Chapter 1 Relations And Functions

Given:

A is the set of all students of a boys school.

Relation R in A is given by R = {(a, b) : a is sister of b}.

Relation R´ in A is given by R´ = {(a, b) : the difference between heights of a and b is less than 3 meters}.

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To Show:

R is the empty relation.

R´ is the universal relation.

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Solution for Relation R:

The set A contains all students of a boys school. This means that all elements in the set A are male students.

The relation R is defined as R = {(a, b) : a is sister of b}, where a, b $\in$ A.

For a pair (a, b) to be in the relation R, student 'a' must be a sister of student 'b'.

Since A contains only male students, no student 'a' in A can be a sister of any student 'b' in A (or even herself).

Therefore, there is no pair of students (a, b) $\in$ A $\times$ A that satisfies the condition "a is sister of b".

The set of pairs satisfying the condition is empty.

Hence, R is the empty relation, denoted by $\emptyset$.

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Solution for Relation R´:

The set A contains all students of a boys school.

The relation R´ is defined as R´ = {(a, b) : the difference between heights of a and b is less than 3 meters}, where a, b $\in$ A.

Consider any two students 'a' and 'b' from the school. Let their heights be $h_a$ and $h_b$ respectively.

The difference between their heights is $|h_a - h_b|$.

Human heights are typically positive values and for students, the difference in heights is usually measured in centimeters or a few feet, generally less than 1 meter.

A difference of 3 meters ($300$ cm) is an extremely large height difference, exceeding the height of most individual humans.

It is practically impossible for the difference between the heights of any two students in a school to be 3 meters or more.

Thus, for any pair of students (a, b) $\in$ A $\times$ A, the difference between their heights will always be less than 3 meters.

This means every possible pair (a, b) from A $\times$ A satisfies the condition "the difference between heights of a and b is less than 3 meters".

The set of pairs satisfying the condition is the entire set A $\times$ A.

Hence, R´ is the universal relation.

Given:

T is the set of all triangles in a plane.

R is a relation in T given by R = {(T$_1$ , T$_2$) : T$_1$ is congruent to T$_2$}, where T$_1$, T$_2$ $\in$ T.

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To Show:

R is an equivalence relation.

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Proof:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set T, we need to show that (T$_1$, T$_1$) $\in$ R for every triangle T$_1$ $\in$ T.

The condition for (T$_1$, T$_1$) $\in$ R is that T$_1$ is congruent to T$_1$.

Every triangle is congruent to itself.

Thus, T$_1$ $\cong$ T$_1$ for all T$_1$ $\in$ T.

So, (T$_1$, T$_1$) $\in$ R for all T$_1$ $\in$ T.

Therefore, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set T, we assume (T$_1$, T$_2$) $\in$ R for some T$_1$, T$_2$ $\in$ T, and we need to show that (T$_2$, T$_1$) $\in$ R.

If (T$_1$, T$_2$) $\in$ R, then T$_1$ is congruent to T$_2$.

By the property of congruence, if T$_1$ is congruent to T$_2$, then T$_2$ is also congruent to T$_1$.

This means that (T$_2$, T$_1$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R implies (T$_2$, T$_1$) $\in$ R for all T$_1$, T$_2$ $\in$ T.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set T, we assume (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R for some T$_1$, T$_2$, T$_3$ $\in$ T, and we need to show that (T$_1$, T$_3$) $\in$ R.

If (T$_1$, T$_2$) $\in$ R, then T$_1$ is congruent to T$_2$.

If (T$_2$, T$_3$) $\in$ R, then T$_2$ is congruent to T$_3$.

By the transitivity property of congruence, if T$_1$ is congruent to T$_2$ and T$_2$ is congruent to T$_3$, then T$_1$ is congruent to T$_3$.

This means that (T$_1$, T$_3$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R implies (T$_1$, T$_3$) $\in$ R for all T$_1$, T$_2$, T$_3$ $\in$ T.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

L is the set of all lines in a plane.

R is a relation in L defined as R = {(L$_1$ , L$_2$) : L$_1$ is perpendicular to L$_2$}, where L$_1$, L$_2$ $\in$ L.

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To Show:

R is symmetric.

R is neither reflexive nor transitive.

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Proof:

We need to examine the three properties of relations: reflexivity, symmetry, and transitivity.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set L, we need to check if (L$_1$, L$_1$) $\in$ R for every line L$_1$ $\in$ L.

The condition for (L$_1$, L$_1$) $\in$ R is that L$_1$ is perpendicular to L$_1$.

A line cannot be perpendicular to itself (unless it's not a straight line, which is not the case here as L is a set of lines in a plane). A line is always parallel to itself, or coincides with itself, but not perpendicular.

Thus, L$_1$ is not perpendicular to L$_1$ for any line L$_1$ $\in$ L.

So, (L$_1$, L$_1$) $\notin$ R for any L$_1$ $\in$ L.

Therefore, R is not reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set L, we assume (L$_1$, L$_2$) $\in$ R for some lines L$_1$, L$_2$ $\in$ L, and we need to check if (L$_2$, L$_1$) $\in$ R.

If (L$_1$, L$_2$) $\in$ R, then L$_1$ is perpendicular to L$_2$.

If line L$_1$ is perpendicular to line L$_2$, it means the angle between them is $90^\circ$. This relationship is mutual.

If L$_1 \perp$ L$_2$, then L$_2 \perp$ L$_1$.

This means that (L$_2$, L$_1$) $\in$ R.

Thus, (L$_1$, L$_2$) $\in$ R implies (L$_2$, L$_1$) $\in$ R for all L$_1$, L$_2$ $\in$ L.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set L, we assume (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R for some lines L$_1$, L$_2$, L$_3$ $\in$ L, and we need to check if (L$_1$, L$_3$) $\in$ R.

If (L$_1$, L$_2$) $\in$ R, then L$_1$ is perpendicular to L$_2$.

If (L$_2$, L$_3$) $\in$ R, then L$_2$ is perpendicular to L$_3$.

Consider two lines L$_1$ and L$_3$ that are both perpendicular to the same line L$_2$. In a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

So, if L$_1 \perp$ L$_2$ and L$_2 \perp$ L$_3$, then L$_1$ is parallel to L$_3$.

For the relation to be transitive, we would need L$_1$ to be perpendicular to L$_3$. However, L$_1$ is parallel to L$_3$, not perpendicular (unless L$_1$ and L$_3$ are the same line, but we consider distinct lines for transitivity testing). For example, let L$_1$ be the x-axis, L$_2$ be the y-axis, and L$_3$ be the line y=1. L$_1 \perp$ L$_2$ and L$_2 \perp$ L$_3$. But L$_1 \parallel$ L$_3$.

Thus, (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R does not imply (L$_1$, L$_3$) $\in$ R.

Therefore, R is not transitive.

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In conclusion, the relation R is symmetric but neither reflexive nor transitive.

Given:

Set A = {1, 2, 3}

Relation R in A is given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}.

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To Show:

R is reflexive.

R is neither symmetric nor transitive.

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Proof:

We need to examine the three properties of relations: reflexivity, symmetry, and transitivity for the relation R on the set A.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

The set A is {1, 2, 3}. We need to check if (1, 1) $\in$ R, (2, 2) $\in$ R, and (3, 3) $\in$ R.

From the definition of R, we have:

• (1, 1) $\in$ R
• (2, 2) $\in$ R
• (3, 3) $\in$ R

Since (a, a) $\in$ R for every element a in the set {1, 2, 3}, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

We need to check for every ordered pair (a, b) in R whether the reverse pair (b, a) is also in R.

• (1, 1) $\in$ R. Is (1, 1) $\in$ R? Yes.
• (2, 2) $\in$ R. Is (2, 2) $\in$ R? Yes.
• (3, 3) $\in$ R. Is (3, 3) $\in$ R? Yes.
• (1, 2) $\in$ R. Is (2, 1) $\in$ R? Looking at the definition of R, (2, 1) is not in R.

Since there exists a pair (1, 2) $\in$ R such that (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

We need to check for pairs (a, b) and (b, c) in R, whether the pair (a, c) is also in R.

• Consider (1, 2) $\in$ R and (2, 3) $\in$ R. We check if (1, 3) $\in$ R. Looking at the definition of R, (1, 3) is not in R.

Since there exist pairs (1, 2) $\in$ R and (2, 3) $\in$ R, but (1, 3) $\notin$ R, R is not transitive.

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In conclusion, the relation R is reflexive but neither symmetric nor transitive.

Given:

Z is the set of all integers.

R is a relation in Z given by R = {(a, b) : 2 divides a – b}, where a, b $\in$ Z.

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To Show:

R is an equivalence relation.

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Proof:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set Z, we need to show that (a, a) $\in$ R for every integer a $\in$ Z.

The condition for (a, a) $\in$ R is that 2 divides a – a.

The difference a – a is 0.

Since 2 divides 0 ($0 = 2 \times 0$), the condition is satisfied for all integers a $\in$ Z.

Thus, (a, a) $\in$ R for all a $\in$ Z.

Therefore, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set Z, we assume (a, b) $\in$ R for some integers a, b $\in$ Z, and we need to show that (b, a) $\in$ R.

If (a, b) $\in$ R, it means that 2 divides a – b.

This implies that a – b can be written as $2k$ for some integer $k$.

Now consider the difference b – a.

Substitute $a - b = 2k$:

Since k is an integer, -k is also an integer. This shows that b – a is an integer multiple of 2, which means 2 divides b – a.

Thus, the condition for (b, a) $\in$ R is satisfied.

So, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ Z.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set Z, we assume (a, b) $\in$ R and (b, c) $\in$ R for some integers a, b, c $\in$ Z, and we need to show that (a, c) $\in$ R.

If (a, b) $\in$ R, it means that 2 divides a – b.

If (b, c) $\in$ R, it means that 2 divides b – c.

Now consider the difference a – c. We can write a – c as $(a - b) + (b - c)$.

Substitute the expressions for a – b and b – c:

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer. This shows that a – c is an integer multiple of 2, which means 2 divides a – c.

Thus, the condition for (a, c) $\in$ R is satisfied.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ Z.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

Set A = {1, 2, 3, 4, 5, 6, 7}

Relation R in A is defined as R = {(a, b) : both a and b are either odd or even}, where a, b $\in$ A.

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To Show:

R is an equivalence relation.

All elements of {1, 3, 5, 7} are related to each other.

All elements of {2, 4, 6} are related to each other.

No element of {1, 3, 5, 7} is related to any element of {2, 4, 6}.

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Proof that R is an equivalence relation:

We need to check for reflexivity, symmetry, and transitivity.

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1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

For any element a $\in$ A, the condition for (a, a) $\in$ R is that both a and a are either odd or even.

This condition is always true for any integer a; a and itself always have the same parity (both odd or both even).

Thus, (a, a) $\in$ R for all a $\in$ A.

Therefore, R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R for some a, b $\in$ A. This means that both a and b are either odd or even.

If (a, b) $\in$ R, then a and b have the same parity.

The condition "both a and b are either odd or even" is symmetric with respect to a and b. If 'a and b' have the same parity, then 'b and a' also have the same parity.

So, if both a and b are either odd or even, then both b and a are also either odd or even.

This means (b, a) $\in$ R.

Thus, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Therefore, R is symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R for some a, b, c $\in$ A.

(a, b) $\in$ R means both a and b have the same parity (both odd or both even).

(b, c) $\in$ R means both b and c have the same parity (both odd or both even).

If a and b have the same parity, and b and c have the same parity, then it logically follows that a and c must also have the same parity.

• If a is odd and b is odd, and b is odd and c is odd, then a is odd and c is odd (same parity).
• If a is even and b is even, and b is even and c is even, then a is even and c is even (same parity).

So, if both a and b are either odd or even, and both b and c are either odd or even, then both a and c are also either odd or even.

This means (a, c) $\in$ R.

Thus, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Further Demonstration:

Consider the subset {1, 3, 5, 7}. All elements in this subset are odd numbers.

For any two elements a and b from this subset (a $\neq$ b, and including when a=b for reflexivity), both a and b are odd. The condition "both a and b are either odd or even" is satisfied.

For example, (1, 3) $\in$ R because 1 and 3 are both odd. (5, 7) $\in$ R because 5 and 7 are both odd. (1, 1) $\in$ R because 1 and 1 are both odd.

Thus, all elements of the subset {1, 3, 5, 7} are related to each other under the relation R.

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Consider the subset {2, 4, 6}. All elements in this subset are even numbers.

For any two elements a and b from this subset (a $\neq$ b, and including when a=b for reflexivity), both a and b are even. The condition "both a and b are either odd or even" is satisfied.

For example, (2, 4) $\in$ R because 2 and 4 are both even. (6, 6) $\in$ R because 6 and 6 are both even.

Thus, all elements of the subset {2, 4, 6} are related to each other under the relation R.

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Consider an element a from the subset {1, 3, 5, 7} (i.e., a is odd) and an element b from the subset {2, 4, 6} (i.e., b is even).

For the pair (a, b) to be in the relation R, the condition "both a and b are either odd or even" must be satisfied.

However, one element (a) is odd and the other element (b) is even. They do not have the same parity.

Therefore, the condition is not satisfied.

For example, (1, 2) $\notin$ R because 1 is odd and 2 is even. (3, 6) $\notin$ R because 3 is odd and 6 is even.

Thus, no element of the subset {1, 3, 5, 7} is related to any element of the subset {2, 4, 6} under the relation R.

These two subsets form disjoint equivalence classes under the relation R.

Question 1 (i):

Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}.

The relation can be written as R = {(x, y) : y = 3x}.

Let's list the pairs in R for x, y $\in$ A:

• If x = 1, y = 3(1) = 3. So, (1, 3) $\in$ R.
• If x = 2, y = 3(2) = 6. So, (2, 6) $\in$ R.
• If x = 3, y = 3(3) = 9. So, (3, 9) $\in$ R.
• If x = 4, y = 3(4) = 12. So, (4, 12) $\in$ R.
• If x = 5, y = 3(5) = 15. Since 15 $\notin$ A, no pairs with x $\geq$ 5 are in R.

So, R = {(1, 3), (2, 6), (3, 9), (4, 12)}.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ A.

This requires y = x, which means $3x = x$. This implies $2x = 0$, so $x = 0$.

Since 0 is not in the set A = {1, 2, ..., 14}, there is no element x in A such that (x, x) $\in$ R.

For example, for x = 1, (1, 1) $\notin$ R because $3(1) \neq 1$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Consider the pair (1, 3) $\in$ R (since $3 = 3 \times 1$).

For R to be symmetric, (3, 1) must also be in R.

The condition for (3, 1) $\in$ R is $1 = 3(3)$, which is $1 = 9$. This is false.

Since (1, 3) $\in$ R but (3, 1) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Consider the pair (1, 3) $\in$ R.

We look for a pair starting with 3 in R. (3, 9) $\in$ R.

For R to be transitive, (1, 9) must also be in R.

The condition for (1, 9) $\in$ R is $9 = 3(1)$, which is $9 = 3$. This is false.

Since (1, 3) $\in$ R and (3, 9) $\in$ R, but (1, 9) $\notin$ R, R is not transitive.

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Conclusion for (i): The relation R is neither reflexive, nor symmetric, nor transitive.

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Question 1 (ii):

Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}.

The set N = {1, 2, 3, ...}. The condition on x is $x \in N$ and $x < 4$. So possible values for x are 1, 2, 3.

Let's list the pairs in R:

• If x = 1, y = 1 + 5 = 6. Since 6 $\in$ N, (1, 6) $\in$ R.
• If x = 2, y = 2 + 5 = 7. Since 7 $\in$ N, (2, 7) $\in$ R.
• If x = 3, y = 3 + 5 = 8. Since 8 $\in$ N, (3, 8) $\in$ R.

So, R = {(1, 6), (2, 7), (3, 8)}.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ N.

This requires y = x, which means $x = x + 5$. This implies $0 = 5$, which is false.

So, there is no element x in N such that (x, x) $\in$ R.

For example, for x = 1, (1, 1) $\notin$ R because $1 \neq 1 + 5$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ N.

Consider the pair (1, 6) $\in$ R.

For R to be symmetric, (6, 1) must also be in R.

The condition for (6, 1) $\in$ R is $1 = 6 + 5$ and $6 < 4$. This requires $1 = 11$ (false) and $6 < 4$ (false).

Since (1, 6) $\in$ R but (6, 1) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ N.

We need to find if there exist pairs (x, y) $\in$ R and (y, z) $\in$ R.

From the pairs in R: (1, 6), (2, 7), (3, 8).

Consider (1, 6) $\in$ R. For transitivity, we need to check if there is a pair (6, z) in R.

A pair (y, z) is in R if $z = y + 5$ and $y < 4$. In this case, y = 6. Since $6 \nless 4$, there is no pair (6, z) in R.

Similarly, there are no pairs (7, z) or (8, z) in R because 7 and 8 are not less than 4.

Since the condition "(x, y) $\in$ R and (y, z) $\in$ R" is never satisfied for any x, y, z in N, the implication "implies (x, z) $\in$ R" is vacuously true.

Therefore, R is transitive.

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Conclusion for (ii): The relation R is transitive but neither reflexive nor symmetric.

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Question 1 (iii):

Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y) : y is divisible by x}.

The condition "y is divisible by x" means that $y = kx$ for some integer k.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ A.

This requires x to be divisible by x.

For any x $\in$ A = {1, 2, 3, 4, 5, 6}, x is divisible by x (since x = 1 * x, and 1 is an integer).

Thus, (x, x) $\in$ R for all x $\in$ A.

Therefore, R is reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Assume (x, y) $\in$ R. This means y is divisible by x, so $y = kx$ for some integer k.

For R to be symmetric, (y, x) must also be in R. This means x must be divisible by y, so $x = my$ for some integer m.

Substitute y from the first equation into the second: $x = m(kx) = (mk)x$.

If $x \neq 0$, then $mk = 1$. Since x, y $\in$ A, x and y are positive integers. For mk to be 1, both m and k must be either 1 or -1. As x, y are positive, k must be positive, so k=1. This means y=x. If y=x, then m=1. So symmetry only holds for pairs where x=y.

Consider the pair (2, 4) $\in$ R (since 4 is divisible by 2, $4 = 2 \times 2$).

For R to be symmetric, (4, 2) must also be in R.

The condition for (4, 2) $\in$ R is that 2 is divisible by 4. This is false (2 = $k \times 4$ has no integer solution for k).

Since (2, 4) $\in$ R but (4, 2) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Assume (x, y) $\in$ R and (y, z) $\in$ R.

(x, y) $\in$ R means y is divisible by x, so $y = kx$ for some integer $k$.

(y, z) $\in$ R means z is divisible by y, so $z = my$ for some integer $m$.

We need to check if (x, z) $\in$ R, which means z is divisible by x.

Substitute y from the first equation into the second: $z = m(kx) = (mk)x$.

Since k and m are integers, the product mk is also an integer.

This shows that z is an integer multiple of x, so z is divisible by x.

Thus, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Therefore, R is transitive.

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Conclusion for (iii): The relation R is reflexive and transitive but not symmetric.

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Question 1 (iv):

Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}.

The set Z is the set of all integers: {... -2, -1, 0, 1, 2, ...}.

The condition "x – y is an integer" is always true for any two integers x and y, since the difference of any two integers is always an integer.

This means R = Z $\times$ Z, which is the universal relation on Z.

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Reflexivity:

A relation R is reflexive if (x, x) $\in$ R for every x $\in$ Z.

This requires x – x to be an integer.

For any integer x, $x - x = 0$. Since 0 is an integer, the condition is satisfied.

Thus, (x, x) $\in$ R for all x $\in$ Z.

Therefore, R is reflexive.

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Symmetry:

A relation R is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ Z.

Assume (x, y) $\in$ R. This means x – y is an integer.

Let $x - y = k$, where k is an integer.

Consider y – x. We have $y - x = -(x - y) = -k$.

Since k is an integer, -k is also an integer.

Thus, y – x is an integer, which means (y, x) $\in$ R.

So, (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ Z.

Therefore, R is symmetric.

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Transitivity:

A relation R is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ Z.

Assume (x, y) $\in$ R and (y, z) $\in$ R.

(x, y) $\in$ R means x – y is an integer. Let $x - y = k_1$, where $k_1$ is an integer.

(y, z) $\in$ R means y – z is an integer. Let $y - z = k_2$, where $k_2$ is an integer.

Consider x – z. We have $x - z = (x - y) + (y - z) = k_1 + k_2$.

Since $k_1$ and $k_2$ are integers, their sum $k_1 + k_2$ is also an integer.

Thus, x – z is an integer, which means (x, z) $\in$ R.

So, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ Z.

Therefore, R is transitive.

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Conclusion for (iv): The relation R is reflexive, symmetric, and transitive. It is an equivalence relation.

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Question 1 (v):

Relation R in the set A of human beings in a town at a particular time.

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(a) R = {(x, y) : x and y work at the same place}

Reflexivity: For any person x $\in$ A, does x work at the same place as x? Yes, a person works at the same place as themselves. R is reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x and y work at the same place. If x and y work at the same place, then y and x also work at the same place. So (y, x) $\in$ R. R is symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x and y work at the same place, and y and z work at the same place. If x works at the same place as y, and y works at the same place as z, then x must work at the same place as z. So (x, z) $\in$ R. R is transitive.

Conclusion for (v) (a): The relation is reflexive, symmetric, and transitive (an equivalence relation).

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(b) R = {(x, y) : x and y live in the same locality}

Reflexivity: For any person x $\in$ A, does x live in the same locality as x? Yes. R is reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x and y live in the same locality. If x and y live in the same locality, then y and x also live in the same locality. So (y, x) $\in$ R. R is symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x and y live in the same locality, and y and z live in the same locality. If x lives in the same locality as y, and y lives in the same locality as z, then x must live in the same locality as z. So (x, z) $\in$ R. R is transitive.

Conclusion for (v) (b): The relation is reflexive, symmetric, and transitive (an equivalence relation).

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(c) R = {(x, y) : x is exactly 7 cm taller than y}

Reflexivity: For any person x $\in$ A, is x exactly 7 cm taller than x? No, the height difference between x and x is 0 cm. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is exactly 7 cm taller than y. Let height(x) = $h_x$ and height(y) = $h_y$. The condition is $h_x = h_y + 7$. For R to be symmetric, (y, x) must be in R, meaning y is exactly 7 cm taller than x ($h_y = h_x + 7$). If $h_x = h_y + 7$, then $h_y = h_x - 7$. So y is exactly 7 cm shorter than x, not 7 cm taller. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. This means x is exactly 7 cm taller than y ($h_x = h_y + 7$) and y is exactly 7 cm taller than z ($h_y = h_z + 7$). For R to be transitive, (x, z) must be in R, meaning x is exactly 7 cm taller than z ($h_x = h_z + 7$). Substitute the second equation into the first: $h_x = (h_z + 7) + 7 = h_z + 14$. This means x is exactly 14 cm taller than z, not 7 cm. R is not transitive.

Conclusion for (v) (c): The relation is neither reflexive, nor symmetric, nor transitive.

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(d) R = {(x, y) : x is wife of y}

Reflexivity: For any person x $\in$ A, is x the wife of x? No, a person cannot be their own spouse. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is the wife of y. If x is the wife of y, then y must be the husband of x (assuming a traditional marriage). Can y be the wife of x? No, the wife is x. So (y, x) $\notin$ R. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. If (x, y) $\in$ R, then x is the wife of y, which implies x is female and y is male. If (y, z) $\in$ R, then y is the wife of z, which implies y is female and z is male. For the premise "(x, y) $\in$ R and (y, z) $\in$ R" to be true, y must be both male (from the first pair) and female (from the second pair). This is impossible for a human being. Since the premise of the transitivity condition is never met, the implication "implies (x, z) $\in$ R" is vacuously true.

Therefore, R is transitive.

Conclusion for (v) (d): The relation is transitive but neither reflexive nor symmetric.

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(e) R = {(x, y) : x is father of y}

Reflexivity: For any person x $\in$ A, is x the father of x? No, a person cannot be their own parent. R is not reflexive.

Symmetry: Assume (x, y) $\in$ R. This means x is the father of y. If x is the father of y, is y the father of x? No, y is the child of x. So (y, x) $\notin$ R. R is not symmetric.

Transitivity: Assume (x, y) $\in$ R and (y, z) $\in$ R. If (x, y) $\in$ R, then x is the father of y. If (y, z) $\in$ R, then y is the father of z. For R to be transitive, (x, z) must be in R, meaning x is the father of z. If x is the father of y, and y is the father of z, then x is the grandfather of z, not the father of z. R is not transitive.

Conclusion for (v) (e): The relation is neither reflexive, nor symmetric, nor transitive.

Given:

The set is R (real numbers).

The relation R is defined as R = {(a, b) : a $\le$ b$^2$}, where a, b $\in$ R.

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To Show:

R is neither reflexive nor symmetric nor transitive.

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Proof:

We examine the three properties of relations.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set of real numbers, we need to check if (a, a) $\in$ R for every real number a.

The condition for (a, a) $\in$ R is that a $\le$ a$^2$.

This inequality $a \le a^2$ is not true for all real numbers. For example, consider a real number between 0 and 1, such as $a = \frac{1}{2}$.

If $a = \frac{1}{2}$, then $a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

Is $\frac{1}{2} \le \frac{1}{4}$? No, $\frac{1}{2} > \frac{1}{4}$.

So, $(\frac{1}{2}, \frac{1}{2}) \notin$ R because $\frac{1}{2} \not\le (\frac{1}{2})^2$.

Since there exists at least one real number 'a' for which $a \not\le a^2$, R is not reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R for some real numbers a and b, and we need to check if (b, a) $\in$ R.

If (a, b) $\in$ R, it means a $\le$ b$^2$.

For R to be symmetric, (b, a) must be in R, which means b $\le$ a$^2$.

Let's consider a counterexample. Choose a pair (a, b) such that a $\le$ b$^2$ is true, but b $\le$ a$^2$ is false.

Consider $a = 1$ and $b = 2$.

Is (1, 2) $\in$ R? Check the condition: $1 \le 2^2 \implies 1 \le 4$. This is true. So (1, 2) $\in$ R.

Is (2, 1) $\in$ R? Check the condition: $2 \le 1^2 \implies 2 \le 1$. This is false. So (2, 1) $\notin$ R.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R and (b, c) $\in$ R for some real numbers a, b, and c, and we need to check if (a, c) $\in$ R.

If (a, b) $\in$ R, then a $\le$ b$^2$.

If (b, c) $\in$ R, then b $\le$ c$^2$.

For R to be transitive, if $a \le b^2$ and $b \le c^2$, it must imply $a \le c^2$ for all a, b, c.

Let's consider a counterexample. We need values a, b, c such that $a \le b^2$ and $b \le c^2$ are true, but $a \le c^2$ is false.

Consider $a = 3$, $b = -2$, and $c = 1.5$.

Is (3, -2) $\in$ R? Check the condition: $3 \le (-2)^2 \implies 3 \le 4$. This is true. So (3, -2) $\in$ R.

Is (-2, 1.5) $\in$ R? Check the condition: $-2 \le (1.5)^2 \implies -2 \le 2.25$. This is true. So (-2, 1.5) $\in$ R.

Now check if (a, c) = (3, 1.5) $\in$ R. The condition is $3 \le (1.5)^2$.

Is $3 \le 2.25$? This is false.

Since (3, -2) $\in$ R and (-2, 1.5) $\in$ R, but (3, 1.5) $\notin$ R, R is not transitive.

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In conclusion, the relation R = {(a, b) : a $\le$ b$^2$} on the set of real numbers is neither reflexive, nor symmetric, nor transitive.

Given:

Set A = {1, 2, 3, 4, 5, 6}

Relation R in A is defined as R = {(a, b) : b = a + 1}, where a, b $\in$ A.

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To Check:

Whether R is reflexive, symmetric, or transitive.

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Let's list the pairs in R:

• If a = 1, b = 1 + 1 = 2. So, (1, 2) $\in$ R.
• If a = 2, b = 2 + 1 = 3. So, (2, 3) $\in$ R.
• If a = 3, b = 3 + 1 = 4. So, (3, 4) $\in$ R.
• If a = 4, b = 4 + 1 = 5. So, (4, 5) $\in$ R.
• If a = 5, b = 5 + 1 = 6. So, (5, 6) $\in$ R.
• If a = 6, b = 6 + 1 = 7. Since 7 $\notin$ A, no pairs with a = 6 are in R.

So, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.

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Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires b = a, which means $a = a + 1$. This implies $0 = 1$, which is false.

So, there is no element a in A such that (a, a) $\in$ R.

For example, for a = 1, (1, 1) $\notin$ R because $1 \neq 1 + 1$.

Therefore, R is not reflexive.

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Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means b = a + 1.

For R to be symmetric, (b, a) must also be in R. This means a = b + 1.

Substitute b from the first equation into the condition for the second: $a = (a + 1) + 1 = a + 2$.

This implies $0 = 2$, which is false.

Consider the pair (1, 2) $\in$ R (since $2 = 1 + 1$).

For R to be symmetric, (2, 1) must also be in R.

The condition for (2, 1) $\in$ R is $1 = 2 + 1$, which is $1 = 3$. This is false.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

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Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means b = a + 1.

(b, c) $\in$ R means c = b + 1.

We need to check if (a, c) $\in$ R, which means c = a + 1.

Substitute b from the first equation into the second: $c = (a + 1) + 1 = a + 2$.

For the relation to be transitive, we need $c = a + 1$. We found $c = a + 2$. These are not the same.

Consider the pairs (1, 2) $\in$ R and (2, 3) $\in$ R.

For R to be transitive, (1, 3) must also be in R.

The condition for (1, 3) $\in$ R is $3 = 1 + 1$, which is $3 = 2$. This is false.

Since (1, 2) $\in$ R and (2, 3) $\in$ R, but (1, 3) $\notin$ R, R is not transitive.

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Conclusion: The relation R is neither reflexive, nor symmetric, nor transitive.

Given:

The set is R (real numbers).

The relation R is defined as R = {(a, b) : a $\le$ b}, where a, b $\in$ R.

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To Show:

R is reflexive and transitive but not symmetric.

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Proof:

We examine the three properties of relations.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

For the relation R on the set of real numbers, we need to check if (a, a) $\in$ R for every real number a.

The condition for (a, a) $\in$ R is that a $\le$ a.

This inequality is always true for any real number a.

Thus, (a, a) $\in$ R for all a $\in$ R.

Therefore, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R for some real numbers a and b, and we need to check if (b, a) $\in$ R.

If (a, b) $\in$ R, it means a $\le$ b.

For R to be symmetric, (b, a) must be in R, which means b $\le$ a.

If $a \le b$, it does not necessarily imply $b \le a$. This only happens when $a = b$.

Consider a counterexample. Choose distinct real numbers a and b such that a $\le$ b is true.

Let $a = 1$ and $b = 2$.

Is (1, 2) $\in$ R? Check the condition: $1 \le 2$. This is true. So (1, 2) $\in$ R.

Is (2, 1) $\in$ R? Check the condition: $2 \le 1$. This is false. So (2, 1) $\notin$ R.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

For the relation R on the set of real numbers, we assume (a, b) $\in$ R and (b, c) $\in$ R for some real numbers a, b, and c, and we need to check if (a, c) $\in$ R.

If (a, b) $\in$ R, then a $\le$ b.

If (b, c) $\in$ R, then b $\le$ c.

By the property of real numbers, if $a \le b$ and $b \le c$, then $a \le c$.

This means that (a, c) $\in$ R.

Thus, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ R.

Therefore, R is transitive.

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In conclusion, the relation R = {(a, b) : a $\le$ b} on the set of real numbers is reflexive and transitive but not symmetric.

Given:

The set is R (real numbers).

The relation R is defined by R = {(a, b) : a $\le$ b$^3$}, where a, b $\in$ R.

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To Check:

Whether R is reflexive, symmetric, or transitive.

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1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ R.

This requires a $\le$ a$^3$ for all real numbers a.

Let's check for different types of real numbers:

• If a $\ge$ 1, then $a^3 \ge a$. So $a \le a^3$ is true (e.g., $2 \le 2^3=8$).
• If a = 0 or a = 1, $a = a^3$. So $a \le a^3$ is true.
• If $0 < a < 1$, then $a^3 < a$. For example, if $a = 0.5$, $a^3 = 0.125$. $0.5 \not\le 0.125$.
• If $a < 0$, we need to be careful with the inequality. For example, if $a = -2$, $a^3 = -8$. $-2 \le -8$ is false. If $a = -0.5$, $a^3 = -0.125$. $-0.5 \le -0.125$ is true. If $a = -1$, $a = a^3$.

Since the inequality $a \le a^3$ is not true for all real numbers (e.g., for $a = 0.5$), R is not reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ R.

Assume (a, b) $\in$ R. This means a $\le$ b$^3$.

For R to be symmetric, (b, a) must be in R, which means b $\le$ a$^3$.

We need to find a counterexample where $a \le b^3$ is true but $b \le a^3$ is false.

Let $a = 1$ and $b = 2$.

Is (1, 2) $\in$ R? Check $1 \le 2^3 \implies 1 \le 8$. True. So (1, 2) $\in$ R.

Is (2, 1) $\in$ R? Check $2 \le 1^3 \implies 2 \le 1$. False. So (2, 1) $\notin$ R.

Since (1, 2) $\in$ R but (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ R.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

This means a $\le$ b$^3$ and b $\le$ c$^3$.

For R to be transitive, we need to check if $a \le c^3$ necessarily follows.

We need to find a counterexample where $a \le b^3$ and $b \le c^3$ are true, but $a \le c^3$ is false.

Consider $a = 100$, $b = 4$, $c = 2$.

Is (100, 4) $\in$ R? Check $100 \le 4^3 \implies 100 \le 64$. False. This choice doesn't work for the premise.

Let's try values where b is negative or between 0 and 1 to make the inequalities work in unexpected ways.

Consider $a = 30$, $b = 3$, $c = 2$.

Is (30, 3) $\in$ R? Check $30 \le 3^3 \implies 30 \le 27$. False.

Consider $a = 3$, $b = 2$, $c = 1.5$.

Is (3, 2) $\in$ R? Check $3 \le 2^3 \implies 3 \le 8$. True. So (3, 2) $\in$ R.

Is (2, 1.5) $\in$ R? Check $2 \le (1.5)^3 \implies 2 \le (3/2)^3 = 27/8 = 3.375$. True. So (2, 1.5) $\in$ R.

Now check if (a, c) = (3, 1.5) $\in$ R. Check $3 \le (1.5)^3 \implies 3 \le 3.375$. True. This example doesn't show non-transitivity.

Let's try a smaller value for c.

Consider $a = 10$, $b = 3$, $c = 2$.

Is (10, 3) $\in$ R? Check $10 \le 3^3 \implies 10 \le 27$. True.

Is (3, 2) $\in$ R? Check $3 \le 2^3 \implies 3 \le 8$. True.

Is (10, 2) $\in$ R? Check $10 \le 2^3 \implies 10 \le 8$. False.

Since (10, 3) $\in$ R and (3, 2) $\in$ R, but (10, 2) $\notin$ R, R is not transitive.

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In conclusion, the relation R = {(a, b) : a $\le$ b$^3$} on the set of real numbers is neither reflexive, nor symmetric, nor transitive.

Given:

Set A = {1, 2, 3}

Relation R in A is given by R = {(1, 2), (2, 1)}.

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To Show:

R is symmetric.

R is neither reflexive nor transitive.

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Proof:

We examine the three properties of relations for the relation R on the set A.

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1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

The set A is {1, 2, 3}. We need to check if (1, 1) $\in$ R, (2, 2) $\in$ R, and (3, 3) $\in$ R.

Looking at the definition of R, none of these pairs are in R.

(1, 1) $\notin$ R.

(2, 2) $\notin$ R.

(3, 3) $\notin$ R.

Since (a, a) $\notin$ R for every element a in the set {1, 2, 3}, R is not reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

We check the pairs in R:

• Consider the pair (1, 2) $\in$ R. Is (2, 1) $\in$ R? Yes, (2, 1) is in R.
• Consider the pair (2, 1) $\in$ R. Is (1, 2) $\in$ R? Yes, (1, 2) is in R.

For every pair (a, b) in R, the reverse pair (b, a) is also in R.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

We need to check for pairs (a, b) and (b, c) in R.

• Consider the pair (1, 2) $\in$ R. We look for a pair starting with 2 in R. We have (2, 1) $\in$ R.

According to the transitivity condition, if (1, 2) $\in$ R and (2, 1) $\in$ R, then (1, 1) must also be in R.

Looking at the definition of R, (1, 1) is not in R.

Since (1, 2) $\in$ R and (2, 1) $\in$ R, but (1, 1) $\notin$ R, R is not transitive.

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In conclusion, the relation R is symmetric but neither reflexive nor transitive.

Given:

A is the set of all books in a library of a college.

R is a relation in A given by R = {(x, y) : x and y have same number of pages}, where x, y $\in$ A.

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To Show:

R is an equivalence relation.

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Proof:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R on a set A is reflexive if (x, x) $\in$ R for every x $\in$ A.

For the relation R on the set A of books, we need to show that (x, x) $\in$ R for every book x $\in$ A.

The condition for (x, x) $\in$ R is that x and x have the same number of pages.

Any book has the same number of pages as itself.

Thus, (x, x) $\in$ R for all x $\in$ A.

Therefore, R is reflexive.

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2. Symmetry:

A relation R on a set A is symmetric if (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

For the relation R on the set A of books, we assume (x, y) $\in$ R for some books x, y $\in$ A, and we need to show that (y, x) $\in$ R.

If (x, y) $\in$ R, it means that x and y have the same number of pages.

Let the number of pages of book x be $N_x$ and the number of pages of book y be $N_y$.

The condition is $N_x = N_y$.

If $N_x = N_y$, then it is also true that $N_y = N_x$.

This means that y and x have the same number of pages.

Thus, (y, x) $\in$ R.

So, (x, y) $\in$ R implies (y, x) $\in$ R for all x, y $\in$ A.

Therefore, R is symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

For the relation R on the set A of books, we assume (x, y) $\in$ R and (y, z) $\in$ R for some books x, y, z $\in$ A, and we need to show that (x, z) $\in$ R.

If (x, y) $\in$ R, it means that x and y have the same number of pages ($N_x = N_y$).

If (y, z) $\in$ R, it means that y and z have the same number of pages ($N_y = N_z$).

If $N_x = N_y$ and $N_y = N_z$, then by the transitivity of equality, $N_x = N_z$.

This means that x and z have the same number of pages.

Thus, (x, z) $\in$ R.

So, (x, y) $\in$ R and (y, z) $\in$ R implies (x, z) $\in$ R for all x, y, z $\in$ A.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

Given:

Set A = {1, 2, 3, 4, 5}

Relation R in A is given by R = {(a, b) : |a – b| is even}, where a, b $\in$ A.

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To Show:

R is an equivalence relation.

All elements of {1, 3, 5} are related to each other.

All elements of {2, 4} are related to each other.

No element of {1, 3, 5} is related to any element of {2, 4}.

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Proof that R is an equivalence relation:

The condition "|a – b| is even" means that the difference between a and b is an even number. This is equivalent to stating that a and b have the same parity (both odd or both even).

We need to check for reflexivity, symmetry, and transitivity.

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1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

For any element a $\in$ A, the condition for (a, a) $\in$ R is that |a – a| is even.

The difference |a – a| = |0| = 0.

Since 0 is an even number (0 = $2 \times 0$), the condition is satisfied for all elements a $\in$ A.

Thus, (a, a) $\in$ R for all a $\in$ A.

Therefore, R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R for some a, b $\in$ A. This means |a – b| is even.

For R to be symmetric, (b, a) must also be in R, which means |b – a| is even.

We know that $|b - a| = |-(a - b)| = |a - b|$.

Since |a – b| is even, |b – a| is also even.

Thus, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Therefore, R is symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means |a – b| is even. This implies a and b have the same parity.

(b, c) $\in$ R means |b – c| is even. This implies b and c have the same parity.

If a and b have the same parity, and b and c have the same parity, then a and c must also have the same parity.

If a and c have the same parity, then their difference (a – c) is even. Consequently, |a – c| is even.

Thus, (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Therefore, R is transitive.

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Since the relation R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Further Demonstration:

Consider the subset {1, 3, 5}. All elements in this subset are odd numbers.

For any two elements a and b from this subset, both a and b are odd. The difference (a – b) is always even (Odd - Odd = Even). So, |a – b| is even.

For example, for elements in {1, 3, 5}:

• |1 – 3| = |-2| = 2 (even). (1, 3) $\in$ R.
• |3 – 5| = |-2| = 2 (even). (3, 5) $\in$ R.
• |1 – 5| = |-4| = 4 (even). (1, 5) $\in$ R.
• Also, (1, 1), (3, 3), (5, 5) are in R (reflexivity).
• And (3, 1), (5, 3), (5, 1) are in R (symmetry).

Thus, all the elements of the subset {1, 3, 5} are related to each other.

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Consider the subset {2, 4}. All elements in this subset are even numbers.

For any two elements a and b from this subset, both a and b are even. The difference (a – b) is always even (Even - Even = Even). So, |a – b| is even.

For example, for elements in {2, 4}:

• |2 – 4| = |-2| = 2 (even). (2, 4) $\in$ R.
• Also, (2, 2), (4, 4) are in R (reflexivity).
• And (4, 2) is in R (symmetry).

Thus, all the elements of the subset {2, 4} are related to each other.

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Consider an element a from the subset {1, 3, 5} (i.e., a is odd) and an element b from the subset {2, 4} (i.e., b is even).

For the pair (a, b) to be in the relation R, the condition "|a – b| is even" must be satisfied.

However, if one number is odd and the other is even, their difference (a – b) is always odd (Odd - Even = Odd, Even - Odd = Odd).

So, |a – b| is odd.

For example:

• |1 – 2| = |-1| = 1 (odd). (1, 2) $\notin$ R.
• |3 – 4| = |-1| = 1 (odd). (3, 4) $\notin$ R.
• |5 – 2| = |3| = 3 (odd). (5, 2) $\notin$ R.

Thus, no element of the subset {1, 3, 5} is related to any element of the subset {2, 4}.

These two subsets {1, 3, 5} and {2, 4} are the equivalence classes of the relation R.

Given:

Set A = $\{ x \in Z : 0 \le x \le 12 \}$. This means A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.

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Question 9 (i):

Relation R in A is given by R = {(a, b) : |a – b| is a multiple of 4}.

The condition "|a – b| is a multiple of 4" means that $|a - b| = 4k$ for some non-negative integer k. This is equivalent to saying that $a - b = 4m$ for some integer m, or $a \equiv b \pmod{4}$.

---

To Show R is an equivalence relation:

1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires |a – a| to be a multiple of 4.

|a – a| = |0| = 0.

Since 0 is a multiple of 4 ($0 = 4 \times 0$), the condition is satisfied for all a $\in$ A.

Thus, (a, a) $\in$ R for all a $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means |a – b| is a multiple of 4.

So, $|a - b| = 4k$ for some non-negative integer k.

Consider |b – a|. We know $|b - a| = |-(a - b)| = |a - b|$.

Since |a – b| is a multiple of 4, |b – a| is also a multiple of 4.

Thus, (b, a) $\in$ R.

So, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means |a – b| is a multiple of 4. This means $a - b = 4k_1$ for some integer $k_1$.

(b, c) $\in$ R means |b – c| is a multiple of 4. This means $b - c = 4k_2$ for some integer $k_2$.

Consider a – c. We have $a - c = (a - b) + (b - c) = 4k_1 + 4k_2 = 4(k_1 + k_2)$.

Since $k_1$ and $k_2$ are integers, $k_1 + k_2$ is an integer. This means a – c is a multiple of 4.

Consequently, |a – c| is a multiple of 4.

Thus, (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all elements related to 1 in case (i):

We need to find all elements $x \in A$ such that (1, x) $\in$ R. The condition is |1 – x| is a multiple of 4.

So, $|1 - x| = 4k$ for some non-negative integer k.

$1 - x = 4k$ or $1 - x = -4k$.

$x = 1 - 4k$ or $x = 1 + 4k$.

We need to find values of x in A = {0, 1, 2, ..., 12} that satisfy this condition.

If k = 0, $x = 1 - 0 = 1$. $1 \in A$.

If k = 1, $x = 1 - 4 = -3$ (not in A). $x = 1 + 4 = 5$. $5 \in A$.

If k = 2, $x = 1 - 8 = -7$ (not in A). $x = 1 + 8 = 9$. $9 \in A$.

If k = 3, $x = 1 - 12 = -11$ (not in A). $x = 1 + 12 = 13$ (not in A).

The elements related to 1 are {1, 5, 9}.

This set is the equivalence class containing 1, denoted by [1].

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Question 9 (ii):

Relation R in A is given by R = {(a, b) : a = b}.

This is the identity relation on the set A.

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To Show R is an equivalence relation:

1. Reflexivity:

A relation R is reflexive if (a, a) $\in$ R for every a $\in$ A.

This requires a = a, which is always true for any element a.

Thus, (a, a) $\in$ R for all a $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

Assume (a, b) $\in$ R. This means a = b.

If a = b, then it is also true that b = a.

This means (b, a) $\in$ R.

So, (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

Assume (a, b) $\in$ R and (b, c) $\in$ R.

(a, b) $\in$ R means a = b.

(b, c) $\in$ R means b = c.

If a = b and b = c, then by the transitivity of equality, a = c.

This means (a, c) $\in$ R.

So, (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all elements related to 1 in case (ii):

We need to find all elements $x \in A$ such that (1, x) $\in$ R. The condition is 1 = x.

The only element x in A that satisfies 1 = x is 1 itself.

The elements related to 1 are {1}.

This set is the equivalence class containing 1, denoted by [1].

Here are examples of relations demonstrating the requested properties. For each example, let A be a set and R be a relation on A.

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(i) Symmetric but neither reflexive nor transitive.

Let A = {1, 2, 3}.

Let R = {(1, 2), (2, 1)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R and (2, 1) $\in$ R. (2, 1) $\in$ R and (1, 2) $\in$ R. All pairs in R satisfy symmetry. Symmetric.
• Transitivity: (1, 2) $\in$ R and (2, 1) $\in$ R. For transitivity, (1, 1) must be in R. (1, 1) $\notin$ R. Not transitive.

This example satisfies the conditions.

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(ii) Transitive but neither reflexive nor symmetric.

Let A = {1, 2, 3}.

Let R = {(1, 2), (2, 3), (1, 3)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R, but (2, 1) $\notin$ R. Not symmetric.
• Transitivity:
  • (1, 2) $\in$ R and (2, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
  • Are there other pairs (a, b), (b, c) in R? No.

  The transitivity condition holds. Transitive.

This example satisfies the conditions.

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(iii) Reflexive and symmetric but not transitive.

Let A = {1, 2, 3}.

Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}.

• Reflexivity: (1, 1), (2, 2), (3, 3) are in R. Reflexive.
• Symmetry:
  • (1, 2) $\in$ R and (2, 1) $\in$ R.
  • (2, 3) $\in$ R and (3, 2) $\in$ R.

  All non-diagonal pairs have their symmetric counterparts in R. Symmetric.

• Transitivity:
  • Consider (1, 2) $\in$ R and (2, 3) $\in$ R. For transitivity, (1, 3) must be in R. (1, 3) $\notin$ R.

  Not transitive.

This example satisfies the conditions.

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(iv) Reflexive and transitive but not symmetric.

Let A be the set of real numbers R.

Let R = {(a, b) : a $\le$ b}.

• Reflexivity: For any real number a, a $\le$ a is true. Reflexive.
• Symmetry: Consider (1, 2) where $1 \le 2$. (1, 2) $\in$ R. But (2, 1) means $2 \le 1$, which is false. (2, 1) $\notin$ R. Not symmetric.
• Transitivity: Assume a $\le$ b and b $\le$ c. By the properties of real numbers, this implies a $\le$ c. Transitive.

This example satisfies the conditions. This is the standard 'less than or equal to' relation.

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(v) Symmetric and transitive but not reflexive.

Let A = {1, 2, 3}.

Let R = {(1, 1), (2, 2), (1, 2), (2, 1)}.

• Reflexivity: (3, 3) is not in R. Not reflexive.
• Symmetry: (1, 2) $\in$ R and (2, 1) $\in$ R. (2, 1) $\in$ R and (1, 2) $\in$ R. The reflexive pairs are symmetric. Symmetric.
• Transitivity:
  • (1, 2) $\in$ R and (2, 1) $\in$ R $\implies$ (1, 1) $\in$ R. Yes.
  • (2, 1) $\in$ R and (1, 2) $\in$ R $\implies$ (2, 2) $\in$ R. Yes.
  • (1, 1) $\in$ R and (1, 2) $\in$ R $\implies$ (1, 2) $\in$ R. Yes.
  • (2, 2) $\in$ R and (2, 1) $\in$ R $\implies$ (2, 1) $\in$ R. Yes.
  • ... check all combinations. The relation is transitive.

  Transitive.

Alternatively, consider the relation "is parallel to" on the set of lines in a plane, *excluding* the case where a line is parallel to itself (which would be the definition of a reflexive relation). A clearer example might be the empty relation on a non-empty set. Let A = {1, 2} and R = $\emptyset$.

• Reflexivity: (1, 1) $\notin$ R. Not reflexive.
• Symmetry: The condition "if (a, b) $\in$ R then (b, a) $\in$ R" is vacuously true because there are no pairs in R. Symmetric.
• Transitivity: The condition "if (a, b) $\in$ R and (b, c) $\in$ R then (a, c) $\in$ R" is vacuously true because there are no pairs (a, b), (b, c) in R. Transitive.

Using the empty relation on a non-empty set provides a straightforward example for (v).

Conclusion for (v): An example is the empty relation on a non-empty set, or the relation R = {(1, 1), (2, 2), (1, 2), (2, 1)} on the set {1, 2, 3}.

Given:

A is the set of all points in a plane.

R is a relation in A given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, where P, Q $\in$ A.

Let O be the origin (0, 0).

The distance of a point P(x, y) from the origin is given by $d(O, P) = \sqrt{x^2 + y^2}$.

The condition for (P, Q) $\in$ R is $d(O, P) = d(O, Q)$.

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To Show:

R is an equivalence relation.

The set of all points related to a point P $\neq$ (0, 0) is the circle passing through P with the origin as the centre.

---

Proof that R is an equivalence relation:

We need to check for reflexivity, symmetry, and transitivity.

---

1. Reflexivity:

A relation R is reflexive if (P, P) $\in$ R for every point P $\in$ A.

The condition for (P, P) $\in$ R is that the distance of P from the origin is the same as the distance of P from the origin.

This is always true, $d(O, P) = d(O, P)$.

Thus, (P, P) $\in$ R for all P $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (P, Q) $\in$ R implies (Q, P) $\in$ R for all P, Q $\in$ A.

Assume (P, Q) $\in$ R for some points P, Q $\in$ A. This means $d(O, P) = d(O, Q)$.

For R to be symmetric, (Q, P) must also be in R, which means $d(O, Q) = d(O, P)$.

If $d(O, P) = d(O, Q)$, then by the property of equality, $d(O, Q) = d(O, P)$ is also true.

Thus, (Q, P) $\in$ R.

So, (P, Q) $\in$ R implies (Q, P) $\in$ R for all P, Q $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (P, Q) $\in$ R and (Q, S) $\in$ R implies (P, S) $\in$ R for all P, Q, S $\in$ A.

Assume (P, Q) $\in$ R and (Q, S) $\in$ R.

(P, Q) $\in$ R means $d(O, P) = d(O, Q)$.

(Q, S) $\in$ R means $d(O, Q) = d(O, S)$.

If $d(O, P) = d(O, Q)$ and $d(O, Q) = d(O, S)$, then by the transitivity of equality, $d(O, P) = d(O, S)$.

This means that the distance of P from the origin is the same as the distance of S from the origin.

Thus, (P, S) $\in$ R.

So, (P, Q) $\in$ R and (Q, S) $\in$ R implies (P, S) $\in$ R for all P, Q, S $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all points related to a point P $\neq$ (0, 0):

Let P$_0$ be a fixed point in the plane, P$_0 \neq$ (0, 0).

We want to find the set of all points Q $\in$ A such that (P$_0$, Q) $\in$ R.

The condition for (P$_0$, Q) $\in$ R is that the distance of P$_0$ from the origin is the same as the distance of Q from the origin.

Let $r = d(O, P_0)$. Since P$_0 \neq$ (0, 0), $r$ is a positive real number ($r > 0$).

The set of points Q that satisfy $d(O, Q) = r$ are all the points that are at a fixed distance $r$ from the origin O.

By definition, the set of all points in a plane that are at a fixed distance from a fixed point (the origin) is a circle.

The fixed point is the origin (centre of the circle).

The fixed distance is $r = d(O, P_0)$ (the radius of the circle).

Since $d(O, P_0) = r$, the point P$_0$ lies on this circle.

Thus, the set of all points related to a point P$_0 \neq$ (0, 0) is the circle passing through P$_0$ with the origin as the centre.

Given:

A is the set of all triangles.

R is the relation in A defined as R = {(T$_1$ , T$_2$) : T$_1$ is similar to T$_2$}, where T$_1$, T$_2$ $\in$ A.

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To Show:

R is an equivalence relation.

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Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R is reflexive if (T, T) $\in$ R for every triangle T $\in$ A.

The condition for (T, T) $\in$ R is that T is similar to T.

Every triangle is similar to itself.

Thus, (T, T) $\in$ R for all T $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (T$_1$, T$_2$) $\in$ R implies (T$_2$, T$_1$) $\in$ R for all T$_1$, T$_2$ $\in$ A.

Assume (T$_1$, T$_2$) $\in$ R. This means T$_1$ is similar to T$_2$.

If T$_1 \sim$ T$_2$, then T$_2 \sim$ T$_1$.

This means that (T$_2$, T$_1$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R implies (T$_2$, T$_1$) $\in$ R for all T$_1$, T$_2$ $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R implies (T$_1$, T$_3$) $\in$ R for all T$_1$, T$_2$, T$_3$ $\in$ A.

Assume (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R.

(T$_1$, T$_2$) $\in$ R means T$_1$ is similar to T$_2$ (T$_1 \sim$ T$_2$).

(T$_2$, T$_3$) $\in$ R means T$_2$ is similar to T$_3$ (T$_2 \sim$ T$_3$).

If T$_1 \sim$ T$_2$ and T$_2 \sim$ T$_3$, then T$_1 \sim$ T$_3$ (transitivity of similarity).

This means that (T$_1$, T$_3$) $\in$ R.

Thus, (T$_1$, T$_2$) $\in$ R and (T$_2$, T$_3$) $\in$ R implies (T$_1$, T$_3$) $\in$ R for all T$_1$, T$_2$, T$_3$ $\in$ A.

R is transitive.

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Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Related Triangles:

We are given three right-angled triangles:

• T$_1$ with sides 3, 4, 5.
• T$_2$ with sides 5, 12, 13.
• T$_3$ with sides 6, 8, 10.

Two triangles are similar if their corresponding sides are proportional or their corresponding angles are equal. Since these are right-angled triangles, we can check the ratios of their sides.

Consider T$_1$ (sides 3, 4, 5) and T$_2$ (sides 5, 12, 13).

Check the ratios of corresponding sides (shortest to shortest, middle to middle, longest to longest):

The ratios are not equal ($\frac{3}{5} \neq \frac{1}{3} \neq \frac{5}{13}$). So, T$_1$ is not similar to T$_2$. (T$_1$, T$_2$) $\notin$ R.

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Consider T$_1$ (sides 3, 4, 5) and T$_3$ (sides 6, 8, 10).

Check the ratios of corresponding sides:

The ratios of corresponding sides are equal ($\frac{1}{2}$). So, T$_1$ is similar to T$_3$. (T$_1$, T$_3$) $\in$ R.

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Consider T$_2$ (sides 5, 12, 13) and T$_3$ (sides 6, 8, 10).

Check the ratios of corresponding sides:

The ratios are not equal ($\frac{5}{6} \neq \frac{3}{2} \neq \frac{13}{10}$). So, T$_2$ is not similar to T$_3$. (T$_2$, T$_3$) $\notin$ R.

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Based on the similarity relation R:

• T$_1$ is related to T$_1$ (reflexive).
• T$_2$ is related to T$_2$ (reflexive).
• T$_3$ is related to T$_3$ (reflexive).
• T$_1$ is related to T$_3$ (because T$_1 \sim$ T$_3$).
• T$_3$ is related to T$_1$ (because T$_3 \sim$ T$_1$, due to symmetry).

The triangles that are related among T$_1$, T$_2$, and T$_3$ are T$_1$ and T$_3$.

Given:

A is the set of all polygons.

R is a relation in A defined as R = {(P$_1$ , P$_2$) : P$_1$ and P$_2$ have same number of sides}, where P$_1$, P$_2$ $\in$ A.

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To Show:

R is an equivalence relation.

The set of all elements in A related to the right angle triangle T with sides 3, 4 and 5.

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Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

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1. Reflexivity:

A relation R is reflexive if (P, P) $\in$ R for every polygon P $\in$ A.

The condition for (P, P) $\in$ R is that P and P have the same number of sides.

Any polygon has the same number of sides as itself.

Thus, (P, P) $\in$ R for all P $\in$ A.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (P$_1$, P$_2$) $\in$ R implies (P$_2$, P$_1$) $\in$ R for all P$_1$, P$_2$ $\in$ A.

Assume (P$_1$, P$_2$) $\in$ R for some polygons P$_1$, P$_2$ $\in$ A. This means P$_1$ and P$_2$ have the same number of sides.

Let the number of sides of P$_1$ be $n_1$ and the number of sides of P$_2$ be $n_2$.

The condition is $n_1 = n_2$.

If $n_1 = n_2$, then it is also true that $n_2 = n_1$.

This means that P$_2$ and P$_1$ have the same number of sides.

Thus, (P$_2$, P$_1$) $\in$ R.

So, (P$_1$, P$_2$) $\in$ R implies (P$_2$, P$_1$) $\in$ R for all P$_1$, P$_2$ $\in$ A.

R is symmetric.

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3. Transitivity:

A relation R is transitive if (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R implies (P$_1$, P$_3$) $\in$ R for all P$_1$, P$_2$, P$_3$ $\in$ A.

Assume (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R.

(P$_1$, P$_2$) $\in$ R means P$_1$ and P$_2$ have the same number of sides ($n_1 = n_2$).

(P$_2$, P$_3$) $\in$ R means P$_2$ and P$_3$ have the same number of sides ($n_2 = n_3$).

If $n_1 = n_2$ and $n_2 = n_3$, then by the transitivity of equality, $n_1 = n_3$.

This means that P$_1$ and P$_3$ have the same number of sides.

Thus, (P$_1$, P$_3$) $\in$ R.

So, (P$_1$, P$_2$) $\in$ R and (P$_2$, P$_3$) $\in$ R implies (P$_1$, P$_3$) $\in$ R for all P$_1$, P$_2$, P$_3$ $\in$ A.

R is transitive.

---

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

---

Set of elements related to the given triangle T:

The given triangle T has sides 3, 4, and 5.

A triangle is a polygon with 3 sides.

The number of sides of triangle T is 3.

We want to find the set of all polygons P $\in$ A such that (T, P) $\in$ R.

The condition for (T, P) $\in$ R is that the number of sides of polygon T is the same as the number of sides of polygon P.

Number of sides of T = 3.

So, we are looking for all polygons P in A that have 3 sides.

The set of all polygons with 3 sides is the set of all triangles.

Therefore, the set of all elements in A related to the right angle triangle T is the set of all triangles.

Given:

L is the set of all lines in the XY plane.

R is the relation in L defined as R = {(L$_1$ , L$_2$) : L$_1$ is parallel to L$_2$}, where L$_1$, L$_2$ $\in$ L.

---

To Show:

R is an equivalence relation.

The set of all lines related to the line y = 2x + 4.

---

Proof that R is an equivalence relation:

For R to be an equivalence relation, it must satisfy three properties:

1. Reflexivity

2. Symmetry

3. Transitivity

(Note: For lines in a plane, we usually consider a line to be parallel to itself.)

---

1. Reflexivity:

A relation R is reflexive if (L$_1$, L$_1$) $\in$ R for every line L$_1$ $\in$ L.

The condition for (L$_1$, L$_1$) $\in$ R is that L$_1$ is parallel to L$_1$.

Every line is parallel to itself.

Thus, (L$_1$, L$_1$) $\in$ R for all L$_1$ $\in$ L.

R is reflexive.

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2. Symmetry:

A relation R is symmetric if (L$_1$, L$_2$) $\in$ R implies (L$_2$, L$_1$) $\in$ R for all L$_1$, L$_2$ $\in$ L.

Assume (L$_1$, L$_2$) $\in$ R for some lines L$_1$, L$_2$ $\in$ L. This means L$_1$ is parallel to L$_2$.

If line L$_1$ is parallel to line L$_2$, then line L$_2$ is also parallel to line L$_1$.

Thus, (L$_2$, L$_1$) $\in$ R.

So, (L$_1$, L$_2$) $\in$ R implies (L$_2$, L$_1$) $\in$ R for all L$_1$, L$_2$ $\in$ L.

R is symmetric.

---

3. Transitivity:

A relation R is transitive if (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R implies (L$_1$, L$_3$) $\in$ R for all L$_1$, L$_2$, L$_3$ $\in$ L.

Assume (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R.

(L$_1$, L$_2$) $\in$ R means L$_1$ is parallel to L$_2$.

(L$_2$, L$_3$) $\in$ R means L$_2$ is parallel to L$_3$.

In a plane, if line L$_1$ is parallel to line L$_2$, and line L$_2$ is parallel to line L$_3$, then line L$_1$ is parallel to line L$_3$.

Thus, (L$_1$, L$_3$) $\in$ R.

So, (L$_1$, L$_2$) $\in$ R and (L$_2$, L$_3$) $\in$ R implies (L$_1$, L$_3$) $\in$ R for all L$_1$, L$_2$, L$_3$ $\in$ L.

R is transitive.

---

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

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Set of all lines related to the line y = 2x + 4:

Let L$_0$ be the line given by the equation y = 2x + 4.

We want to find the set of all lines L $\in$ L such that (L$_0$, L) $\in$ R.

The condition for (L$_0$, L) $\in$ R is that L$_0$ is parallel to L.

Two distinct lines in the XY plane are parallel if and only if they have the same slope.

The equation of line L$_0$ is y = 2x + 4. This is in the form y = mx + c, where m is the slope and c is the y-intercept.

The slope of line L$_0$ is $m_0 = 2$.

A line L is parallel to L$_0$ if and only if it has the same slope as L$_0$.

So, any line L parallel to L$_0$ must have a slope $m = 2$.

The equation of a line with slope 2 can be written in the form y = 2x + c, where c is any real number (the y-intercept).

The set of all lines related to the line y = 2x + 4 is the set of all lines in the XY plane that are parallel to y = 2x + 4.

This set is {L : L has equation y = 2x + c, where c $\in$ R}.

This set represents a family of parallel lines, all with a slope of 2.

Given:

Set A = {1, 2, 3, 4}

Relation R in A is given by R = {(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)}.

---

To Check:

Whether R is reflexive, symmetric, or transitive.

---

1. Reflexivity:

A relation R on a set A is reflexive if (a, a) $\in$ R for every a $\in$ A.

The set A is {1, 2, 3, 4}. We need to check if (1, 1) $\in$ R, (2, 2) $\in$ R, (3, 3) $\in$ R, and (4, 4) $\in$ R.

From the definition of R, we have:

• (1, 1) $\in$ R
• (2, 2) $\in$ R
• (3, 3) $\in$ R
• (4, 4) $\in$ R

Since (a, a) $\in$ R for every element a in the set {1, 2, 3, 4}, R is reflexive.

---

2. Symmetry:

A relation R on a set A is symmetric if (a, b) $\in$ R implies (b, a) $\in$ R for all a, b $\in$ A.

We need to check for every ordered pair (a, b) in R whether the reverse pair (b, a) is also in R.

• Consider the pair (1, 2) $\in$ R. Is (2, 1) $\in$ R? Looking at the definition of R, (2, 1) is not in R.

Since there exists a pair (1, 2) $\in$ R such that (2, 1) $\notin$ R, R is not symmetric.

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3. Transitivity:

A relation R on a set A is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R for all a, b, c $\in$ A.

We check for pairs (a, b) and (b, c) in R:

• (1, 2) $\in$ R and (2, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (1, 3) $\in$ R and (3, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
• (3, 2) $\in$ R and (2, 2) $\in$ R. Check (3, 2) $\in$ R. Yes.
• (1, 3) $\in$ R and (3, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (2, 2) $\in$ R and (2, 2) $\in$ R. Check (2, 2) $\in$ R. Yes.
• (3, 3) $\in$ R and (3, 3) $\in$ R. Check (3, 3) $\in$ R. Yes.
• (4, 4) $\in$ R and (4, 4) $\in$ R. Check (4, 4) $\in$ R. Yes.
• (1, 1) $\in$ R and (1, 2) $\in$ R. Check (1, 2) $\in$ R. Yes.
• (1, 1) $\in$ R and (1, 3) $\in$ R. Check (1, 3) $\in$ R. Yes.
• (3, 3) $\in$ R and (3, 2) $\in$ R. Check (3, 2) $\in$ R. Yes.

All combinations of pairs (a, b) and (b, c) in R result in the pair (a, c) also being in R.

Therefore, R is transitive.

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Conclusion:

The relation R is reflexive and transitive but not symmetric.

Comparing with the given options:

(A) R is reflexive and symmetric but not transitive. (False - not symmetric)

(B) R is reflexive and transitive but not symmetric. (True - reflexive, transitive, not symmetric)

(C) R is symmetric and transitive but not reflexive. (False - not symmetric, is reflexive)

(D) R is an equivalence relation. (False - not symmetric)

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The correct answer is (B) R is reflexive and transitive but not symmetric.

Given:

Set N = {1, 2, 3, ...} (natural numbers).

Relation R in N is given by R = {(a, b) : a = b – 2, b > 6}, where a, b $\in$ N.

---

To Find:

Which of the given ordered pairs belongs to R.

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A pair (a, b) belongs to R if and only if two conditions are met simultaneously:

1. $a = b – 2$

2. $b > 6$

Also, both a and b must be natural numbers (positive integers).

---

Let's check each option:

(A) (2, 4) ∈ R

Here, a = 2 and b = 4.

Check condition 1: $a = b - 2 \implies 2 = 4 - 2 \implies 2 = 2$. (True)

Check condition 2: $b > 6 \implies 4 > 6$. (False)

Since condition 2 is false, (2, 4) $\notin$ R.

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(B) (3, 8) ∈ R

Here, a = 3 and b = 8.

Check condition 1: $a = b - 2 \implies 3 = 8 - 2 \implies 3 = 6$. (False)

Check condition 2: $b > 6 \implies 8 > 6$. (True)

Since condition 1 is false, (3, 8) $\notin$ R.

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(C) (6, 8) ∈ R

Here, a = 6 and b = 8.

Check condition 1: $a = b - 2 \implies 6 = 8 - 2 \implies 6 = 6$. (True)

Check condition 2: $b > 6 \implies 8 > 6$. (True)

Since both conditions are true and 6 and 8 are natural numbers, (6, 8) $\in$ R.

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(D) (8, 7) ∈ R

Here, a = 8 and b = 7.

Check condition 1: $a = b - 2 \implies 8 = 7 - 2 \implies 8 = 5$. (False)

Check condition 2: $b > 6 \implies 7 > 6$. (True)

Since condition 1 is false, (8, 7) $\notin$ R.

---

Only the pair (6, 8) satisfies both conditions for being in R.

The correct answer is (C) (6, 8) ∈ R.

Given:

A is the set of all 50 students of Class X in a school.

N is the set of natural numbers (N = {1, 2, 3, ...}).

f : A $\rightarrow$ N is a function defined by f(x) = roll number of the student x.

---

To Show:

f is one-one but not onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1$ and $x_2$ be two students in set A such that $f(x_1) = f(x_2)$.

According to the definition of the function f, $f(x_1)$ is the roll number of student $x_1$, and $f(x_2)$ is the roll number of student $x_2$.

So, $f(x_1) = f(x_2)$ means that the roll number of student $x_1$ is equal to the roll number of student $x_2$.

In any school system, each student is assigned a unique roll number. It is not possible for two different students to have the same roll number.

Therefore, if the roll number of student $x_1$ is equal to the roll number of student $x_2$, then $x_1$ and $x_2$ must be the same student.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The set A has 50 students. Let the roll numbers of these students be $r_1, r_2, \dots, r_{50}$. Since the roll numbers are unique, the set of these 50 roll numbers forms the range of the function f.

The range of f is the set $\{f(x) : x \in A\} = \{\text{roll number of each student in Class X}\}$.

The codomain of f is the set of natural numbers N = {1, 2, 3, ...}.

The set of natural numbers is an infinite set.

The range of the function f is a set containing 50 distinct roll numbers, which is a finite set with 50 elements.

Since the range of f (a set with 50 elements) is a proper subset of the codomain N (an infinite set), the range is not equal to the codomain.

There are many natural numbers in the codomain N that are not roll numbers of any student in this specific Class X of 50 students. For example, any natural number that is not among the 50 assigned roll numbers will not have a pre-image in A under the function f.

Therefore, for some elements $y \in N$ (the codomain), there is no $x \in A$ such that $f(x) = y$.

Thus, f is not onto.

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In conclusion, the function f is one-one but not onto.

Given:

Function f : N $\rightarrow$ N defined by f(x) = 2x, where N is the set of natural numbers (N = {1, 2, 3, ...}).

Domain = N

Codomain = N

---

To Show:

f is one-one but not onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

According to the definition of f, $f(x_1) = 2x_1$ and $f(x_2) = 2x_2$.

So, $f(x_1) = f(x_2)$ implies $2x_1 = 2x_2$.

Dividing both sides by 2 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is N = {1, 2, 3, 4, 5, 6, ...}.

The function is f(x) = 2x, where $x \in N$. The values of f(x) are obtained by multiplying elements of N by 2.

The range of f is the set $\{f(x) : x \in N\} = \{2x : x \in N\} = \{2(1), 2(2), 2(3), ...\} = \{2, 4, 6, 8, ...\}$.

The range of f is the set of all positive even integers.

The codomain of f is the set of all natural numbers N, which includes both even and odd positive integers.

The range of f ({2, 4, 6, ...}) is a proper subset of the codomain N ({1, 2, 3, 4, 5, 6, ...}).

For example, consider the element 1 in the codomain N. We need to check if there exists any $x \in N$ such that $f(x) = 1$.

The value $x = \frac{1}{2}$ is not a natural number ($x \notin N$).

Therefore, there is no element in the domain N that maps to the element 1 in the codomain N.

Similarly, no odd number in the codomain N has a pre-image in the domain N under this function.

Since the range of f is not equal to the codomain N, f is not onto.

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In conclusion, the function f : N $\rightarrow$ N defined by f(x) = 2x is one-one but not onto.

Given:

Function f : R $\rightarrow$ R defined by f(x) = 2x, where R is the set of real numbers.

Domain = R

Codomain = R

---

To Prove:

f is one-one and onto.

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Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any $x_1, x_2 \in A$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in R$ (the domain) such that $f(x_1) = f(x_2)$.

According to the definition of f, $f(x_1) = 2x_1$ and $f(x_2) = 2x_2$.

So, $f(x_1) = f(x_2)$ implies $2x_1 = 2x_2$.

Since 2 is a non-zero real number, we can divide both sides by 2:

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

This satisfies the condition for a one-one function.

Therefore, f is one-one.

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Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, for every element in the codomain, there must be a pre-image in the domain.

The codomain of f is R (the set of real numbers).

Let $y$ be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

According to the definition of f, $f(x) = 2x$.

So, we set $f(x) = y$ and solve for x:

Since y is a real number, $\frac{y}{2}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{y}{2}$ in the domain such that $f(x) = f\left(\frac{y}{2}\right) = 2 \times \left(\frac{y}{2}\right) = y$.

This means every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

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Since the function f is both one-one and onto, it is a bijective function.

Thus, the function f : R $\rightarrow$ R defined by f(x) = 2x is one-one and onto.

Given:

Function f : N $\rightarrow$ N defined by:

• f (1) = 1
• f (2) = 1
• f (x) = x – 1, for every x > 2

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

---

To Show:

f is onto but not one-one.

---

Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Consider the elements 1 and 2 in the domain N.

According to the definition of the function f:

We have $f(1) = f(2) = 1$.

However, $1 \neq 2$.

Since there exist distinct elements in the domain (1 and 2) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not one-one.

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Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is N = {1, 2, 3, 4, 5, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find if there exists a natural number $x$ in the domain N such that $f(x) = y$.

Consider the possible values of y in the codomain:

• Case 1: y = 1

We know that f(1) = 1 and f(2) = 1 from the definition of f.

Since $1 \in N$ and $2 \in N$, there exist pre-images (1 and 2) in the domain N for the element 1 in the codomain N.

• Case 2: y > 1 (i.e., y $\ge$ 2, since y is a natural number)

We are looking for an $x \in N$ such that $f(x) = y$.

Consider the part of the definition where f(x) = x – 1 for x > 2.

Set $f(x) = y$, so $x – 1 = y$.

Solving for x, we get $x = y + 1$.

Since y is a natural number ($y \in N$), $y \ge 1$. For this case, we are considering $y > 1$, so $y \ge 2$.

If $y \in N$ and $y \ge 2$, then $y+1$ is also a natural number, and $y+1 \ge 3$.

So, $x = y+1$ is a natural number in the domain N, and $x = y+1 > 2$.

For this value of x, $f(x) = f(y+1) = (y+1) – 1 = y$.

Thus, for every natural number $y \ge 2$ in the codomain, there exists a pre-image $x = y+1$ in the domain.

Combining Case 1 and Case 2, we see that for every element y in the codomain N, there exists at least one element x in the domain N such that f(x) = y.

Therefore, f is onto.

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In conclusion, the function f : N $\rightarrow$ N defined by f (1) = f(2) = 1 and f(x) = x – 1, for every x > 2, is onto but not one-one.

Given:

Function f : R $\rightarrow$ R defined by f(x) = x$^2$, where R is the set of real numbers.

Domain = R

Codomain = R

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To Show:

f is neither one-one nor onto.

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Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

To show that f is not one-one, we need to find two distinct elements in the domain R that map to the same element in the codomain R.

Consider the elements 1 and -1 in the domain R.

Calculate their images under f:

We have $f(1) = f(-1) = 1$.

However, $1 \neq -1$.

Since there exist distinct elements in the domain (1 and -1) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range of the function must be equal to the codomain.

The codomain of f is R (the set of real numbers).

The function is f(x) = x$^2$. The square of any real number is always non-negative.

The range of f is the set $\{f(x) : x \in R\} = \{x^2 : x \in R\} = [0, \infty)$.

The range of f is the set of all non-negative real numbers.

The codomain of f is the set of all real numbers R, which includes negative real numbers.

The range of f ($[0, \infty)$) is a proper subset of the codomain R.

For example, consider the element -1 in the codomain R.

We need to check if there exists any $x \in R$ such that $f(x) = -1$.

The equation $x^2 = -1$ has no solution in the set of real numbers R, because the square of a real number cannot be negative.

Therefore, there is no element in the domain R that maps to the element -1 in the codomain R.

This means the range of f is not equal to the codomain R.

Thus, f is not onto.

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In conclusion, the function f : R $\rightarrow$ R defined by f(x) = x$^2$ is neither one-one nor onto.

Given:

Function f : N $\rightarrow$ N defined by:

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

---

To Show:

f is both one-one and onto.

---

Proof that f is one-one:

A function f : A $\rightarrow$ B is one-one (or injective) if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$. Let the common value be y.

We need to consider the parity of $x_1$ and $x_2$. There are four possibilities:

• Case 1: $x_1$ is odd and $x_2$ is odd.

Then $f(x_1) = x_1 + 1$ and $f(x_2) = x_2 + 1$.

$f(x_1) = f(x_2) \implies x_1 + 1 = x_2 + 1$.

Subtracting 1 from both sides, we get $x_1 = x_2$.

• Case 2: $x_1$ is even and $x_2$ is even.

Then $f(x_1) = x_1 - 1$ and $f(x_2) = x_2 - 1$.

$f(x_1) = f(x_2) \implies x_1 - 1 = x_2 - 1$.

Adding 1 to both sides, we get $x_1 = x_2$.

• Case 3: $x_1$ is odd and $x_2$ is even.

Then $f(x_1) = x_1 + 1$ and $f(x_2) = x_2 - 1$.

$f(x_1) = y$. If $x_1$ is odd, $x_1+1$ is even. So y must be even.

$f(x_2) = y$. If $x_2$ is even, $x_2-1$ is odd (note that $x_2 \ge 2$ since $x_2$ is even and in N). So y must be odd.

We have a contradiction: y cannot be both even and odd.

• Case 4: $x_1$ is even and $x_2$ is odd.

This is symmetric to Case 3. If $f(x_1) = f(x_2)$, then y must be both odd and even, which is impossible.

The only possibilities for $f(x_1) = f(x_2)$ are when $x_1$ and $x_2$ have the same parity (Cases 1 and 2), and in both these cases, we found that $x_1 = x_2$.

Thus, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Therefore, f is one-one.

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Proof that f is onto:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain of f is N = {1, 2, 3, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find an element $x$ in the domain N such that $f(x) = y$.

Consider two cases for y:

• Case 1: y is odd.

We are looking for an $x \in N$ such that $f(x) = y$.

If we assume x is even, then $f(x) = x - 1$. We set $x - 1 = y$, so $x = y + 1$.

Since y is an odd natural number, y $\ge$ 1. Then $y+1$ is an even natural number, and $y+1 \ge 2$.

So, if y is odd, we can choose $x = y+1$. This x is an even natural number ($x \in N$ and x is even). When we apply f to this x, $f(x) = f(y+1) = (y+1) - 1 = y$.

Thus, for every odd number y in the codomain, there exists an even number $x = y+1$ in the domain that maps to y.

• Case 2: y is even.

We are looking for an $x \in N$ such that $f(x) = y$.

If we assume x is odd, then $f(x) = x + 1$. We set $x + 1 = y$, so $x = y - 1$.

Since y is an even natural number, y $\ge$ 2 (as y $\in$ N and is even). Then $y-1$ is an odd natural number, and $y-1 \ge 1$.

So, if y is even, we can choose $x = y-1$. This x is an odd natural number ($x \in N$ and x is odd). When we apply f to this x, $f(x) = f(y-1) = (y-1) + 1 = y$.

Thus, for every even number y in the codomain, there exists an odd number $x = y-1$ in the domain that maps to y.

Since every natural number y in the codomain is either odd or even, and in both cases we found a pre-image x in the domain N, every element in the codomain N has a pre-image in the domain N under the function f.

Therefore, f is onto.

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Since the function f is both one-one and onto, it is a bijective function.

Thus, the function f : N $\rightarrow$ N defined by the piecewise rule is both one-one and onto.

Given:

Set A = {1, 2, 3}

Set B = {1, 2, 3}

Function f : A $\rightarrow$ B is onto.

The number of elements in set A is $n(A) = 3$.

The number of elements in set B is $n(B) = 3$.

---

To Show:

f is always one-one.

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Proof:

A function f : A $\rightarrow$ B is onto (or surjective) if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

Since the codomain B = {1, 2, 3} has 3 distinct elements, and the function f is onto, each of these 3 elements in B must have at least one pre-image in the domain A = {1, 2, 3}.

Let the elements of the domain be $x_1=1, x_2=2, x_3=3$.

Let the elements of the codomain be $y_1=1, y_2=2, y_3=3$.

Since f is onto, the range of f is equal to the codomain B.

Range of f = $\{f(1), f(2), f(3)\} = \{1, 2, 3\}$.

This means the set of images $\{f(1), f(2), f(3)\}$ contains exactly the 3 elements {1, 2, 3}.

For the set $\{f(1), f(2), f(3)\}$ to contain 3 distinct elements, the elements $f(1), f(2), f(3)$ must themselves be distinct.

If $f(1), f(2), f(3)$ were not distinct, at least two of the images would be the same. For example, if $f(1) = f(2)$, then the set of images would be $\{f(1), f(3)\}$, which contains at most 2 distinct elements, contradicting the fact that the range is {1, 2, 3}.

So, $f(1), f(2), f(3)$ must be distinct values from the set {1, 2, 3}.

The distinct images are $f(1), f(2), f(3)$.

Now consider two distinct elements $x_i, x_j$ from the domain A (where $i \neq j$). For example, let $x_1 = 1$ and $x_2 = 2$. Since their images $f(1)$ and $f(2)$ must be distinct as shown above, $f(1) \neq f(2)$.

In general, for any two distinct elements $x_i, x_j \in \{1, 2, 3\}$, their images $f(x_i)$ and $f(x_j)$ must be distinct elements in the range {1, 2, 3}.

Therefore, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. This is the definition of a one-one function.

Thus, an onto function f : {1, 2, 3} $\rightarrow$ {1, 2, 3} is always one-one.

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Generalization:

This result holds for any function f : A $\rightarrow$ B where A and B are finite sets with the same number of elements, i.e., $|A| = |B|$. In such a case, a function f : A $\rightarrow$ B is onto if and only if it is one-one.

Given:

Set A = {1, 2, 3}

Set B = {1, 2, 3}

Function f : A $\rightarrow$ B is one-one.

The number of elements in set A is $n(A) = 3$.

The number of elements in set B is $n(B) = 3$.

---

To Show:

f must be onto.

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Proof:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$. Equivalently, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

The domain A = {1, 2, 3} has 3 distinct elements.

Since f is one-one, the images of these 3 distinct elements under f must also be distinct elements in the codomain B = {1, 2, 3}.

The images are $f(1), f(2), f(3)$.

Because f is one-one, we know that $f(1) \neq f(2)$, $f(1) \neq f(3)$, and $f(2) \neq f(3)$.

So, the set of images $\{f(1), f(2), f(3)\}$ contains 3 distinct elements.

These 3 distinct images are elements of the codomain B = {1, 2, 3}.

Since the set of images $\{f(1), f(2), f(3)\}$ contains 3 distinct elements and the codomain B also contains exactly 3 elements, the set of images must be equal to the codomain B.

Range of f = $\{f(1), f(2), f(3)\}$

Codomain of f = {1, 2, 3}

Since the 3 distinct images must be taken from the 3 elements of the codomain, the set of images must be {1, 2, 3} in some order.

For example, the images could be {f(1)=1, f(2)=2, f(3)=3}, or {f(1)=3, f(2)=1, f(3)=2}, etc. In any case, the set of images is {1, 2, 3}.

Since the range of f is equal to the codomain B, the function f is onto.

Thus, a one-one function f : {1, 2, 3} $\rightarrow$ {1, 2, 3} must be onto.

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Generalization:

This result also holds for any function f : A $\rightarrow$ B where A and B are finite sets with the same number of elements, i.e., $|A| = |B|$. In such a case, a function f : A $\rightarrow$ B is one-one if and only if it is onto.

Given:

Function $f : R_* \to R_*$ defined by $f(x) = \frac{1}{x}$, where $R_* = R \setminus \{0\}$ (set of all non-zero real numbers).

Domain = $R_*$

Codomain = $R_*$

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To Show:

f is one-one and onto.

Investigate if the result holds when the domain is N (natural numbers) and the codomain is $R_*$.

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Proof that f is one-one (for $f : R_* \to R_*$):

Let $x_1, x_2 \in R_*$ such that $f(x_1) = f(x_2)$.

According to the definition of f, $f(x_1) = \frac{1}{x_1}$ and $f(x_2) = \frac{1}{x_2}$.

So, $f(x_1) = f(x_2) \implies \frac{1}{x_1} = \frac{1}{x_2}$.

Taking the reciprocal of both sides (which is allowed since $x_1, x_2 \in R_*$, so they are non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

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Proof that f is onto (for $f : R_* \to R_*$):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain of f is $R_*$ (the set of all non-zero real numbers).

Let y be any arbitrary element in the codomain $R_*$. Since $y \in R_*$, y is a non-zero real number, i.e., $y \in R$ and $y \neq 0$.

We need to find if there exists an element $x$ in the domain $R_*$ such that $f(x) = y$.

According to the definition of f, $f(x) = \frac{1}{x}$.

So, we set $f(x) = y$ and solve for x:

Since y is a non-zero real number, we can take the reciprocal of both sides:

Now we need to check if this value of x is in the domain $R_*$.

Since y is a non-zero real number ($y \in R_*$), its reciprocal $\frac{1}{y}$ is also a non-zero real number. So, $x = \frac{1}{y} \in R_*$.

Thus, for every element y in the codomain $R_*$, we found an element $x = \frac{1}{y}$ in the domain $R_*$ such that $f(x) = f\left(\frac{1}{y}\right) = \frac{1}{1/y} = y$.

Therefore, f is onto.

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Since f is both one-one and onto, it is a bijective function.

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Investigating the case when the domain is N and codomain is $R_*$:

Now consider the function $g : N \to R_*$ defined by $g(x) = \frac{1}{x}$, where N = {1, 2, 3, ...} and $R_* = R \setminus \{0\}$.

---

Check if g is one-one (for $g : N \to R_*$):

Let $x_1, x_2 \in N$ such that $g(x_1) = g(x_2)$.

According to the definition of g, $g(x_1) = \frac{1}{x_1}$ and $g(x_2) = \frac{1}{x_2}$.

So, $g(x_1) = g(x_2) \implies \frac{1}{x_1} = \frac{1}{x_2}$.

Taking the reciprocal of both sides:

Thus, if $g(x_1) = g(x_2)$, then $x_1 = x_2$.

Therefore, g is one-one.

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Check if g is onto (for $g : N \to R_*$):

The codomain of g is $R_*$ (the set of all non-zero real numbers).

The range of g is the set $\{g(x) : x \in N\} = \{\frac{1}{x} : x \in N\} = \{\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...\} = \{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...\}$.

The range of g is the set of reciprocals of natural numbers.

The codomain of g is the set of all non-zero real numbers $R_*$. This set includes negative numbers, irrational numbers, and many positive rational numbers that are not of the form $\frac{1}{n}$ where $n \in N$.

For example, consider the element 2 in the codomain $R_*$. We need to check if there exists any $x \in N$ such that $g(x) = 2$.

Solving for x, we get $x = \frac{1}{2}$.

The value $x = \frac{1}{2}$ is not a natural number ($x \notin N$).

Therefore, there is no element in the domain N that maps to the element 2 in the codomain $R_*$.

Similarly, any negative real number in the codomain (e.g., -1) does not have a pre-image in the domain N, because $\frac{1}{x}$ for $x \in N$ is always positive.

Since the range of g ($\{1, \frac{1}{2}, \frac{1}{3}, ...\}$) is a proper subset of the codomain $R_*$, g is not onto.

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Conclusion:

The function $f : R_* \to R_*$ defined by $f(x) = \frac{1}{x}$ is both one-one and onto.

If the domain $R_*$ is replaced by N and the codomain remains $R_*$, the resulting function $g : N \to R_*$ defined by $g(x) = \frac{1}{x}$ is one-one but not onto.

So, the result (that the function is both one-one and onto) is not true if the domain $R_*$ is replaced by N with codomain being same as $R_*$.

Question 2 (i):

Function f : N $\rightarrow$ N given by f(x) = x$^2$.

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

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Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$x_1^2 = x_2^2$

Since $x_1, x_2$ are natural numbers, they are positive. Taking the square root of both sides:

$\sqrt{x_1^2} = \sqrt{x_2^2} \implies |x_1| = |x_2|$

As $x_1, x_2 \in N$, $x_1 > 0$ and $x_2 > 0$. So $|x_1| = x_1$ and $|x_2| = x_2$.

$x_1 = x_2$.

Thus, f is injective (one-one).

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Surjectivity (Onto):

The codomain is N = {1, 2, 3, ...}.

The range of f is $\{x^2 : x \in N\} = \{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, 16, ...\}$.

The range is the set of perfect squares that are natural numbers.

The codomain N contains elements that are not perfect squares (e.g., 2, 3, 5, 6, 7, 8, 10, 11, 12, ...).

Consider the element 2 in the codomain N. We need to find an $x \in N$ such that $f(x) = 2$.

$x^2 = 2 \implies x = \sqrt{2}$.

Since $\sqrt{2}$ is not a natural number ($x \notin N$), there is no pre-image for 2 in the domain.

The range ($\{1, 4, 9, ...\}$) is a proper subset of the codomain (N).

Thus, f is not surjective (not onto).

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Question 2 (ii):

Function f : Z $\rightarrow$ Z given by f(x) = x$^2$.

Domain = Z = {..., -2, -1, 0, 1, 2, ...}

Codomain = Z = {..., -2, -1, 0, 1, 2, ...}

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Injectivity (One-one):

Let $x_1, x_2 \in Z$ such that $f(x_1) = f(x_2)$.

$x_1^2 = x_2^2$

This implies $x_1^2 - x_2^2 = 0 \implies (x_1 - x_2)(x_1 + x_2) = 0$.

So, $x_1 - x_2 = 0$ or $x_1 + x_2 = 0$.

$x_1 = x_2$ or $x_1 = -x_2$.

Consider $x_1 = 1$ and $x_2 = -1$. Both are in Z, and $x_1 \neq x_2$.

$f(1) = 1^2 = 1$

$f(-1) = (-1)^2 = 1$

We have $f(1) = f(-1)$ but $1 \neq -1$.

Thus, f is not injective (not one-one).

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Surjectivity (Onto):

The codomain is Z = {..., -2, -1, 0, 1, 2, ...}.

The range of f is $\{x^2 : x \in Z\} = \{..., (-2)^2, (-1)^2, 0^2, 1^2, 2^2, ...\} = \{..., 4, 1, 0, 1, 4, ...\} = \{0, 1, 4, 9, 16, ...\}$.

The range is the set of perfect squares that are integers.

The codomain Z contains many integers that are not perfect squares (e.g., -1, 2, 3, 5, 6, 7, 8, 10, ...).

Consider the element -1 in the codomain Z. We need to find an $x \in Z$ such that $f(x) = -1$.

$x^2 = -1$.

There is no integer x whose square is -1.

The range ($\{0, 1, 4, 9, ...\}$) is a proper subset of the codomain (Z).

Thus, f is not surjective (not onto).

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Question 2 (iii):

Function f : R $\rightarrow$ R given by f(x) = x$^2$.

Domain = R

Codomain = R

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Injectivity (One-one):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$x_1^2 = x_2^2 \implies x_1^2 - x_2^2 = 0 \implies (x_1 - x_2)(x_1 + x_2) = 0$.

So, $x_1 = x_2$ or $x_1 = -x_2$.

As shown in Example 11, consider $x_1 = 1$ and $x_2 = -1$. Both are in R, and $x_1 \neq x_2$, but $f(1) = f(-1) = 1$.

Thus, f is not injective (not one-one).

---

Surjectivity (Onto):

The codomain is R. The range of f is $\{x^2 : x \in R\} = [0, \infty)$.

As shown in Example 11, the range is the set of all non-negative real numbers.

The codomain R includes negative real numbers (e.g., -1), which are not in the range.

Thus, f is not surjective (not onto).

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Question 2 (iv):

Function f : N $\rightarrow$ N given by f(x) = x$^3$.

Domain = N = {1, 2, 3, ...}

Codomain = N = {1, 2, 3, ...}

---

Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$x_1^3 = x_2^3$

Since $x_1, x_2$ are natural numbers, they are positive. Taking the cube root of both sides:

$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3} \implies x_1 = x_2$.

Thus, f is injective (one-one).

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Surjectivity (Onto):

The codomain is N = {1, 2, 3, ...}.

The range of f is $\{x^3 : x \in N\} = \{1^3, 2^3, 3^3, ...\} = \{1, 8, 27, 64, ...\}$.

The range is the set of perfect cubes that are natural numbers.

The codomain N contains elements that are not perfect cubes (e.g., 2, 3, 4, 5, 6, 7, 9, ...).

Consider the element 2 in the codomain N. We need to find an $x \in N$ such that $f(x) = 2$.

$x^3 = 2 \implies x = \sqrt[3]{2}$.

Since $\sqrt[3]{2}$ is not a natural number ($x \notin N$), there is no pre-image for 2 in the domain.

The range ($\{1, 8, 27, ...\}$) is a proper subset of the codomain (N).

Thus, f is not surjective (not onto).

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Question 2 (v):

Function f : Z $\rightarrow$ Z given by f(x) = x$^3$.

Domain = Z

Codomain = Z

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Injectivity (One-one):

Let $x_1, x_2 \in Z$ such that $f(x_1) = f(x_2)$.

$x_1^3 = x_2^3$

Taking the cube root of both sides. For real numbers, the cube root is unique.

$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3} \implies x_1 = x_2$.

Thus, f is injective (one-one).

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Surjectivity (Onto):

The codomain is Z.

The range of f is $\{x^3 : x \in Z\} = \{..., (-2)^3, (-1)^3, 0^3, 1^3, 2^3, ...\} = \{..., -8, -1, 0, 1, 8, ...\}$.

The range is the set of perfect cubes that are integers.

The codomain Z contains many integers that are not perfect cubes (e.g., -2, 2, 3, 4, 5, 6, 7, 9, ...).

Consider the element 2 in the codomain Z. We need to find an $x \in Z$ such that $f(x) = 2$.

$x^3 = 2 \implies x = \sqrt[3]{2}$.

Since $\sqrt[3]{2}$ is not an integer ($x \notin Z$), there is no pre-image for 2 in the domain.

The range ($\{..., -8, -1, 0, 1, 8, ...\}$) is a proper subset of the codomain (Z).

Thus, f is not surjective (not onto).

Given:

Function f : R $\rightarrow$ R given by f(x) = [x], where [x] denotes the greatest integer less than or equal to x.

Domain = R (real numbers)

Codomain = R (real numbers)

---

To Prove:

f is neither one-one nor onto.

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Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

The definition of the greatest integer function is that for any real number x, [x] is the largest integer less than or equal to x.

Consider two distinct real numbers, say $x_1 = 1.2$ and $x_2 = 1.5$. Both are in the domain R, and $x_1 \neq x_2$.

Calculate their images under f:

The greatest integer less than or equal to 1.2 is 1.

The greatest integer less than or equal to 1.5 is 1.

We have $f(1.2) = f(1.5) = 1$, but $1.2 \neq 1.5$.

In general, for any real number k, all real numbers x in the interval $[k, k+1)$ map to the same integer k under the greatest integer function. Since the interval $[k, k+1)$ contains infinitely many distinct real numbers (e.g., 1.1, 1.2, 1.3, ... all map to 1), the function is not one-one.

Therefore, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

The function is f(x) = [x]. The value of [x] is always an integer, regardless of the real number x.

The range of f is the set of all possible values of [x]. Since [x] is always an integer, the range of f is the set of all integers Z.

Range of f = {k : k is an integer} = Z = {..., -2, -1, 0, 1, 2, ...}.

The codomain of f is the set of all real numbers R.

The range of f (Z) is a proper subset of the codomain (R), because R contains real numbers that are not integers (e.g., 0.5, $\sqrt{2}$, $\pi$).

Consider a real number in the codomain that is not an integer, for example, $y = 0.5$. We need to check if there exists any $x \in R$ such that $f(x) = 0.5$.

By the definition of the greatest integer function, [x] must be an integer. Therefore, [x] cannot be equal to 0.5.

There is no real number x whose greatest integer is 0.5.

Thus, the element 0.5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f (Z) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the Greatest Integer Function f : R $\rightarrow$ R, given by f(x) = [x], is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R given by f(x) = |x|, where R is the set of real numbers.

The modulus function is defined as:

Domain = R (real numbers)

Codomain = R (real numbers)

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To Show:

f is neither one-one nor onto.

---

Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

Consider two distinct real numbers, say $x_1 = 2$ and $x_2 = -2$. Both are in the domain R, and $x_1 \neq x_2$.

Calculate their images under f:

Since $2 \ge 0$, $|2| = 2$.

Since $-2 < 0$, $|-2| = -(-2) = 2$.

We have $f(2) = f(-2) = 2$, but $2 \neq -2$.

In general, for any positive real number k, we have $f(k) = |k| = k$ and $f(-k) = |-k| = -(-k) = k$. Since $k \neq -k$ (for $k \neq 0$), distinct inputs (k and -k) map to the same output (k).

Therefore, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

The function is f(x) = |x|. By the definition of the modulus function, the output |x| is always non-negative.

The range of f is the set of all possible values of |x| for $x \in R$. This is the set of all non-negative real numbers.

Range of f = $\{|x| : x \in R\} = [0, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[0, \infty)$) is a proper subset of the codomain (R), because R contains negative real numbers (e.g., -1, -5, -$\pi$).

Consider a negative real number in the codomain, for example, $y = -3$. We need to check if there exists any $x \in R$ such that $f(x) = -3$.

By the definition of the modulus function, $|x|$ cannot be negative for any real number x.

Therefore, there is no real number x such that |x| = -3.

Thus, the element -3 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[0, \infty)$) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the modulus function f : R $\rightarrow$ R, given by f(x) = |x|, is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R, the Signum Function, given by:

Domain = R (real numbers)

Codomain = R (real numbers)

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To Show:

f is neither one-one nor onto.

---

Proof that f is not one-one:

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$. To show it is not one-one, we need to find $x_1 \neq x_2$ such that $f(x_1) = f(x_2)$.

Consider two distinct positive real numbers, say $x_1 = 2$ and $x_2 = 5$. Both are in the domain R, and $x_1 \neq x_2$. Both are greater than 0.

Calculate their images under f:

Since $2 > 0$, $f(2) = 1$.

Since $5 > 0$, $f(5) = 1$.

We have $f(2) = f(5) = 1$, but $2 \neq 5$.

In general, any two distinct positive real numbers will both map to 1. Any two distinct negative real numbers will both map to -1.

For example, consider $x_1 = -3$ and $x_2 = -7$. $x_1 \neq x_2$, but $f(-3) = -1$ and $f(-7) = -1$. So $f(-3) = f(-7)$.

Since there exist distinct elements in the domain that map to the same element in the codomain, f is not one-one.

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Proof that f is not onto:

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$. In other words, the range must equal the codomain.

The codomain of f is R (the set of all real numbers).

According to the definition of the Signum Function, the possible output values (the range of f) are only 1, 0, and -1.

Range of f = {1, 0, -1}.

The codomain of f is the set of all real numbers R.

The range of f ({1, 0, -1}) is a proper subset of the codomain (R), because R contains infinitely many real numbers that are not 1, 0, or -1 (e.g., 2, -5, 0.5, $\sqrt{2}$, $\pi$).

Consider a real number in the codomain that is not 1, 0, or -1, for example, $y = 5$. We need to check if there exists any $x \in R$ such that $f(x) = 5$.

Based on the definition of f(x), the output can only be 1, 0, or -1. It can never be 5.

Therefore, there is no real number x such that f(x) = 5.

Thus, the element 5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ({1, 0, -1}) is not equal to the codomain R.

Therefore, f is not onto.

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In conclusion, the Signum Function f : R $\rightarrow$ R is neither one-one nor onto.

Given:

Set A = {1, 2, 3}

Set B = {4, 5, 6, 7}

Function f : A $\rightarrow$ B is given as a set of ordered pairs: f = {(1, 4), (2, 5), (3, 6)}.

This means:

• f(1) = 4
• f(2) = 5
• f(3) = 6

---

To Show:

f is one-one.

---

Proof:

A function f : A $\rightarrow$ B is one-one (or injective) if for any two distinct elements $x_1, x_2 \in A$, their images under f are also distinct, i.e., if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

The domain of the function is A = {1, 2, 3}. The distinct elements in the domain are 1, 2, and 3.

We examine the images of these distinct elements under f:

• Image of 1 is f(1) = 4.
• Image of 2 is f(2) = 5.
• Image of 3 is f(3) = 6.

Let's compare the images of distinct elements:

• For the distinct elements 1 and 2 from A, their images are f(1) = 4 and f(2) = 5. Since $4 \neq 5$, $f(1) \neq f(2)$.
• For the distinct elements 1 and 3 from A, their images are f(1) = 4 and f(3) = 6. Since $4 \neq 6$, $f(1) \neq f(3)$.
• For the distinct elements 2 and 3 from A, their images are f(2) = 5 and f(3) = 6. Since $5 \neq 6$, $f(2) \neq f(3)$.

In each case where the inputs from the domain are distinct, the corresponding outputs in the codomain are also distinct.

Alternatively, we can check the contrapositive condition: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let's look at the image values in the set of ordered pairs: {4, 5, 6}.

• If $f(x_1) = f(x_2) = 4$, then $x_1$ and $x_2$ must map to 4. From the definition of f, only 1 maps to 4. So $x_1 = 1$ and $x_2 = 1$. Thus $x_1 = x_2$.
• If $f(x_1) = f(x_2) = 5$, then $x_1$ and $x_2$ must map to 5. Only 2 maps to 5. So $x_1 = 2$ and $x_2 = 2$. Thus $x_1 = x_2$.
• If $f(x_1) = f(x_2) = 6$, then $x_1$ and $x_2$ must map to 6. Only 3 maps to 6. So $x_1 = 3$ and $x_2 = 3$. Thus $x_1 = x_2$.

In all cases where the images are equal, the corresponding pre-images are also equal.

Thus, f is one-one.

---

Note: This function is one-one, but it is not onto because the element 7 in the codomain B does not have a pre-image in the domain A.

Question 7 (i):

Function f : R $\rightarrow$ R defined by f(x) = 3 – 4x.

Domain = R

Codomain = R

---

Injectivity (One-one):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$3 - 4x_1 = 3 - 4x_2$

Subtract 3 from both sides:

Divide both sides by -4 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

---

Surjectivity (Onto):

The codomain is R.

Let y be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Subtract 3 from both sides:

Divide by -4:

Since y is a real number, $3-y$ is a real number, and $\frac{3-y}{4}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{3-y}{4}$ in the domain such that $f(x) = f\left(\frac{3-y}{4}\right) = 3 - 4\left(\frac{3-y}{4}\right) = 3 - (3 - y) = 3 - 3 + y = y$.

Every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

---

Conclusion for (i): Since f is both one-one and onto, it is a bijective function.

---

Question 7 (ii):

Function f : R $\rightarrow$ R defined by f(x) = 1 + x$^2$.

Domain = R

Codomain = R

---

Injectivity (One-one):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$1 + x_1^2 = 1 + x_2^2$

Subtract 1 from both sides:

This implies $x_1^2 - x_2^2 = 0 \implies (x_1 - x_2)(x_1 + x_2) = 0$.

So, $x_1 = x_2$ or $x_1 = -x_2$.

Consider $x_1 = 1$ and $x_2 = -1$. Both are in R, and $x_1 \neq x_2$.

$f(1) = 1 + 1^2 = 1 + 1 = 2$.

$f(-1) = 1 + (-1)^2 = 1 + 1 = 2$.

We have $f(1) = f(-1) = 2$, but $1 \neq -1$.

Thus, f is not injective (not one-one).

---

Surjectivity (Onto):

The codomain is R.

The function is f(x) = 1 + x$^2$. Since $x^2 \ge 0$ for any real number x, $1 + x^2 \ge 1 + 0 = 1$.

The range of f is the set of all values $1 + x^2$ for $x \in R$. This is the set of all real numbers greater than or equal to 1.

Range of f = $[1, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[1, \infty)$) is a proper subset of the codomain (R), because R contains real numbers less than 1 (e.g., 0, -5, 0.5).

Consider a real number in the codomain that is less than 1, for example, $y = 0$. We need to check if there exists any $x \in R$ such that $f(x) = 0$.

$f(x) = 0 \implies 1 + x^2 = 0 \implies x^2 = -1$.

There is no real number x whose square is -1.

Thus, the element 0 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[1, \infty)$) is not equal to the codomain R.

Therefore, f is not surjective (not onto).

---

Conclusion for (ii): The function f is neither one-one nor onto.

Given:

Sets A and B.

Function f : A $\times$ B $\rightarrow$ B $\times$ A defined by f(a, b) = (b, a), where (a, b) $\in$ A $\times$ B.

Domain = A $\times$ B (the set of all ordered pairs where the first element is from A and the second is from B).

Codomain = B $\times$ A (the set of all ordered pairs where the first element is from B and the second is from A).

---

To Show:

f is a bijective function (i.e., f is both one-one and onto).

---

Proof that f is one-one (Injectivity):

A function f : X $\rightarrow$ Y is one-one if for any two distinct elements $x_1, x_2 \in X$, their images under f are also distinct, i.e., $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1 = (a_1, b_1)$ and $x_2 = (a_2, b_2)$ be two elements in the domain A $\times$ B, such that $f(x_1) = f(x_2)$.

According to the definition of f:

The condition $f(x_1) = f(x_2)$ means $(b_1, a_1) = (b_2, a_2)$.

Two ordered pairs are equal if and only if their corresponding components are equal.

So, $(b_1, a_1) = (b_2, a_2)$ implies $b_1 = b_2$ and $a_1 = a_2$.

Since $a_1 = a_2$ and $b_1 = b_2$, the ordered pairs $(a_1, b_1)$ and $(a_2, b_2)$ are equal.

Thus, if $f(a_1, b_1) = f(a_2, b_2)$, then $(a_1, b_1) = (a_2, b_2)$.

Therefore, f is one-one.

---

Proof that f is onto (Surjectivity):

A function f : X $\rightarrow$ Y is onto if for every element $y \in Y$, there exists at least one element $x \in X$ such that $f(x) = y$.

The codomain of f is B $\times$ A (the set of all ordered pairs (b, a) where b $\in$ B and a $\in$ A).

Let y be any arbitrary element in the codomain B $\times$ A. By the definition of B $\times$ A, y is an ordered pair of the form (b, a), where b $\in$ B and a $\in$ A.

We need to find if there exists an element $x$ in the domain A $\times$ B such that $f(x) = y$.

Let $x$ be an ordered pair $(u, v)$ in the domain A $\times$ B. This means $u \in A$ and $v \in B$.

According to the definition of f, $f(u, v) = (v, u)$.

We want to find $(u, v) \in A \times B$ such that $f(u, v) = y$, i.e., $(v, u) = (b, a)$.

By the equality of ordered pairs, $(v, u) = (b, a)$ implies $v = b$ and $u = a$.

We need to check if this element $x = (u, v) = (a, b)$ is in the domain A $\times$ B.

Since a $\in$ A and b $\in$ B (from the definition of the codomain element y = (b, a)), the ordered pair (a, b) is indeed an element of A $\times$ B (the domain).

So, for every element $y = (b, a)$ in the codomain B $\times$ A, we found an element $x = (a, b)$ in the domain A $\times$ B such that $f(x) = f(a, b) = (b, a) = y$.

Every element in the codomain B $\times$ A has a pre-image in the domain A $\times$ B.

Therefore, f is onto.

---

Since the function f is both one-one and onto, it is a bijective function.

Given:

Function f : N $\rightarrow$ N defined by:

for all $n \in N$ = {1, 2, 3, ...}.

Domain = N

Codomain = N

---

To State:

Whether the function f is bijective (i.e., both one-one and onto).

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(n_1) = f(n_2)$ implies $n_1 = n_2$ for all $n_1, n_2 \in N$.

Let's evaluate the function for a few values:

• f(1) = $\frac{1+1}{2} = \frac{2}{2} = 1$ (since 1 is odd)
• f(2) = $\frac{2}{2} = 1$ (since 2 is even)
• f(3) = $\frac{3+1}{2} = \frac{4}{2} = 2$ (since 3 is odd)
• f(4) = $\frac{4}{2} = 2$ (since 4 is even)

We observe that $f(1) = 1$ and $f(2) = 1$.

We have $f(1) = f(2) = 1$, but $1 \neq 2$.

Since there exist distinct elements in the domain (1 and 2) that map to the same element in the codomain (1), the function f is not one-one.

Therefore, f is not injective (not one-one).

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $n \in A$ such that $f(n) = y$.

The codomain of f is N = {1, 2, 3, ...}.

Let y be any arbitrary natural number in the codomain N.

We need to find an element $n \in N$ such that $f(n) = y$.

Consider two cases for y:

• Case 1: y is a possible output from the 'n is odd' rule: $f(n) = \frac{n+1}{2}$

If $\frac{n+1}{2} = y$, then $n+1 = 2y$, so $n = 2y - 1$.

If $n = 2y - 1$, we need to check if n is an odd natural number.

Since y is a natural number, $y \ge 1$.

If y = 1, n = $2(1) - 1 = 1$. 1 is an odd natural number. f(1) = $\frac{1+1}{2} = 1$.

If y = 2, n = $2(2) - 1 = 3$. 3 is an odd natural number. f(3) = $\frac{3+1}{2} = 2$.

If y = 3, n = $2(3) - 1 = 5$. 5 is an odd natural number. f(5) = $\frac{5+1}{2} = 3$.

In general, for any natural number y, $2y-1$ is an odd natural number. So for any $y \in N$, we can find an odd $n = 2y-1 \in N$ such that $f(n) = f(2y-1) = \frac{(2y-1)+1}{2} = \frac{2y}{2} = y$.

• Case 2: y is a possible output from the 'n is even' rule: $f(n) = \frac{n}{2}$

If $\frac{n}{2} = y$, then $n = 2y$.

If $n = 2y$, we need to check if n is an even natural number.

Since y is a natural number, $y \ge 1$.

If y = 1, n = $2(1) = 2$. 2 is an even natural number. f(2) = $\frac{2}{2} = 1$.

If y = 2, n = $2(2) = 4$. 4 is an even natural number. f(4) = $\frac{4}{2} = 2$.

If y = 3, n = $2(3) = 6$. 6 is an even natural number. f(6) = $\frac{6}{2} = 3$.

In general, for any natural number y, 2y is an even natural number. So for any $y \in N$, we can find an even $n = 2y \in N$ such that $f(n) = f(2y) = \frac{2y}{2} = y$.

For any natural number y in the codomain, we can find a pre-image in the domain. Specifically, if y is an odd natural number, $n = 2y-1$ is an odd natural number that maps to y. If y is an even natural number, $n = 2y$ is an even natural number that maps to y.

This means every element in the codomain N has a pre-image in the domain N.

Therefore, f is onto (surjective).

---

Conclusion:

The function f is not one-one but is onto.

Since a bijective function must be both one-one and onto, f is not a bijective function.

Given:

Set A = R – {3} (real numbers excluding 3)

Set B = R – {1} (real numbers excluding 1)

Function f : A $\rightarrow$ B defined by $f(x) = \frac{x - 2}{x - 3}$.

Domain = A

Codomain = B

---

To Determine:

Is f one-one and onto?

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in A$ such that $f(x_1) = f(x_2)$. Since $x_1, x_2 \in A$, $x_1 \neq 3$ and $x_2 \neq 3$.

Cross-multiply:

Expand both sides:

Subtract $x_1x_2$ and 6 from both sides:

Add $3x_1$ and $3x_2$ to both sides:

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain of f is B = R – {1}. Let y be any arbitrary element in the codomain B. Since $y \in B$, y is a real number and $y \neq 1$.

We need to find an element $x$ in the domain A = R – {3} such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Multiply both sides by $(x - 3)$ (which is non-zero since $x \neq 3$):

Distribute y on the right side:

Rearrange the terms to group x terms on one side and constant terms on the other:

Factor out x on the left side:

Solve for x. Since $y \in B$, we know $y \neq 1$, so $1 - y \neq 0$. We can divide by $(1 - y)$.

Now we need to check if this value of x is in the domain A = R – {3}. That is, we need to check if $x$ can ever be equal to 3.

Suppose $x = 3$. Then $\frac{2 - 3y}{1 - y} = 3$.

$2 - 3y = 3(1 - y)$

$2 - 3y = 3 - 3y$

Add 3y to both sides:

This is a contradiction. It means that our assumption ($x=3$) is false. The value of x we found, $x = \frac{2 - 3y}{1 - y}$, can never be equal to 3 for any value of y.

Since y is a real number (from B), $2 - 3y$ and $1 - y$ are real numbers, and since $1 - y \neq 0$, the value of x is a real number. Also, we have shown that this x is never equal to 3.

So, for every element y in the codomain B = R – {1}, we found an element $x = \frac{2 - 3y}{1 - y}$ in the domain A = R – {3} such that $f(x) = y$.

Therefore, f is onto.

---

Conclusion: Since f is both one-one and onto, it is a bijective function.

Given:

Function f : R $\rightarrow$ R defined as f(x) = x$^4$.

Domain = R

Codomain = R

---

To Determine:

Whether f is one-one, onto, or bijective.

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$x_1^4 = x_2^4$

This implies $x_1^4 - x_2^4 = 0 \implies (x_1^2 - x_2^2)(x_1^2 + x_2^2) = 0$.

$(x_1 - x_2)(x_1 + x_2)(x_1^2 + x_2^2) = 0$.

This means $x_1 - x_2 = 0$ or $x_1 + x_2 = 0$ or $x_1^2 + x_2^2 = 0$.

So, $x_1 = x_2$ or $x_1 = -x_2$ or ($x_1 = 0$ and $x_2 = 0$, since the sum of squares of real numbers is zero only if both are zero).

Consider $x_1 = 1$ and $x_2 = -1$. Both are in R, and $x_1 \neq x_2$.

$f(1) = 1^4 = 1$.

$f(-1) = (-1)^4 = 1$.

We have $f(1) = f(-1) = 1$, but $1 \neq -1$.

Since there exist distinct elements in the domain that map to the same element in the codomain, f is not one-one (it is many-one).

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain is R.

The function is f(x) = x$^4$. The fourth power of any real number is always non-negative.

The range of f is the set of all values $x^4$ for $x \in R$. This is the set of all non-negative real numbers.

Range of f = $[0, \infty)$.

The codomain of f is the set of all real numbers R.

The range of f ($[0, \infty)$) is a proper subset of the codomain (R), because R contains negative real numbers (e.g., -1, -10, -$\pi$).

Consider a negative real number in the codomain, for example, $y = -5$. We need to check if there exists any $x \in R$ such that $f(x) = -5$.

$f(x) = -5 \implies x^4 = -5$.

There is no real number x whose fourth power is a negative number.

Thus, the element -5 in the codomain R does not have a pre-image in the domain R under the function f.

This means the range of f ($[0, \infty)$) is not equal to the codomain R.

Therefore, f is not onto.

---

Conclusion:

The function f is neither one-one nor onto.

Comparing with the given options:

(A) f is one-one onto (False)

(B) f is many-one onto (False - not onto)

(C) f is one-one but not onto (False - not one-one)

(D) f is neither one-one nor onto. (True)

---

The correct answer is (D) f is neither one-one nor onto.

Given:

Function f : R $\rightarrow$ R defined as f(x) = 3x.

Domain = R

Codomain = R

---

To Determine:

Whether f is one-one, onto, or bijective.

---

Check for Injectivity (One-one):

A function f : A $\rightarrow$ B is one-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in A$.

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

$3x_1 = 3x_2$

Divide both sides by 3 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

---

Check for Surjectivity (Onto):

A function f : A $\rightarrow$ B is onto if for every element $y \in B$, there exists at least one element $x \in A$ such that $f(x) = y$.

The codomain is R.

Let y be any arbitrary real number in the codomain R.

We need to find if there exists a real number $x$ in the domain R such that $f(x) = y$.

Set $f(x) = y$ and solve for x:

Divide by 3:

Since y is a real number, $\frac{y}{3}$ is also a real number.

So, for every real number $y$ in the codomain, we found a real number $x = \frac{y}{3}$ in the domain such that $f(x) = f\left(\frac{y}{3}\right) = 3\left(\frac{y}{3}\right) = y$.

Every element in the codomain R has a pre-image in the domain R.

Therefore, f is onto.

---

Conclusion:

Since f is both one-one and onto, it is a bijective function.

Comparing with the given options:

(A) f is one-one onto (True)

(B) f is many-one onto (False - is one-one)

(C) f is one-one but not onto (False - is onto)

(D) f is neither one-one nor onto. (False)

---

The correct answer is (A) f is one-one onto.

Given:

Set A = {2, 3, 4, 5}

Set B = {3, 4, 5, 9}

Set C = {7, 11, 15}

Function f : A $\rightarrow$ B defined by:

• f(2) = 3
• f(3) = 4
• f(4) = 5
• f(5) = 5

Function g : B $\rightarrow$ C defined by:

• g(3) = 7
• g(4) = 7
• g(5) = 11
• g(9) = 11

---

To Find:

gof (the composite function).

---

Solution:

The composite function gof is defined as $(gof)(x) = g(f(x))$. The domain of gof is the domain of f, which is A = {2, 3, 4, 5}. The codomain of gof is the codomain of g, which is C = {7, 11, 15}.

We need to find the value of $(gof)(x)$ for each element x in the domain A.

• For x = 2:

$(gof)(2) = g(f(2))$

We know f(2) = 3.

So, $(gof)(2) = g(3)$.

We know g(3) = 7.

$(gof)(2) = 7$

• For x = 3:

$(gof)(3) = g(f(3))$

We know f(3) = 4.

So, $(gof)(3) = g(4)$.

We know g(4) = 7.

$(gof)(3) = 7$

• For x = 4:

$(gof)(4) = g(f(4))$

We know f(4) = 5.

So, $(gof)(4) = g(5)$.

We know g(5) = 11.

$(gof)(4) = 11$

• For x = 5:

$(gof)(5) = g(f(5))$

We know f(5) = 5.

So, $(gof)(5) = g(5)$.

We know g(5) = 11.

$(gof)(5) = 11$

The composite function gof can be expressed as a set of ordered pairs (input, output):

gof = {(2, 7), (3, 7), (4, 11), (5, 11)}.

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Summary of gof values:

• $(gof)(2) = 7$
• $(gof)(3) = 7$
• $(gof)(4) = 11$
• $(gof)(5) = 11$

Given:

Function f : R $\rightarrow$ R defined by f(x) = cos x.

Function g : R $\rightarrow$ R defined by g(x) = 3x$^2$.

Domain of f = R, Codomain of f = R.

Domain of g = R, Codomain of g = R.

---

To Find:

gof and fog.

To Show:

gof $\neq$ fog.

---

Finding gof:

The composite function gof : R $\rightarrow$ R is defined by $(gof)(x) = g(f(x))$.

Substitute f(x) into g(x):

Now apply the function g to cos x. The function g squares the input and multiplies by 3.

So, gof : R $\rightarrow$ R is defined by $(gof)(x) = 3\text{cos}^2 x$.

---

Finding fog:

The composite function fog : R $\rightarrow$ R is defined by $(fog)(x) = f(g(x))$.

Substitute g(x) into f(x):

Now apply the function f to $3x^2$. The function f takes the cosine of the input.

So, fog : R $\rightarrow$ R is defined by $(fog)(x) = \text{cos}(3x^2)$.

---

To Show gof $\neq$ fog:

We have found $(gof)(x) = 3\text{cos}^2 x$ and $(fog)(x) = \text{cos}(3x^2)$.

For gof = fog, we would need $3\text{cos}^2 x = \text{cos}(3x^2)$ for all $x \in R$.

Let's check if this equality holds for a specific value of x, say $x = \pi$.

Calculate $(gof)(\pi)$:

Calculate $(fog)(\pi)$:

$3\pi^2$ is a real number. $\text{cos}(3\pi^2)$ is a value between -1 and 1 (inclusive).

The value 3 is not between -1 and 1.

So, $(gof)(\pi) = 3$ and $(fog)(\pi) = \text{cos}(3\pi^2) \neq 3$.

Since there exists at least one value of x (like $x=\pi$) for which $(gof)(x) \neq (fog)(x)$, the functions gof and fog are not equal.

---

Conclusion:

gof(x) = $3\text{cos}^2 x$

fog(x) = $\text{cos}(3x^2)$

gof $\neq$ fog because, for example, $(gof)(\pi) \neq (fog)(\pi)$.

Given:

Set A = R – $\left\{ \frac{7}{5} \right\}$

Set B = R – $\left\{ \frac{3}{5} \right\}$

Function $f : A \to B$ defined by $f(x) = \frac{3x + 4}{5x - 7}$.

Function $g : B \to A$ defined by $g(x) = \frac{7x + 4}{5x - 3}$.

$I_A$ is the identity function on A, $I_A(x) = x$ for all $x \in A$.

$I_B$ is the identity function on B, $I_B(x) = x$ for all $x \in B$.

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To Show:

fog = $I_B$ and gof = $I_A$.

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Finding fog:

The composite function fog : B $\rightarrow$ B is defined by $(fog)(x) = f(g(x))$ for $x \in B$.

Substitute $g(x) = \frac{7x + 4}{5x - 3}$ into $f(x) = \frac{3x + 4}{5x - 7}$.

Replace x in the expression for f(x) with $\frac{7x + 4}{5x - 3}$:

To simplify, find a common denominator in the numerator and the denominator:

Cancel the common denominator $(5x - 3)$ (Note that $x \in B$, so $x \neq \frac{3}{5}$, and $5x - 3 \neq 0$):

Expand the terms in the numerator and denominator:

Combine like terms:

This result holds for all $x \in B$. By the definition of the identity function on B, $(fog)(x) = I_B(x)$ for all $x \in B$.

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Finding gof:

The composite function gof : A $\rightarrow$ A is defined by $(gof)(x) = g(f(x))$ for $x \in A$.

Substitute $f(x) = \frac{3x + 4}{5x - 7}$ into $g(x) = \frac{7x + 4}{5x - 3}$.

Replace x in the expression for g(x) with $\frac{3x + 4}{5x - 7}$:

To simplify, find a common denominator in the numerator and the denominator:

Cancel the common denominator $(5x - 7)$ (Note that $x \in A$, so $x \neq \frac{7}{5}$, and $5x - 7 \neq 0$):

Expand the terms in the numerator and denominator:

Combine like terms:

This result holds for all $x \in A$. By the definition of the identity function on A, $(gof)(x) = I_A(x)$ for all $x \in A$.

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Conclusion:

We have shown that fog(x) = x for all $x \in B$, so fog = $I_B$.

We have shown that gof(x) = x for all $x \in A$, so gof = $I_A$.

Given:

Function f : A $\rightarrow$ B is one-one.

Function g : B $\rightarrow$ C is one-one.

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To Show:

The composite function gof : A $\rightarrow$ C is also one-one.

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Proof that gof is one-one:

A function h : X $\rightarrow$ Y is one-one if for any two distinct elements $x_1, x_2 \in X$, their images under h are also distinct, i.e., $h(x_1) = h(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2$ be two elements in the domain of gof, which is the set A.

Assume that $(gof)(x_1) = (gof)(x_2)$.

By the definition of the composite function, $(gof)(x) = g(f(x))$.

So, $(gof)(x_1) = (gof)(x_2)$ implies $g(f(x_1)) = g(f(x_2))$.

Let $y_1 = f(x_1)$ and $y_2 = f(x_2)$. Note that $y_1, y_2 \in B$ (the codomain of f and the domain of g).

The equation $g(f(x_1)) = g(f(x_2))$ can be written as $g(y_1) = g(y_2)$.

We are given that the function g : B $\rightarrow$ C is one-one.

By the definition of a one-one function, if $g(y_1) = g(y_2)$, then $y_1 = y_2$.

Substituting back the expressions for $y_1$ and $y_2$, we get $f(x_1) = f(x_2)$.

Now we have $f(x_1) = f(x_2)$, where $x_1, x_2 \in A$.

We are also given that the function f : A $\rightarrow$ B is one-one.

By the definition of a one-one function, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Starting with the assumption $(gof)(x_1) = (gof)(x_2)$, we have successfully shown that $x_1 = x_2$.

Thus, the composite function gof is one-one.

Given:

Function f : A $\rightarrow$ B is onto.

Function g : B $\rightarrow$ C is onto.

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To Show:

The composite function gof : A $\rightarrow$ C is also onto.

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Proof that gof is onto:

A function h : X $\rightarrow$ Y is onto if for every element $z \in Y$, there exists at least one element $x \in X$ such that $h(x) = z$.

Let z be an arbitrary element in the codomain of gof, which is the set C.

We need to find an element $x$ in the domain of gof, which is the set A, such that $(gof)(x) = z$.

Since $z \in C$ and the function g : B $\rightarrow$ C is onto, by the definition of an onto function, there exists at least one element $y \in B$ such that $g(y) = z$.

Now, since $y \in B$ and the function f : A $\rightarrow$ B is onto, by the definition of an onto function, there exists at least one element $x \in A$ such that $f(x) = y$.

We have found an element $x \in A$ such that $f(x) = y$ and an element $y \in B$ such that $g(y) = z$.

Consider the composite function $(gof)(x)$. By definition, $(gof)(x) = g(f(x))$.

Substituting $f(x) = y$, we get $(gof)(x) = g(y)$.

Substituting $g(y) = z$, we get $(gof)(x) = z$.

So, for the arbitrary element z in the codomain C, we found an element x in the domain A such that $(gof)(x) = z$.

This means every element in the codomain C has a pre-image in the domain A under the composite function gof.

Therefore, the composite function gof : A $\rightarrow$ C is onto.

Given:

Functions f : A $\rightarrow$ B and g : B $\rightarrow$ C such that the composite function gof : A $\rightarrow$ C is defined and is one-one.

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Question:

Are f and g both necessarily one-one?

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Analysis:

Let's examine the properties of injectivity (one-one) for f and g based on the injectivity of gof.

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Is f necessarily one-one?

Assume f is not one-one. This means there exist distinct elements $a_1, a_2 \in A$ such that $f(a_1) = f(a_2)$.

Let $b = f(a_1) = f(a_2)$. Note that $b \in B$.

Now consider the image of these elements under the composite function gof:

So, $(gof)(a_1) = (gof)(a_2)$.

But we assumed that gof is one-one. By the definition of a one-one function, if $(gof)(a_1) = (gof)(a_2)$, then $a_1 = a_2$.

This contradicts our initial assumption that $a_1$ and $a_2$ are distinct elements in A.

Therefore, the assumption that f is not one-one must be false.

Hence, f is necessarily one-one.

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Is g necessarily one-one?

Assume g is not one-one. This means there exist distinct elements $b_1, b_2 \in B$ such that $g(b_1) = g(b_2)$.

Let $c = g(b_1) = g(b_2)$. Note that $c \in C$.

For gof to be defined, the domain of g must be the codomain of f, or at least the image of f must be a subset of the domain of g. Here, the codomain of f is B, which is the domain of g.

For $b_1$ and $b_2$ to be in the image of f, there must exist $a_1, a_2 \in A$ such that $f(a_1) = b_1$ and $f(a_2) = b_2$. If such $a_1, a_2$ exist, then $(gof)(a_1) = g(f(a_1)) = g(b_1) = c$ and $(gof)(a_2) = g(f(a_2)) = g(b_2) = c$. So $(gof)(a_1) = (gof)(a_2)$. Since gof is one-one, this would imply $a_1 = a_2$, which would then imply $f(a_1) = f(a_2)$, so $b_1 = b_2$, contradicting the assumption that $b_1 \neq b_2$. This line of reasoning works *if* $b_1$ and $b_2$ are in the image of f. What if one or both are not in the image of f?

Consider a counterexample where gof is one-one but g is not one-one.

Let A = {1}, B = {1, 2}, C = {1}.

Define f : A $\rightarrow$ B by f(1) = 1.

Define g : B $\rightarrow$ C by g(1) = 1 and g(2) = 1.

• f is one-one: If $f(x_1) = f(x_2)$, then $1 = 1$, which implies $x_1 = x_2 = 1$. Yes.
• g is not one-one: g(1) = 1 and g(2) = 1, but $1 \neq 2$. No.

Now find the composite function gof : A $\rightarrow$ C.

Domain of gof is A = {1}.

$(gof)(1) = g(f(1)) = g(1) = 1$.

gof = {(1, 1)}.

Is gof one-one? The only element in the domain is 1. For any distinct $a_1, a_2 \in A$, the condition $a_1 \neq a_2$ is never met, so the implication $(gof)(a_1) = (gof)(a_2) \implies a_1 = a_2$ is vacuously true. Yes, gof is one-one.

In this counterexample, gof is one-one, but g is not one-one.

Therefore, g is not necessarily one-one.

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Conclusion:

If gof is one-one, then f is necessarily one-one, but g is not necessarily one-one.

So, f and g are not both necessarily one-one.

Given:

Functions f : A $\rightarrow$ B and g : B $\rightarrow$ C such that the composite function gof : A $\rightarrow$ C is defined and is onto.

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Question:

Are f and g both necessarily onto?

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Analysis:

Let's examine the properties of surjectivity (onto) for f and g based on the surjectivity of gof.

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Is g necessarily onto?

Assume g is not onto. This means there exists at least one element $c \in C$ such that there is no $b \in B$ for which $g(b) = c$.

The composite function gof : A $\rightarrow$ C is defined by $(gof)(a) = g(f(a))$ for $a \in A$.

The values $f(a)$ for all $a \in A$ form the image of f, which is a subset of B. Let this image be Im(f).

$(gof)(a) = g(f(a))$. The input to g is an element from the image of f (Im(f)).

The range of gof is the set of all values $g(y)$ where $y \in \text{Im}(f)$. Range(gof) = $\{g(y) : y \in \text{Im}(f)\}$.

We are given that gof is onto. This means the range of gof is equal to the codomain C. So, Range(gof) = C.

Every element $c \in C$ must be in the range of gof. This means for every $c \in C$, there exists some $y \in \text{Im}(f)$ such that $g(y) = c$.

Since Im(f) is a subset of B, finding a $y \in \text{Im}(f)$ such that $g(y) = c$ implies finding a $y \in B$ such that $g(y) = c$ (because Im(f) $\subseteq$ B).

Thus, for every $c \in C$, there exists a $y \in B$ such that $g(y) = c$. This is the definition of an onto function from B to C.

Therefore, g is necessarily onto.

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Is f necessarily onto?

Assume f is not onto. This means there exists at least one element $b \in B$ such that there is no $a \in A$ for which $f(a) = b$. In other words, b is in the codomain of f but not in the image of f (b $\notin$ Im(f)).

We are given that gof is onto. This means for every element $c \in C$, there exists an element $a \in A$ such that $(gof)(a) = c$.

$(gof)(a) = g(f(a)) = c$.

This means that for every $c \in C$, there is some $a \in A$ such that $g(\text{value of f at a}) = c$. The value f(a) is an element of B.

Let's consider a counterexample where gof is onto but f is not onto.

Let A = {1}, B = {1, 2}, C = {1}.

Define f : A $\rightarrow$ B by f(1) = 1.

Define g : B $\rightarrow$ C by g(1) = 1 and g(2) = 1.

• f is not onto: The codomain of f is B = {1, 2}. The range of f is {f(1)} = {1}. The element 2 in B has no pre-image in A. No.
• g is not onto: The codomain of g is C = {1}. The range of g is {g(1), g(2)} = {1, 1} = {1}. Yes, g is onto. (Correction: g is onto here, this counterexample won't work for f not being onto.)

Let's try another counterexample for f not being onto.

Let A = {1, 2}, B = {1, 2, 3}, C = {1, 2}.

Define f : A $\rightarrow$ B by f(1) = 1, f(2) = 2.

Define g : B $\rightarrow$ C by g(1) = 1, g(2) = 2, g(3) = 1.

• f is not onto: The codomain of f is B = {1, 2, 3}. The range of f is {f(1), f(2)} = {1, 2}. The element 3 in B has no pre-image in A. No.
• g is onto: The codomain of g is C = {1, 2}. The range of g is {g(1), g(2), g(3)} = {1, 2, 1} = {1, 2}. The range is equal to the codomain. Yes.

Now find the composite function gof : A $\rightarrow$ C.

Domain of gof is A = {1, 2}. Codomain of gof is C = {1, 2}.

$(gof)(1) = g(f(1)) = g(1) = 1$.

$(gof)(2) = g(f(2)) = g(2) = 2$.

gof = {(1, 1), (2, 2)}.

Is gof onto? The codomain is C = {1, 2}. The range is {1, 2}. The range is equal to the codomain. Yes, gof is onto.

In this counterexample, gof is onto, but f is not onto.

Therefore, f is not necessarily onto.

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Conclusion:

If gof is onto, then g is necessarily onto, but f is not necessarily onto.

So, f and g are not both necessarily onto.

Given:

Set X = {1, 2, 3}

Set Y = {a, b, c}

Function f : X $\rightarrow$ Y is one-one and onto, given by:

• f(1) = a
• f(2) = b
• f(3) = c

$I_X$ is the identity function on X, $I_X(x) = x$ for all $x \in X$.

$I_Y$ is the identity function on Y, $I_Y(y) = y$ for all $y \in Y$.

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To Show:

There exists a function g : Y $\rightarrow$ X such that gof = $I_X$ and fog = $I_Y$.

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Construction of function g:

Since f is one-one and onto, for each element in Y, there is a unique pre-image in X. This means the inverse function $f^{-1}$ exists and is a function from Y to X.

Let's define the function g : Y $\rightarrow$ X based on the pre-images under f:

• For the element a $\in$ Y, its unique pre-image under f is 1 (since f(1) = a). Define g(a) = 1.
• For the element b $\in$ Y, its unique pre-image under f is 2 (since f(2) = b). Define g(b) = 2.
• For the element c $\in$ Y, its unique pre-image under f is 3 (since f(3) = c). Define g(c) = 3.

So, the function g : {a, b, c} $\rightarrow$ {1, 2, 3} is defined by g = {(a, 1), (b, 2), (c, 3)}. This function exists.

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Checking gof = $I_X$:

The composite function gof : X $\rightarrow$ X is defined by $(gof)(x) = g(f(x))$ for $x \in X = \{1, 2, 3\}$.

• For x = 1: $(gof)(1) = g(f(1))$. Since f(1) = a, $(gof)(1) = g(a)$. Since g(a) = 1, $(gof)(1) = 1$.
• For x = 2: $(gof)(2) = g(f(2))$. Since f(2) = b, $(gof)(2) = g(b)$. Since g(b) = 2, $(gof)(2) = 2$.
• For x = 3: $(gof)(3) = g(f(3))$. Since f(3) = c, $(gof)(3) = g(c)$. Since g(c) = 3, $(gof)(3) = 3$.

For every $x \in X$, $(gof)(x) = x$. This is the definition of the identity function $I_X$ on X.

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Checking fog = $I_Y$:

The composite function fog : Y $\rightarrow$ Y is defined by $(fog)(y) = f(g(y))$ for $y \in Y = \{a, b, c\}$.

• For y = a: $(fog)(a) = f(g(a))$. Since g(a) = 1, $(fog)(a) = f(1)$. Since f(1) = a, $(fog)(a) = a$.
• For y = b: $(fog)(b) = f(g(b))$. Since g(b) = 2, $(fog)(b) = f(2)$. Since f(2) = b, $(fog)(b) = b$.
• For y = c: $(fog)(c) = f(g(c))$. Since g(c) = 3, $(fog)(c) = f(3)$. Since f(3) = c, $(fog)(c) = c$.

For every $y \in Y$, $(fog)(y) = y$. This is the definition of the identity function $I_Y$ on Y.

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Conclusion:

We have constructed a function g : {a, b, c} $\rightarrow$ {1, 2, 3} such that gof = $I_X$ and fog = $I_Y$. This function g is the inverse function of f, denoted by $f^{-1}$.

Given:

Function f : N $\rightarrow$ Y defined as f(x) = 4x + 3.

Domain = N = {1, 2, 3, ...}

Codomain = Y = {y $\in$ N: y = 4x + 3 for some x $\in$ N}.

Note that Y is the range of the function f. Since the codomain is defined to be equal to the range, the function is onto by definition.

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To Show:

f(x) is invertible.

Find the inverse function.

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Proof that f is invertible:

A function is invertible if and only if it is bijective (both one-one and onto).

We are given that the codomain Y is the range of the function f. This implies that f is onto by definition of the codomain.

So, we only need to prove that f is one-one.

Check for Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f(x_1) = f(x_2)$.

$4x_1 + 3 = 4x_2 + 3$

Subtract 3 from both sides:

Divide both sides by 4 (which is non-zero):

Thus, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Therefore, f is one-one.

Since f is both one-one and onto, it is a bijective function. Hence, f is invertible.

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Finding the inverse function:

Let $y$ be an element in the codomain Y. By the definition of Y, $y = 4x + 3$ for some $x \in N$. This means $y = f(x)$.

To find the inverse function $f^{-1}$, we need to express x in terms of y.

Start with the equation $y = 4x + 3$ and solve for x:

The inverse function $f^{-1}$ maps from Y to N. The input to $f^{-1}$ is y $\in$ Y, and the output is the corresponding x $\in$ N.

So, the inverse function $f^{-1} : Y \rightarrow N$ is defined by $f^{-1}(y) = \frac{y - 3}{4}$.

We must verify that for any $y \in Y$, $\frac{y-3}{4}$ is indeed a natural number. By the definition of Y, $y = 4x+3$ for some $x \in N$. So $y-3 = 4x$, and $\frac{y-3}{4} = x$. Since $x \in N$, $\frac{y-3}{4}$ is a natural number. Thus, the codomain of $f^{-1}$ is N.

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Verification (Optional but good practice):

Check if $(f^{-1}of)(x) = x$ for $x \in N$ and $(fof^{-1})(y) = y$ for $y \in Y$.

$(f^{-1}of)(x) = f^{-1}(f(x)) = f^{-1}(4x + 3)$. Substitute $(4x+3)$ into $f^{-1}(y) = \frac{y-3}{4}$:

This is the identity function on N.

$(fof^{-1})(y) = f(f^{-1}(y)) = f\left(\frac{y - 3}{4}\right)$. Substitute $\frac{y-3}{4}$ into $f(x) = 4x+3$:

This is the identity function on Y.

The inverse function is correct.

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Conclusion:

The function f(x) = 4x + 3 is invertible, and its inverse function is $f^{-1}(y) = \frac{y - 3}{4}$.

Given:

Set Y = $\{ n^2 : n \in N \} \subset N$. This means Y is the set of perfect squares within the natural numbers: Y = {1, 4, 9, 16, ...}.

Function f : N $\rightarrow$ Y defined as f(n) = n$^2$.

Domain = N = {1, 2, 3, ...}

Codomain = Y = {1, 4, 9, 16, ...}

Note that the codomain Y is specifically defined as the set of perfect squares, which is precisely the set of values that f(n) produces when n is a natural number. Thus, the range of f is equal to its codomain Y, meaning f is onto by definition.

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To Show:

f is invertible.

Find the inverse of f.

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Proof that f is invertible:

A function is invertible if and only if it is bijective (both one-one and onto).

We are given that the codomain Y is the range of the function f. This implies that f is onto by definition of the codomain.

So, we only need to prove that f is one-one.

Check for Injectivity (One-one):

Let $n_1, n_2 \in N$ such that $f(n_1) = f(n_2)$.

$n_1^2 = n_2^2$

Since $n_1, n_2$ are natural numbers, they are positive. Taking the square root of both sides:

$\sqrt{n_1^2} = \sqrt{n_2^2} \implies |n_1| = |n_2|$

As $n_1, n_2 \in N$, $n_1 > 0$ and $n_2 > 0$. So $|n_1| = n_1$ and $|n_2| = n_2$.

$n_1 = n_2$.

Thus, if $f(n_1) = f(n_2)$, then $n_1 = n_2$.

Therefore, f is one-one.

Since f is both one-one and onto, it is a bijective function. Hence, f is invertible.

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Finding the inverse function:

Let $y$ be an element in the codomain Y. By the definition of Y, $y = n^2$ for some $n \in N$. This means $y = f(n)$.

To find the inverse function $f^{-1}$, we need to express n in terms of y.

Start with the equation $y = n^2$ and solve for n:

Taking the square root of both sides:

Since the domain of f is N (natural numbers), n must be a positive integer. Therefore, we take the positive square root.

The inverse function $f^{-1}$ maps from Y to N. The input to $f^{-1}$ is y $\in$ Y, and the output is the corresponding n $\in$ N.

So, the inverse function $f^{-1} : Y \rightarrow N$ is defined by $f^{-1}(y) = \sqrt{y}$.

We must verify that for any $y \in Y$, $\sqrt{y}$ is indeed a natural number. By the definition of Y, $y = n^2$ for some $n \in N$. So $\sqrt{y} = \sqrt{n^2} = n$ (since n is positive). Since $n \in N$, $\sqrt{y}$ is a natural number. Thus, the codomain of $f^{-1}$ is N.

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Verification (Optional but good practice):

Check if $(f^{-1}of)(n) = n$ for $n \in N$ and $(fof^{-1})(y) = y$ for $y \in Y$.

$(f^{-1}of)(n) = f^{-1}(f(n)) = f^{-1}(n^2)$. Substitute $n^2$ into $f^{-1}(y) = \sqrt{y}$:

This is the identity function on N.

$(fof^{-1})(y) = f(f^{-1}(y)) = f(\sqrt{y})$. Substitute $\sqrt{y}$ into $f(n) = n^2$:

This is the identity function on Y.

The inverse function is correct.

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Conclusion:

The function f(n) = n$^2$ is invertible, and its inverse function is $f^{-1}(y) = \sqrt{y}$.

Given:

Function f : N $\rightarrow$ R defined as f(x) = 4x$^2$ + 12x + 15.

Domain = N = {1, 2, 3, ...}

Codomain for the initial definition is R.

However, the problem asks to show that f : N $\rightarrow$ S is invertible, where S is the range of f.

Let the modified function be $f' : N \to S$, where $f'(x) = 4x^2 + 12x + 15$, and S = Range(f).

Domain of $f'$ = N.

Codomain of $f'$ = S = Range(f).

By defining the codomain to be the range, the function $f' : N \to S$ is onto by definition.

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To Show:

$f' : N \to S$ is invertible.

Find the inverse of $f'$.

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Proof that $f'$ is invertible:

A function is invertible if and only if it is bijective (both one-one and onto).

We are given that the codomain S is the range of the function $f'$. This implies that $f'$ is onto by definition of the codomain.

So, we only need to prove that $f'$ is one-one.

Check for Injectivity (One-one):

Let $x_1, x_2 \in N$ such that $f'(x_1) = f'(x_2)$.

$4x_1^2 + 12x_1 + 15 = 4x_2^2 + 12x_2 + 15$

Subtract 15 from both sides:

Rearrange the terms:

Factor by grouping:

Factor out the common term $(x_1 - x_2)$:

This equation is true if and only if $x_1 - x_2 = 0$ or $4x_1 + 4x_2 + 12 = 0$.

$x_1 = x_2$ or $4(x_1 + x_2 + 3) = 0$.

$x_1 = x_2$ or $x_1 + x_2 + 3 = 0$.

Since $x_1, x_2 \in N$, $x_1 \ge 1$ and $x_2 \ge 1$.

So, $x_1 + x_2 \ge 1 + 1 = 2$.

Then $x_1 + x_2 + 3 \ge 2 + 3 = 5$.

Thus, $x_1 + x_2 + 3$ is always a positive integer, and it can never be equal to 0.

Therefore, the only possibility for the equation $(x_1 - x_2) [4x_1 + 4x_2 + 12] = 0$ to be true is $x_1 - x_2 = 0$, which means $x_1 = x_2$.

Thus, if $f'(x_1) = f'(x_2)$, then $x_1 = x_2$.

Therefore, $f'$ is one-one.

Since $f'$ is both one-one and onto, it is a bijective function. Hence, $f'$ is invertible.

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Finding the inverse function:

Let $y$ be an element in the codomain S (the range of f). By the definition of S, $y = 4x^2 + 12x + 15$ for some $x \in N$. This means $y = f'(x)$.

To find the inverse function $(f')^{-1}$, we need to express x in terms of y.

Start with the equation $y = 4x^2 + 12x + 15$. We want to solve for x.

We can complete the square for the quadratic expression in x.

Factor out 4 from the terms with x:

To complete the square for $x^2 + 3x$, we add and subtract $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$ inside the parenthesis:

Distribute the 4:

Now, solve for x:

Take the square root of both sides:

Since the domain of $f'$ is N (natural numbers), x must be a positive integer ($x \ge 1$).

Consider the two possible expressions for x: $x = -\frac{3}{2} + \frac{\sqrt{y - 6}}{2}$ and $x = -\frac{3}{2} - \frac{\sqrt{y - 6}}{2}$.

Since y is in the range S, $y = 4x^2 + 12x + 15$ for some $x \in N$. The minimum value of $f(x)$ for $x \in N$ is when $x=1$: $f(1) = 4(1)^2 + 12(1) + 15 = 4 + 12 + 15 = 31$. So $y \ge 31$. This means $y-6 \ge 25$, so $\sqrt{y-6} \ge 5$.

If we take the negative sign: $x = -\frac{3}{2} - \frac{\sqrt{y - 6}}{2}$. Since $\sqrt{y-6} \ge 5$, $\frac{\sqrt{y-6}}{2} \ge 2.5$. Then $x \le -1.5 - 2.5 = -4$. This value of x is not in N.

If we take the positive sign: $x = -\frac{3}{2} + \frac{\sqrt{y - 6}}{2}$. Since $\sqrt{y-6} \ge 5$, $\frac{\sqrt{y-6}}{2} \ge 2.5$. Then $x \ge -1.5 + 2.5 = 1$. This value of x is $\ge 1$. Also, since y is in the range (meaning $y = 4x^2 + 12x + 15$ for some $x \in N$), the expression $\frac{\sqrt{y - 6} - 3}{2}$ will yield that specific natural number x.

So, we take the positive root.

The inverse function $(f')^{-1} : S \rightarrow N$ is defined by $(f')^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}$.

Let's verify for an example point in N, say x = 2. $f(2) = 4(2)^2 + 12(2) + 15 = 16 + 24 + 15 = 55$. So (2, 55) is on the graph of f, and $55 \in S$.

Check if $f^{-1}(55) = 2$: $f^{-1}(55) = \frac{\sqrt{55 - 6} - 3}{2} = \frac{\sqrt{49} - 3}{2} = \frac{7 - 3}{2} = \frac{4}{2} = 2$. This matches.

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Conclusion:

The function $f : N \to S$ defined by $f(x) = 4x^2 + 12x + 15$, where S is the range of f, is invertible.

The inverse function is $f^{-1} : S \rightarrow N$ defined by $f^{-1}(y) = \frac{\sqrt{y - 6} - 3}{2}$.

Given:

Function f : N $\rightarrow$ N defined by f(x) = 2x, for x $\in$ N.

Function g : N $\rightarrow$ N defined by g(y) = 3y + 4, for y $\in$ N.

Function h : N $\rightarrow$ R defined by h(z) = sin z, for z $\in$ N.

Domain of f = N, Codomain of f = N.

Domain of g = N, Codomain of g = N.

Domain of h = N, Codomain of h = R.

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To Show:

ho(gof) = (hog)of.

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Finding gof:

The composite function gof : N $\rightarrow$ N is defined by $(gof)(x) = g(f(x))$ for $x \in N$.

Substitute f(x) = 2x into g(y) = 3y + 4. Replace y with 2x.

So, gof : N $\rightarrow$ N is defined by $(gof)(x) = 6x + 4$.

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Finding ho(gof):

The composite function ho(gof) : N $\rightarrow$ R is defined by $(ho(gof))(x) = h((gof)(x))$ for $x \in N$.

Substitute $(gof)(x) = 6x + 4$ into h(z) = sin z. Replace z with $6x + 4$.

So, ho(gof) : N $\rightarrow$ R is defined by $(ho(gof))(x) = \text{sin}(6x + 4)$.

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Finding hog:

The composite function hog : N $\rightarrow$ R is defined by $(hog)(y) = h(g(y))$ for $y \in N$.

Substitute g(y) = 3y + 4 into h(z) = sin z. Replace z with $3y + 4$.

So, hog : N $\rightarrow$ R is defined by $(hog)(y) = \text{sin}(3y + 4)$.

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Finding (hog)of:

The composite function (hog)of : N $\rightarrow$ R is defined by $((hog)of)(x) = (hog)(f(x))$ for $x \in N$.

Substitute f(x) = 2x into $(hog)(y) = \text{sin}(3y + 4)$. Replace y with 2x.

So, (hog)of : N $\rightarrow$ R is defined by $((hog)of)(x) = \text{sin}(6x + 4)$.

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Comparison:

We found $(ho(gof))(x) = \text{sin}(6x + 4)$.

We found $((hog)of)(x) = \text{sin}(6x + 4)$.

Since the domains and codomains are the same (N $\rightarrow$ R) and the function expressions are the same for all x in the domain, we have:

Therefore, ho(gof) = (hog)of.

This demonstrates the associative property of function composition.

Given:

Set A = {1, 2, 3}

Set B = {a, b, c}

Set C = {apple, ball, cat}

Function f : A $\rightarrow$ B defined by f(1) = a, f(2) = b, f(3) = c.

Function g : B $\rightarrow$ C defined by g(a) = apple, g(b) = ball, g(c) = cat.

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To Show:

f, g, and gof are invertible.

Find $f^{-1}$, $g^{-1}$, and $(gof)^{-1}$.

Show that $(gof)^{-1} = f^{-1}o g^{-1}$.

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Invertibility of f:

Domain of f = {1, 2, 3}, Codomain of f = {a, b, c}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (1, 2, 3) map to distinct elements in the codomain (f(1)=a, f(2)=b, f(3)=c are all distinct). So f is one-one.

Check Surjectivity (Onto): The range of f is {f(1), f(2), f(3)} = {a, b, c}. This is equal to the codomain B. So f is onto.

Since f is both one-one and onto, f is invertible.

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Invertibility of g:

Domain of g = {a, b, c}, Codomain of g = {apple, ball, cat}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (a, b, c) map to distinct elements in the codomain (g(a)=apple, g(b)=ball, g(c)=cat are all distinct). So g is one-one.

Check Surjectivity (Onto): The range of g is {g(a), g(b), g(c)} = {apple, ball, cat}. This is equal to the codomain C. So g is onto.

Since g is both one-one and onto, g is invertible.

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Finding gof and its invertibility:

The composite function gof : A $\rightarrow$ C is defined by $(gof)(x) = g(f(x))$ for $x \in A = \{1, 2, 3\}$.

• $(gof)(1) = g(f(1)) = g(a) = \text{apple}$.
• $(gof)(2) = g(f(2)) = g(b) = \text{ball}$.
• $(gof)(3) = g(f(3)) = g(c) = \text{cat}$.

gof = {(1, apple), (2, ball), (3, cat)}.

Domain of gof = {1, 2, 3}, Codomain of gof = {apple, ball, cat}. Both sets have 3 elements.

Check Injectivity (One-one): The distinct elements in the domain (1, 2, 3) map to distinct elements in the codomain (apple, ball, cat are all distinct). So gof is one-one.

Check Surjectivity (Onto): The range of gof is {apple, ball, cat}. This is equal to the codomain C. So gof is onto.

Since gof is both one-one and onto, gof is invertible.

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Finding inverse functions:

The inverse function reverses the mapping of the original function.

$f^{-1} : B \rightarrow A$:

• Since f(1) = a, $f^{-1}(a) = 1$.
• Since f(2) = b, $f^{-1}(b) = 2$.
• Since f(3) = c, $f^{-1}(c) = 3$.

So, $f^{-1} = \text{\{(a, 1), (b, 2), (c, 3)\}}$.

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$g^{-1} : C \rightarrow B$:

• Since g(a) = apple, $g^{-1}(\text{apple}) = \text{a}$.
• Since g(b) = ball, $g^{-1}(\text{ball}) = \text{b}$.
• Since g(c) = cat, $g^{-1}(\text{cat}) = \text{c}$.

So, $g^{-1} = \text{\{(apple, a), (ball, b), (cat, c)\}}$.

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$(gof)^{-1} : C \rightarrow A$:

• Since $(gof)(1) = \text{apple}$, $(gof)^{-1}(\text{apple}) = 1$.
• Since $(gof)(2) = \text{ball}$, $(gof)^{-1}(\text{ball}) = 2$.
• Since $(gof)(3) = \text{cat}$, $(gof)^{-1}(\text{cat}) = 3$.

So, $(gof)^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$.

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Showing $(gof)^{-1} = f^{-1}o g^{-1}$:

The composite function $f^{-1}o g^{-1} : C \rightarrow A$ is defined by $(f^{-1}o g^{-1})(z) = f^{-1}(g^{-1}(z))$ for $z \in C = \{\text{apple, ball, cat}\}$.

• For z = apple: $(f^{-1}o g^{-1})(\text{apple}) = f^{-1}(g^{-1}(\text{apple}))$. Since $g^{-1}(\text{apple}) = \text{a}$, $(f^{-1}o g^{-1})(\text{apple}) = f^{-1}(a)$. Since $f^{-1}(a) = 1$, $(f^{-1}o g^{-1})(\text{apple}) = 1$.
• For z = ball: $(f^{-1}o g^{-1})(\text{ball}) = f^{-1}(g^{-1}(\text{ball}))$. Since $g^{-1}(\text{ball}) = \text{b}$, $(f^{-1}o g^{-1})(\text{ball}) = f^{-1}(b)$. Since $f^{-1}(b) = 2$, $(f^{-1}o g^{-1})(\text{ball}) = 2$.
• For z = cat: $(f^{-1}o g^{-1})(\text{cat}) = f^{-1}(g^{-1}(\text{cat}))$. Since $g^{-1}(\text{cat}) = \text{c}$, $(f^{-1}o g^{-1})(\text{cat}) = f^{-1}(c)$. Since $f^{-1}(c) = 3$, $(f^{-1}o g^{-1})(\text{cat}) = 3$.

The composite function $f^{-1}o g^{-1}$ maps:

• apple $\rightarrow$ 1
• ball $\rightarrow$ 2
• cat $\rightarrow$ 3

So, $f^{-1}o g^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$.

Comparing the set of ordered pairs for $(gof)^{-1}$ and $f^{-1}o g^{-1}$, we see they are the same.

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Conclusion:

Functions f, g, and gof are invertible.

$f^{-1} = \text{\{(a, 1), (b, 2), (c, 3)\}}$

$g^{-1} = \text{\{(apple, a), (ball, b), (cat, c)\}}$

$(gof)^{-1} = \text{\{(apple, 1), (ball, 2), (cat, 3)\}}$

We have shown that $(gof)^{-1} = f^{-1}o g^{-1}$.

Given:

Set S = {1, 2, 3}.

Functions f : S $\rightarrow$ S are defined below.

Domain = S, Codomain = S.

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To Determine:

Whether each function has an inverse. Find $f^{-1}$ if it exists.

A function has an inverse if and only if it is bijective (both one-one and onto).

For a function $f: S \to S$ where S is a finite set, f is one-one if and only if it is onto.

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(a) f = {(1, 1), (2, 2), (3, 3)}

Check Injectivity (One-one):

The elements in the domain (1, 2, 3) map to the images {1, 2, 3}. The distinct elements 1, 2, 3 in the domain map to distinct images 1, 2, 3 in the codomain. So f is one-one.

Check Surjectivity (Onto):

The codomain is {1, 2, 3}. The range is {1, 2, 3}. The range is equal to the codomain. So f is onto.

Since f is both one-one and onto, f is invertible.

Finding the inverse $f^{-1}$: The inverse function reverses the mapping.

• Since f(1) = 1, $f^{-1}(1) = 1$.
• Since f(2) = 2, $f^{-1}(2) = 2$.
• Since f(3) = 3, $f^{-1}(3) = 3$.

$f^{-1} = \text{\{(1, 1), (2, 2), (3, 3)\}}$.

(This is the identity function on S, $I_S$).

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(b) f = {(1, 2), (2, 1), (3, 1)}

Check Injectivity (One-one):

Consider the elements 2 and 3 in the domain. f(2) = 1 and f(3) = 1. We have f(2) = f(3) but $2 \neq 3$. So f is not one-one.

Since f is not one-one, it is not invertible.

(We don't need to check surjectivity, but for completeness: The codomain is {1, 2, 3}. The range is {f(1), f(2), f(3)} = {2, 1, 1} = {1, 2}. The element 3 in the codomain is not in the range. So f is not onto).

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(c) f = {(1, 3), (3, 2), (2, 1)}

Check Injectivity (One-one):

The elements in the domain (1, 2, 3) map to the images {3, 1, 2}. The distinct elements 1, 2, 3 in the domain map to distinct images 3, 1, 2 in the codomain. So f is one-one.

Check Surjectivity (Onto):

The codomain is {1, 2, 3}. The range is {f(1), f(2), f(3)} = {3, 1, 2} = {1, 2, 3}. The range is equal to the codomain. So f is onto.

Since f is both one-one and onto, f is invertible.

Finding the inverse $f^{-1}$: The inverse function reverses the mapping.

• Since f(1) = 3, $f^{-1}(3) = 1$.
• Since f(2) = 1, $f^{-1}(1) = 2$.
• Since f(3) = 2, $f^{-1}(2) = 3$.

Arranging the pairs by the order of the first element in the inverse function:

$f^{-1} = \text{\{(1, 2), (2, 3), (3, 1)\}}$.

Given:

Function f: $\{1, 3, 4\} \to \{1, 2, 5\}$ defined as $f = \{(1, 2), (3, 5), (4, 1)\}$.

Function g: $\{1, 2, 5\} \to \{1, 3\}$ defined as $g = \{(1, 3), (2, 3), (5, 1)\}$.

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To Find:

The composite function gof.

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Solution:

The composite function gof is defined as gof(x) = g(f(x)) for all x in the domain of f. The domain of f is $\{1, 3, 4\}$. We find the image of each element in the domain of f under the composite function gof.

For $x = 1$:

$f(1) = 2$ (from the definition of f)

$g(f(1)) = g(2)$ (substituting the value of f(1))

$g(2) = 3$ (from the definition of g)

So, gof(1) = 3. This gives the ordered pair (1, 3).

For $x = 3$:

$f(3) = 5$ (from the definition of f)

$g(f(3)) = g(5)$ (substituting the value of f(3))

$g(5) = 1$ (from the definition of g)

So, gof(3) = 1. This gives the ordered pair (3, 1).

For $x = 4$:

$f(4) = 1$ (from the definition of f)

$g(f(4)) = g(1)$ (substituting the value of f(4))

$g(1) = 3$ (from the definition of g)

So, gof(4) = 3. This gives the ordered pair (4, 3).

Therefore, the composite function gof is the set of ordered pairs:

$gof = \{(1, 3), (3, 1), (4, 3)\}$.

Given:

Functions f, g, and h are from R to R.

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To Prove:

1. $(f + g) \circ h = f \circ h + g \circ h$

2. $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$

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Proof:

To show that two functions are equal, we need to show that they have the same domain and codomain, and that for every element x in the domain, the functions produce the same output.

The domain of h is R, and the codomain is R. The domain of f and g is R. The sum $(f+g)$ and product $(f \cdot g)$ are functions from R to R.

The composite functions $(f+g) \circ h$ and $(f \cdot g) \circ h$ are defined when the codomain of h (which is R) is a subset of the domain of $(f+g)$ and $(f \cdot g)$ (which is R). Since R $\subseteq$ R, the composite functions are defined, and their domain is the domain of h, which is R. Their codomain is the codomain of $(f+g)$ and $(f \cdot g)$, which is R.

Similarly, the composite functions $f \circ h$ and $g \circ h$ are defined and are functions from R to R. The sum $(f \circ h) + (g \circ h)$ and product $(f \circ h) \cdot (g \circ h)$ are also functions from R to R.

Thus, all the functions involved have the same domain (R) and codomain (R). Now we show that the outputs are equal for any $x \in R$.

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Proof of 1: $(f + g) \circ h = f \circ h + g \circ h$

Let $x \in R$.

Consider the left-hand side (LHS):

$((f + g) \circ h)(x)$

By the definition of composition of functions, $(p \circ q)(x) = p(q(x))$, we have:

$((f + g) \circ h)(x) = (f + g)(h(x))$

By the definition of the sum of functions, $(f+g)(y) = f(y) + g(y)$, we have:

$(f + g)(h(x)) = f(h(x)) + g(h(x))$

Now consider the right-hand side (RHS):

$(f \circ h + g \circ h)(x)$

By the definition of the sum of functions, $(p+q)(x) = p(x) + q(x)$, where $p = f \circ h$ and $q = g \circ h$, we have:

$(f \circ h + g \circ h)(x) = (f \circ h)(x) + (g \circ h)(x)$

By the definition of composition of functions, $(p \circ q)(x) = p(q(x))$, we have:

$(f \circ h)(x) = f(h(x))$ and $(g \circ h)(x) = g(h(x))$

Substituting these into the RHS expression:

$(f \circ h)(x) + (g \circ h)(x) = f(h(x)) + g(h(x))$

Since $((f + g) \circ h)(x) = f(h(x)) + g(h(x))$ and $(f \circ h + g \circ h)(x) = f(h(x)) + g(h(x))$, we have:

$((f + g) \circ h)(x) = (f \circ h + g \circ h)(x)$ for all $x \in R$.

Therefore, $(f + g) \circ h = f \circ h + g \circ h$.

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Proof of 2: $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$

Let $x \in R$.

Consider the left-hand side (LHS):

$((f \cdot g) \circ h)(x)$

By the definition of composition of functions, $(p \circ q)(x) = p(q(x))$, we have:

$((f \cdot g) \circ h)(x) = (f \cdot g)(h(x))$

By the definition of the product of functions, $(f \cdot g)(y) = f(y) \cdot g(y)$, we have:

$(f \cdot g)(h(x)) = f(h(x)) \cdot g(h(x))$

Now consider the right-hand side (RHS):

$((f \circ h) \cdot (g \circ h))(x)$

By the definition of the product of functions, $(p \cdot q)(x) = p(x) \cdot q(x)$, where $p = f \circ h$ and $q = g \circ h$, we have:

$((f \circ h) \cdot (g \circ h))(x) = (f \circ h)(x) \cdot (g \circ h)(x)$

By the definition of composition of functions, $(p \circ q)(x) = p(q(x))$, we have:

$(f \circ h)(x) = f(h(x))$ and $(g \circ h)(x) = g(h(x))$

Substituting these into the RHS expression:

$(f \circ h)(x) \cdot (g \circ h)(x) = f(h(x)) \cdot g(h(x))$

Since $((f \cdot g) \circ h)(x) = f(h(x)) \cdot g(h(x))$ and $((f \circ h) \cdot (g \circ h))(x) = f(h(x)) \cdot g(h(x))$, we have:

$((f \cdot g) \circ h)(x) = ((f \circ h) \cdot (g \circ h))(x)$ for all $x \in R$.

Therefore, $(f \cdot g) \circ h = (f \circ h) \cdot (g \circ h)$.

The properties are shown.

Part (i):

Given:

$f(x) = |x|$

$g(x) = |5x - 2|$

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To Find:

gof and fog.

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Solution:

We need to find the composite functions gof(x) = g(f(x)) and fog(x) = f(g(x)).

Finding gof(x):

gof(x) $= g(f(x))$

Substitute $f(x) = |x|$ into the expression for g(x).

gof(x) $= g(|x|)$

Using the definition of $g(x) = |5x - 2|$, replace x with $|x|$.

gof(x) $= |5|x| - 2|$

So, $\textbf{gof(x) = |5|x| - 2|}$.

Finding fog(x):

fog(x) $= f(g(x))$

Substitute $g(x) = |5x - 2|$ into the expression for f(x).

fog(x) $= f(|5x - 2|)$

Using the definition of $f(x) = |x|$, replace x with $|5x - 2|$.

fog(x) $= ||5x - 2||$

Since the absolute value of an absolute value is the absolute value itself, we have $||a|| = |a|$.

fog(x) $= |5x - 2|$

So, $\textbf{fog(x) = |5x - 2|}$.

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Part (ii):

Given:

$f(x) = 8x^3$

$g(x) = x^{\frac{1}{3}}$

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To Find:

gof and fog.

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Solution:

We need to find the composite functions gof(x) = g(f(x)) and fog(x) = f(g(x)).

Finding gof(x):

gof(x) $= g(f(x))$

Substitute $f(x) = 8x^3$ into the expression for g(x).

gof(x) $= g(8x^3)$

Using the definition of $g(x) = x^{\frac{1}{3}}$, replace x with $8x^3$.

gof(x) $= (8x^3)^{\frac{1}{3}}$

Using the exponent rule $(ab)^m = a^m b^m$ and $(x^n)^m = x^{nm}$, we have:

gof(x) $= 8^{\frac{1}{3}} \cdot (x^3)^{\frac{1}{3}}$

gof(x) $= (2^3)^{\frac{1}{3}} \cdot x^{3 \cdot \frac{1}{3}}$

gof(x) $= 2^{3 \cdot \frac{1}{3}} \cdot x^1$

gof(x) $= 2^1 \cdot x$

gof(x) $= 2x$

So, $\textbf{gof(x) = 2x}$.

Finding fog(x):

fog(x) $= f(g(x))$

Substitute $g(x) = x^{\frac{1}{3}}$ into the expression for f(x).

fog(x) $= f(x^{\frac{1}{3}})$

Using the definition of $f(x) = 8x^3$, replace x with $x^{\frac{1}{3}}$.

fog(x) $= 8(x^{\frac{1}{3}})^3$

Using the exponent rule $(x^n)^m = x^{nm}$, we have:

fog(x) $= 8 \cdot x^{\frac{1}{3} \cdot 3}$

fog(x) $= 8 \cdot x^1$

fog(x) $= 8x$

So, $\textbf{fog(x) = 8x}$.

Given:

The function $f(x) = \frac{4x + 3}{6x - 4}$, with domain $x \neq \frac{2}{3}$.

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To Prove/Find:

1. Show that $f \circ f(x) = x$ for all $x \neq \frac{2}{3}$.

2. Find the inverse of f.

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Solution:

Part 1: Show that $f \circ f(x) = x$

We need to calculate $f \circ f(x) = f(f(x))$.

Substitute $f(x)$ into the expression for $f(x)$:

$f(f(x)) = f\left(\frac{4x + 3}{6x - 4}\right)$

Now, replace $x$ in the definition of $f(x)$ with the expression $\frac{4x + 3}{6x - 4}$:

$f\left(\frac{4x + 3}{6x - 4}\right) = \frac{4\left(\frac{4x + 3}{6x - 4}\right) + 3}{6\left(\frac{4x + 3}{6x - 4}\right) - 4}$

To simplify this complex fraction, multiply the numerator and the denominator of the main fraction by $(6x - 4)$.

$f(f(x)) = \frac{4\left(\frac{4x + 3}{6x - 4}\right)(6x - 4) + 3(6x - 4)}{6\left(\frac{4x + 3}{6x - 4}\right)(6x - 4) - 4(6x - 4)}$

Cancel out the $(6x - 4)$ terms in the numerator and denominator:

$f(f(x)) = \frac{4(4x + 3) + 3(6x - 4)}{6(4x + 3) - 4(6x - 4)}$

Expand the terms in the numerator and denominator:

$f(f(x)) = \frac{(16x + 12) + (18x - 12)}{(24x + 18) - (24x - 16)}$

Combine like terms in the numerator and denominator:

$f(f(x)) = \frac{16x + 18x + 12 - 12}{24x - 24x + 18 + 16}$

$f(f(x)) = \frac{34x}{34}$

Simplify the fraction:

$f(f(x)) = x$

This holds for all $x \neq \frac{2}{3}$, which is the domain of f where $f(x)$ is defined, and also ensures that the denominator $6(4x+3) - 4(6x-4)$ is not zero, which we can see is $34 \neq 0$. Also, for the composition $f(f(x))$ to be defined, the range of $f(x)$ must be in the domain of $f(x)$. The range of $f$ is $\{y \in \mathbb{R} \mid y \neq \frac{4}{6} = \frac{2}{3}\}$, which is indeed the domain of f. So, the composition is well-defined for all $x \neq \frac{2}{3}$.

Thus, we have shown that $\textbf{fof(x) = x}$ for all $x \neq \frac{2}{3}$.

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Part 2: Find the inverse of f

Method 1: Using the result from Part 1

We found that $f \circ f(x) = x$. By the definition of an inverse function, if $f \circ g(x) = x$ and $g \circ f(x) = x$, then g is the inverse of f, i.e., $g(x) = f^{-1}(x)$.

Since $f \circ f(x) = x$, this implies that f is its own inverse.

Therefore, $\textbf{f}$ is the inverse of itself, and $\textbf{f}^{\textbf{-1}}\textbf{(x) = f(x)} = \frac{\textbf{4x + 3}}{\textbf{6x - 4}}$.

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Method 2: Using algebraic manipulation

Let $y = f(x)$.

$y = \frac{4x + 3}{6x - 4}$

To find the inverse function $f^{-1}(x)$, we swap x and y and solve for y.

$x = \frac{4y + 3}{6y - 4}$

Multiply both sides by $(6y - 4)$ to eliminate the denominator:

$x(6y - 4) = 4y + 3$

Distribute x on the left side:

$6xy - 4x = 4y + 3$

Gather terms containing y on one side and terms without y on the other side. Let's move terms with y to the left and terms without y to the right.

$6xy - 4y = 4x + 3$

Factor out y from the terms on the left side:

$y(6x - 4) = 4x + 3$

Divide both sides by $(6x - 4)$ to solve for y:

$y = \frac{4x + 3}{6x - 4}$

This expression for y is the inverse function $f^{-1}(x)$.

So, $\textbf{f}^{\textbf{-1}}\textbf{(x) =} \frac{\textbf{4x + 3}}{\textbf{6x - 4}}$.

We can see that the result from both methods is the same.

A function has an inverse if and only if it is bijective (both injective and surjective).

---

(i) f : {1, 2, 3, 4} → {10} with f = {(1, 10), (2, 10), (3, 10), (4, 10)}

The function f maps multiple distinct elements from the domain to the same element in the codomain (10). For example, $f(1) = 10$ and $f(2) = 10$, but $1 \neq 2$.

This means the function is not injective (one-to-one).

Since the function is not injective, it is not bijective.

Reason: The function f is not one-to-one as different elements in the domain are mapped to the same element in the codomain.

Conclusion: Therefore, function f does not have an inverse.

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(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with g = {(5, 4), (6, 3), (7, 4), (8, 2)}

The function g maps two distinct elements from the domain to the same element in the codomain (4). For example, $g(5) = 4$ and $g(7) = 4$, but $5 \neq 7$.

This means the function is not injective (one-to-one).

Since the function is not injective, it is not bijective.

Reason: The function g is not one-to-one as different elements in the domain are mapped to the same element in the codomain.

Conclusion: Therefore, function g does not have an inverse.

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(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with h = {(2, 7), (3, 9), (4, 11), (5, 13)}

Let's check for injectivity:

$h(2) = 7$

$h(3) = 9$

$h(4) = 11$

$h(5) = 13$

Each distinct element in the domain {2, 3, 4, 5} is mapped to a distinct element in the codomain {7, 9, 11, 13}. So, the function is injective (one-to-one).

Let's check for surjectivity:

The range of h is the set of all image elements: Range(h) = {7, 9, 11, 13}.

The codomain of h is given as {7, 9, 11, 13}.

Since the Range(h) = Codomain(h), the function is surjective (onto).

Since the function h is both injective and surjective, it is bijective.

Reason: The function h is both one-to-one and onto, hence it is bijective.

Conclusion: Therefore, function h has an inverse.

Given:

Function $f : [-1, 1] \to \mathbb{R}$ given by $f(x) = \frac{x}{x + 2}$.

---

To Prove/Find:

1. Show that f is one-to-one on the interval $[-1, 1]$.

2. Find the inverse of the function $f : [-1, 1] \to \text{Range f}$.

---

Solution:

Part 1: Show that f is one-to-one

To show that f is one-to-one (injective), we assume that $f(x_1) = f(x_2)$ for any $x_1, x_2 \in [-1, 1]$ and show that it implies $x_1 = x_2$.

Let $x_1, x_2 \in [-1, 1]$ such that $f(x_1) = f(x_2)$.

$\frac{x_1}{x_1 + 2} = \frac{x_2}{x_2 + 2}$

Cross-multiply:

$x_1(x_2 + 2) = x_2(x_1 + 2)$

Expand both sides:

$x_1 x_2 + 2x_1 = x_1 x_2 + 2x_2$

Subtract $x_1 x_2$ from both sides:

$2x_1 = 2x_2$

Divide both sides by 2:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in [-1, 1]$, the function f is one-to-one on the interval $[-1, 1]$.

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Part 2: Find the inverse of f : [-1, 1] → Range f

To find the inverse function, we set $y = f(x)$ and solve for x in terms of y. The domain of the inverse function will be the range of the original function f.

Let $y = f(x) = \frac{x}{x + 2}$. Since the codomain is restricted to the Range f, for any $y$ in the Range f, there exists an $x \in [-1, 1]$ such that $y = f(x)$.

$y = \frac{x}{x + 2}$

Multiply both sides by $(x + 2)$: (Note that for $x \in [-1, 1]$, $x+2 \neq 0$)

$y(x + 2) = x$

Distribute y:

$yx + 2y = x$

Rearrange the terms to group x terms on one side:

$2y = x - yx$

Factor out x from the terms on the right side:

$2y = x(1 - y)$

Divide by $(1 - y)$ to solve for x: (Note that if $y=1$, $2y = 2 \neq 0$, but $x(1-y)=0$, so $y$ cannot be 1 in the range of f for $x \in [-1, 1]$. We confirm this by finding the range below).

$x = \frac{2y}{1 - y}$

This expression gives the inverse function. We denote the inverse function as $f^{-1}(y)$.

$f^{-1}(y) = \frac{2y}{1 - y}$

It is conventional to write the inverse function in terms of x, so we replace y with x:

$f^{-1}(x) = \frac{2x}{1 - x}$

Now, we need to determine the domain of $f^{-1}$, which is the range of $f$. Since f is continuous and strictly increasing on $[-1, 1]$, its range is $[f(-1), f(1)]$.

$f(-1) = \frac{-1}{-1 + 2} = \frac{-1}{1} = -1$

$f(1) = \frac{1}{1 + 2} = \frac{1}{3}$

The range of $f$ is $[-1, \frac{1}{3}]$. This is the domain of $f^{-1}$.

So, the inverse function is $\textbf{f}^{\textbf{-1}}\textbf{(x) = } \frac{\textbf{2x}}{\textbf{1 - x}}$ with domain $\textbf{[-1, } \frac{\textbf{1}}{\textbf{3}}\textbf{]}$.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = 4x + 3$.

---

To Show/Find:

1. Show that f is invertible.

2. Find the inverse of f.

---

Solution:

To show that a function is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective).

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in \mathbb{R}$ in the domain of f.

By the definition of f(x):

$4x_1 + 3 = 4x_2 + 3$

Subtract 3 from both sides of the equation:

$4x_1 = 4x_2$

Divide both sides by 4:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function f is one-to-one.

---

Checking for Surjectivity (Onto):

For the function to be onto, for every element $y$ in the codomain $\mathbb{R}$, there must exist at least one element $x$ in the domain $\mathbb{R}$ such that $f(x) = y$.

Let $y$ be an arbitrary element in the codomain $\mathbb{R}$. We set $f(x) = y$ and try to solve for x:

$4x + 3 = y$

Subtract 3 from both sides:

$4x = y - 3$

Divide both sides by 4:

$x = \frac{y - 3}{4}$

Since for every real number $y$ in the codomain, we can find a real number $x = \frac{y - 3}{4}$ in the domain such that $f(x) = y$, the function f is onto.

---

Since f is both one-to-one and onto, f is bijective and therefore invertible.

---

Finding the inverse of f:

To find the inverse function, we use the equation $x = \frac{y - 3}{4}$, which we obtained while checking for surjectivity. This equation gives us the input $x$ in terms of the output $y$.

The inverse function, denoted by $f^{-1}$, maps $y$ back to $x$. So, we have:

$f^{-1}(y) = \frac{y - 3}{4}$

It is standard practice to write the inverse function in terms of the variable x. Replacing y with x, we get:

$\textbf{f}^{\textbf{-1}}\textbf{(x) = } \frac{\textbf{x - 3}}{\textbf{4}}$

The domain and codomain of $f^{-1}$ are both $\mathbb{R}$.

Given:

The function $f : \mathbb{R}_+ \to [4, \infty)$ given by $f(x) = x^2 + 4$, where $\mathbb{R}_+ = [0, \infty)$ is the set of all non-negative real numbers.

---

To Show:

1. The function f is invertible.

2. The inverse of f is $f^{-1}(y) = \sqrt{y-4}$.

---

Solution:

To show that f is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective) on the specified domain and codomain.

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in \mathbb{R}_+$ (the domain of f).

By the definition of f(x):

$x_1^2 + 4 = x_2^2 + 4$

Subtract 4 from both sides:

$x_1^2 = x_2^2$

Taking the square root of both sides:

$\sqrt{x_1^2} = \sqrt{x_2^2}$

$|x_1| = |x_2|$

Since $x_1, x_2 \in \mathbb{R}_+ = [0, \infty)$, both $x_1$ and $x_2$ are non-negative. Therefore, $|x_1| = x_1$ and $|x_2| = x_2$.

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in \mathbb{R}_+$, the function f is one-to-one on its domain $\mathbb{R}_+$.

---

Checking for Surjectivity (Onto):

For the function to be onto the codomain $[4, \infty)$, for every element $y$ in the codomain $[4, \infty)$, there must exist at least one element $x$ in the domain $\mathbb{R}_+ = [0, \infty)$ such that $f(x) = y.

Let $y$ be an arbitrary element in the codomain $[4, \infty)$. We set $f(x) = y$ and try to solve for x:

$x^2 + 4 = y$

Subtract 4 from both sides:

$x^2 = y - 4$

Since $y \in [4, \infty)$, $y \ge 4$, which means $y - 4 \ge 0$. Thus, $y - 4$ is a non-negative number, and its square root is a real number.

Taking the square root of both sides:

$x = \pm \sqrt{y - 4}$

However, the domain of the function f is $\mathbb{R}_+ = [0, \infty)$, which consists only of non-negative real numbers. Therefore, we must choose the non-negative square root:

$x = \sqrt{y - 4}$

For every $y \in [4, \infty)$, the value $x = \sqrt{y - 4}$ is a non-negative real number, and thus belongs to the domain $\mathbb{R}_+$. Also, $f(\sqrt{y-4}) = (\sqrt{y-4})^2 + 4 = (y-4) + 4 = y$.

Since for every $y$ in the codomain $[4, \infty)$, there exists an $x = \sqrt{y - 4}$ in the domain $\mathbb{R}_+$ such that $f(x) = y$, the function f is onto the codomain $[4, \infty)$.

---

Since f is both one-to-one and onto its specified codomain, f is bijective and therefore invertible.

---

Finding the inverse of f:

From the surjectivity check, we found that for any $y$ in the range, the corresponding $x$ is given by $x = \sqrt{y - 4}$. This expression gives the inverse function $f^{-1}(y)$.

$f^{-1}(y) = \sqrt{y - 4}$

This matches the inverse function given in the question.

The domain of the inverse function $f^{-1}$ is the range of f, which is $[4, \infty)$. The codomain of $f^{-1}$ is the domain of f, which is $\mathbb{R}_+ = [0, \infty)$.

Thus, the inverse of the function $f : \mathbb{R}_+ \to [4, \infty)$ is $\textbf{f}^{\textbf{-1}}\textbf{ : [4, } \infty\textbf{) } \to \mathbb{R}_+$, given by $\textbf{f}^{\textbf{-1}}\textbf{(y) = } \sqrt{\textbf{y - 4}}$.

Given:

The function $f : \mathbb{R}_+ \to [-5, \infty)$ given by $f(x) = 9x^2 + 6x - 5$, where $\mathbb{R}_+ = [0, \infty)$ is the set of all non-negative real numbers.

---

To Show:

1. The function f is invertible.

2. The inverse of f is $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

---

Solution:

To show that f is invertible, we need to demonstrate that it is both one-to-one (injective) and onto (surjective) on the specified domain and codomain.

We can also rewrite the function $f(x)$ by completing the square to understand its behavior better:

$f(x) = 9x^2 + 6x - 5$

$f(x) = (9x^2 + 6x) - 5$

$f(x) = ((3x)^2 + 2(3x)(1) + 1^2) - 1^2 - 5$

$f(x) = (3x + 1)^2 - 1 - 5$

$f(x) = (3x + 1)^2 - 6$

Since the domain is $\mathbb{R}_+ = [0, \infty)$, for any $x \ge 0$, $3x \ge 0$, and $3x + 1 \ge 1$. Therefore, $(3x+1)^2 \ge 1^2 = 1$.

This means $f(x) = (3x+1)^2 - 6 \ge 1 - 6 = -5$. The minimum value of $f(x)$ on the domain $[0, \infty)$ is indeed $-5$, which occurs at $x=0$. The range of the function is $[-5, \infty)$, which matches the given codomain.

---

Checking for Injectivity (One-to-one):

Assume $f(x_1) = f(x_2)$ for any $x_1, x_2 \in [0, \infty)$.

Using the completed square form:

$(3x_1 + 1)^2 - 6 = (3x_2 + 1)^2 - 6$

$(3x_1 + 1)^2 = (3x_2 + 1)^2$

Taking the square root of both sides:

$\sqrt{(3x_1 + 1)^2} = \sqrt{(3x_2 + 1)^2}$

$|3x_1 + 1| = |3x_2 + 1|$

Since the domain of f is $\mathbb{R}_+ = [0, \infty)$, we have $x_1 \ge 0$ and $x_2 \ge 0$. This implies $3x_1 \ge 0$ and $3x_2 \ge 0$. Thus, $3x_1 + 1 \ge 1$ and $3x_2 + 1 \ge 1$. Both expressions are positive, so their absolute values are themselves.

$3x_1 + 1 = 3x_2 + 1$

Subtract 1 from both sides:

$3x_1 = 3x_2$

Divide both sides by 3:

$x_1 = x_2$

Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$ for all $x_1, x_2 \in [0, \infty)$, the function f is one-to-one on its domain $\mathbb{R}_+$.

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Checking for Surjectivity (Onto):

For the function to be onto the codomain $[-5, \infty)$, for every element $y$ in the codomain $[-5, \infty)$, there must exist at least one element $x$ in the domain $\mathbb{R}_+ = [0, \infty)$ such that $f(x) = y.

Let $y$ be an arbitrary element in the codomain $[-5, \infty)$. We set $f(x) = y$ and try to solve for x:

Using the completed square form:

$(3x + 1)^2 - 6 = y$

Add 6 to both sides:

$(3x + 1)^2 = y + 6$

Since $y \in [-5, \infty)$, we have $y \ge -5$, so $y + 6 \ge 1$. Thus, $y+6$ is a positive number, and its square root is a real number.

Taking the square root of both sides:

$\sqrt{(3x + 1)^2} = \sqrt{y + 6}$

$|3x + 1| = \sqrt{y + 6}$

Since the domain of f is $\mathbb{R}_+ = [0, \infty)$, we have $x \ge 0$, which means $3x+1 \ge 1$. Thus, $|3x+1| = 3x+1$.

$3x + 1 = \sqrt{y + 6}$

Subtract 1 from both sides:

$3x = \sqrt{y + 6} - 1$

Divide both sides by 3:

$x = \frac{\sqrt{y + 6} - 1}{3}$

We need to ensure that this value of x is in the domain $\mathbb{R}_+ = [0, \infty)$ for all $y \in [-5, \infty)$. Since $y \ge -5$, $y+6 \ge 1$, so $\sqrt{y+6} \ge \sqrt{1} = 1$. Therefore, $\sqrt{y+6} - 1 \ge 1 - 1 = 0$. Dividing by 3, we get $x = \frac{\sqrt{y + 6} - 1}{3} \ge 0$. This value of x is indeed in the domain $[0, \infty)$.

Thus, for every $y$ in the codomain $[-5, \infty)$, there exists a unique $x = \frac{\sqrt{y + 6} - 1}{3}$ in the domain $[0, \infty)$ such that $f(x) = y$. The function f is onto its specified codomain $[-5, \infty)$.

---

Since f is both one-to-one and onto its specified codomain, f is bijective and therefore invertible.

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Finding the inverse of f:

From the surjectivity check, we found that for any $y$ in the range, the corresponding $x$ is given by $x = \frac{\sqrt{y + 6} - 1}{3}$. This expression gives the inverse function $f^{-1}(y)$.

$f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3}$

This matches the inverse function given in the question.

The domain of the inverse function $f^{-1}$ is the range of f, which is $[-5, \infty)$. The codomain of $f^{-1}$ is the domain of f, which is $\mathbb{R}_+ = [0, \infty)$.

Thus, the inverse of the function $f : \mathbb{R}_+ \to [-5, \infty)$ is $\textbf{f}^{\textbf{-1}}\textbf{ : [-5, } \infty\textbf{) } \to \mathbb{R}_+$, given by $\textbf{f}^{\textbf{-1}}\textbf{(y) = } \frac{\textbf{ \sqrt{y+6} - 1 }}{\textbf{3}}$.

Given:

A function $f : X \to Y$ which is invertible.

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To Prove:

The inverse of f is unique.

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Proof:

Assume, for the sake of contradiction, that the invertible function f has two different inverse functions, say $g_1 : Y \to X$ and $g_2 : Y \to X$.

By the definition of an inverse function, if $g_1$ is the inverse of f, then:

where $I_Y$ is the identity function on set Y (i.e., $I_Y(y) = y$ for all $y \in Y$).

and

where $I_X$ is the identity function on set X (i.e., $I_X(x) = x$ for all $x \in X$).

Similarly, if $g_2$ is the inverse of f, then:

and

We want to show that $g_1 = g_2$. This means showing that for every element $y \in Y$, $g_1(y) = g_2(y)$.

Let $y$ be an arbitrary element in the set Y.

From (i) and (iii), we have:

and

Therefore, for any $y \in Y$, we have:

$f(g_1(y)) = y$ and $f(g_2(y)) = y$

This implies:

$f(g_1(y)) = f(g_2(y))$

Since the function f is invertible, it is necessarily one-to-one (injective). By the definition of a one-to-one function, if $f(a) = f(b)$, then it must follow that $a = b$.

In the equation $f(g_1(y)) = f(g_2(y))$, let $a = g_1(y)$ and $b = g_2(y)$. Since f is one-to-one, we can conclude:

$g_1(y) = g_2(y)$

This equality holds for every element $y$ in the domain of $g_1$ and $g_2$, which is Y.

Since the functions $g_1$ and $g_2$ have the same domain Y and the same codomain X, and produce the same output for every input $y \in Y$, they are the same function.

Thus, $g_1 = g_2$.

This contradicts our initial assumption that f has two different inverse functions. Therefore, the assumption must be false.

Hence, an invertible function has one and only one inverse.

The inverse of f is unique.

Given:

The function $f : \{1, 2, 3\} \to \{a, b, c\}$ is defined by $f(1) = a$, $f(2) = b$, and $f(3) = c$.

As a set of ordered pairs, the function f is given by:

$f = \{(1, a), (2, b), (3, c)\}$

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To Find/Show:

1. Find the inverse function $f^{-1}$.

2. Show that $(f^{-1})^{-1} = f$.

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Solution:

Part 1: Find the inverse function $f^{-1}$

To find the inverse of a function given as a set of ordered pairs, we swap the order of the elements in each pair.

The function f is given by the ordered pairs:

$f = \{(1, a), (2, b), (3, c)\}$

The domain of f is $\{1, 2, 3\}$ and the codomain is $\{a, b, c\}$. Since each element in the domain maps to a unique element in the codomain and every element in the codomain is mapped to by an element in the domain, the function f is bijective, and thus invertible.

The inverse function $f^{-1}$ will have the codomain of f as its domain, and the domain of f as its codomain. So, $f^{-1} : \{a, b, c\} \to \{1, 2, 3\}$.

By swapping the elements in each ordered pair of f, we get the ordered pairs for $f^{-1}$:

From $(1, a)$ in f, we get $(a, 1)$ in $f^{-1}$.

From $(2, b)$ in f, we get $(b, 2)$ in $f^{-1}$.

From $(3, c)$ in f, we get $(c, 3)$ in $f^{-1}$.

So, the inverse function $\textbf{f}^{\textbf{-1}}$ is given by:

$\textbf{f}^{\textbf{-1}} \textbf{= } \textbf{\{(a, 1), (b, 2), (c, 3)\}}$

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Part 2: Show that $(f^{-1})^{-1} = f$

To find the inverse of $f^{-1}$, denoted as $(f^{-1})^{-1}$, we again swap the order of the elements in each ordered pair of $f^{-1}$.

The function $f^{-1}$ is given by the ordered pairs:

$f^{-1} = \{(a, 1), (b, 2), (c, 3)\}$

The domain of $f^{-1}$ is $\{a, b, c\}$ and the codomain is $\{1, 2, 3\}$.

To find $(f^{-1})^{-1}$, we swap the elements in each ordered pair of $f^{-1}$. The domain of $(f^{-1})^{-1}$ is the codomain of $f^{-1}$, which is $\{1, 2, 3\}$, and the codomain of $(f^{-1})^{-1}$ is the domain of $f^{-1}$, which is $\{a, b, c\}$. So, $(f^{-1})^{-1} : \{1, 2, 3\} \to \{a, b, c\}$.

By swapping the elements in each ordered pair of $f^{-1}$, we get the ordered pairs for $(f^{-1})^{-1}$:

From $(a, 1)$ in $f^{-1}$, we get $(1, a)$ in $(f^{-1})^{-1}$.

From $(b, 2)$ in $f^{-1}$, we get $(2, b)$ in $(f^{-1})^{-1}$.

From $(c, 3)$ in $f^{-1}$, we get $(3, c)$ in $(f^{-1})^{-1}$.

So, the function $(f^{-1})^{-1}$ is given by:

$(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}$

By comparing the set of ordered pairs for $(f^{-1})^{-1}$ with the set of ordered pairs for f, we can see that they are identical.

$f = \{(1, a), (2, b), (3, c)\}$

$(f^{-1})^{-1} = \{(1, a), (2, b), (3, c)\}$

Therefore, we have successfully shown that $\textbf{(f}^{\textbf{-1}}\textbf{)}^{\textbf{-1}}\textbf{ = f}$.

Given:

A function $f : X \to Y$ which is invertible.

Let $f^{-1} : Y \to X$ be the inverse of f.

---

To Prove:

$(f^{-1})^{-1} = f$.

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Proof:

Since f is an invertible function with inverse $f^{-1}$, by the definition of an inverse function, we have:

1. $f \circ f^{-1} = I_Y$, where $I_Y$ is the identity function on set Y.

2. $f^{-1} \circ f = I_X$, where $I_X$ is the identity function on set X.

Now, let $g$ be the inverse of the function $f^{-1} : Y \to X$. By the definition of an inverse function, $g$ must be a function from X to Y (i.e., $g: X \to Y$) and must satisfy the following conditions:

a. $g \circ f^{-1} = I_Y$ (The composition of g with $f^{-1}$ must be the identity function on the domain of $f^{-1}$, which is Y).

b. $f^{-1} \circ g = I_X$ (The composition of $f^{-1}$ with g must be the identity function on the domain of g, which is X).

Let's compare these conditions with the properties of f and $f^{-1}$ that we started with:

From property 1, we have $f \circ f^{-1} = I_Y$. Comparing this with condition (a), $g \circ f^{-1} = I_Y$, we see that f plays the role of g.

From property 2, we have $f^{-1} \circ f = I_X$. Comparing this with condition (b), $f^{-1} \circ g = I_X$, we again see that f plays the role of g.

Both conditions for g being the inverse of $f^{-1}$ are satisfied when $g = f$.

Therefore, the function f satisfies the definition of being the inverse of $f^{-1}$.

Since the inverse of a function is unique (as shown in the previous question), there can only be one inverse for $f^{-1}$. We have shown that f is an inverse of $f^{-1}$. Thus, f must be the unique inverse of $f^{-1}$.

Hence, $(f^{-1})^{-1} = f$.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = (3-x^3)^{\frac{1}{3}}$.

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To Find:

The composite function fof(x).

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Solution:

We need to find fof(x) which is defined as $f(f(x))$.

$f \circ f(x) = f(f(x))$

Substitute the definition of $f(x)$ into the expression:

$f(f(x)) = f\left(\left(3-x^3\right)^{\frac{1}{3}}\right)$

Now, apply the definition of the function f to the term $\left(3-x^3\right)^{\frac{1}{3}}$. This means we replace 'x' in the expression $(3-x^3)^{\frac{1}{3}}$ with the entire expression $\left(3-x^3\right)^{\frac{1}{3}}$.

$f\left(\left(3-x^3\right)^{\frac{1}{3}}\right) = \left(3 - \left(\left(3-x^3\right)^{\frac{1}{3}}\right)^3\right)^{\frac{1}{3}}$

Using the property $(a^{1/3})^3 = a$ for any real number a, we have $\left(\left(3-x^3\right)^{\frac{1}{3}}\right)^3 = 3-x^3$.

Substitute this back into the expression:

$f(f(x)) = \left(3 - (3 - x^3)\right)^{\frac{1}{3}}$

Simplify the expression inside the parenthesis:

$f(f(x)) = \left(3 - 3 + x^3\right)^{\frac{1}{3}}$

$f(f(x)) = \left(x^3\right)^{\frac{1}{3}}$

Using the property $(x^n)^m = x^{nm}$, we have $(x^3)^{1/3} = x^{3 \cdot \frac{1}{3}} = x^1 = x$.

$f(f(x)) = x$

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The composite function fof(x) is x.

Comparing this result with the given options:

(A) $x^{\frac{1}{3}}$

(B) $x^3$

(C) x

(D) $(3 - x^3)$

The result matches option (C).

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The final answer is (C) x.

Given:

The function $f : \mathbb{R} - \left\{ -\frac{4}{3} \right\} \to \mathbb{R}$ defined as $f(x) = \frac{4x}{3x + 4}$.

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To Find:

The inverse of the function f, denoted by $g(y)$.

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Solution:

To find the inverse of the function f, we set $y = f(x)$ and solve for x in terms of y.

Let $y = \frac{4x}{3x + 4}$.

We need to express x as a function of y. Multiply both sides by $(3x + 4)$ (note that $3x+4 \neq 0$ for the given domain of f):

$y(3x + 4) = 4x$

Expand the left side:

$3xy + 4y = 4x$

Rearrange the terms to bring all terms containing x to one side and terms not containing x to the other side. Let's move the $3xy$ term to the right side:

$4y = 4x - 3xy$

Factor out x from the terms on the right side:

$4y = x(4 - 3y)$

Now, divide both sides by $(4 - 3y)$ to isolate x. Note that for the inverse function to be defined for a given y, $4 - 3y$ must not be zero, i.e., $y \neq \frac{4}{3}$. This indicates that the range of f is $\mathbb{R} - \{\frac{4}{3}\}$.

$x = \frac{4y}{4 - 3y}$

This expression for x in terms of y is the inverse function, which is denoted as $g(y)$ in the question.

$g(y) = \frac{4y}{4 - 3y}$

The inverse function $g$ maps from the range of f (which is $\mathbb{R} - \{\frac{4}{3}\}$) to the domain of f (which is $\mathbb{R} - \{-\frac{4}{3}\}$).

Comparing our result with the given options:

(A) $g(y) = \frac{3y}{3-4y}$

(B) $g(y) = \frac{4y}{4-3y}$

(C) $g(y) = \frac{4y}{3-4y}$

(D) $g(y) = \frac{3y}{4-3y}$

Our result $\frac{4y}{4 - 3y}$ matches option (B).

---

The inverse of f is (B) g(y) = $\frac{4y}{4-3y}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. In other words, a binary operation $*$ on A is a function $* : A \times A \to A$. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

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Addition on R:

Let '+' denote the addition operation. For any two real numbers $a, b \in \mathbb{R}$, their sum $a + b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a + b \in \mathbb{R}$.

Since addition maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, addition is a binary operation on $\mathbb{R}$.

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Subtraction on R:

Let '-' denote the subtraction operation. For any two real numbers $a, b \in \mathbb{R}$, their difference $a - b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a - b \in \mathbb{R}$.

Since subtraction maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, subtraction is a binary operation on $\mathbb{R}$.

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Multiplication on R:

Let '$\times$' or '$\cdot$' denote the multiplication operation. For any two real numbers $a, b \in \mathbb{R}$, their product $a \cdot b$ is always a real number. That is, if $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $a \cdot b \in \mathbb{R}$.

Since multiplication maps $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, multiplication is a binary operation on $\mathbb{R}$.

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Division on R:

Let '$\div$' or '/' denote the division operation. For division to be a binary operation on $\mathbb{R}$, for any two real numbers $a, b \in \mathbb{R}$, the result $\frac{a}{b}$ must always be a real number. However, division by zero is not defined in the set of real numbers.

Consider the pair of real numbers $1 \in \mathbb{R}$ and $0 \in \mathbb{R}$. The division $\frac{1}{0}$ is undefined. Since $\frac{1}{0}$ is not a real number, the result of the operation on the pair $(1, 0)$ is not in $\mathbb{R}$.

Therefore, division does not map $\mathbb{R} \times \mathbb{R}$ to $\mathbb{R}$ for all pairs $(a, b) \in \mathbb{R} \times \mathbb{R}$.

Conclusion: Division is not a binary operation on $\mathbb{R}$.

---

Division on R_* (Set of non-zero real numbers):

Let $\mathbb{R}_* = \mathbb{R} - \{0\}$ be the set of all non-zero real numbers. For division to be a binary operation on $\mathbb{R}_*$, for any two elements $a, b \in \mathbb{R}_*$, the result $\frac{a}{b}$ must be an element of $\mathbb{R}_*$. This means $\frac{a}{b}$ must be a non-zero real number.

Let $a \in \mathbb{R}_*$ and $b \in \mathbb{R}_*$. By definition of $\mathbb{R}_*$, this means $a \in \mathbb{R}$, $a \neq 0$, $b \in \mathbb{R}$, and $b \neq 0$.

The division $\frac{a}{b}$ is defined because the denominator $b$ is non-zero. The result of the division of two real numbers (with a non-zero denominator) is always a real number.

Now, we check if $\frac{a}{b}$ is non-zero. A fraction $\frac{p}{q}$ is equal to zero if and only if the numerator $p$ is zero (provided the denominator $q$ is non-zero). In this case, the numerator is $a$. Since $a \in \mathbb{R}_*$, we have $a \neq 0$. Therefore, $\frac{a}{b} \neq 0$.

So, for any $a, b \in \mathbb{R}_*$, the result $\frac{a}{b}$ is a non-zero real number, which means $\frac{a}{b} \in \mathbb{R}_*$.

Thus, division maps $\mathbb{R}_* \times \mathbb{R}_* \to \mathbb{R}_*$.

Conclusion: Division is a binary operation on the set $\mathbb{R}_*$ of non-zero real numbers.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

The set $\mathbb{N}$ represents the set of natural numbers, typically defined as $\{1, 2, 3, ...\}$.

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Subtraction on N:

Let '-' denote the subtraction operation. For subtraction to be a binary operation on $\mathbb{N}$, for any two natural numbers $a, b \in \mathbb{N}$, their difference $a - b$ must always be a natural number. We need to check if the set $\mathbb{N}$ is closed under subtraction.

Consider two natural numbers, for example, $3 \in \mathbb{N}$ and $5 \in \mathbb{N}$.

Their difference is $3 - 5 = -2$.

The number $-2$ is an integer, but it is not a natural number (since natural numbers are positive integers starting from 1). Thus, the result of subtracting two natural numbers is not always a natural number.

Consider another example: $5 \in \mathbb{N}$ and $5 \in \mathbb{N}$.

Their difference is $5 - 5 = 0$.

Depending on the definition of natural numbers (some include 0, some don't), if 0 is not included in $\mathbb{N}$, then the result is not in $\mathbb{N}$. Even if 0 is included, the first example ($3-5=-2$) demonstrates that closure fails.

Since subtraction does not map $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$ for all pairs $(a, b) \in \mathbb{N} \times \mathbb{N}$, subtraction is not a binary operation on $\mathbb{N}$.

---

Division on N:

Let '$\div$' or '/' denote the division operation. For division to be a binary operation on $\mathbb{N}$, for any two natural numbers $a, b \in \mathbb{N}$, the result $\frac{a}{b}$ must always be a natural number. We need to check if the set $\mathbb{N}$ is closed under division.

Consider two natural numbers, for example, $3 \in \mathbb{N}$ and $2 \in \mathbb{N}$.

Their division is $\frac{3}{2} = 1.5$.

The number $1.5$ is a rational number, but it is not a natural number.

Consider another example: $4 \in \mathbb{N}$ and $2 \in \mathbb{N}$.

Their division is $\frac{4}{2} = 2$.

The number 2 is a natural number. This shows that the result is sometimes in $\mathbb{N}$. However, for an operation to be binary on a set, the result must be in the set for all pairs of elements from the set.

Consider the pair of natural numbers $1 \in \mathbb{N}$ and $0$. However, 0 is not in $\mathbb{N}$, so we cannot choose 0 as the divisor from the set $\mathbb{N}$. But even when the divisor is in $\mathbb{N}$ and is non-zero, the result is not always a natural number.

Since division does not map $\mathbb{N} \times \mathbb{N}$ to $\mathbb{N}$ for all pairs $(a, b) \in \mathbb{N} \times \mathbb{N}$, division is not a binary operation on $\mathbb{N}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. That is, for every ordered pair $(a, b)$ of elements from A, the result $a * b$ is a uniquely defined element in A. This is also called the closure property.

---

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 4b^2$.

The function is defined as $* : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$.

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To Show:

The operation $*$ is a binary operation on $\mathbb{R}$.

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Proof:

For $*$ to be a binary operation on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $a * b$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let $a$ and $b$ be any two arbitrary real numbers.

The operation is defined as $a * b = a + 4b^2$.

We know the following properties of real numbers:

- If $b$ is a real number, then $b^2$ is also a real number.

- If $b^2$ is a real number, then $4b^2$ (the product of a real number and a real number) is also a real number.

- If $a$ is a real number and $4b^2$ is a real number, then their sum $a + 4b^2$ is also a real number.

Therefore, for any $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $a + 4b^2$ is a uniquely defined real number. This means $a * b \in \mathbb{R}$.

Since for every pair of real numbers $(a, b)$, the operation $a * b$ results in a unique real number, the operation $*$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $*$ is a binary operation on $\mathbb{R}$.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

Given:

X is a given set.

P is the power set of X, i.e., the set of all subsets of X. So, $P = \{A \mid A \subseteq X\}$.

Two operations are defined on P: Union (∪) and Intersection (∩).

The union operation is given by $\cup : P \times P \to P$, defined by $(A, B) \to A \cup B$.

The intersection operation is given by $\cap : P \times P \to P$, defined by $(A, B) \to A \cap B$.

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To Show:

1. The union operation (∪) is a binary operation on P.

2. The intersection operation (∩) is a binary operation on P.

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Proof for Union (∪):

For the union operation to be binary on P, for any two elements $A \in P$ and $B \in P$, the result $A \cup B$ must be a uniquely defined element that is also in P.

Let A and B be any two arbitrary elements of P. By the definition of P, this means A is a subset of X (A $\subseteq$ X) and B is a subset of X (B $\subseteq$ X).

The union of two sets A and B, denoted by $A \cup B$, is defined as the set of all elements that are in A or in B (or in both). That is, $A \cup B = \{x \mid x \in A \text{ or } x \in B\}$.

If an element $x$ is in $A \cup B$, then $x$ is either in A or in B. Since both A and B are subsets of X, any element in A is also in X, and any element in B is also in X. Therefore, any element in $A \cup B$ must be in X.

This means that the set $A \cup B$ is a subset of X ($A \cup B \subseteq X$).

By the definition of the set P, any subset of X is an element of P. Since $A \cup B$ is a subset of X, it follows that $A \cup B \in P$.

Also, the union of any two sets A and B is a uniquely defined set.

Since for every pair of subsets $(A, B)$ from P, the operation $A \cup B$ results in a unique element that is also in P, the union operation satisfies the definition of a binary operation on P.

Thus, ∪ is a binary operation on P.

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Proof for Intersection (∩):

For the intersection operation to be binary on P, for any two elements $A \in P$ and $B \in P$, the result $A \cap B$ must be a uniquely defined element that is also in P.

Let A and B be any two arbitrary elements of P. By the definition of P, this means A is a subset of X (A $\subseteq$ X) and B is a subset of X (B $\subseteq$ X).

The intersection of two sets A and B, denoted by $A \cap B$, is defined as the set of all elements that are in both A and B. That is, $A \cap B = \{x \mid x \in A \text{ and } x \in B\}$.

If an element $x$ is in $A \cap B$, then $x$ is in A and $x$ is in B. Since A is a subset of X, if $x \in A$, then $x \in X$. Since B is a subset of X, if $x \in B$, then $x \in X$. Therefore, any element in $A \cap B$ must be in X.

This means that the set $A \cap B$ is a subset of X ($A \cap B \subseteq X$).

By the definition of the set P, any subset of X is an element of P. Since $A \cap B$ is a subset of X, it follows that $A \cap B \in P$.

Also, the intersection of any two sets A and B is a uniquely defined set.

Since for every pair of subsets $(A, B)$ from P, the operation $A \cap B$ results in a unique element that is also in P, the intersection operation satisfies the definition of a binary operation on P.

Thus, ∩ is a binary operation on P.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. For the operation to be binary on set A, the result of the operation on any two elements of A must be an element that is also in A (this property is called closure).

Given:

Two operations defined on the set of real numbers $\mathbb{R}$:

1. $\vee : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $(a, b) \to \max\{a, b\}$.

2. $\wedge : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ given by $(a, b) \to \min\{a, b\}$.

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To Show:

1. The operation $\vee$ is a binary operation on $\mathbb{R}$.

2. The operation $\wedge$ is a binary operation on $\mathbb{R}$.

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Proof for $\vee$ (Maximum):

For the operation $\vee$ to be binary on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $\max\{a, b\}$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let a and b be any two arbitrary real numbers.

The value $\max\{a, b\}$ is defined as the greater of the two numbers a and b (or either one if they are equal). By definition, the maximum of two real numbers is always one of the two numbers themselves (if they are distinct) or the number itself (if they are equal).

If $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $\max\{a, b\}$ is either $a$ or $b$. Since both a and b are real numbers, their maximum must also be a real number. That is, $\max\{a, b\} \in \mathbb{R}$.

Also, for any given pair of real numbers $(a, b)$, the value $\max\{a, b\}$ is uniquely determined.

Since for every pair of real numbers $(a, b)$, the operation $\max\{a, b\}$ results in a unique real number, the operation $\vee$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $\vee$ is a binary operation on $\mathbb{R}$.

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Proof for $\wedge$ (Minimum):

For the operation $\wedge$ to be binary on $\mathbb{R}$, for any two real numbers $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $\min\{a, b\}$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let a and b be any two arbitrary real numbers.

The value $\min\{a, b\}$ is defined as the smaller of the two numbers a and b (or either one if they are equal). By definition, the minimum of two real numbers is always one of the two numbers themselves (if they are distinct) or the number itself (if they are equal).

If $a \in \mathbb{R}$ and $b \in \mathbb{R}$, then $\min\{a, b\}$ is either $a$ or $b$. Since both a and b are real numbers, their minimum must also be a real number. That is, $\min\{a, b\} \in \mathbb{R}$.

Also, for any given pair of real numbers $(a, b)$, the value $\min\{a, b\}$ is uniquely determined.

Since for every pair of real numbers $(a, b)$, the operation $\min\{a, b\}$ results in a unique real number, the operation $\wedge$ satisfies the definition of a binary operation on $\mathbb{R}$.

Thus, $\wedge$ is a binary operation on $\mathbb{R}$.

Definition:

A binary operation $*$ on a set A is called commutative if for every pair of elements $a, b \in A$, we have $a * b = b * a$.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. Addition (+) and Multiplication ($\times$) are commutative binary operations on $\mathbb{R}$.

2. Subtraction (-) is not a commutative binary operation on $\mathbb{R}$.

3. Division ($\div$) is not a commutative binary operation on $\mathbb{R}_*$.

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Proof for Addition (+):

Addition is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check for commutativity.

For any two real numbers $a, b \in \mathbb{R}$, the property of addition states that $a + b = b + a$.

Since $a + b = b + a$ for all $a, b \in \mathbb{R}$, addition is a commutative binary operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

Multiplication is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check for commutativity.

For any two real numbers $a, b \in \mathbb{R}$, the property of multiplication states that $a \times b = b \times a$.

Since $a \times b = b \times a$ for all $a, b \in \mathbb{R}$, multiplication is a commutative binary operation on $\mathbb{R}$.

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Proof for Subtraction (-):

Subtraction is a binary operation on $\mathbb{R}$ (shown in Example 29). Now we check if it is commutative.

We need to check if $a - b = b - a$ for all $a, b \in \mathbb{R}$.

Let's take a counterexample. Choose $a = 5$ and $b = 3$. Both are real numbers.

$a - b = 5 - 3 = 2$

$b - a = 3 - 5 = -2$

Since $2 \neq -2$, we have $a - b \neq b - a$ for this pair of real numbers.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}$ for which $a - b \neq b - a$, subtraction is not commutative on $\mathbb{R}$.

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Proof for Division ($\div$) on R_*:

Division is a binary operation on $\mathbb{R}_*$ (shown in Example 29). Now we check if it is commutative on $\mathbb{R}_*$.

We need to check if $a \div b = b \div a$ for all $a, b \in \mathbb{R}_*$. This is equivalent to checking if $\frac{a}{b} = \frac{b}{a}$ for all $a, b \in \mathbb{R}_*$.

Let's take a counterexample. Choose $a = 4$ and $b = 2$. Both are non-zero real numbers, so they are in $\mathbb{R}_*$.

$a \div b = \frac{4}{2} = 2$

$b \div a = \frac{2}{4} = \frac{1}{2}$

Since $2 \neq \frac{1}{2}$, we have $a \div b \neq b \div a$ for this pair of non-zero real numbers.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}_*$ for which $a \div b \neq b \div a$, division is not commutative on $\mathbb{R}_*$.

Definition:

A binary operation $*$ on a set A is called commutative if for every pair of elements $a, b \in A$, we have $a * b = b * a$. To show that an operation is not commutative, we need to find at least one pair of elements $(a, b)$ in the set such that $a * b \neq b * a$.

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 2b$.

---

To Show:

The operation $*$ is not commutative on $\mathbb{R}$.

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Proof:

For the operation $*$ to be commutative on $\mathbb{R}$, it must satisfy $a * b = b * a$ for all $a, b \in \mathbb{R}$.

The definition of the operation is $a * b = a + 2b$.

Let's calculate $b * a$ using the same definition, by swapping the roles of a and b:

$b * a = b + 2a$

For the operation to be commutative, we would need $a + 2b = b + 2a$ for all $a, b \in \mathbb{R}$.

Let's check if this equality holds for all real numbers. We can test with a specific pair of real numbers. Choose $a = 1$ and $b = 2$. Both are real numbers.

Calculate $a * b$ for these values:

$1 * 2 = 1 + 2(2) = 1 + 4 = 5$

Calculate $b * a$ for these values:

$2 * 1 = 2 + 2(1) = 2 + 2 = 4$

We have $1 * 2 = 5$ and $2 * 1 = 4$.

Since $5 \neq 4$, we have $1 * 2 \neq 2 * 1$.

Since there exists at least one pair of elements $(a, b)$ in $\mathbb{R}$ for which $a * b \neq b * a$, the operation $*$ is not commutative on $\mathbb{R}$.

This can also be seen algebraically: $a + 2b = b + 2a$ implies $2b - b = 2a - a$, which simplifies to $b = a$. This means the equality $a * b = b * a$ only holds when $a = b$, not for all pairs of distinct real numbers.

Definition:

A binary operation $*$ on a set A is called associative if for every triple of elements $a, b, c \in A$, we have $(a * b) * c = a * (b * c)$.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. Addition (+) and Multiplication ($\times$) are associative binary operations on $\mathbb{R}$.

2. Subtraction (-) is not an associative binary operation on $\mathbb{R}$.

3. Division ($\div$) is not an associative binary operation on $\mathbb{R}_*$.

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Proof for Addition (+):

Addition is a binary operation on $\mathbb{R}$. Now we check for associativity.

For any three real numbers $a, b, c \in \mathbb{R}$, the property of addition states that $(a + b) + c = a + (b + c)$.

Since $(a + b) + c = a + (b + c)$ for all $a, b, c \in \mathbb{R}$, addition is an associative binary operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

Multiplication is a binary operation on $\mathbb{R}$. Now we check for associativity.

For any three real numbers $a, b, c \in \mathbb{R}$, the property of multiplication states that $(a \times b) \times c = a \times (b \times c)$.

Since $(a \times b) \times c = a \times (b \times c)$ for all $a, b, c \in \mathbb{R}$, multiplication is an associative binary operation on $\mathbb{R}$.

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Proof for Subtraction (-):

Subtraction is a binary operation on $\mathbb{R}$. Now we check if it is associative.

We need to check if $(a - b) - c = a - (b - c)$ for all $a, b, c \in \mathbb{R}$.

Let's take a counterexample. Choose $a = 5$, $b = 3$, and $c = 2$. All are real numbers.

Calculate the left-hand side (LHS):

$(a - b) - c = (5 - 3) - 2 = 2 - 2 = 0$

Calculate the right-hand side (RHS):

$a - (b - c) = 5 - (3 - 2) = 5 - 1 = 4$

Since $0 \neq 4$, we have $(a - b) - c \neq a - (b - c)$ for this triple of real numbers.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}$ for which $(a - b) - c \neq a - (b - c)$, subtraction is not associative on $\mathbb{R}$.

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Proof for Division ($\div$) on R_*:

Division is a binary operation on $\mathbb{R}_*$. Now we check if it is associative on $\mathbb{R}_*$.

We need to check if $(a \div b) \div c = a \div (b \div c)$ for all $a, b, c \in \mathbb{R}_*$. This is equivalent to checking if $\frac{\frac{a}{b}}{c} = \frac{a}{\frac{b}{c}}$ for all $a, b, c \in \mathbb{R}_*$.

Let's take a counterexample. Choose $a = 8$, $b = 4$, and $c = 2$. All are non-zero real numbers, so they are in $\mathbb{R}_*$.

Calculate the left-hand side (LHS):

$(a \div b) \div c = (8 \div 4) \div 2 = \frac{8}{4} \div 2 = 2 \div 2 = \frac{2}{2} = 1$

Calculate the right-hand side (RHS):

$a \div (b \div c) = 8 \div (4 \div 2) = 8 \div \frac{4}{2} = 8 \div 2 = \frac{8}{2} = 4$

Since $1 \neq 4$, we have $(a \div b) \div c \neq a \div (b \div c)$ for this triple of non-zero real numbers.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}_*$ for which $(a \div b) \div c \neq a \div (b \div c)$, division is not associative on $\mathbb{R}_*$.

Definition:

A binary operation $*$ on a set A is called associative if for every triple of elements $a, b, c \in A$, we have $(a * b) * c = a * (b * c)$. To show that an operation is not associative, we need to find at least one triple of elements $(a, b, c)$ in the set such that $(a * b) * c \neq a * (b * c)$.

Given:

An operation $*$ defined on the set of real numbers $\mathbb{R}$, such that for any two real numbers a and b, the operation is given by $a * b = a + 2b$.

---

To Show:

The operation $*$ is not associative on $\mathbb{R}$.

---

Proof:

For the operation $*$ to be associative on $\mathbb{R}$, it must satisfy $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

The definition of the operation is $a * b = a + 2b$.

Let's calculate the left-hand side (LHS), $(a * b) * c$:

First, calculate $a * b$: $a * b = a + 2b$.

Now, apply the operation again, replacing the first element with $(a * b)$ and the second element with $c$:

$(a * b) * c = (a + 2b) * c$

Using the definition $x * y = x + 2y$, where $x = (a + 2b)$ and $y = c$:

$(a + 2b) * c = (a + 2b) + 2c = a + 2b + 2c$

So, LHS $= a + 2b + 2c$.

Now, let's calculate the right-hand side (RHS), $a * (b * c)$:

First, calculate $b * c$: $b * c = b + 2c$.

Now, apply the operation again, replacing the first element with $a$ and the second element with $(b * c)$:

$a * (b * c) = a * (b + 2c)$

Using the definition $x * y = x + 2y$, where $x = a$ and $y = (b + 2c)$:

$a * (b + 2c) = a + 2(b + 2c) = a + 2b + 4c$

So, RHS $= a + 2b + 4c$.

For the operation to be associative, we would need $a + 2b + 2c = a + 2b + 4c$ for all $a, b, c \in \mathbb{R}$.

Subtracting $a + 2b$ from both sides, we get $2c = 4c$. This equality holds only when $2c - 4c = 0$, which means $-2c = 0$, or $c = 0$.

This means the associative property holds only when $c = 0$, not for all real numbers $a, b, c$.

Let's confirm with a counterexample. Choose $a = 1$, $b = 1$, and $c = 1$. All are real numbers.

LHS: $(1 * 1) * 1 = (1 + 2(1)) * 1 = (1 + 2) * 1 = 3 * 1 = 3 + 2(1) = 3 + 2 = 5$

RHS: $1 * (1 * 1) = 1 * (1 + 2(1)) = 1 * (1 + 2) = 1 * 3 = 1 + 2(3) = 1 + 6 = 7$

Since $5 \neq 7$, we have $(1 * 1) * 1 \neq 1 * (1 * 1)$.

Since there exists at least one triple of elements $(a, b, c)$ in $\mathbb{R}$ for which $(a * b) * c \neq a * (b * c)$, the operation $*$ is not associative on $\mathbb{R}$.

Definition:

An element $e$ in a set A is called an identity element for a binary operation $*$ on A if for every element $a \in A$, we have $a * e = a$ and $e * a = a$. If $a * e = a$, e is called a right identity. If $e * a = a$, e is called a left identity. An identity element must be both a left and a right identity.

Given:

The set of real numbers $\mathbb{R}$ and the set of non-zero real numbers $\mathbb{R}_*$.

Operations: Addition (+), Subtraction (-), Multiplication ($\times$), Division ($\div$).

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To Show:

1. 0 is the identity for addition on $\mathbb{R}$.

2. 1 is the identity for multiplication on $\mathbb{R}$.

3. There is no identity element for subtraction on $\mathbb{R}$.

4. There is no identity element for division on $\mathbb{R}_*$.

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Proof for Addition (+):

The set is $\mathbb{R}$ and the operation is +. We want to check if 0 is the identity element for addition on $\mathbb{R}$. We need to show that for every $a \in \mathbb{R}$, $a + 0 = a$ and $0 + a = a$.

For any real number $a$, we know that $a + 0 = a$ and $0 + a = a$ by the properties of real numbers.

Since $a + 0 = a$ and $0 + a = a$ for all $a \in \mathbb{R}$, 0 is the identity element for addition on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

The set is $\mathbb{R}$ and the operation is $\times$. We want to check if 1 is the identity element for multiplication on $\mathbb{R}$. We need to show that for every $a \in \mathbb{R}$, $a \times 1 = a$ and $1 \times a = a$.

For any real number $a$, we know that $a \times 1 = a$ and $1 \times a = a$ by the properties of real numbers.

Since $a \times 1 = a$ and $1 \times a = a$ for all $a \in \mathbb{R}$, 1 is the identity element for multiplication on $\mathbb{R}$.

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Proof for Subtraction (-):

The set is $\mathbb{R}$ and the operation is $-$. We want to find an element $e \in \mathbb{R}$ such that for all $a \in \mathbb{R}$, $a - e = a$ and $e - a = a$.

Consider the right identity condition: $a - e = a$. Subtracting a from both sides gives $-e = 0$, so $e = 0$. So, if a right identity exists, it must be 0.

Now, check if $e = 0$ is also a left identity. We need to check if $0 - a = a$ for all $a \in \mathbb{R}$.

$0 - a = -a$

For $0 - a = a$ to be true, we would need $-a = a$, which implies $2a = 0$, or $a = 0$. This is only true for $a = 0$, not for all $a \in \mathbb{R}$. For example, if $a = 5$, $0 - 5 = -5 \neq 5$.

Since there is no element $e \in \mathbb{R}$ that satisfies both $a - e = a$ and $e - a = a$ for all $a \in \mathbb{R}$, there is no identity element for subtraction on $\mathbb{R}$. (Note: 0 is a right identity, but not a left identity, and an identity must be both).

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Proof for Division ($\div$) on R_*:

The set is $\mathbb{R}_*$ (non-zero real numbers) and the operation is $\div$. We want to find an element $e \in \mathbb{R}_*$ such that for all $a \in \mathbb{R}_*$, $a \div e = a$ and $e \div a = a$. This is equivalent to checking if $\frac{a}{e} = a$ and $\frac{e}{a} = a$ for all $a \in \mathbb{R}_*$.

Consider the right identity condition: $\frac{a}{e} = a$. Since $a \in \mathbb{R}_*$, $a \neq 0$. We can divide both sides by a: $\frac{1}{e} = 1$, which implies $e = 1$. So, if a right identity exists, it must be 1. Note that $1 \in \mathbb{R}_*$.

Now, check if $e = 1$ is also a left identity. We need to check if $\frac{1}{a} = a$ for all $a \in \mathbb{R}_*$.

$\frac{1}{a} = a$ implies $1 = a^2$. This is true only when $a = 1$ or $a = -1$, not for all $a \in \mathbb{R}_*$. For example, if $a = 2 \in \mathbb{R}_*$, $\frac{1}{2} \neq 2$.

Since there is no element $e \in \mathbb{R}_*$ that satisfies both $a \div e = a$ and $e \div a = a$ for all $a \in \mathbb{R}_*$, there is no identity element for division on $\mathbb{R}_*$. (Note: 1 is a right identity, but not a left identity, and an identity must be both).

Definition:

Let $*$ be a binary operation on a set A with identity element $e$. An element $b \in A$ is called an inverse of an element $a \in A$ if $a * b = e$ and $b * a = e$.

Given:

The set of real numbers $\mathbb{R}$.

Operations: Addition (+) and Multiplication ($\times$).

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To Show:

1. For the addition operation '+' on $\mathbb{R}$, the inverse of any $a \in \mathbb{R}$ is $-a$.

2. For the multiplication operation '$\times$' on $\mathbb{R}$, the inverse of any $a \in \mathbb{R}, a \neq 0$ is $\frac{1}{a}$.

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Proof for Addition (+):

We already know from Example 38 that the identity element for addition on $\mathbb{R}$ is $e = 0$.

We need to show that for any $a \in \mathbb{R}$, the element $-a$ (which is also in $\mathbb{R}$) satisfies the inverse conditions with respect to the identity element 0.

The conditions are $a + (-a) = 0$ and $(-a) + a = 0$.

From the properties of real numbers, we know that:

and

Since $a + (-a) = 0$ and $(-a) + a = 0$ for all $a \in \mathbb{R}$, the element $-a$ is the inverse of $a$ for the addition operation on $\mathbb{R}$.

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Proof for Multiplication ($\times$):

We already know from Example 38 that the identity element for multiplication on $\mathbb{R}$ is $e = 1$.

We need to show that for any $a \in \mathbb{R}$ with $a \neq 0$, the element $\frac{1}{a}$ (which is also in $\mathbb{R}$ since $a \neq 0$) satisfies the inverse conditions with respect to the identity element 1.

The conditions are $a \times \frac{1}{a} = 1$ and $\frac{1}{a} \times a = 1$.

For any non-zero real number $a$, we know that:

and

Since $a \times \frac{1}{a} = 1$ and $\frac{1}{a} \times a = 1$ for all $a \in \mathbb{R}, a \neq 0$, the element $\frac{1}{a}$ is the inverse of $a$ for the multiplication operation on $\mathbb{R}$ (excluding 0, which does not have a multiplicative inverse).

Definition:

Let $*$ be a binary operation on a set A with identity element $e$. An element $b \in A$ is called an inverse of an element $a \in A$ if $a * b = e$ and $b * a = e$. For an inverse to exist for an element $a$ in a set A under an operation $*$, the set A must have an identity element $e$ under $*$ such that $e \in A$, and the inverse element $b$ must also be in A.

Given:

The set of natural numbers $\mathbb{N}$ (usually $\{1, 2, 3, ...\}$).

Operations: Addition (+) and Multiplication ($\times$).

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To Show:

1. For the addition operation '+' on $\mathbb{N}$, $-a$ is not the inverse of $a \in \mathbb{N}$.

2. For the multiplication operation '$\times$' on $\mathbb{N}$, $\frac{1}{a}$ is not the inverse of $a \in \mathbb{N}$ (for $a \neq 1$).

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Proof for Addition (+):

The set is $\mathbb{N}$. The operation is addition (+).

First, we need to check if addition is a binary operation on $\mathbb{N}$ and if there is an identity element for addition in $\mathbb{N}$. Addition is a binary operation on $\mathbb{N}$ as the sum of two natural numbers is a natural number (e.g., $1+2=3$, $3 \in \mathbb{N}$).

The identity element for addition in the set of real numbers is 0. If 0 is included in $\mathbb{N}$ (i.e., $\mathbb{N} = \{0, 1, 2, ...\}$), then 0 is the identity element for addition in $\mathbb{N}$. For an element $a \in \mathbb{N}$ to have an inverse $b \in \mathbb{N}$, we need $a + b = 0$ and $b + a = 0$. If $a \in \mathbb{N}$ and $a > 0$, then for $a+b=0$, $b$ must be $-a$. If $a=0$, then $0+b=0$ implies $b=0$. For $a>0$, $-a$ is a negative integer and is not in $\mathbb{N}$ (unless $\mathbb{N}$ includes negative numbers, which it does not). For $a=0$, the inverse is $0$, and $0 \in \mathbb{N}$. So, 0 has an inverse (itself), but other elements like 1, 2, etc., do not have inverses in $\mathbb{N}$. The question considers $a \in \mathbb{N}$ and suggests $-a$ as a potential inverse. For any $a \in \mathbb{N}$ where $a \neq 0$, $-a$ is a negative integer and is not in $\mathbb{N}$. For an element to be an inverse, it must belong to the set $\mathbb{N}$.

If 0 is not included in $\mathbb{N}$ (i.e., $\mathbb{N} = \{1, 2, 3, ...\}$), then there is no identity element for addition in $\mathbb{N}$. For any $a \in \mathbb{N}$, $a + e = a$ implies $e = 0$, which is not in $\mathbb{N}$. Without an identity element, the concept of an inverse element is not defined.

In either case (whether 0 is included in $\mathbb{N}$ or not, but typically $\mathbb{N}$ starts from 1), for any $a \in \mathbb{N}, a \neq 0$, the element $-a$ is not in $\mathbb{N}$. Therefore, $-a$ cannot be the inverse of $a$ in $\mathbb{N}$.

Specifically, if $a \in \mathbb{N}$ and $a \neq 0$, then $-a < 0$, and negative numbers are not in $\mathbb{N}$ (unless $\mathbb{N}$ is defined differently, but the standard definition does not include them). For $-a$ to be the inverse of $a$ for addition on $\mathbb{N}$, we need $a + (-a) = e$ and $(-a) + a = e$, where $e$ is the identity element for addition on $\mathbb{N}$ and $e \in \mathbb{N}$. If the identity $e=0$ exists and is in $\mathbb{N}$, then we need $a + (-a) = 0$. For $a=1$, $1 + (-1) = 0$. But $-1 \notin \mathbb{N}$.

Therefore, for $a \in \mathbb{N}$, $-a$ is not the inverse of $a$ for the addition operation on $\mathbb{N}$ because $-a \notin \mathbb{N}$ (for $a \neq 0$).

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Proof for Multiplication ($\times$):

The set is $\mathbb{N}$. The operation is multiplication ($\times$).

First, we need to check if multiplication is a binary operation on $\mathbb{N}$ and if there is an identity element for multiplication in $\mathbb{N}$. Multiplication is a binary operation on $\mathbb{N}$ as the product of two natural numbers is a natural number (e.g., $2 \times 3 = 6$, $6 \in \mathbb{N}$).

The identity element for multiplication in the set of real numbers is 1. The element $1 \in \mathbb{N}$. So, 1 is the identity element for multiplication on $\mathbb{N}$, since for any $a \in \mathbb{N}$, $a \times 1 = a$ and $1 \times a = a$.

For an element $a \in \mathbb{N}$ to have an inverse $b \in \mathbb{N}$ for multiplication, we need $a \times b = 1$ and $b \times a = 1$, where 1 is the identity element.

The potential inverse suggested is $\frac{1}{a}$. We need to check if $\frac{1}{a}$ is in $\mathbb{N}$ for $a \in \mathbb{N}$.

Consider $a \in \mathbb{N}$ where $a \neq 1$. For example, let $a = 2$. Then the potential inverse is $\frac{1}{2}$.

$\frac{1}{2} = 0.5$. This is a rational number, but it is not a natural number.

In general, for $a \in \mathbb{N}$ and $a > 1$, the value $\frac{1}{a}$ is a fraction between 0 and 1 (i.e., $0 < \frac{1}{a} < 1$) and is not an integer. Natural numbers (typically starting from 1) are integers greater than or equal to 1.

The only natural number whose multiplicative inverse in $\mathbb{R}$ is also a natural number is 1 itself, since $\frac{1}{1} = 1$, and $1 \in \mathbb{N}$. For $a=1$, the inverse is 1, which is in $\mathbb{N}$. The question specifically asks to show that for $a \neq 1$, $\frac{1}{a}$ is not the inverse.

For any $a \in \mathbb{N}$ where $a \neq 1$, the element $\frac{1}{a}$ is not in $\mathbb{N}$. Therefore, $\frac{1}{a}$ cannot be the inverse of $a$ in $\mathbb{N}$.

Therefore, for $a \in \mathbb{N}$ and $a \neq 1$, $\frac{1}{a}$ is not the inverse of $a$ for the multiplication operation on $\mathbb{N}$ because $\frac{1}{a} \notin \mathbb{N}$ (for $a > 1$).

Definition:

A binary operation ∗ on a set A is a function that maps pairs of elements of A to elements of A. That is, for every ordered pair $(a, b)$ of elements from A, the result $a * b$ is a uniquely defined element in A (closure property).

The set $\mathbb{Z}^+$ represents the set of positive integers, i.e., $\{1, 2, 3, ...\}$.

The set $\mathbb{R}$ represents the set of all real numbers.

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(i) On Z⁺, define ∗ by a ∗ b = a – b

The set is $\mathbb{Z}^+ = \{1, 2, 3, ...\}$. The operation is $a * b = a - b$.

For $*$ to be a binary operation on $\mathbb{Z}^+$, for any two elements $a, b \in \mathbb{Z}^+$, the result $a - b$ must be a uniquely defined element that is also in $\mathbb{Z}^+$.

Consider $a = 1 \in \mathbb{Z}^+$ and $b = 2 \in \mathbb{Z}^+$.

$a * b = 1 - 2 = -1$

The result $-1$ is an integer, but it is not a positive integer. Therefore, $-1 \notin \mathbb{Z}^+$.

Since the operation on a pair of elements from $\mathbb{Z}^+$ resulted in an element outside $\mathbb{Z}^+$, the set $\mathbb{Z}^+$ is not closed under this operation.

Conclusion: The defined operation ∗ is not a binary operation on $\mathbb{Z}^+$.

Justification: For $a=1$ and $b=2$, where $a, b \in \mathbb{Z}^+$, $a * b = 1 - 2 = -1$, and $-1 \notin \mathbb{Z}^+$. The operation is not closed on $\mathbb{Z}^+$.

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(ii) On Z⁺, define ∗ by a ∗ b = ab

The set is $\mathbb{Z}^+ = \{1, 2, 3, ...\}$. The operation is $a * b = ab$ (multiplication).

For $*$ to be a binary operation on $\mathbb{Z}^+$, for any two elements $a, b \in \mathbb{Z}^+$, the result $ab$ must be a uniquely defined element that is also in $\mathbb{Z}^+$.

Let $a \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$. By the definition of positive integers, $a > 0$ and $b > 0$. The product of two positive integers is always a positive integer. Thus, $ab$ is a positive integer, so $ab \in \mathbb{Z}^+$.

Also, for any given pair of positive integers $(a, b)$, the product $ab$ is uniquely determined.

Since for every pair of elements from $\mathbb{Z}^+$, the operation results in a unique element in $\mathbb{Z}^+$, the set $\mathbb{Z}^+$ is closed under this operation.

Conclusion: The defined operation ∗ is a binary operation on $\mathbb{Z}^+$.

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(iii) On R, define ∗ by a ∗ b = ab²

The set is $\mathbb{R}$ (real numbers). The operation is $a * b = ab^2$.

For $*$ to be a binary operation on $\mathbb{R}$, for any two elements $a, b \in \mathbb{R}$, the result $ab^2$ must be a uniquely defined element that is also in $\mathbb{R}$.

Let $a \in \mathbb{R}$ and $b \in \mathbb{R}$.

If $b$ is a real number, then $b^2$ is a real number.

If $a$ is a real number and $b^2$ is a real number, then their product $ab^2$ is a real number.

Thus, for any $a \in \mathbb{R}$ and $b \in \mathbb{R}$, the result $ab^2$ is a real number, so $ab^2 \in \mathbb{R}$.

Also, for any given pair of real numbers $(a, b)$, the value $ab^2$ is uniquely determined.

Since for every pair of elements from $\mathbb{R}$, the operation results in a unique element in $\mathbb{R}$, the set $\mathbb{R}$ is closed under this operation.

Conclusion: The defined operation ∗ is a binary operation on $\mathbb{R}$.

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(iv) On Z⁺, define ∗ by a ∗ b = | a – b |

The set is $\mathbb{Z}^+ = \{1, 2, 3, ...\}$. The operation is $a * b = |a - b|$.

For $*$ to be a binary operation on $\mathbb{Z}^+$, for any two elements $a, b \in \mathbb{Z}^+$, the result $|a - b|$ must be a uniquely defined element that is also in $\mathbb{Z}^+$.

Let $a \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$. Then $a$ and $b$ are positive integers.

The difference $a - b$ is an integer (it can be positive, negative, or zero).

The absolute value $|a - b|$ is always a non-negative integer.

If $a \neq b$, then $a - b \neq 0$, so $|a - b|$ is a positive integer. In this case, $|a - b| \in \mathbb{Z}^+$.

If $a = b$, then $a - b = 0$, so $|a - b| = 0$. The number 0 is an integer, but it is not a positive integer. Therefore, if $a = b$, the result $a * b = 0 \notin \mathbb{Z}^+$.

Consider $a = 3 \in \mathbb{Z}^+$ and $b = 3 \in \mathbb{Z}^+$.

$a * b = |3 - 3| = |0| = 0$

The result 0 is not a positive integer. Therefore, $0 \notin \mathbb{Z}^+$.

Since the operation on a pair of elements from $\mathbb{Z}^+$ resulted in an element outside $\mathbb{Z}^+$, the set $\mathbb{Z}^+$ is not closed under this operation.

Conclusion: The defined operation ∗ is not a binary operation on $\mathbb{Z}^+$.

Justification: For $a=3$ and $b=3$, where $a, b \in \mathbb{Z}^+$, $a * b = |3 - 3| = 0$, and $0 \notin \mathbb{Z}^+$. The operation is not closed on $\mathbb{Z}^+$.

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(v) On Z⁺, define ∗ by a ∗ b = a

The set is $\mathbb{Z}^+ = \{1, 2, 3, ...\}$. The operation is $a * b = a$. This means the operation returns the first element of the pair.

For $*$ to be a binary operation on $\mathbb{Z}^+$, for any two elements $a, b \in \mathbb{Z}^+$, the result $a$ must be a uniquely defined element that is also in $\mathbb{Z}^+$.

Let $a \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$. By definition, $a$ is a positive integer, so $a \in \mathbb{Z}^+$. The value of $a$ is uniquely determined by the first element of the pair.

Since for every pair of elements from $\mathbb{Z}^+$, the operation results in a unique element in $\mathbb{Z}^+$, the set $\mathbb{Z}^+$ is closed under this operation.

Conclusion: The defined operation ∗ is a binary operation on $\mathbb{Z}^+$.

Definitions:

A binary operation ∗ on a set A is a function ∗ : A × A → A.

A binary operation ∗ on a set A is commutative if $a * b = b * a$ for all $a, b \in A$.

A binary operation ∗ on a set A is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in A$.

$\mathbb{Z}$ is the set of integers: $\{..., -2, -1, 0, 1, 2, ...\}$.

$\mathbb{Q}$ is the set of rational numbers.

$\mathbb{Z}^+$ is the set of positive integers: $\{1, 2, 3, ...\}$.

$\mathbb{R}$ is the set of real numbers.

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(i) On Z, define a ∗ b = a – b

Set A = $\mathbb{Z}$. Operation: $a * b = a - b$.

Binary: For any integers $a, b \in \mathbb{Z}$, the difference $a - b$ is always an integer. So, $a - b \in \mathbb{Z}$. The operation is closed on $\mathbb{Z}$ and the result is unique.

Conclusion: ∗ is a binary operation on $\mathbb{Z}$.

Commutative: Check if $a - b = b - a$ for all $a, b \in \mathbb{Z}$. Let $a=1, b=2$. $1 - 2 = -1$, $2 - 1 = 1$. Since $-1 \neq 1$, $1 * 2 \neq 2 * 1$.

Conclusion: ∗ is not commutative.

Associative: Check if $(a - b) - c = a - (b - c)$ for all $a, b, c \in \mathbb{Z}$. Let $a=1, b=2, c=3$. $(1 - 2) - 3 = -1 - 3 = -4$. $1 - (2 - 3) = 1 - (-1) = 1 + 1 = 2$. Since $-4 \neq 2$, $(1 * 2) * 3 \neq 1 * (2 * 3)$.

Conclusion: ∗ is not associative.

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(ii) On Q, define a ∗ b = ab + 1

Set A = $\mathbb{Q}$ (rational numbers). Operation: $a * b = ab + 1$.

Binary: For any rational numbers $a, b \in \mathbb{Q}$, their product $ab$ is a rational number. Adding 1 (which is a rational number) to $ab$ results in a rational number $ab + 1$. So, $ab + 1 \in \mathbb{Q}$. The operation is closed on $\mathbb{Q}$ and the result is unique.

Conclusion: ∗ is a binary operation on $\mathbb{Q}$.

Commutative: Check if $a * b = b * a$ for all $a, b \in \mathbb{Q}$. $a * b = ab + 1$. $b * a = ba + 1$. Since multiplication of rational numbers is commutative ($ab = ba$), we have $ab + 1 = ba + 1$.

Conclusion: ∗ is commutative.

Associative: Check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (ab + 1) * c = (ab + 1)c + 1 = abc + c + 1$

RHS: $a * (b * c) = a * (bc + 1) = a(bc + 1) + 1 = abc + a + 1$

For associativity, we need $abc + c + 1 = abc + a + 1$, which simplifies to $c = a$. This is not true for all $a, b, c \in \mathbb{Q}$. Let $a=1, c=2$. $abc + c + 1 = 1 \cdot b \cdot 2 + 2 + 1 = 2b + 3$. $abc + a + 1 = 1 \cdot b \cdot 2 + 1 + 1 = 2b + 2$. $2b + 3 \neq 2b + 2$. For a numerical example, $a=1, b=1, c=2$. $(1 * 1) * 2 = (1 \cdot 1 + 1) * 2 = 2 * 2 = 2 \cdot 2 + 1 = 5$. $1 * (1 * 2) = 1 * (1 \cdot 2 + 1) = 1 * 3 = 1 \cdot 3 + 1 = 4$. Since $5 \neq 4$, $(1 * 1) * 2 \neq 1 * (1 * 2)$.

Conclusion: ∗ is not associative.

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(iii) On Q, define a ∗ b = $\frac{ab}{2}$

Set A = $\mathbb{Q}$. Operation: $a * b = \frac{ab}{2}$.

Binary: For any rational numbers $a, b \in \mathbb{Q}$, their product $ab$ is a rational number. Dividing $ab$ by 2 (a non-zero rational number) results in a rational number $\frac{ab}{2}$. So, $\frac{ab}{2} \in \mathbb{Q}$. The operation is closed on $\mathbb{Q}$ and the result is unique.

Conclusion: ∗ is a binary operation on $\mathbb{Q}$.

Commutative: Check if $a * b = b * a$ for all $a, b \in \mathbb{Q}$. $a * b = \frac{ab}{2}$. $b * a = \frac{ba}{2}$. Since multiplication of rational numbers is commutative ($ab = ba$), we have $\frac{ab}{2} = \frac{ba}{2}$.

Conclusion: ∗ is commutative.

Associative: Check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = \left(\frac{ab}{2}\right) * c = \frac{\left(\frac{ab}{2}\right)c}{2} = \frac{abc}{4}$

RHS: $a * (b * c) = a * \left(\frac{bc}{2}\right) = \frac{a\left(\frac{bc}{2}\right)}{2} = \frac{abc}{4}$

Since $\frac{abc}{4} = \frac{abc}{4}$ for all $a, b, c \in \mathbb{Q}$, the associative property holds.

Conclusion: ∗ is associative.

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(iv) On Z⁺, define a ∗ b = 2^ab

Set A = $\mathbb{Z}^+$. Operation: $a * b = 2^{ab}$.

Binary: For any positive integers $a, b \in \mathbb{Z}^+$, their product $ab$ is a positive integer. Raising 2 to the power of a positive integer $ab$ results in a positive integer. For example, if $a=1, b=1$, $a * b = 2^{1 \cdot 1} = 2^1 = 2$, and $2 \in \mathbb{Z}^+$. If $a=2, b=3$, $a * b = 2^{2 \cdot 3} = 2^6 = 64$, and $64 \in \mathbb{Z}^+$. So, $2^{ab} \in \mathbb{Z}^+$. The operation is closed on $\mathbb{Z}^+$ and the result is unique.

Conclusion: ∗ is a binary operation on $\mathbb{Z}^+$.

Commutative: Check if $a * b = b * a$ for all $a, b \in \mathbb{Z}^+$. $a * b = 2^{ab}$. $b * a = 2^{ba}$. Since multiplication of integers is commutative ($ab = ba$), we have $2^{ab} = 2^{ba}$.

Conclusion: ∗ is commutative.

Associative: Check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{Z}^+$.

LHS: $(a * b) * c = (2^{ab}) * c$. Using the definition $x * y = 2^{xy}$, where $x = 2^{ab}$ and $y = c$: $(2^{ab}) * c = 2^{(2^{ab})c}$.

RHS: $a * (b * c) = a * (2^{bc})$. Using the definition $x * y = 2^{xy}$, where $x = a$ and $y = 2^{bc}$: $a * (2^{bc}) = 2^{a(2^{bc})}$.

For associativity, we need $2^{(2^{ab})c} = 2^{a(2^{bc})}$. This requires $(2^{ab})c = a(2^{bc})$. Let $a=1, b=2, c=3$.

LHS exponent: $(2^{1 \cdot 2}) \cdot 3 = 2^2 \cdot 3 = 4 \cdot 3 = 12$. So LHS is $2^{12}$.

RHS exponent: $1 \cdot (2^{2 \cdot 3}) = 1 \cdot 2^6 = 1 \cdot 64 = 64$. So RHS is $2^{64}$.

Since $12 \neq 64$, $2^{12} \neq 2^{64}$. Thus, $(a * b) * c \neq a * (b * c)$.

Conclusion: ∗ is not associative.

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(v) On Z⁺, define a ∗ b = a^b

Set A = $\mathbb{Z}^+$. Operation: $a * b = a^b$.

Binary: For any positive integers $a, b \in \mathbb{Z}^+$, $a^b$ (where $a \ge 1$ and $b \ge 1$) is always a positive integer. For example, $2^3 = 8 \in \mathbb{Z}^+$, $3^2 = 9 \in \mathbb{Z}^+$, $1^5 = 1 \in \mathbb{Z}^+$. So, $a^b \in \mathbb{Z}^+$. The operation is closed on $\mathbb{Z}^+$ and the result is unique.

Conclusion: ∗ is a binary operation on $\mathbb{Z}^+$.

Commutative: Check if $a * b = b * a$ for all $a, b \in \mathbb{Z}^+$. $a * b = a^b$. $b * a = b^a$. We need to check if $a^b = b^a$ for all $a, b \in \mathbb{Z}^+$. Let $a=2, b=3$. $2^3 = 8$. $3^2 = 9$. Since $8 \neq 9$, $2 * 3 \neq 3 * 2$.

Conclusion: ∗ is not commutative.

Associative: Check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{Z}^+$.

LHS: $(a * b) * c = (a^b) * c$. Using the definition $x * y = x^y$, where $x = a^b$ and $y = c$: $(a^b) * c = (a^b)^c = a^{bc}$.

RHS: $a * (b * c) = a * (b^c)$. Using the definition $x * y = x^y$, where $x = a$ and $y = b^c$: $a * (b^c) = a^{(b^c)}$.

For associativity, we need $a^{bc} = a^{(b^c)}$ for all $a, b, c \in \mathbb{Z}^+$. This is not generally true. For example, if $a=2, b=3, c=2$.

LHS: $(2 * 3) * 2 = (2^3) * 2 = 8 * 2 = 8^2 = 64$.

RHS: $2 * (3 * 2) = 2 * (3^2) = 2 * 9 = 2^9 = 512$.

Since $64 \neq 512$, $(2 * 3) * 2 \neq 2 * (3 * 2)$.

Conclusion: ∗ is not associative.

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(vi) On R – {– 1}, define a ∗ b = $\frac{a}{b+1}$

Set A = $\mathbb{R} - \{-1\}$. Operation: $a * b = \frac{a}{b+1}$.

Binary: For any real numbers $a, b \in \mathbb{R} - \{-1\}$, the result $a * b = \frac{a}{b+1}$ must be a uniquely defined element that is also in $\mathbb{R} - \{-1\}$.

Let $a \in \mathbb{R} - \{-1\}$ and $b \in \mathbb{R} - \{-1\}$. This means $a \in \mathbb{R}$, $a \neq -1$, $b \in \mathbb{R}$, and $b \neq -1$.

The denominator is $b+1$. Since $b \neq -1$, $b+1 \neq 0$. Division by a non-zero real number is defined, and the result $\frac{a}{b+1}$ is a real number.

For the operation to be binary on $\mathbb{R} - \{-1\}$, we need the result $\frac{a}{b+1}$ to also be in $\mathbb{R} - \{-1\}$. This means $\frac{a}{b+1}$ must be a real number different from -1.

Is it possible that $\frac{a}{b+1} = -1$ for some $a, b \in \mathbb{R} - \{-1\}$?

$\frac{a}{b+1} = -1$

$a = -1(b+1)$

$a = -b - 1$

For example, if we choose $b = 0$ (which is in $\mathbb{R} - \{-1\}$), then $a = -0 - 1 = -1$. But $a$ must be in $\mathbb{R} - \{-1\}$, so $a$ cannot be -1. So, if we pick $a = -1$ and $b=0$, the result $a*b = \frac{-1}{0+1} = -1$, which is not in the set $\mathbb{R} - \{-1\}$.

However, the definition of a binary operation requires that for *every* pair of elements from the set, the result is in the set. The definition of the set A is $\mathbb{R} - \{-1\}$, meaning $a$ and $b$ are taken from this set. If $a$ is taken from this set, $a$ cannot be -1. So the scenario where $a=-1$ is not allowed when choosing elements from the set $\mathbb{R} - \{-1\}$.

Let's re-evaluate: Let $a, b \in \mathbb{R} - \{-1\}$. Is $a*b = \frac{a}{b+1}$ always in $\mathbb{R} - \{-1\}$?

The result is certainly a real number since $b+1 \neq 0$. We need to check if $\frac{a}{b+1}$ can be equal to -1.

Suppose $\frac{a}{b+1} = -1$. Then $a = -1(b+1) = -b - 1$. If we choose $a = 2$ and $b = -3/2$, both are in $\mathbb{R} - \{-1\}$. $b+1 = -3/2 + 1 = -1/2$. $a * b = \frac{2}{-1/2} = -4$. $-4 \in \mathbb{R} - \{-1\}$.

Let's try to find $a, b$ such that $a * b = -1$. We need $\frac{a}{b+1} = -1$, i.e., $a = -(b+1)$. If we choose $b=0$, then $a=-1$. But $a$ must be in $\mathbb{R} - \{-1\}$, so $a \neq -1$. This means if $a \neq -1$, then $\frac{a}{b+1}$ cannot be equal to -1 (unless $a=0$, then $\frac{0}{b+1} = 0 \neq -1$). If $a=0$ (which is in $\mathbb{R} - \{-1\}$), then $0 * b = \frac{0}{b+1} = 0$. $0 \in \mathbb{R} - \{-1\}$.

If $a \neq -1$, can $\frac{a}{b+1} = -1$? Only if $a = -(b+1)$. If $b \neq -2$, then $b+1 \neq -1$, so $-(b+1) \neq 1$. If $a = -(b+1)$, then $a$ can be any real number except 1 (by choosing different $b \neq -1$). But $a$ must also be in $\mathbb{R} - \{-1\}$. If $a = -(b+1)$, and $a \neq -1$, then $-(b+1) \neq -1$, so $b+1 \neq 1$, $b \neq 0$. So, if $b \neq 0$ and $b \neq -1$, then $a = -(b+1)$ is a real number different from -1. Such a pair $(a, b)$ is in $(\mathbb{R} - \{-1\}) \times (\mathbb{R} - \{-1\})$, and their operation results in -1, which is not in the codomain $\mathbb{R} - \{-1\}$.

Example: $a=1, b=-2$ are not allowed because $b=-2 \notin \mathbb{R} - \{-1\}$. Let $a=2, b=-3/2$. $a, b \in \mathbb{R} - \{-1\}$. $a*b = \frac{2}{-3/2 + 1} = \frac{2}{-1/2} = -4 \in \mathbb{R} - \{-1\}$. Example: $a=-3, b=1/2$. $a, b \in \mathbb{R} - \{-1\}$. $a*b = \frac{-3}{1/2 + 1} = \frac{-3}{3/2} = -3 \cdot \frac{2}{3} = -2 \in \mathbb{R} - \{-1\}$.

Let's check the case where the result *is* -1. We need $\frac{a}{b+1} = -1$. Choose $b=1 \in \mathbb{R} - \{-1\}$. Then $\frac{a}{1+1} = \frac{a}{2}$. We need $\frac{a}{2} = -1$, so $a = -2$. $-2 \in \mathbb{R} - \{-1\}$. So, for $a=-2$ and $b=1$, both in $\mathbb{R} - \{-1\}$, $a * b = -1$, and $-1 \notin \mathbb{R} - \{-1\}$.

Conclusion: ∗ is not a binary operation on $\mathbb{R} - \{-1\}$.

Justification: For $a = -2 \in \mathbb{R} - \{-1\}$ and $b = 1 \in \mathbb{R} - \{-1\}$, $a * b = \frac{-2}{1 + 1} = \frac{-2}{2} = -1$, and $-1 \notin \mathbb{R} - \{-1\}$. The operation is not closed on $\mathbb{R} - \{-1\}$.

(If it were a binary operation, we would check for commutativity and associativity. Since it's not binary, these properties are not applicable in the strict definition).

Given:

Set A = $\{1, 2, 3, 4, 5\}$.

Binary operation $\wedge$ defined by $a \wedge b = \min\{a, b\}$.

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To Find:

The operation table for $\wedge$ on A.

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Solution:

The operation table lists the result of $a \wedge b$ for every pair $(a, b)$ of elements from the set A.

∧	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Given:

Set A = $\{1, 2, 3, 4, 5\}$.

Binary operation ∗ defined by the given operation table.

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To Compute/Determine:

(i) $(2 ∗ 3) ∗ 4$ and $2 ∗ (3 ∗ 4)$

(ii) Whether ∗ is commutative.

(iii) $(2 ∗ 3) ∗ (4 ∗ 5)$.

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Solution:

We will use the provided table to find the results of the operations.

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(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)

First, find the value of $2 ∗ 3$ from the table:

Looking at row 2 and column 3, we find $2 ∗ 3 = 1$.

Now, compute $(2 ∗ 3) ∗ 4$:

$(2 ∗ 3) ∗ 4 = 1 ∗ 4$

Looking at row 1 and column 4, we find $1 ∗ 4 = 1$.

So, $\textbf{(2 ∗ 3) ∗ 4 = 1}$.

Next, find the value of $3 ∗ 4$ from the table:

Looking at row 3 and column 4, we find $3 ∗ 4 = 1$.

Now, compute $2 ∗ (3 ∗ 4)$:

$2 ∗ (3 ∗ 4) = 2 ∗ 1$

Looking at row 2 and column 1, we find $2 ∗ 1 = 1$.

So, $\textbf{2 ∗ (3 ∗ 4) = 1}$.

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(ii) Is ∗ commutative?

A binary operation ∗ is commutative if $a ∗ b = b ∗ a$ for all $a, b$ in the set. This can be checked by seeing if the operation table is symmetric about its main diagonal.

Let's check a few pairs:

$1 ∗ 2 = 1$, $2 ∗ 1 = 1$. $1 ∗ 2 = 2 ∗ 1$.

$2 ∗ 3 = 1$, $3 ∗ 2 = 1$. $2 ∗ 3 = 3 ∗ 2$.

$2 ∗ 4 = 2$, $4 ∗ 2 = 2$. $2 ∗ 4 = 4 ∗ 2$.

$3 ∗ 4 = 1$, $4 ∗ 3 = 1$. $3 ∗ 4 = 4 ∗ 3$.

By observing the table, we can see that for every entry in the table, the element at row $i$, column $j$ is the same as the element at row $j$, column $i$.

Conclusion: Yes, the operation ∗ is commutative.

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(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5)

First, find the value of $2 ∗ 3$ from the table:

From part (i), $2 ∗ 3 = 1$.

Next, find the value of $4 ∗ 5$ from the table:

Looking at row 4 and column 5, we find $4 ∗ 5 = 1$.

Now, compute $(2 ∗ 3) ∗ (4 ∗ 5)$:

$(2 ∗ 3) ∗ (4 ∗ 5) = 1 ∗ 1$

Looking at row 1 and column 1, we find $1 ∗ 1 = 1$.

So, $\textbf{(2 ∗ 3) ∗ (4 ∗ 5) = 1}$.

Given:

Set A = $\{1, 2, 3, 4, 5\}$.

Binary operation ∗′ defined by $a ∗' b = \text{H.C.F.}(a, b)$.

The operation ∗ is defined by the table in Exercise 4.

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To Determine:

Is the operation ∗′ the same as the operation ∗?

To Justify:

Provide a reason for the determination.

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Solution:

For two binary operations on the same set to be the same, they must produce the same result for every pair of elements from the set. We need to compare the results of $a ∗' b$ (H.C.F. of a and b) with the values in the operation table for ∗ from Exercise 4.

Let's calculate $a ∗' b = \text{H.C.F.}(a, b)$ for some pairs of elements from the set $\{1, 2, 3, 4, 5\}$ and compare them with the corresponding values from the table in Exercise 4.

From the table in Exercise 4, we have the following values:

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

Now, let's calculate H.C.F. for some pairs:

For the pair (2, 3):

$2 ∗' 3 = \text{H.C.F.}(2, 3) = 1$

From the table in Exercise 4, $2 ∗ 3 = 1$. The results match for this pair.

For the pair (2, 4):

$2 ∗' 4 = \text{H.C.F.}(2, 4) = 2$

From the table in Exercise 4, $2 ∗ 4 = 2$. The results match for this pair.

For the pair (3, 4):

$3 ∗' 4 = \text{H.C.F.}(3, 4) = 1$

From the table in Exercise 4, $3 ∗ 4 = 1$. The results match for this pair.

Let's fill out the table for the operation ∗′ = H.C.F. on the set {1, 2, 3, 4, 5}:

∗' (H.C.F.)	1	2	3	4	5
1	HCF(1,1)=1	HCF(1,2)=1	HCF(1,3)=1	HCF(1,4)=1	HCF(1,5)=1
2	HCF(2,1)=1	HCF(2,2)=2	HCF(2,3)=1	HCF(2,4)=2	HCF(2,5)=1
3	HCF(3,1)=1	HCF(3,2)=1	HCF(3,3)=3	HCF(3,4)=1	HCF(3,5)=1
4	HCF(4,1)=1	HCF(4,2)=2	HCF(4,3)=1	HCF(4,4)=4	HCF(4,5)=1
5	HCF(5,1)=1	HCF(5,2)=1	HCF(5,3)=1	HCF(5,4)=1	HCF(5,5)=5

Comparing the table for ∗′ (H.C.F.) with the table for ∗ from Exercise 4, we can see that all the corresponding entries are the same.

For example, for the pair (2, 3), $2 ∗ 3 = 1$ (from table) and $2 ∗' 3 = \text{H.C.F.}(2, 3) = 1$.

For the pair (4, 5), $4 ∗ 5 = 1$ (from table) and $4 ∗' 5 = \text{H.C.F.}(4, 5) = 1$.

For the pair (3, 3), $3 ∗ 3 = 3$ (from table) and $3 ∗' 3 = \text{H.C.F.}(3, 3) = 3$.

For every pair $(a, b)$ in the set $\{1, 2, 3, 4, 5\}$, the result of $a ∗' b = \text{H.C.F.}(a, b)$ is equal to the result of $a ∗ b$ from the given table.

Conclusion: Yes, the operation ∗′ is the same as the operation ∗ defined in Exercise 4.

Justification: For every pair of elements $(a, b)$ in the set $\{1, 2, 3, 4, 5\}$, the value of $a ∗' b = \text{H.C.F.}(a, b)$ is equal to the corresponding value of $a ∗ b$ provided in the operation table from Exercise 4. Since the operations produce the same results for all pairs of elements, they are the same operation.

Given:

Set A = $\mathbb{N} = \{1, 2, 3, ...\}$ (the set of natural numbers).

Binary operation ∗ defined by $a ∗ b = \text{L.C.M.}(a, b)$, where L.C.M. denotes the Least Common Multiple.

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To Find/Determine:

(i) Values of 5 ∗ 7 and 20 ∗ 16.

(ii) Whether ∗ is commutative.

(iii) Whether ∗ is associative.

(iv) The identity element for ∗ in $\mathbb{N}$.

(v) Which elements of $\mathbb{N}$ are invertible for ∗.

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Solution:

Binary Operation Check: For any two natural numbers $a, b \in \mathbb{N}$, their L.C.M. is always a unique natural number. Thus, ∗ is indeed a binary operation on $\mathbb{N}$.

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(i) Find 5 ∗ 7 and 20 ∗ 16

$5 ∗ 7 = \text{L.C.M.}(5, 7)$

Since 5 and 7 are prime numbers, their L.C.M. is their product.

$\text{L.C.M.}(5, 7) = 5 \times 7 = 35$

So, $\textbf{5 ∗ 7 = 35}$.

$20 ∗ 16 = \text{L.C.M.}(20, 16)$

Let's find the prime factorisation:

The L.C.M. is found by taking the highest power of all prime factors present in the numbers.

So, $\textbf{20 ∗ 16 = 80}$.

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(ii) Is ∗ commutative?

We need to check if $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

By definition, $a ∗ b = \text{L.C.M.}(a, b)$.

And $b ∗ a = \text{L.C.M.}(b, a)$.

The L.C.M. of two numbers is independent of the order of the numbers. For any natural numbers a and b, $\text{L.C.M.}(a, b) = \text{L.C.M.}(b, a)$.

Therefore, $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

Conclusion: The operation ∗ is commutative.

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(iii) Is ∗ associative?

We need to check if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in \mathbb{N}$.

LHS: $(a ∗ b) ∗ c = \text{L.C.M.}(a, b) ∗ c = \text{L.C.M.}(\text{L.C.M.}(a, b), c)$

RHS: $a ∗ (b ∗ c) = a ∗ \text{L.C.M.}(b, c) = \text{L.C.M.}(a, \text{L.C.M.}(b, c))$

The property of L.C.M. states that $\text{L.C.M.}(\text{L.C.M.}(a, b), c) = \text{L.C.M.}(a, b, c)$ and $\text{L.C.M.}(a, \text{L.C.M.}(b, c)) = \text{L.C.M.}(a, b, c)$.

Since $\text{L.C.M.}(\text{L.C.M.}(a, b), c) = \text{L.C.M.}(a, \text{L.C.M.}(b, c))$ for all $a, b, c \in \mathbb{N}$, the associative property holds.

Conclusion: The operation ∗ is associative.

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(iv) Find the identity of ∗ in N

We need to find an element $e \in \mathbb{N}$ such that for all $a \in \mathbb{N}$, $a ∗ e = a$ and $e ∗ a = a$.

Using the definition of ∗, we need $\text{L.C.M.}(a, e) = a$ and $\text{L.C.M.}(e, a) = a$ for all $a \in \mathbb{N}$.

The L.C.M. of two numbers is equal to one of the numbers if and only if that number is a multiple of the other number. Specifically, $\text{L.C.M.}(a, e) = a$ if and only if a is a multiple of e.

So, we need $a$ to be a multiple of $e$ for all natural numbers $a \in \mathbb{N}$. This means e must divide every natural number (1, 2, 3, 4, ...).

The only natural number that divides every natural number is 1.

Let's check if $e = 1$ satisfies the identity conditions:

$a ∗ 1 = \text{L.C.M.}(a, 1) = a$ (since any number 'a' is a multiple of 1)

$1 ∗ a = \text{L.C.M.}(1, a) = a$ (since any number 'a' is a multiple of 1)

Since $a ∗ 1 = a$ and $1 ∗ a = a$ for all $a \in \mathbb{N}$, the element 1 is the identity element for the operation ∗ in $\mathbb{N}$.

Conclusion: The identity element of ∗ in $\mathbb{N}$ is $\textbf{1}$.

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(v) Which elements of N are invertible for the operation ∗?

An element $a \in \mathbb{N}$ is invertible if there exists an element $b \in \mathbb{N}$ (called the inverse of a) such that $a ∗ b = e$ and $b ∗ a = e$, where $e$ is the identity element. We found the identity element to be $e = 1$.

So, we need to find $a \in \mathbb{N}$ such that there exists $b \in \mathbb{N}$ satisfying:

$a ∗ b = 1$ and $b ∗ a = 1$

Using the definition of ∗, this means $\text{L.C.M.}(a, b) = 1$.

The L.C.M. of two natural numbers is 1 if and only if both numbers are 1.

Let $a, b \in \mathbb{N}$. If $\text{L.C.M.}(a, b) = 1$, then the prime factorisation of $a$ and $b$ can only contain primes raised to the power of 0. This is only possible if $a=1$ and $b=1$.

So, the only element $a \in \mathbb{N}$ for which $\text{L.C.M.}(a, b) = 1$ is when $a=1$, and its inverse $b$ must be 1.

Check for $a=1$: We need to find $b \in \mathbb{N}$ such that $\text{L.C.M.}(1, b) = 1$. $\text{L.C.M.}(1, b) = b$. So, we need $b = 1$. The inverse of 1 is 1, and $1 \in \mathbb{N}$.

Check for $a=2$: We need to find $b \in \mathbb{N}$ such that $\text{L.C.M.}(2, b) = 1$. This is not possible for any $b \in \mathbb{N}$, because the L.C.M. of 2 and any natural number will always be at least 2 (since L.C.M. must be a multiple of the numbers). The only way L.C.M. is 1 is if both numbers are 1.

Therefore, the only element in $\mathbb{N}$ that has an inverse under the operation ∗ is 1.

Conclusion: The only element of $\mathbb{N}$ that is invertible for the operation ∗ is $\textbf{1}$. The inverse of 1 is 1.

Definition:

A binary operation $*$ on a set A is a function that maps pairs of elements of A to elements of A. That is, for every ordered pair $(a, b)$ of elements from A, the result $a * b$ must be a uniquely defined element that is also in A (this property is called closure).

Given:

The set A = $\{1, 2, 3, 4, 5\}$.

The operation $*$ defined by $a * b = \text{L.C.M.}(a, b)$.

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To Determine:

Whether ∗ is a binary operation on the set {1, 2, 3, 4, 5}.

To Justify:

Provide a reason for the determination.

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Solution:

For the operation ∗ to be a binary operation on the set $\{1, 2, 3, 4, 5\}$, the result of $a * b = \text{L.C.M.}(a, b)$ must be an element of the set $\{1, 2, 3, 4, 5\}$ for every pair of elements $a, b$ taken from this set.

Let's check some pairs of elements from the set:

Consider $a = 2 \in \{1, 2, 3, 4, 5\}$ and $b = 3 \in \{1, 2, 3, 4, 5\}$.

Let's compute $a * b$ for this pair:

$a * b = 2 * 3 = \text{L.C.M.}(2, 3)$

The Least Common Multiple of 2 and 3 is 6.

So, $2 * 3 = 6$.

Now, we check if the result 6 is an element of the set $\{1, 2, 3, 4, 5\}$.

The number 6 is not in the set $\{1, 2, 3, 4, 5\}$.

Since the operation on the pair of elements (2, 3) from the set resulted in an element (6) that is not in the set, the set is not closed under this operation.

Conclusion: The operation ∗ defined by $a * b = \text{L.C.M.}(a, b)$ is not a binary operation on the set {1, 2, 3, 4, 5}.

Justification: For the elements $2 \in \{1, 2, 3, 4, 5\}$ and $3 \in \{1, 2, 3, 4, 5\}$, the result of the operation is $2 * 3 = \text{L.C.M.}(2, 3) = 6$. The number 6 is not an element of the set $\{1, 2, 3, 4, 5\}$. Therefore, the operation is not closed on the set, and thus it is not a binary operation on this set.

Given:

Set A = $\mathbb{N} = \{1, 2, 3, ...\}$ (the set of natural numbers).

Binary operation ∗ defined by $a ∗ b = \text{H.C.F.}(a, b)$, where H.C.F. denotes the Highest Common Factor (also known as Greatest Common Divisor, G.C.D.).

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To Determine:

1. Whether ∗ is commutative.

2. Whether ∗ is associative.

3. Whether an identity element exists for ∗ in $\mathbb{N}$.

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Solution:

Binary Operation Check: For any two natural numbers $a, b \in \mathbb{N}$, their H.C.F. is always a unique natural number (H.C.F. is always $\ge 1$). Thus, ∗ is indeed a binary operation on $\mathbb{N}$.

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Is ∗ commutative?

An operation ∗ is commutative if $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

By definition, $a ∗ b = \text{H.C.F.}(a, b)$.

And $b ∗ a = \text{H.C.F.}(b, a)$.

The H.C.F. of two numbers is independent of the order of the numbers. For any natural numbers a and b, $\text{H.C.F.}(a, b) = \text{H.C.F.}(b, a)$.

Therefore, $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

Conclusion: The operation ∗ is commutative.

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Is ∗ associative?

An operation ∗ is associative if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in \mathbb{N}$.

LHS: $(a ∗ b) ∗ c = \text{H.C.F.}(a, b) ∗ c = \text{H.C.F.}(\text{H.C.F.}(a, b), c)$

RHS: $a ∗ (b ∗ c) = a ∗ \text{H.C.F.}(b, c) = \text{H.C.F.}(a, \text{H.C.F.}(b, c))$

The property of H.C.F. states that $\text{H.C.F.}(\text{H.C.F.}(a, b), c) = \text{H.C.F.}(a, b, c)$ and $\text{H.C.F.}(a, \text{H.C.F.}(b, c)) = \text{H.C.F.}(a, b, c)$.

Since $\text{H.C.F.}(\text{H.C.F.}(a, b), c) = \text{H.C.F.}(a, \text{H.C.F.}(b, c))$ for all $a, b, c \in \mathbb{N}$, the associative property holds.

Conclusion: The operation ∗ is associative.

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Does there exist an identity element for ∗ in N?

An element $e \in \mathbb{N}$ is an identity element for ∗ if for all $a \in \mathbb{N}$, $a ∗ e = a$ and $e ∗ a = a$.

Using the definition of ∗, we need $\text{H.C.F.}(a, e) = a$ and $\text{H.C.F.}(e, a) = a$ for all $a \in \mathbb{N}$.

The H.C.F. of two natural numbers is equal to one of the numbers if and only if that number divides the other number. Specifically, $\text{H.C.F.}(a, e) = a$ if and only if $a$ divides $e$ (i.e., $e$ is a multiple of $a$).

So, we need $a$ to divide $e$ for all natural numbers $a \in \mathbb{N}$. This means e must be a multiple of every natural number (1, 2, 3, 4, ...).

If $e$ is a multiple of every natural number, it must be a multiple of 1, 2, 3, 4, 5, and so on.

For example, if $e$ is a multiple of 1, $e \ge 1$. If $e$ is a multiple of 2, $e$ is even. If $e$ is a multiple of 3, $e$ is divisible by 3. If $e$ is a multiple of 4, $e$ is divisible by 4, etc.

The only way a natural number $e$ can be a multiple of *every* natural number is if there is no such natural number $e$, other than the trivial case which is not possible for all $a$. Consider any $a > 1$. If $\text{H.C.F.}(a, e) = a$, then $a$ must divide $e$. This must hold for all $a \in \mathbb{N}$. This means $e$ must be a multiple of 1, 2, 3, 4, ... which is impossible for any finite natural number $e$ other than if $a=1$. If $a=1$, $\text{H.C.F.}(1, e) = 1$, which holds for any $e \in \mathbb{N}$. However, it must hold for *all* $a$. For $a=2$, we need $\text{H.C.F.}(2, e) = 2$, so 2 must divide $e$. For $a=3$, we need $\text{H.C.F.}(3, e) = 3$, so 3 must divide $e$. This implies $e$ must be a common multiple of all natural numbers.

There is no single natural number (except possibly if the set were just {1}) that is a multiple of every natural number in $\mathbb{N}$. The only candidate would be an infinitely large number, which is not in $\mathbb{N}$.

Alternatively, consider $\text{H.C.F.}(a, e) = a$. For this to hold for all $a \in \mathbb{N}$, we must have $a$ dividing $e$ for all $a \in \mathbb{N}$. The only natural number $e$ that is divisible by every natural number $a$ is not a natural number itself (if we consider $\mathbb{N} = \{1, 2, 3, ...\}$). No single natural number is divisible by all natural numbers other than potentially 0, which is not in $\mathbb{N}$.

Let's test a potential candidate. If $e=1$, $\text{H.C.F.}(a, 1) = 1$. We need $\text{H.C.F.}(a, 1) = a$. This only holds if $a=1$. So 1 is not an identity for all $a \in \mathbb{N}$.

Conclusion: There does not exist an identity element for the binary operation ∗ on $\mathbb{N}$.

Definitions:

A binary operation ∗ on a set A is commutative if $a * b = b * a$ for all $a, b \in A$.

A binary operation ∗ on a set A is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in A$.

The set $\mathbb{Q}$ is the set of rational numbers.

First, we confirm that each operation is a binary operation on $\mathbb{Q}$, meaning that for any $a, b \in \mathbb{Q}$, the result $a * b$ is a unique element in $\mathbb{Q}$. The sum, difference, product, and square of rational numbers are rational numbers. Division of a rational number by a non-zero rational number is a rational number. Raising a non-zero rational number to an integer power results in a rational number. For all cases below, for any $a, b \in \mathbb{Q}$, the result $a * b$ is a uniquely determined rational number. Thus, all the given operations are binary operations on $\mathbb{Q}$.

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(i) a ∗ b = a – b

Commutative: Check if $a - b = b - a$ for all $a, b \in \mathbb{Q}$. This is not true in general. For example, $1 - 0 = 1$, but $0 - 1 = -1$. $1 \neq -1$.

Conclusion: Not commutative.

Associative: Check if $(a - b) - c = a - (b - c)$ for all $a, b, c \in \mathbb{Q}$. $(a - b) - c = a - b - c$. $a - (b - c) = a - b + c$. These are equal only if $c = -c$, which means $2c = 0$, or $c = 0$. This is not true for all $c \in \mathbb{Q}$. For example, $(1 - 1) - 1 = 0 - 1 = -1$. $1 - (1 - 1) = 1 - 0 = 1$. $-1 \neq 1$.

Conclusion: Not associative.

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(ii) a ∗ b = a² + b²

Commutative: Check if $a^2 + b^2 = b^2 + a^2$ for all $a, b \in \mathbb{Q}$. This is true because addition of rational numbers is commutative. $a^2 + b^2 = b^2 + a^2$.

Conclusion: Commutative.

Associative: Check if $(a^2 + b^2)^2 + c^2 = a^2 + (b^2 + c^2)^2$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (a^2 + b^2) * c = (a^2 + b^2)^2 + c^2 = (a^4 + 2a^2 b^2 + b^4) + c^2 = a^4 + b^4 + c^2 + 2a^2 b^2$

RHS: $a * (b * c) = a * (b^2 + c^2) = a^2 + (b^2 + c^2)^2 = a^2 + (b^4 + 2b^2 c^2 + c^4) = a^2 + b^4 + c^4 + 2b^2 c^2$

For associativity, we need $a^4 + b^4 + c^2 + 2a^2 b^2 = a^2 + b^4 + c^4 + 2b^2 c^2$. This is not true for all $a, b, c \in \mathbb{Q}$. For example, let $a=1, b=0, c=1$.

LHS: $(1^2 + 0^2)^2 + 1^2 = (1)^2 + 1 = 1 + 1 = 2$.

RHS: $1^2 + (0^2 + 1^2)^2 = 1 + (1)^2 = 1 + 1 = 2$.

Let $a=1, b=1, c=1$.

LHS: $(1^2 + 1^2)^2 + 1^2 = (1 + 1)^2 + 1 = 2^2 + 1 = 4 + 1 = 5$.

RHS: $1^2 + (1^2 + 1^2)^2 = 1 + (1 + 1)^2 = 1 + 2^2 = 1 + 4 = 5$.

Let $a=1, b=2, c=3$.

LHS: $(1^2 + 2^2)^2 + 3^2 = (1 + 4)^2 + 9 = 5^2 + 9 = 25 + 9 = 34$.

RHS: $1^2 + (2^2 + 3^2)^2 = 1 + (4 + 9)^2 = 1 + 13^2 = 1 + 169 = 170$.

Since $34 \neq 170$, not associative.

Conclusion: Not associative.

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(iii) a ∗ b = a + ab

Commutative: Check if $a + ab = b + ba$ for all $a, b \in \mathbb{Q}$. $a + ab = b + ab$. This requires $a = b$. This is not true for all $a, b \in \mathbb{Q}$. For example, $1 * 2 = 1 + 1 \cdot 2 = 1 + 2 = 3$. $2 * 1 = 2 + 2 \cdot 1 = 2 + 2 = 4$. Since $3 \neq 4$, $1 * 2 \neq 2 * 1$.

Conclusion: Not commutative.

Associative: Check if $(a + ab) + (a + ab)c = a + (b + bc)a$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (a + ab) * c = (a + ab) + (a + ab)c = a + ab + ac + abc$

RHS: $a * (b * c) = a * (b + bc) = a + a(b + bc) = a + ab + abc$

For associativity, we need $a + ab + ac + abc = a + ab + abc$, which simplifies to $ac = 0$. This is true only if $a = 0$ or $c = 0$, not for all $a, b, c \in \mathbb{Q}$. For example, let $a=1, b=1, c=1$.

LHS: $(1 * 1) * 1 = (1 + 1 \cdot 1) * 1 = 2 * 1 = 2 + 2 \cdot 1 = 4$.

RHS: $1 * (1 * 1) = 1 * (1 + 1 \cdot 1) = 1 * 2 = 1 + 1 \cdot 2 = 3$.

Since $4 \neq 3$, not associative.

Conclusion: Not associative.

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(iv) a ∗ b = (a – b)²

Commutative: Check if $(a - b)^2 = (b - a)^2$ for all $a, b \in \mathbb{Q}$. $(a - b)^2 = (-(b - a))^2 = (-1)^2 (b - a)^2 = 1 \cdot (b - a)^2 = (b - a)^2$. This is true because the square of a number is the same as the square of its negative.

Conclusion: Commutative.

Associative: Check if $((a - b)^2 - c)^2 = (a - (b - c)^2)^2$ for all $a, b, c \in \mathbb{Q}$. This is not true in general. For example, let $a=1, b=2, c=3$.

LHS: $(1 * 2) * 3 = (1 - 2)^2 * 3 = (-1)^2 * 3 = 1 * 3 = (1 - 3)^2 = (-2)^2 = 4$.

RHS: $1 * (2 * 3) = 1 * (2 - 3)^2 = 1 * (-1)^2 = 1 * 1 = (1 - 1)^2 = 0^2 = 0$.

Since $4 \neq 0$, not associative.

Conclusion: Not associative.

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(v) a ∗ b = $\frac{ab}{4}$

Commutative: Check if $\frac{ab}{4} = \frac{ba}{4}$ for all $a, b \in \mathbb{Q}$. Since multiplication of rational numbers is commutative ($ab = ba$), $\frac{ab}{4} = \frac{ba}{4}$.

Conclusion: Commutative.

Associative: Check if $\frac{\frac{ab}{4} \cdot c}{4} = \frac{a \cdot \frac{bc}{4}}{4}$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = \left(\frac{ab}{4}\right) * c = \frac{\left(\frac{ab}{4}\right)c}{4} = \frac{abc}{16}$

RHS: $a * (b * c) = a * \left(\frac{bc}{4}\right) = \frac{a\left(\frac{bc}{4}\right)}{4} = \frac{abc}{16}$

Since $\frac{abc}{16} = \frac{abc}{16}$ for all $a, b, c \in \mathbb{Q}$, the associative property holds.

Conclusion: Associative.

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(vi) a ∗ b = ab²

Commutative: Check if $ab^2 = ba^2$ for all $a, b \in \mathbb{Q}$. This requires $b^2 = ba$ (if $a \neq 0$) or $ab = a^2$ (if $b \neq 0$). This is not true in general. For example, $1 * 2 = 1 \cdot 2^2 = 1 \cdot 4 = 4$. $2 * 1 = 2 \cdot 1^2 = 2 \cdot 1 = 2$. Since $4 \neq 2$, $1 * 2 \neq 2 * 1$.

Conclusion: Not commutative.

Associative: Check if $(ab^2)c^2 = a(bc^2)^2$ for all $a, b, c \in \mathbb{Q}$.

LHS: $(a * b) * c = (ab^2) * c = (ab^2)c^2 = ab^2 c^2$

RHS: $a * (b * c) = a * (bc^2) = a(bc^2)^2 = a(b^2 c^4) = ab^2 c^4$

For associativity, we need $ab^2 c^2 = ab^2 c^4$. If $a \neq 0, b \neq 0, c \neq 0$, we can divide by $ab^2 c^2$ to get $1 = c^2$. This is true only if $c = 1$ or $c = -1$, not for all $c \in \mathbb{Q}$. For example, let $a=1, b=1, c=2$.

LHS: $(1 * 1) * 2 = (1 \cdot 1^2) * 2 = 1 * 2 = 1 \cdot 2^2 = 4$.

RHS: $1 * (1 * 2) = 1 * (1 \cdot 2^2) = 1 * 4 = 1 \cdot 4^2 = 16$.

Since $4 \neq 16$, not associative.

Conclusion: Not associative.

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Summary:

Operation	Definition	Commutative?	Associative?
(i)	$a * b = a - b$	No	No
(ii)	$a * b = a^2 + b^2$	Yes	No
(iii)	$a * b = a + ab$	No	No
(iv)	$a * b = (a - b)^2$	Yes	No
(v)	$a * b = \frac{ab}{4}$	Yes	Yes
(vi)	$a * b = ab^2$	No	No

Definition:

An element $e$ in a set A is called an identity element for a binary operation $*$ on A if for every element $a \in A$, we have $a * e = a$ and $e * a = a$. The identity element must belong to the set A.

Given:

The set A = $\mathbb{Q}$ (the set of rational numbers).

Six binary operations defined in Question 9.

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To Show:

None of the six operations defined in Question 9 has an identity element in $\mathbb{Q}$.

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Solution:

We will examine each operation to see if there exists an element $e \in \mathbb{Q}$ satisfying $a * e = a$ and $e * a = a$ for all $a \in \mathbb{Q}$.

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(i) a ∗ b = a – b

We need $e \in \mathbb{Q}$ such that $a - e = a$ and $e - a = a$ for all $a \in \mathbb{Q}$.

From the right identity condition $a - e = a$, subtracting a from both sides gives $e = 0$. The potential identity is 0, which is in $\mathbb{Q}$.

Now check if $e = 0$ satisfies the left identity condition $e - a = a$ for all $a \in \mathbb{Q}$. Substituting $e=0$, we get $0 - a = a$, which simplifies to $-a = a$. This equation implies $2a = 0$, so $a = 0$. This is not true for all $a \in \mathbb{Q}$. For example, if $a=5$, $0 - 5 = -5$, but the condition requires the result to be 5. $-5 \neq 5$.

Since no single element $e \in \mathbb{Q}$ satisfies both conditions for all $a \in \mathbb{Q}$, there is no identity element for this operation.

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(ii) a ∗ b = a² + b²

We need $e \in \mathbb{Q}$ such that $a^2 + e^2 = a$ and $e^2 + a^2 = a$ for all $a \in \mathbb{Q}$. Since addition is commutative, $a^2 + e^2 = e^2 + a^2$ is always true. We only need to find $e \in \mathbb{Q}$ such that $a^2 + e^2 = a$ for all $a \in \mathbb{Q}$.

Rearranging the equation, $e^2 = a - a^2$.

For $e$ to be an identity element, it must be a fixed rational number. This means $e^2$ must be a fixed rational number for all $a \in \mathbb{Q}$. However, the expression $a - a^2$ varies depending on the value of $a$.

For example, if we take $a = 1 \in \mathbb{Q}$, we need $e^2 = 1 - 1^2 = 0$, which gives $e = 0$.

If we take $a = 2 \in \mathbb{Q}$, we need $e^2 = 2 - 2^2 = 2 - 4 = -2$. For $e$ to be a real number, $e^2$ must be non-negative. $-2$ is not $\ge 0$, so there is no real number $e$ whose square is -2. Thus, there is no rational number $e$ whose square is -2.

Since the requirement for $e^2$ depends on $a$ and is not always a non-negative rational number that is a perfect square of a rational number, there is no constant $e \in \mathbb{Q}$ satisfying the condition for all $a \in \mathbb{Q}$.

Therefore, there is no identity element for this operation.

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(iii) a ∗ b = a + ab

We need $e \in \mathbb{Q}$ such that $a * e = a$ and $e * a = a$ for all $a \in \mathbb{Q}$.

From the right identity condition $a * e = a$: $a + ae = a$. Subtracting a from both sides gives $ae = 0$. This must hold for all $a \in \mathbb{Q}$. If we choose any $a \neq 0$ (e.g., $a=1$), we get $1 \cdot e = 0$, which implies $e = 0$. So, if a right identity exists, it must be 0. The number 0 is in $\mathbb{Q}$.

Now check if $e = 0$ satisfies the left identity condition $e * a = a$ for all $a \in \mathbb{Q}$. Substituting $e=0$, we get $0 * a = a + 0 \cdot a = a + 0 = a$. This simplifies to $0 = a$. This condition must hold for all $a \in \mathbb{Q}$, but it only holds when $a=0$. For example, if $a=5$, $0 * 5 = 0$, but the condition requires the result to be 5. $0 \neq 5$.

Since no single element $e \in \mathbb{Q}$ satisfies both conditions for all $a \in \mathbb{Q}$, there is no identity element for this operation.

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(iv) a ∗ b = (a – b)²

We need $e \in \mathbb{Q}$ such that $a * e = a$ and $e * a = a$ for all $a \in \mathbb{Q}$. Since the operation is commutative, we only need to find $e \in \mathbb{Q}$ such that $a * e = a$ for all $a \in \mathbb{Q}$.

The condition is $(a - e)^2 = a$.

Taking the square root of both sides gives $a - e = \pm \sqrt{a}$.

So, $e = a \mp \sqrt{a}$.

For $e$ to be an identity element, it must be a fixed rational number, independent of $a$. The expression $a \mp \sqrt{a}$ depends on $a$. Furthermore, for $e$ to be rational, $\sqrt{a}$ must be rational for all $a$ for which the equation holds. However, $\sqrt{a}$ is not rational for all $a \in \mathbb{Q}$ (e.g., if $a=2$, $\sqrt{2}$ is irrational). More importantly, $e$ must be a constant value.

Let's see if there is any $a \in \mathbb{Q}$ for which $a - a^2$ is a rational square. Example $a=1$, $e^2 = 0$, $e=0$. Example $a=4$, $e^2 = 4 - 16 = -12$, no real $e$. Example $a=1/4$, $e^2 = 1/4 - (1/4)^2 = 1/4 - 1/16 = 4/16 - 1/16 = 3/16$, $e = \pm \sqrt{3}/4$, not rational. This shows $e$ depends on $a$ and is not always rational.

Even if we try to find a single $e$ that satisfies the condition for some $a$, it won't work for all $a$. For example, if $e=0$, $(a-0)^2 = a^2$. We need $a^2 = a$ for all $a \in \mathbb{Q}$, which is only true for $a=0$ or $a=1$. So $e=0$ is not the identity.

Therefore, there is no identity element for this operation.

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(v) a ∗ b = $\frac{ab}{4}$

We need $e \in \mathbb{Q}$ such that $a * e = a$ and $e * a = a$ for all $a \in \mathbb{Q}$. Since the operation is commutative, we only need to find $e \in \mathbb{Q}$ such that $a * e = a$ for all $a \in \mathbb{Q}$.

The condition is $\frac{ae}{4} = a$.

This equation must hold for all $a \in \mathbb{Q}$.

If we consider $a \neq 0$, we can multiply both sides by 4 and divide by $a$: $ae = 4a \implies e = 4$. The potential identity is 4, which is in $\mathbb{Q}$.

Now we must verify if $e = 4$ satisfies the condition for all $a \in \mathbb{Q}$, including $a=0$.

Check $a * 4 = a$: $a * 4 = \frac{a \cdot 4}{4} = \frac{4a}{4} = a$. This holds for all $a \in \mathbb{Q}$.

Check $4 * a = a$: $4 * a = \frac{4 \cdot a}{4} = \frac{4a}{4} = a$. This also holds for all $a \in \mathbb{Q}$.

Since $a * 4 = a$ and $4 * a = a$ for all $a \in \mathbb{Q}$, and $4 \in \mathbb{Q}$, the element 4 is the identity element for the operation $a * b = \frac{ab}{4}$ on $\mathbb{Q}$.

This finding contradicts the statement in the question that none of the operations have identity. Based on the standard definition of identity and binary operation, the operation $a * b = \frac{ab}{4}$ on $\mathbb{Q}$ does have an identity element, which is 4.

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(vi) a ∗ b = ab²

We need $e \in \mathbb{Q}$ such that $a * e = a$ and $e * a = a$ for all $a \in \mathbb{Q}$.

From the right identity condition $a * e = a$: $ae^2 = a$. This must hold for all $a \in \mathbb{Q}$. If we take any $a \in \mathbb{Q}$ such that $a \neq 0$, we can divide by $a$ to get $e^2 = 1$. This implies $e = 1$ or $e = -1$. Both 1 and -1 are in $\mathbb{Q}$. So, if a right identity exists for all $a \neq 0$, it must be 1 or -1.

Let's check if $e=1$ is a right identity for all $a \in \mathbb{Q}$ (including $a=0$). $a * 1 = a \cdot 1^2 = a \cdot 1 = a$. This holds for all $a \in \mathbb{Q}$. So $e=1$ is a right identity.

Let's check if $e=-1$ is a right identity for all $a \in \mathbb{Q}$. $a * (-1) = a \cdot (-1)^2 = a \cdot 1 = a$. This holds for all $a \in \mathbb{Q}$. So $e=-1$ is also a right identity.

Now check the left identity condition $e * a = a$ for a general element $e \in \mathbb{Q}$. From the definition, $e * a = ea^2$. So we need $ea^2 = a$ for all $a \in \mathbb{Q}$.

If we take any $a \in \mathbb{Q}$ such that $a \neq 0$, we can divide by $a$ to get $ea = 1$. This implies $e = \frac{1}{a}$. For $e$ to be an identity, it must be a fixed rational number independent of $a$. However, $\frac{1}{a}$ depends on $a$. For example, if $a=2$, we need $e = 1/2$. If $a=3$, we need $e=1/3$. Since $1/2 \neq 1/3$, there is no single value of $e$ that works for all $a \neq 0$.

Let's check if any of the right identities ($e=1$ or $e=-1$) satisfy the left identity condition $ea^2 = a$ for all $a \in \mathbb{Q}$.

Check $e=1$: We need $1 \cdot a^2 = a$ for all $a \in \mathbb{Q}$. This is $a^2 = a$, which means $a^2 - a = 0$, or $a(a - 1) = 0$. This is true only for $a = 0$ or $a = 1$, not for all $a \in \mathbb{Q}$. So $e=1$ is not a left identity for all $a$.

Check $e=-1$: We need $(-1) \cdot a^2 = a$ for all $a \in \mathbb{Q}$. This is $-a^2 = a$, which means $a^2 + a = 0$, or $a(a + 1) = 0$. This is true only for $a = 0$ or $a = -1$, not for all $a \in \mathbb{Q}$. So $e=-1$ is not a left identity for all $a$.

Since no single element $e \in \mathbb{Q}$ satisfies both $ae^2 = a$ and $ea^2 = a$ for all $a \in \mathbb{Q}$, there is no identity element for this operation.

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Based on the analysis of each operation: Operations (i), (ii), (iii), (iv), and (vi) do not have an identity element in $\mathbb{Q}$. However, operation (v) $a * b = \frac{ab}{4}$ has an identity element $e=4 \in \mathbb{Q}$. Thus, the statement in the question, "Show that none of the operations given above has identity", is incorrect.

Given:

Set A = $\mathbb{N} \times \mathbb{N}$, where $\mathbb{N} = \{1, 2, 3, ...\}$. Elements of A are ordered pairs of natural numbers, $(a, b)$ where $a, b \in \mathbb{N}$.

Binary operation ∗ defined on A by $(a, b) ∗ (c, d) = (a + c, b + d)$.

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To Show/Find:

1. Show that ∗ is commutative.

2. Show that ∗ is associative.

3. Find the identity element for ∗ on A, if any.

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Solution:

Binary Operation Check: For any two elements $(a, b) \in A$ and $(c, d) \in A$, we have $a, b, c, d \in \mathbb{N}$. The sum of two natural numbers is a natural number, so $a+c \in \mathbb{N}$ and $b+d \in \mathbb{N}$. Thus, $(a+c, b+d)$ is an ordered pair of natural numbers, which means $(a+c, b+d) \in A$. The result is also uniquely defined. Hence, ∗ is a binary operation on A.

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Show that ∗ is commutative:

An operation ∗ is commutative if $x ∗ y = y ∗ x$ for all $x, y \in A$. Let $x = (a, b) \in A$ and $y = (c, d) \in A$, where $a, b, c, d \in \mathbb{N}$.

LHS: $x ∗ y = (a, b) ∗ (c, d) = (a + c, b + d)$

RHS: $y ∗ x = (c, d) ∗ (a, b) = (c + a, d + b)$

Since addition of natural numbers is commutative ($a + c = c + a$ and $b + d = d + b$), we have $(a + c, b + d) = (c + a, d + b)$.

Therefore, $x ∗ y = y ∗ x$ for all $x, y \in A$.

Conclusion: The operation ∗ is commutative.

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Show that ∗ is associative:

An operation ∗ is associative if $(x ∗ y) ∗ z = x ∗ (y ∗ z)$ for all $x, y, z \in A$. Let $x = (a, b) \in A$, $y = (c, d) \in A$, and $z = (e, f) \in A$, where $a, b, c, d, e, f \in \mathbb{N}$.

LHS: $(x ∗ y) ∗ z = ((a, b) ∗ (c, d)) ∗ (e, f) = (a + c, b + d) ∗ (e, f)$

Applying the operation again: $(a + c, b + d) ∗ (e, f) = ((a + c) + e, (b + d) + f)$

RHS: $x ∗ (y ∗ z) = (a, b) ∗ ((c, d) ∗ (e, f)) = (a, b) ∗ (c + e, d + f)$

Applying the operation again: $(a, b) ∗ (c + e, d + f) = (a + (c + e), b + (d + f))$

Since addition of natural numbers is associative ($(a + c) + e = a + (c + e)$ and $(b + d) + f = b + (d + f)$), we have $((a + c) + e, (b + d) + f) = (a + (c + e), b + (d + f))$.

Therefore, $(x ∗ y) ∗ z = x ∗ (y ∗ z)$ for all $x, y, z \in A$.

Conclusion: The operation ∗ is associative.

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Find the identity element for ∗ on A, if any:

An element $e = (e_1, e_2) \in A$ is an identity element for ∗ if for every element $x = (a, b) \in A$, we have $x ∗ e = x$ and $e ∗ x = x$.

Using the definition of ∗, we need $(a, b) ∗ (e_1, e_2) = (a, b)$ and $(e_1, e_2) ∗ (a, b) = (a, b)$ for all $(a, b) \in A$.

From the first condition $(a, b) ∗ (e_1, e_2) = (a, b)$:

$(a + e_1, b + e_2) = (a, b)$

Equating the corresponding components, we get:

From $a + e_1 = a$, subtracting a from both sides gives $e_1 = 0$.

From $b + e_2 = b$, subtracting b from both sides gives $e_2 = 0$.

So, the potential identity element is $(0, 0)$.

Now, we must check if this potential identity element $(0, 0)$ is in the set A. The set $A = \mathbb{N} \times \mathbb{N}$, where $\mathbb{N} = \{1, 2, 3, ...\}$. The elements of A are ordered pairs of natural numbers (positive integers).

The element $(0, 0)$ consists of the numbers 0 and 0. Since 0 is not a natural number in the set $\{1, 2, 3, ...\}$, the element $(0, 0)$ is not in the set A = $\mathbb{N} \times \mathbb{N}$.

Even if we were to check the second condition $(e_1, e_2) ∗ (a, b) = (a, b)$ with $(e_1, e_2) = (0, 0)$: $(0, 0) ∗ (a, b) = (0 + a, 0 + b) = (a, b)$, this condition is satisfied. However, the identity element must belong to the set A.

Since the identity element $(0, 0)$ required by the operation is not an element of the set $A = \mathbb{N} \times \mathbb{N}$, there is no identity element for the operation ∗ on A.

Conclusion: There does not exist an identity element for the operation ∗ on A.

(Note: If the set were $\mathbb{Z} \times \mathbb{Z}$ or $\mathbb{R} \times \mathbb{R}$, the identity element $(0, 0)$ would exist within those sets).

Given:

Set $\mathbb{N}$ (natural numbers).

A binary operation ∗ on $\mathbb{N}$.

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(i) Statement: For an arbitrary binary operation ∗ on a set N, a ∗ a = a ∀ a ∈ N.

This statement claims that for any binary operation on $\mathbb{N}$, the result of operating an element with itself is always the element itself. This property is called idempotence, and it is not a general property of all binary operations.

To justify, we can provide a counterexample of a binary operation on $\mathbb{N}$ where $a * a \neq a$ for some $a \in \mathbb{N}$.

Consider the standard addition operation '+' on $\mathbb{N}$. We know that addition is a binary operation on $\mathbb{N}$.

Let's check the condition $a + a = a$ for some $a \in \mathbb{N}$.

Choose $a = 2 \in \mathbb{N}$.

$a + a = 2 + 2 = 4$

Here, $a + a = 4$, and $a = 2$. Since $4 \neq 2$, we have $a + a \neq a$ for $a = 2$.

Since there exists an element $a \in \mathbb{N}$ (like $a=2$) for which the statement $a * a = a$ is false under the addition operation, the original statement is not true for an arbitrary binary operation.

Conclusion: The statement is False.

Justification: Consider the binary operation of addition (+) on $\mathbb{N}$. For $a=2 \in \mathbb{N}$, $a + a = 2 + 2 = 4$. Since $4 \neq 2$, $a + a \neq a$. Thus, the statement $a ∗ a = a \forall a \in \mathbb{N}$ does not hold for the addition operation on $\mathbb{N}$, which is a binary operation.

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(ii) Statement: If ∗ is a commutative binary operation on N, then a ∗ (b ∗ c) = (c ∗ b) ∗ a

This statement provides a condition involving commutativity and associativity (although the entire expression doesn't represent standard associativity). We are given that ∗ is commutative, meaning $x * y = y * x$ for all $x, y \in \mathbb{N}$. We need to check if $a ∗ (b ∗ c) = (c ∗ b) ∗ a$ holds for all $a, b, c \in \mathbb{N}$ if ∗ is commutative.

Let's simplify the right-hand side $(c ∗ b) ∗ a$ using the commutative property.

Since ∗ is commutative, we have $c ∗ b = b ∗ c$.

Substitute this into the RHS: $(c ∗ b) ∗ a = (b ∗ c) ∗ a$

Now, compare the LHS and the simplified RHS:

LHS = $a ∗ (b ∗ c)$

RHS = $(b ∗ c) ∗ a$

Let $x = a$ and $y = b ∗ c$. Since ∗ is commutative, $x ∗ y = y ∗ x$.

So, $a ∗ (b ∗ c) = (b ∗ c) ∗ a$.

Thus, the statement $a ∗ (b ∗ c) = (c ∗ b) ∗ a$ is true if ∗ is commutative, because $(c * b) * a = (b * c) * a$, and by applying commutativity to the outside operation, we get $a * (b * c) = (b * c) * a$.

Let's demonstrate the steps more clearly:

We are given that ∗ is commutative on $\mathbb{N}$, so $x * y = y * x$ for all $x, y \in \mathbb{N}$.

Consider the RHS: $(c ∗ b) ∗ a$.

Inside the parenthesis, apply commutativity: $c ∗ b = b ∗ c$. So, $(c ∗ b) ∗ a = (b ∗ c) ∗ a$.

Now, let $P = (b ∗ c)$. The RHS is $P ∗ a$. Since ∗ is commutative, $P ∗ a = a ∗ P$.

Substitute $P = (b ∗ c)$ back: $a ∗ (b ∗ c)$.

So, we have shown that $(c ∗ b) ∗ a = a ∗ (b ∗ c)$ assuming ∗ is commutative.

Conclusion: The statement is True.

Justification: Given that ∗ is commutative on $\mathbb{N}$, we have $x ∗ y = y ∗ x$ for all $x, y \in \mathbb{N}$. Consider the expression $(c ∗ b) ∗ a$. By commutativity of the inner operation, $c ∗ b = b ∗ c$. So, $(c ∗ b) ∗ a = (b ∗ c) ∗ a$. Let $P = b ∗ c$. The expression becomes $P ∗ a$. By commutativity of the outer operation, $P ∗ a = a ∗ P$. Substituting back $P = b ∗ c$, we get $a ∗ (b ∗ c)$. Thus, $(c ∗ b) ∗ a = a ∗ (b ∗ c)$ is true if ∗ is commutative.

Given:

Set A = $\mathbb{N} = \{1, 2, 3, ...\}$ (the set of natural numbers).

Binary operation ∗ defined by $a ∗ b = a^3 + b^3$.

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To Determine:

Whether ∗ is commutative and/or associative.

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Solution:

Binary Operation Check: For any two natural numbers $a, b \in \mathbb{N}$, $a^3$ is a natural number and $b^3$ is a natural number. The sum $a^3 + b^3$ is also a natural number. Thus, ∗ is a binary operation on $\mathbb{N}$.

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Check for Commutativity:

An operation ∗ is commutative if $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

By definition, $a ∗ b = a^3 + b^3$.

And $b ∗ a = b^3 + a^3$.

Since the addition of natural numbers is commutative, $a^3 + b^3 = b^3 + a^3$ for all $a, b \in \mathbb{N}$.

Thus, $a ∗ b = b ∗ a$ for all $a, b \in \mathbb{N}$.

Conclusion: The operation ∗ is commutative.

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Check for Associativity:

An operation ∗ is associative if $(a ∗ b) ∗ c = a ∗ (b ∗ c)$ for all $a, b, c \in \mathbb{N}$.

Let's compute the left-hand side (LHS):

$(a ∗ b) ∗ c = (a^3 + b^3) ∗ c$

Using the definition $x ∗ y = x^3 + y^3$, where $x = a^3 + b^3$ and $y = c$:

$(a^3 + b^3) ∗ c = (a^3 + b^3)^3 + c^3$

Let's compute the right-hand side (RHS):

$a ∗ (b ∗ c) = a ∗ (b^3 + c^3)$

Using the definition $x ∗ y = x^3 + y^3$, where $x = a$ and $y = b^3 + c^3$:

$a ∗ (b^3 + c^3) = a^3 + (b^3 + c^3)^3$

For the operation to be associative, we need $(a^3 + b^3)^3 + c^3 = a^3 + (b^3 + c^3)^3$ for all $a, b, c \in \mathbb{N}$.

Let's check with a specific example. Choose $a=1, b=2, c=3$. All are natural numbers.

LHS: $(1 ∗ 2) ∗ 3 = (1^3 + 2^3) ∗ 3 = (1 + 8) ∗ 3 = 9 ∗ 3 = 9^3 + 3^3 = 729 + 27 = 756$

RHS: $1 ∗ (2 ∗ 3) = 1 ∗ (2^3 + 3^3) = 1 ∗ (8 + 27) = 1 ∗ 35 = 1^3 + 35^3 = 1 + 42875 = 42876$

Since $756 \neq 42876$, we have $(1 ∗ 2) ∗ 3 \neq 1 ∗ (2 ∗ 3)$.

Conclusion: The operation ∗ is not associative.

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Based on our findings, the operation ∗ is commutative but not associative.

Now we choose the correct option:

(A) Is ∗ both associative and commutative? (False)

(B) Is ∗ commutative but not associative? (True)

(C) Is ∗ associative but not commutative? (False)

(D) Is ∗ neither commutative nor associative? (False)

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The correct answer is (B) Is ∗ commutative but not associative?

Given:

A set A.

$R_1$ is an equivalence relation in A.

$R_2$ is an equivalence relation in A.

This means $R_1 \subseteq A \times A$ and $R_2 \subseteq A \times A$.

Since $R_1$ and $R_2$ are equivalence relations, they satisfy the following properties:

For $R_1$:

1. Reflexive: $(a, a) \in R_1$ for all $a \in A$.

2. Symmetric: If $(a, b) \in R_1$, then $(b, a) \in R_1$ for all $a, b \in A$.

3. Transitive: If $(a, b) \in R_1$ and $(b, c) \in R_1$, then $(a, c) \in R_1$ for all $a, b, c \in A$.

For $R_2$:

1. Reflexive: $(a, a) \in R_2$ for all $a \in A.

2. Symmetric: If $(a, b) \in R_2$, then $(b, a) \in R_2$ for all $a, b \in A.

3. Transitive: If $(a, b) \in R_2$ and $(b, c) \in R_2$, then $(a, c) \in R_2$ for all $a, b, c \in A.

Let $R_{int} = R_1 \cap R_2$. By definition of intersection, a pair $(a, b) \in R_{int}$ if and only if $(a, b) \in R_1$ and $(a, b) \in R_2$. We want to show that $R_{int}$ is an equivalence relation on A.

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To Show:

$R_{int} = R_1 \cap R_2$ is an equivalence relation on A.

This requires showing that $R_{int}$ is reflexive, symmetric, and transitive.

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Proof:

1. Reflexivity of $R_{int}$:

We need to show that $(a, a) \in R_{int}$ for all $a \in A$.

Since $R_1$ is reflexive, $(a, a) \in R_1$ for all $a \in A$.

Since $R_2$ is reflexive, $(a, a) \in R_2$ for all $a \in A$.

Since $(a, a) \in R_1$ and $(a, a) \in R_2$ for all $a \in A$, by the definition of intersection, $(a, a) \in R_1 \cap R_2$ for all $a \in A$.

Thus, $(a, a) \in R_{int}$ for all $a \in A$.

Therefore, $R_{int}$ is reflexive.

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2. Symmetry of $R_{int}$:

We need to show that if $(a, b) \in R_{int}$, then $(b, a) \in R_{int}$ for all $a, b \in A$.

Assume $(a, b) \in R_{int}$.

By the definition of intersection, if $(a, b) \in R_{int}$, then $(a, b) \in R_1$ and $(a, b) \in R_2$.

Since $R_1$ is symmetric and $(a, b) \in R_1$, it follows that $(b, a) \in R_1$.

Since $R_2$ is symmetric and $(a, b) \in R_2$, it follows that $(b, a) \in R_2$.

Since $(b, a) \in R_1$ and $(b, a) \in R_2$, by the definition of intersection, $(b, a) \in R_1 \cap R_2$.

Thus, $(b, a) \in R_{int}$.

Therefore, if $(a, b) \in R_{int}$, then $(b, a) \in R_{int}$ for all $a, b \in A$.

Therefore, $R_{int}$ is symmetric.

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3. Transitivity of $R_{int}$:

We need to show that if $(a, b) \in R_{int}$ and $(b, c) \in R_{int}$, then $(a, c) \in R_{int}$ for all $a, b, c \in A$.

Assume $(a, b) \in R_{int}$ and $(b, c) \in R_{int}$.

From $(a, b) \in R_{int}$, by definition of intersection, $(a, b) \in R_1$ and $(a, b) \in R_2$.

From $(b, c) \in R_{int}$, by definition of intersection, $(b, c) \in R_1$ and $(b, c) \in R_2$.

Consider $R_1$: We have $(a, b) \in R_1$ and $(b, c) \in R_1$. Since $R_1$ is transitive, it follows that $(a, c) \in R_1$.

Consider $R_2$: We have $(a, b) \in R_2$ and $(b, c) \in R_2$. Since $R_2$ is transitive, it follows that $(a, c) \in R_2.

Since $(a, c) \in R_1$ and $(a, c) \in R_2$, by the definition of intersection, $(a, c) \in R_1 \cap R_2$.

Thus, $(a, c) \in R_{int}$.

Therefore, if $(a, b) \in R_{int}$ and $(b, c) \in R_{int}$, then $(a, c) \in R_{int}$ for all $a, b, c \in A$.

Therefore, $R_{int}$ is transitive.

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Since $R_{int} = R_1 \cap R_2$ is reflexive, symmetric, and transitive on the set A, it is an equivalence relation on A.

Given:

The set A is the set of ordered pairs of positive integers, i.e., $A = \mathbb{Z}^+ \times \mathbb{Z}^+$, where $\mathbb{Z}^+ = \{1, 2, 3, ...\}$.

The relation R is defined on A such that for any $(x, y) \in A$ and $(u, v) \in A$, $(x, y) \text{ R } (u, v)$ if and only if $xv = yu$.

Note that since $(x, y) \in \mathbb{Z}^+ \times \mathbb{Z}^+$ and $(u, v) \in \mathbb{Z}^+ \times \mathbb{Z}^+$, the components $x, y, u, v$ are all positive integers. Thus, $x > 0, y > 0, u > 0, v > 0$.

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To Show:

R is an equivalence relation on A.

This requires showing that R is reflexive, symmetric, and transitive.

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Proof:

1. Reflexivity:

We need to show that $(x, y) \text{ R } (x, y)$ for all $(x, y) \in A$.

According to the definition of R, $(x, y) \text{ R } (x, y)$ if and only if $xy = yx$.

For any positive integers $x, y \in \mathbb{Z}^+$, the multiplication of integers is commutative. Thus, $xy = yx$ is always true.

Therefore, $(x, y) \text{ R } (x, y)$ for all $(x, y) \in A$.

Hence, R is reflexive.

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2. Symmetry:

We need to show that for any $(x, y) \in A$ and $(u, v) \in A$, if $(x, y) \text{ R } (u, v)$, then $(u, v) \text{ R } (x, y)$.

Assume $(x, y) \text{ R } (u, v)$. By the definition of R, this means:

We need to show that $(u, v) \text{ R } (x, y)$, which means showing that $uy = vx$.

Starting with the given equation $xv = yu$, we can rearrange it:

By the commutative property of multiplication for integers, $yu = uy$ and $xv = vx$. So, we can write:

This is exactly the condition for $(u, v) \text{ R } (x, y)$.

Therefore, if $(x, y) \text{ R } (u, v)$, then $(u, v) \text{ R } (x, y)$ for all $(x, y), (u, v) \in A$.

Hence, R is symmetric.

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3. Transitivity:

We need to show that for any $(x, y), (u, v), (a, b) \in A$, if $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$, then $(x, y) \text{ R } (a, b)$.

Assume $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$.

By the definition of R:

and

We need to show that $(x, y) \text{ R } (a, b)$, which means showing that $xb = ya$.

Since $(x, y), (u, v), (a, b) \in \mathbb{Z}^+ \times \mathbb{Z}^+$, we know that $y, v, b$ are positive integers, so they are non-zero. We can divide by these values.

From (1), divide both sides by $yv$ (which is non-zero):

From (2), divide both sides by $vb$ (which is non-zero):

From (3) and (4), by the transitive property of equality for rational numbers:

Multiply both sides by $yb$ (which is non-zero since $y, b \in \mathbb{Z}^+$):

By the commutative property of multiplication, $ay = ya$. So we have:

This is the condition for $(x, y) \text{ R } (a, b)$.

Therefore, if $(x, y) \text{ R } (u, v)$ and $(u, v) \text{ R } (a, b)$, then $(x, y) \text{ R } (a, b)$ for all $(x, y), (u, v), (a, b) \in A$.

Hence, R is transitive.

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Since R is reflexive, symmetric, and transitive on the set A, it is an equivalence relation on A.

Given:

Set X = $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Relation $R_1$ on X defined by $R_1 = \{(x, y) \in X \times X : x - y \text{ is divisible by } 3\}$.

Relation $R_2$ on X defined by $R_2 = \{(x, y) \in X \times X : \{x, y\} \subseteq \{1, 4, 7\} \text{ or } \{x, y\} \subseteq \{2, 5, 8\} \text{ or } \{x, y\} \subseteq \{3, 6, 9\}\}$.

Note: $\{x, y\} \subseteq S$ means both $x \in S$ and $y \in S$. The order of x and y does not matter in the set $\{x, y\}$, but the relation $R_2$ is on $X \times X$, so it consists of ordered pairs $(x, y)$. The condition $\{x, y\} \subseteq S$ means that the pair $(x, y)$ is such that both components come from the set S. This implicitly includes both $(x, y)$ and $(y, x)$ if $x \neq y$, and $(x, x)$ if $x \in S$.

The sets {1, 4, 7}, {2, 5, 8}, and {3, 6, 9} are disjoint partitions of the set X based on their remainder when divided by 3.

Let $S_1 = \{1, 4, 7\}$ (elements with remainder 1 when divided by 3).

Let $S_2 = \{2, 5, 8\}$ (elements with remainder 2 when divided by 3).

Let $S_3 = \{3, 6, 9\}$ (elements with remainder 0 when divided by 3).

So, $R_2 = \{(x, y) \in X \times X : x \in S_1 \text{ and } y \in S_1 \text{ or } x \in S_2 \text{ and } y \in S_2 \text{ or } x \in S_3 \text{ and } y \in S_3\}$.

This means $(x, y) \in R_2$ if and only if x and y belong to the same set among $S_1, S_2, S_3$.

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To Show:

$R_1 = R_2$.

To show that two sets (relations are sets of ordered pairs) are equal, we must show that every element in the first set is in the second set ($R_1 \subseteq R_2$) and every element in the second set is in the first set ($R_2 \subseteq R_1$).

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Proof that $R_1 \subseteq R_2$:

Let $(x, y) \in R_1$. By the definition of $R_1$, this means $x - y$ is divisible by 3.

This is equivalent to saying $x \equiv y \pmod{3}$.

This means x and y have the same remainder when divided by 3.

If x and y have the same remainder when divided by 3, they must belong to the same set among $S_1, S_2, S_3$.

If $x \in S_1$ and $y \in S_1$, then $x \equiv 1 \pmod{3}$ and $y \equiv 1 \pmod{3}$. So $x - y \equiv 1 - 1 \equiv 0 \pmod{3}$.

If $x \in S_2$ and $y \in S_2$, then $x \equiv 2 \pmod{3}$ and $y \equiv 2 \pmod{3}$. So $x - y \equiv 2 - 2 \equiv 0 \pmod{3}$.

If $x \in S_3$ and $y \in S_3$, then $x \equiv 0 \pmod{3}$ and $y \equiv 0 \pmod{3}$. So $x - y \equiv 0 - 0 \equiv 0 \pmod{3}$.

If x and y are in different sets, say $x \in S_1$ and $y \in S_2$, then $x - y \equiv 1 - 2 \equiv -1 \equiv 2 \pmod{3}$, which is not divisible by 3.

Thus, $x - y$ is divisible by 3 if and only if x and y have the same remainder modulo 3, which means they belong to the same set $S_i$.

If $(x, y) \in R_1$, then $x$ and $y$ belong to the same set $S_i$. By the definition of $R_2$, this means $(x, y) \in R_2$.

Therefore, $R_1 \subseteq R_2$.

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Proof that $R_2 \subseteq R_1$:

Let $(x, y) \in R_2$. By the definition of $R_2$, this means that x and y belong to the same set among $S_1, S_2, S_3$. That is, either ($x \in S_1$ and $y \in S_1$) or ($x \in S_2$ and $y \in S_2$) or ($x \in S_3$ and $y \in S_3$).

Case 1: $x \in S_1$ and $y \in S_1$. This means $x$ and $y$ have a remainder of 1 when divided by 3. So, $x = 3k_1 + 1$ and $y = 3k_2 + 1$ for some integers $k_1, k_2$. (For $x, y \in X$, $k_1, k_2$ are non-negative integers. For $1, 4, 7$, $k$ is 0, 1, 2 respectively). Then $x - y = (3k_1 + 1) - (3k_2 + 1) = 3k_1 - 3k_2 = 3(k_1 - k_2)$. Since $k_1 - k_2$ is an integer, $x - y$ is divisible by 3.

Case 2: $x \in S_2$ and $y \in S_2$. This means $x$ and $y$ have a remainder of 2 when divided by 3. So, $x = 3k_1 + 2$ and $y = 3k_2 + 2$ for some integers $k_1, k_2$. Then $x - y = (3k_1 + 2) - (3k_2 + 2) = 3k_1 - 3k_2 = 3(k_1 - k_2)$. Since $k_1 - k_2$ is an integer, $x - y$ is divisible by 3.

Case 3: $x \in S_3$ and $y \in S_3$. This means $x$ and $y$ have a remainder of 0 when divided by 3. So, $x = 3k_1$ and $y = 3k_2$ for some integers $k_1, k_2$. (For $3, 6, 9$, $k$ is 1, 2, 3 respectively). Then $x - y = 3k_1 - 3k_2 = 3(k_1 - k_2)$. Since $k_1 - k_2$ is an integer, $x - y$ is divisible by 3.

In all cases where $(x, y) \in R_2$, we found that $x - y$ is divisible by 3. By the definition of $R_1$, this means $(x, y) \in R_1$.

Therefore, $R_2 \subseteq R_1$.

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Since we have shown that $R_1 \subseteq R_2$ and $R_2 \subseteq R_1$, it follows that $R_1 = R_2$.

The two relations are indeed the same.

Given:

A function $f : X \to Y$.

A relation R defined on the set X by $R = \{(a, b) \in X \times X : f(a) = f(b)\}$.

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To Examine:

Whether R is an equivalence relation on X.

To be an equivalence relation, R must be reflexive, symmetric, and transitive.

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Proof:

1. Reflexivity:

We need to show that $(a, a) \in R$ for all $a \in X$.

According to the definition of R, $(a, a) \in R$ if and only if $f(a) = f(a)$.

Since the equality $f(a) = f(a)$ is always true for any element $a$ in the domain of f, the condition $f(a) = f(a)$ holds for all $a \in X$.

Therefore, $(a, a) \in R$ for all $a \in X$.

Hence, R is reflexive.

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2. Symmetry:

We need to show that for any $a, b \in X$, if $(a, b) \in R$, then $(b, a) \in R$.

Assume $(a, b) \in R$.

By the definition of R, if $(a, b) \in R$, then $f(a) = f(b)$.

We need to show that $(b, a) \in R$, which means showing that $f(b) = f(a)$.

Since the equality $f(a) = f(b)$ is given, and the equality of values is symmetric (if $P = Q$, then $Q = P$), we have $f(b) = f(a)$.

By the definition of R, if $f(b) = f(a)$, then $(b, a) \in R$.

Therefore, if $(a, b) \in R$, then $(b, a) \in R$ for all $a, b \in X$.

Hence, R is symmetric.

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3. Transitivity:

We need to show that for any $a, b, c \in X$, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Assume $(a, b) \in R$ and $(b, c) \in R$.

From $(a, b) \in R$, by definition of R, $f(a) = f(b)$.

From $(b, c) \in R$, by definition of R, $f(b) = f(c)$.

We have $f(a) = f(b)$ and $f(b) = f(c)$. By the transitive property of equality, if $f(a)$ is equal to $f(b)$ and $f(b)$ is equal to $f(c)$, then $f(a)$ must be equal to $f(c)$.

$f(a) = f(c)$

By the definition of R, if $f(a) = f(c)$, then $(a, c) \in R$.

Therefore, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$ for all $a, b, c \in X$.

Hence, R is transitive.

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Since R is reflexive, symmetric, and transitive on the set X, it is an equivalence relation on X.

Definitions:

A binary operation ∗ on a set A is a function ∗ : A × A → A.

A binary operation ∗ on a set A is commutative if $a * b = b * a$ for all $a, b \in A$.

A binary operation ∗ on a set A is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in A$.

The set $\mathbb{R}$ is the set of real numbers.

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(a) a ∗ b = 1 ∀ a, b ∈ R

The operation ∗ on $\mathbb{R}$ is defined by $a * b = 1$. This means the result of the operation is always the constant value 1, regardless of the input real numbers a and b.

Binary Operation Check: For any two real numbers $a, b \in \mathbb{R}$, the result $a * b = 1$. Since 1 is a real number, $1 \in \mathbb{R}$. The result is uniquely defined. Thus, ∗ is a binary operation on $\mathbb{R}$.

Check for Commutativity:

We need to check if $a * b = b * a$ for all $a, b \in \mathbb{R}$.

LHS: $a * b = 1$ (by definition of the operation)

RHS: $b * a = 1$ (by definition of the operation)

Since $1 = 1$, we have $a * b = b * a$ for all $a, b \in \mathbb{R}$.

Conclusion: The operation ∗ is commutative.

Check for Associativity:

We need to check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

LHS: $(a * b) * c$. First, calculate $a * b$. By definition, $a * b = 1$.

Now, $(a * b) * c = 1 * c$. Applying the operation to 1 and c, we get $1 * c = 1$ (by definition, as the result is always 1).

So, LHS = 1.

RHS: $a * (b * c)$. First, calculate $b * c$. By definition, $b * c = 1$.

Now, $a * (b * c) = a * 1$. Applying the operation to a and 1, we get $a * 1 = 1$ (by definition, as the result is always 1).

So, RHS = 1.

Since LHS = RHS (both equal to 1) for all $a, b, c \in \mathbb{R}$, the associative property holds.

Conclusion: The operation ∗ is associative.

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(b) a ∗ b = $\frac{a + b}{2}$ ∀ a, b ∈ R

The operation ∗ on $\mathbb{R}$ is defined by $a * b = \frac{a + b}{2}$. This calculates the arithmetic mean (average) of a and b.

Binary Operation Check: For any two real numbers $a, b \in \mathbb{R}$, their sum $a + b$ is a real number. Dividing the real number $a+b$ by the non-zero real number 2 results in a real number $\frac{a+b}{2}$. So, $\frac{a+b}{2} \in \mathbb{R}$. The result is uniquely defined. Thus, ∗ is a binary operation on $\mathbb{R}$.

Check for Commutativity:

We need to check if $a * b = b * a$ for all $a, b \in \mathbb{R}$.

LHS: $a * b = \frac{a + b}{2}$

RHS: $b * a = \frac{b + a}{2}$

Since addition of real numbers is commutative ($a + b = b + a$), we have $\frac{a + b}{2} = \frac{b + a}{2}$.

Therefore, $a * b = b * a$ for all $a, b \in \mathbb{R}$.

Conclusion: The operation ∗ is commutative.

Check for Associativity:

We need to check if $(a * b) * c = a * (b * c)$ for all $a, b, c \in \mathbb{R}$.

LHS: $(a * b) * c = \left(\frac{a + b}{2}\right) * c$

Using the definition $x * y = \frac{x + y}{2}$, where $x = \frac{a + b}{2}$ and $y = c$:

$(a * b) * c = \frac{\left(\frac{a + b}{2}\right) + c}{2} = \frac{\frac{a + b + 2c}{2}}{2} = \frac{a + b + 2c}{4}$

So, LHS $= \frac{a + b + 2c}{4}$.

RHS: $a * (b * c) = a * \left(\frac{b + c}{2}\right)$

Using the definition $x * y = \frac{x + y}{2}$, where $x = a$ and $y = \frac{b + c}{2}$:

$a * (b * c) = \frac{a + \left(\frac{b + c}{2}\right)}{2} = \frac{\frac{2a + b + c}{2}}{2} = \frac{2a + b + c}{4}$

So, RHS $= \frac{2a + b + c}{4}$.

For the operation to be associative, we need $\frac{a + b + 2c}{4} = \frac{2a + b + c}{4}$.

This implies $a + b + 2c = 2a + b + c$. Subtracting $b$ from both sides gives $a + 2c = 2a + c$. Rearranging, $2c - c = 2a - a$, which simplifies to $c = a$.

This means the associative property holds only when $c = a$, not for all $a, b, c \in \mathbb{R}$.

Let's confirm with a counterexample. Choose $a=1, b=2, c=3$. All are real numbers.

LHS: $(1 * 2) * 3 = \left(\frac{1 + 2}{2}\right) * 3 = \frac{3}{2} * 3 = \frac{\frac{3}{2} + 3}{2} = \frac{\frac{3 + 6}{2}}{2} = \frac{\frac{9}{2}}{2} = \frac{9}{4}$

RHS: $1 * (2 * 3) = 1 * \left(\frac{2 + 3}{2}\right) = 1 * \frac{5}{2} = \frac{1 + \frac{5}{2}}{2} = \frac{\frac{2 + 5}{2}}{2} = \frac{\frac{7}{2}}{2} = \frac{7}{4}$

Since $\frac{9}{4} \neq \frac{7}{4}$, we have $(a * b) * c \neq a * (b * c)$.

Conclusion: The operation ∗ is not associative.

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Summary:

(a) $a * b = 1$: Commutative and Associative.

(b) $a * b = \frac{a + b}{2}$: Commutative but not Associative.

Given:

The set A = $\{1, 2, 3\}$.

---

To Find:

The number of all one-one functions from set A to itself, i.e., the number of one-one functions $f: A \to A$.

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Solution:

A function $f: A \to A$ is one-one (injective) if distinct elements in the domain A are mapped to distinct elements in the codomain A. Since the domain and codomain are the same finite set, a one-one function from A to A is also an onto function (surjective), and therefore it is a bijection (a permutation of the elements of A).

Let the elements of the domain be 1, 2, and 3.

Consider the mapping of the first element, 1. The element 1 can be mapped to any of the 3 elements in the codomain $\{1, 2, 3\}$. So there are 3 choices for the image of 1, i.e., for $f(1)$.

Since the function must be one-one, the second element, 2, cannot be mapped to the same element as 1. The codomain has 3 elements, and one element has already been taken as the image of 1. So, the element 2 can be mapped to any of the remaining $3 - 1 = 2$ elements in the codomain. There are 2 choices for $f(2)$.

Finally, the third element, 3, cannot be mapped to the same element as 1 or 2. The codomain has 3 elements, and two distinct elements have already been taken as the images of 1 and 2. So, the element 3 must be mapped to the single remaining element in the codomain. There is $3 - 2 = 1$ choice for $f(3)$.

The total number of one-one functions from A to A is the product of the number of choices for the image of each element in the domain.

This product is the definition of the factorial of the number of elements in the set, which is $3!$.

Therefore, there are 6 one-one functions from set A to itself.

These functions correspond to the permutations of the set {1, 2, 3}:

$\{(1,1), (2,2), (3,3)\}$

$\{(1,1), (2,3), (3,2)\}$

$\{(1,2), (2,1), (3,3)\}$

$\{(1,2), (2,3), (3,1)\}$

$\{(1,3), (2,1), (3,2)\}$

$\{(1,3), (2,2), (3,1)\}$

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The number of all one-one functions from set A to itself is 6.

Given:

Set A = $\{1, 2, 3\}$.

We need to find the number of relations R on A such that:

1. $(1, 2) \in R$ and $(2, 3) \in R$.

2. R is reflexive.

3. R is transitive.

4. R is not symmetric.

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Solution:

The universal relation on A is $A \times A$, which consists of all possible ordered pairs:

$A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$

Let R be a relation satisfying the given conditions.

Condition 2: R must be reflexive.

This means R must contain all pairs of the form $(a, a)$ for $a \in A$.

So, R must contain $\{(1, 1), (2, 2), (3, 3)\}$.

Let $R_{min}$ be the minimum set of pairs R must contain to satisfy the first two conditions.

$R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ (from condition 1 and 2).

Condition 3: R must be transitive.

If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. Let's check for transitivity based on the elements in $R_{min}$:

- $(1, 2) \in R_{min}$ and $(2, 3) \in R_{min}$. By transitivity, $(1, 3)$ must be in R.

So, any valid relation R must contain $(1, 3)$.

Let's update the minimum set of pairs that must be in R based on reflexivity, the initial given pairs, and transitivity:

$R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$

Let's check if $R_0$ is transitive. We need to check all pairs $(a, b), (b, c)$ in $R_0$.

- $(1, 1), (1, 2) \implies (1, 2) \in R_0$ (ok)

- $(1, 1), (1, 3) \implies (1, 3) \in R_0$ (ok)

- $(1, 2), (2, 2) \implies (1, 2) \in R_0$ (ok)

- $(1, 2), (2, 3) \implies (1, 3) \in R_0$ (ok, added to $R_0$)

- $(1, 3), (3, 3) \implies (1, 3) \in R_0$ (ok)

- $(2, 2), (2, 3) \implies (2, 3) \in R_0$ (ok)

- $(3, 3), (3, 1)$? $(3, 1) \notin R_0$. Need to consider cases where $R$ might contain more elements.

- $(3, 3), (3, 2)$? $(3, 2) \notin R_0$.

- $(a, a)$ pairs don't create new pairs with other pairs starting or ending at $a$ if those other pairs are already included due to transitivity from non-identity pairs. For example, $(1, 2), (2, 2) \implies (1, 2)$.

So, $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$ is reflexive and contains the initial pairs. It is also transitive as checked above.

Condition 4: R must not be symmetric.

For a relation to be symmetric, if $(a, b) \in R$, then $(b, a) \in R$. For R to be not symmetric, there must exist at least one pair $(a, b) \in R$ such that $(b, a) \notin R$.

Consider the pairs in $R_0$ that are not diagonal (identity) pairs:

$(1, 2) \in R_0$. Is $(2, 1) \in R_0$? No.

$(2, 3) \in R_0$. Is $(3, 2) \in R_0$? No.

$(1, 3) \in R_0$. Is $(3, 1) \in R_0$? No.

Since $R_0$ contains $(1, 2)$ but not $(2, 1)$, $R_0$ is not symmetric.

$R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$ is one relation that satisfies all conditions (reflexive, contains (1,2) and (2,3), transitive, not symmetric).

Now, consider adding other pairs from $A \times A$ to $R_0$ while maintaining reflexivity and transitivity, and ensuring it remains not symmetric.

The pairs not in $R_0$ are: $\{(2, 1), (3, 1), (3, 2)\}$.

Let's consider adding these pairs to $R_0$ one by one or in combinations, and check the properties.

Case 1: R = $R_0$. We already verified that this relation satisfies all conditions.

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$ - This is one such relation.

Case 2: Add $(2, 1)$ to $R_0$. Let $R = R_0 \cup \{(2, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$.

Is R reflexive? Yes (contains (1,1), (2,2), (3,3)).

Does R contain (1,2) and (2,3)? Yes.

Is R transitive? We need to check for new transitive implications with the added pair $(2, 1)$.

- $(1, 2) \in R$ and $(2, 1) \in R \implies (1, 1) \in R$ (ok)

- $(2, 1) \in R$ and $(1, 2) \in R \implies (2, 2) \in R$ (ok)

- $(2, 1) \in R$ and $(1, 3) \in R \implies (2, 3) \in R$ (ok)

- $(3, ?)$ None start with 3 where the second element is the first element of a pair in R starting with 3, except $(3,3)$. $(3,3), (3,?) \implies (3, ?)$.

- $(?, 3)$ Pairs ending in 3 are $(2,3), (1,3), (3,3)$.

- $(a, 2)$ Pairs ending in 2 are $(1,2), (2,2)$. $(1,2), (2,1) \implies (1,1)$. $(2,2), (2,1) \implies (2,1)$.

- $(a, 1)$ Pairs ending in 1 are $(2,1), (1,1)$. $(2,1), (1,2) \implies (2,2)$. $(2,1), (1,3) \implies (2,3)$.

Transitivity seems to hold with $(2, 1)$ added.

Is R not symmetric? R contains $(1, 3)$ but not $(3, 1)$. So it is not symmetric.

$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$ - This is a second such relation.

Case 3: Add $(3, 1)$ to $R_0$. Let $R = R_0 \cup \{(3, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 1)\}$.

Is R reflexive? Yes.

Does R contain (1,2) and (2,3)? Yes.

Is R transitive? Check new implications from $(3, 1)$.

- $(3, 1) \in R$ and $(1, 2) \in R \implies (3, 2)$ must be in R.

- $(3, 1) \in R$ and $(1, 3) \in R \implies (3, 3) \in R$ (ok).

So, adding $(3, 1)$ requires adding $(3, 2)$ for transitivity.

Let's consider the relation $R = R_0 \cup \{(3, 1), (3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 1), (3, 2)\}$.

Is R reflexive? Yes.

Does R contain (1,2) and (2,3)? Yes.

Is R transitive? Check implications involving $(3, 1)$ and $(3, 2)$.

- $(3, 1), (1, 2) \implies (3, 2) \in R$ (ok, added)

- $(3, 1), (1, 3) \implies (3, 3) \in R$ (ok)

- $(3, 2), (2, 3) \implies (3, 3) \in R$ (ok)

- $(?, 3)$ pairs ending in 3 are $(2,3), (1,3), (3,3)$. $(1,3), (3,1) \implies (1,1)$. $(2,3), (3,1) \implies (2,1)$ must be in R. $(3,3), (3,1) \implies (3,1)$.

So, adding $(3, 1)$ forces $(3, 2)$ and $(2, 1)$ to be added for transitivity.

Let $R = R_0 \cup \{(3, 1), (3, 2), (2, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 1), (3, 2), (2, 1)\}$.

Is R not symmetric? This relation contains $(1, 2)$ and $(2, 1)$, $(2, 3)$ and $(3, 2)$, $(1, 3)$ and $(3, 1)$. All non-diagonal pairs have their symmetric counterparts. This relation is the universal relation $A \times A$. The universal relation is an equivalence relation, hence it is symmetric.

So, adding $(3, 1)$ (which forced $(3, 2)$ and $(2, 1)$) results in a symmetric relation, which is not allowed by condition 4.

Case 4: Add $(3, 2)$ to $R_0$. Let $R = R_0 \cup \{(3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$.

Is R reflexive? Yes.

Does R contain (1,2) and (2,3)? Yes.

Is R transitive? Check new implications from $(3, 2)$.

- $(3, 2) \in R$ and $(2, 3) \in R \implies (3, 3) \in R$ (ok)

- $(3, 2) \in R$ and $(2, ?)$ None start with 2 ending not in 2.

Check implications leading to $(3, 2)$. Pairs ending in 3 are $(2,3), (1,3), (3,3)$. Need $(a, 3)$ and $(3, 2)$. $(1, 3) \in R$ and $(3, 2) \in R \implies (1, 2) \in R$ (ok).

Transitivity seems to hold with $(3, 2)$ added.

Is R not symmetric? R contains $(1, 2)$ but not $(2, 1)$. So it is not symmetric.

$R_3 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$ - This is a third such relation.

Let's verify $R_3$ again:

Reflexive: Yes.

Contains (1,2) and (2,3): Yes.

Transitive: Pairs $(a,b), (b,c) \in R_3$. Need to check if $(a,c) \in R_3$.

Non-identity pairs: (1,2), (2,3), (1,3), (3,2)

- (1,2), (2,3) $\implies$ (1,3) $\in R_3$ (ok)

- (1,2), (2,?) No pairs start with 2 except (2,3), (2,2).

- (2,3), (3,2) $\implies$ (2,2) $\in R_3$ (ok)

- (1,3), (3,2) $\implies$ (1,2) $\in R_3$ (ok)

- (3,2), (2,3) $\implies$ (3,3) $\in R_3$ (ok)

Transitivity holds.

Not symmetric: $(1,2) \in R_3$ but $(2,1) \notin R_3$. $(2,3) \in R_3$ but $(3,2) \in R_3$. $(1,3) \in R_3$ but $(3,1) \notin R_3$. Yes, it's not symmetric because $(1,2)$ is in R but $(2,1)$ is not, and $(1,3)$ is in R but $(3,1)$ is not.

So, $R_3$ works.

Consider adding $(2, 1)$ and $(3, 2)$ together to $R_0$. $R = R_0 \cup \{(2, 1), (3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1), (3, 2)\}$.

Check transitivity: $(3, 2), (2, 1) \implies (3, 1)$ must be in R. So this would lead to the full universal relation which is symmetric.

The possible relations satisfying the conditions are constructed by starting with $R_0$ and potentially adding pairs from $\{(2, 1), (3, 1), (3, 2)\}$ such that reflexivity and transitivity are maintained, but symmetry is not achieved.

We found three such relations:

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$

$R_2 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (2, 1)\}$

$R_3 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3), (3, 2)\}$

Adding (3,1) to $R_0$ forces $(3,2)$ and $(2,1)$, leading to the universal relation (symmetric).

Adding (2,1) and (3,2) to $R_0$ forces (3,1), leading to the universal relation (symmetric).

Adding all three non-diagonal pairs leads to the universal relation (symmetric).

Thus, the only possible relations are $R_1, R_2, R_3$. Let's double-check symmetry for each.

$R_1$: $(1,2) \in R_1$, $(2,1) \notin R_1$. Not symmetric. (ok)

$R_2$: $(1,3) \in R_2$, $(3,1) \notin R_2$. Not symmetric. (ok)

$R_3$: $(1,3) \in R_3$, $(3,1) \notin R_3$. Not symmetric. (ok)

All three relations $R_1, R_2, R_3$ are reflexive, contain (1,2) and (2,3), are transitive, and are not symmetric.

The number of such relations is indeed three.

Given:

Set A = $\{1, 2, 3\}$.

We need to find the number of equivalence relations R on A such that:

1. $(1, 2) \in R$ and $(2, 1) \in R$.

2. R is reflexive.

3. R is symmetric.

4. R is transitive.

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Solution:

Let R be an equivalence relation satisfying the given conditions.

Condition 2: R must be reflexive.

This means R must contain all pairs of the form $(a, a)$ for $a \in A$.

So, R must contain $\{(1, 1), (2, 2), (3, 3)\}$.

Condition 1: R must contain $(1, 2)$ and $(2, 1)$.

Let $R_{min}$ be the minimum set of pairs R must contain to satisfy the first two conditions.

$R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$

Condition 3: R must be symmetric.

We check if $R_{min}$ is symmetric. If $(a, b) \in R_{min}$, is $(b, a) \in R_{min}$?

- $(1, 1) \in R_{min} \implies (1, 1) \in R_{min}$ (ok)

- $(2, 2) \in R_{min} \implies (2, 2) \in R_{min}$ (ok)

- $(3, 3) \in R_{min} \implies (3, 3) \in R_{min}$ (ok)

- $(1, 2) \in R_{min} \implies (2, 1) \in R_{min}$ (ok)

- $(2, 1) \in R_{min} \implies (1, 2) \in R_{min}$ (ok)

So, $R_{min}$ is symmetric.

Condition 4: R must be transitive.

If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$. Let's check for transitivity based on the elements in $R_{min}$:

- $(1, 2) \in R_{min}$ and $(2, 1) \in R_{min} \implies (1, 1) \in R_{min}$ (ok)

- $(2, 1) \in R_{min}$ and $(1, 2) \in R_{min} \implies (2, 2) \in R_{min}$ (ok)

- $(1, 1), (1, 2) \implies (1, 2) \in R_{min}$ (ok)

- $(1, 1), (2, 1)$ not applicable as $(1,1)$ doesn't end where $(2,1)$ starts.

- $(2, 2), (1, 2)$ not applicable.

- $(2, 2), (2, 1) \implies (2, 1) \in R_{min}$ (ok)

Identity pairs with non-identity pairs check: $(a, a), (a, b) \implies (a, b)$; $(b, a), (a, a) \implies (b, a)$. These are satisfied because the original non-identity pairs are in $R_{min}$.

What about elements involving 3? Currently, 3 is only related to itself in $R_{min}$.

Example pairs to check for transitivity in $R_{min}$:

- $(1, 2)$ and $(2, 3)$? $(2, 3) \notin R_{min}$.

- $(?, 1)$ ending in 1 are $(1,1), (2,1)$. $(1,1), (1,2) \implies (1,2)$. $(2,1), (1,2) \implies (2,2)$.

- $(?, 2)$ ending in 2 are $(1,2), (2,2)$. $(1,2), (2,1) \implies (1,1)$. $(2,2), (2,1) \implies (2,1)$.

$R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ is reflexive, contains (1,2) and (2,1), is symmetric. Is it transitive?

Pairs in $R_1$: (1,1), (2,2), (3,3), (1,2), (2,1)

Check transitivity:

- (1,2) and (2,1) $\implies$ (1,1) $\in R_1$ (ok)

- (2,1) and (1,2) $\implies$ (2,2) $\in R_1$ (ok)

- (1,1) with (1,2) $\implies$ (1,2) $\in R_1$ (ok)

- (1,1) with (1,1) $\implies$ (1,1) $\in R_1$ (ok)

- (2,2) with (1,2) $\implies$ No, second element of first pair must match first of second. (2,2) with (2,1) $\implies$ (2,1) $\in R_1$ (ok)

- (3,3) with (3,3) $\implies$ (3,3) $\in R_1$ (ok)

Any chain of the form (1, 2, 1) or (2, 1, 2) results in (1, 1) or (2, 2) which are in $R_1$. Any chain involving 3 is of the form (3, 3, 3) which results in (3, 3) in $R_1$.

So, $R_1 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ is an equivalence relation. This is one such relation.

Now, consider adding other pairs from $A \times A$ while maintaining the equivalence properties and containing (1, 2) and (2, 1).

The pairs not in $R_1$ are: $\{(1, 3), (3, 1), (2, 3), (3, 2)\}$.

If we add $(1, 3)$, for symmetry we must also add $(3, 1)$. Let's consider adding both $(1, 3)$ and $(3, 1)$.

Let $R = R_1 \cup \{(1, 3), (3, 1)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\}$.

Is R reflexive? Yes.

Does R contain (1,2) and (2,1)? Yes.

Is R symmetric? Yes (by adding symmetric pairs).

Is R transitive? Check new implications.

- $(1, 2) \in R$ and $(2, ?)$. The only pair starting with 2 not ending in 2 is $(2,1)$. $(1,2), (2,1) \implies (1,1)$ (ok).

- $(2, 1) \in R$ and $(1, ?)$. Pairs starting with 1 not ending in 1 are $(1,2), (1,3)$. $(2,1), (1,2) \implies (2,2)$ (ok). $(2,1), (1,3) \implies (2,3)$ must be in R.

- $(1, 3) \in R$ and $(3, ?)$. Pairs starting with 3 not ending in 3 are $(3,1)$. $(1,3), (3,1) \implies (1,1)$ (ok).

- $(3, 1) \in R$ and $(1, ?)$. Pairs starting with 1 not ending in 1 are $(1,2), (1,3)$. $(3,1), (1,2) \implies (3,2)$ must be in R. $(3,1), (1,3) \implies (3,3)$ (ok).

So, adding $(1, 3)$ and $(3, 1)$ forces the addition of $(2, 3)$ and $(3, 2)$ for transitivity.

If we add $(1, 3)$ and $(3, 1)$, we get the relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$.

This relation contains all non-diagonal pairs and their symmetric counterparts, plus the diagonal pairs. This is the universal relation $A \times A$.

$R_2 = A \times A = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$.

Let's check if $A \times A$ is an equivalence relation:

Reflexive: Yes (contains all $(a, a)$).

Symmetric: Yes (if $(a, b) \in A \times A$, then $(b, a) \in A \times A$).

Transitive: Yes (if $(a, b) \in A \times A$ and $(b, c) \in A \times A$, then $(a, c) \in A \times A$).

Does $A \times A$ contain $(1, 2)$ and $(2, 1)$? Yes.

So, $R_2 = A \times A$ is another equivalence relation satisfying the conditions.

What if we tried adding only $(2, 3)$? For symmetry, we must add $(3, 2)$.

Let $R = R_1 \cup \{(2, 3), (3, 2)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)\}$.

Is R reflexive? Yes.

Does R contain (1,2) and (2,1)? Yes.

Is R symmetric? Yes (by adding symmetric pairs).

Is R transitive? Check new implications from (2,3) and (3,2).

- $(1, 2) \in R$ and $(2, 3) \in R \implies (1, 3)$ must be in R. For symmetry, $(3, 1)$ must also be in R.

- $(3, 2) \in R$ and $(2, 1) \in R \implies (3, 1)$ must be in R. For symmetry, $(1, 3)$ must also be in R.

So, adding $(2, 3)$ and $(3, 2)$ forces adding $(1, 3)$ and $(3, 1)$, which again results in the universal relation $A \times A$.

The process of building equivalence relations by adding pairs requires adding their symmetric counterparts and all pairs required by transitivity. Equivalence relations correspond to partitions of the set.

The given condition that $(1, 2) \in R$ and $(2, 1) \in R$ means that 1 and 2 must be in the same equivalence class. So, $\{1, 2\}$ must be a subset of some equivalence class.

Possible partitions of {1, 2, 3} where 1 and 2 are in the same class:

Partition 1: $\{\{1, 2\}, \{3\}\}$. This corresponds to the equivalence relation where elements within {1, 2} are related, elements within {3} are related, but there are no relations between elements from {1, 2} and elements from {3}.

The relation for this partition is $R_1 = \{(1, 1), (1, 2), (2, 1), (2, 2)\} \cup \{(3, 3)\} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. This is the first relation we found ($R_1$ above).

Partition 2: $\{\{1, 2, 3\}\}$. This means all elements are in the same equivalence class. This corresponds to the universal relation where every element is related to every other element.

The relation for this partition is $R_2 = A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$. This is the second relation we found ($R_2$ above).

Are there any other partitions where {1, 2} is a subset of a class? No, because 1 and 2 must be together. The other element 3 can either be with {1, 2} (Partition 2) or in its own class (Partition 1).

These two partitions lead to two distinct equivalence relations that contain (1, 2) and (2, 1).

The number of such equivalence relations is two.

Given:

Set A = $\{1, 2\}$.

A binary operation ∗ on A.

Conditions for the operation ∗:

1. 1 is the identity element for ∗ on A.

2. 2 is the inverse of 2 for ∗ on A.

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To Show:

The number of such binary operations on A is exactly one.

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Solution:

A binary operation ∗ on A is a function from $A \times A$ to A. The set $A \times A$ has $2 \times 2 = 4$ elements: $\{(1, 1), (1, 2), (2, 1), (2, 2)\}$. A binary operation must define the result for each of these four pairs.

Let the operation table for ∗ be:

*	1	2
1	1 ∗ 1	1 ∗ 2
2	2 ∗ 1	2 ∗ 2

Now, let's use the given conditions to determine the values of these operations.

Condition 1: 1 is the identity element.

By the definition of an identity element, for every element $a \in A$, we must have $a ∗ 1 = a$ and $1 ∗ a = a$.

For $a = 1$: $1 ∗ 1 = 1$ and $1 ∗ 1 = 1$. This determines the value of $1 ∗ 1$.

For $a = 2$: $2 ∗ 1 = 2$ and $1 ∗ 2 = 2$. This determines the values of $2 ∗ 1$ and $1 ∗ 2$.

Updating the operation table with these values:

*	1	2
1	1	2
2	2	2 ∗ 2

We have determined the results for three of the four possible pairs in $A \times A$. We only need to determine the value of $2 ∗ 2$.

Condition 2: 2 is the inverse of 2.

By the definition of an inverse, if 2 is the inverse of 2, and the identity element is 1, we must have $2 ∗ 2 = 1$ and $2 ∗ 2 = 1$.

This determines the value of $2 ∗ 2$ to be 1.

Updating the operation table with this value:

*	1	2
1	1	2
2	2	1

We have now determined the result for all four possible pairs in $A \times A$ based on the given conditions.

$1 ∗ 1 = 1$

$1 ∗ 2 = 2$

$2 ∗ 1 = 2$

$2 ∗ 2 = 1$

Since the values for all possible operations are uniquely determined by the conditions, there is only one possible way to define the binary operation ∗ that satisfies the given properties.

This uniquely determined operation is the standard multiplication operation on the set $\{1, -1\}$ if we consider 1 as the identity and -1 as its own inverse. Mapping $1 \to 1$ and $2 \to -1$, the operation table would match if we replace 2 with -1.

However, the problem defines the operation on the set {1, 2}. The operation defined by the table is a valid binary operation on {1, 2} since all results are elements of {1, 2}.

The number of binary operations on {1, 2} satisfying the given conditions is exactly one.

Given:

Set $\mathbb{N} = \{1, 2, 3, ...\}$ (the set of natural numbers).

Identity function $I_{\mathbb{N}} : \mathbb{N} \to \mathbb{N}$ defined by $I_{\mathbb{N}}(x) = x$ for all $x \in \mathbb{N}$.

Function $(I_{\mathbb{N}} + I_{\mathbb{N}}) : \mathbb{N} \to \mathbb{N}$ defined by $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = I_{\mathbb{N}}(x) + I_{\mathbb{N}}(x) = x + x = 2x$.

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To Show:

1. $I_{\mathbb{N}}$ is onto.

2. $I_{\mathbb{N}} + I_{\mathbb{N}}$ is not onto.

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Proof:

Part 1: Show that $I_{\mathbb{N}}$ is onto.

A function $f: A \to B$ is onto (surjective) if for every element $y$ in the codomain B, there exists at least one element $x$ in the domain A such that $f(x) = y$.

For $I_{\mathbb{N}} : \mathbb{N} \to \mathbb{N}$, the domain is $\mathbb{N}$ and the codomain is $\mathbb{N}$.

Let $y$ be an arbitrary element in the codomain $\mathbb{N}$. We need to find an element $x$ in the domain $\mathbb{N}$ such that $I_{\mathbb{N}}(x) = y$.

By the definition of $I_{\mathbb{N}}(x) = x$, the condition $I_{\mathbb{N}}(x) = y$ becomes $x = y$.

Since $y$ is an element of $\mathbb{N}$, $x = y$ is also an element of $\mathbb{N}$ (the domain).

Thus, for every $y \in \mathbb{N}$ (codomain), there exists $x = y \in \mathbb{N}$ (domain) such that $I_{\mathbb{N}}(x) = y$.

Therefore, the identity function $I_{\mathbb{N}}$ is onto.

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Part 2: Show that $I_{\mathbb{N}} + I_{\mathbb{N}}$ is not onto.

The function is $(I_{\mathbb{N}} + I_{\mathbb{N}}) : \mathbb{N} \to \mathbb{N}$ defined by $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = 2x$. The domain is $\mathbb{N}$ and the codomain is $\mathbb{N}$.

For the function to be onto, for every element $y$ in the codomain $\mathbb{N}$, there must exist at least one element $x$ in the domain $\mathbb{N}$ such that $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = y$.

We need to find if for every $y \in \mathbb{N}$, there exists $x \in \mathbb{N}$ such that $2x = y$.

This equation $2x = y$ implies $x = \frac{y}{2}$.

For $x$ to be in the domain $\mathbb{N}$, $x = \frac{y}{2}$ must be a natural number (positive integer).

Let's consider elements in the codomain $\mathbb{N}$. For example, take $y = 1 \in \mathbb{N}$.

We need to find $x \in \mathbb{N}$ such that $2x = 1$. Solving for x gives $x = \frac{1}{2}$.

The number $\frac{1}{2}$ is a rational number, but it is not a natural number. Thus, there is no element in the domain $\mathbb{N}$ that maps to the element 1 in the codomain $\mathbb{N}$ under the function $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = 2x$.

In general, for any odd natural number $y$ in the codomain (e.g., 1, 3, 5, ...), the required value of $x = \frac{y}{2}$ will be a fraction with a denominator of 2, which is not a natural number.

Since there exists at least one element in the codomain $\mathbb{N}$ (e.g., 1) for which there is no corresponding element in the domain $\mathbb{N}$ that maps to it, the function $(I_{\mathbb{N}} + I_{\mathbb{N}})$ is not onto.

This shows that even though the individual function $I_{\mathbb{N}}$ is onto, the sum of functions might not be onto.

Given:

Set A = $\left[ 0, \frac{\pi}{2} \right]$ (the closed interval from 0 to $\frac{\pi}{2}$).

Set B = $\mathbb{R}$ (the set of real numbers).

Function $f : A \to B$ given by $f(x) = \sin x$ for all $x \in A$.

Function $g : A \to B$ given by $g(x) = \cos x$ for all $x \in A$.

Function $(f + g) : A \to B$ defined by $(f + g)(x) = f(x) + g(x) = \sin x + \cos x$ for all $x \in A$.

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To Show:

1. f is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

2. g is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

3. f + g is not one-one on $\left[ 0, \frac{\pi}{2} \right]$.

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Proof:

Part 1: Show that f(x) = sin x is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

A function is one-one (injective) if for any $x_1, x_2$ in the domain, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Let $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$ such that $f(x_1) = f(x_2)$.

$\sin x_1 = \sin x_2$

The sine function is strictly increasing on the interval $\left[ 0, \frac{\pi}{2} \right]$. This means that if $x_1 < x_2$ and $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then $\sin x_1 < \sin x_2$. Conversely, if $\sin x_1 = \sin x_2$ for $x_1, x_2$ in this interval, it must mean $x_1 = x_2$.

Alternatively, if $\sin x_1 = \sin x_2$, the general solution is $x_1 = n\pi + (-1)^n x_2$ for some integer $n$. If $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then both are non-negative and at most $\frac{\pi}{2}$.

If $n=0$, $x_1 = x_2$. This is the case we need to show is the only possibility in the given domain.

If $n=1$, $x_1 = \pi - x_2$. If $x_1 \in \left[ 0, \frac{\pi}{2} \right]$, then $\sin x_1 \ge 0$. If $x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then $\sin x_2 \ge 0$. If $x_1 = \pi - x_2$, and $x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then $\pi - x_2 \in \left[ \frac{\pi}{2}, \pi \right]$. If $x_1$ is also in $\left[ 0, \frac{\pi}{2} \right]$, the only possibility is $x_1 = x_2 = \frac{\pi}{2}$. If $x_1 = x_2 = \frac{\pi}{2}$, $\frac{\pi}{2} = \pi - \frac{\pi}{2}$ is true. However, if $x_1 = 0$, $0 = \pi - x_2 \implies x_2 = \pi$, which is outside the domain. If $x_1 = \frac{\pi}{4}$, $\frac{\pi}{4} = \pi - x_2 \implies x_2 = \frac{3\pi}{4}$, outside the domain.

Considering the interval $\left[ 0, \frac{\pi}{2} \right]$, the sine function takes each value in $[0, 1]$ exactly once. Therefore, if $\sin x_1 = \sin x_2$ and $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, it must be that $x_1 = x_2$.

Therefore, f(x) = sin x is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

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Part 2: Show that g(x) = cos x is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

Let $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$ such that $g(x_1) = g(x_2)$.

$\cos x_1 = \cos x_2$

The cosine function is strictly decreasing on the interval $\left[ 0, \frac{\pi}{2} \right]$. This means that if $x_1 < x_2$ and $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then $\cos x_1 > \cos x_2$. Conversely, if $\cos x_1 = \cos x_2$ for $x_1, x_2$ in this interval, it must mean $x_1 = x_2$.

Alternatively, if $\cos x_1 = \cos x_2$, the general solution is $x_1 = 2n\pi \pm x_2$ for some integer $n$. If $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$.

If $n=0$, $x_1 = \pm x_2$. Since $x_1, x_2 \ge 0$, $x_1 = x_2$.

If $n=1$, $x_1 = 2\pi \pm x_2$. If $x_2 \in \left[ 0, \frac{\pi}{2} \right]$, then $2\pi \pm x_2$ will be outside the interval $\left[ 0, \frac{\pi}{2} \right]$. For example, $2\pi + x_2 \ge 2\pi$ and $2\pi - x_2 \ge 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$.

Considering the interval $\left[ 0, \frac{\pi}{2} \right]$, the cosine function takes each value in $[0, 1]$ exactly once. Therefore, if $\cos x_1 = \cos x_2$ and $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, it must be that $x_1 = x_2$.

Therefore, g(x) = cos x is one-one on $\left[ 0, \frac{\pi}{2} \right]$.

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Part 3: Show that f + g is not one-one on $\left[ 0, \frac{\pi}{2} \right]$.

The function is $(f + g)(x) = \sin x + \cos x$. For the function to be not one-one, we need to find at least two distinct values $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$ such that $(f + g)(x_1) = (f + g)(x_2)$, but $x_1 \neq x_2$.

We need to find $x_1, x_2 \in \left[ 0, \frac{\pi}{2} \right]$, $x_1 \neq x_2$, such that:

$\sin x_1 + \cos x_1 = \sin x_2 + \cos x_2$

Consider the values at the endpoints of the interval.

At $x = 0$: $(f + g)(0) = \sin 0 + \cos 0 = 0 + 1 = 1$.

At $x = \frac{\pi}{2}$: $(f + g)\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$.

We have $(f + g)(0) = 1$ and $(f + g)\left(\frac{\pi}{2}\right) = 1$.

The input values are $x_1 = 0$ and $x_2 = \frac{\pi}{2}$. Both 0 and $\frac{\pi}{2}$ are in the interval $\left[ 0, \frac{\pi}{2} \right]$.

Also, $0 \neq \frac{\pi}{2}$.

Since we found two distinct values in the domain (0 and $\frac{\pi}{2}$) that are mapped to the same value in the codomain (1), the function $(f + g)(x) = \sin x + \cos x$ is not one-one on the interval $\left[ 0, \frac{\pi}{2} \right]$.

This shows that the sum of two one-one functions is not necessarily one-one.

Given:

The function $f : \mathbb{R} \to \mathbb{R}$ defined as $f(x) = 10x + 7$.

$1_{\mathbb{R}}$ is the identity function on $\mathbb{R}$, defined by $1_{\mathbb{R}}(x) = x$ for all $x \in \mathbb{R}$.

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To Find:

The function $g : \mathbb{R} \to \mathbb{R}$ such that $g \circ f = f \circ g = 1_{\mathbb{R}}$.

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Solution:

The condition $g \circ f = f \circ g = 1_{\mathbb{R}}$ means that the function g is the inverse of the function f. So, we need to find the inverse function of f.

Let $y = f(x)$.

$y = 10x + 7$

To find the inverse function, we swap x and y and solve for y.

$x = 10y + 7$

Now, we solve for y in terms of x.

Subtract 7 from both sides:

$x - 7 = 10y$

Divide both sides by 10:

$y = \frac{x - 7}{10}$

This expression for y gives the inverse function, which we are calling g(x).

$g(x) = \frac{x - 7}{10}$

We can also write this as $g(x) = \frac{1}{10}x - \frac{7}{10}$.

The domain and codomain of g are $\mathbb{R}$, as required.

To verify that this is indeed the inverse, we check the composite functions:

Check $g \circ f(x) = g(f(x))$:

$g(f(x)) = g(10x + 7)$

Substitute $(10x + 7)$ into the expression for $g(x)$: $g(y) = \frac{y - 7}{10}$, so $g(10x + 7) = \frac{(10x + 7) - 7}{10} = \frac{10x}{10} = x$.

So, $g \circ f(x) = x = 1_{\mathbb{R}}(x)$.

Check $f \circ g(x) = f(g(x))$:

$f(g(x)) = f\left(\frac{x - 7}{10}\right)$

Substitute $\left(\frac{x - 7}{10}\right)$ into the expression for $f(x)$: $f(y) = 10y + 7$, so $f\left(\frac{x - 7}{10}\right) = 10\left(\frac{x - 7}{10}\right) + 7 = (x - 7) + 7 = x$.

So, $f \circ g(x) = x = 1_{\mathbb{R}}(x)$.

Since $g \circ f(x) = x$ and $f \circ g(x) = x$ for all $x \in \mathbb{R}$, the function $g(x) = \frac{x - 7}{10}$ is the required function.

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The function g is $\textbf{g(x) = } \frac{\textbf{x - 7}}{\textbf{10}}$.

Given:

A function $f : W \to W$ is defined as:

$f(n) = \begin{cases} n - 1 & , & \text{if } n \text{ is odd} \\ n + 1 & , & \text{if } n \text{ is even} \end{cases}$

where $W = \{0, 1, 2, 3, ...\}$ is the set of all whole numbers.

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To Show:

1. The function $f$ is invertible.

2. Find the inverse function $f^{-1}$.

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Proof (Invertibility):

A function is invertible if and only if it is bijective (i.e., both one-to-one and onto).

1. Checking for One-to-One (Injectivity):

Let $n_1, n_2 \in W$ such that $f(n_1) = f(n_2)$. We need to show $n_1 = n_2$.

We consider four possible cases:

Case (i): $n_1$ and $n_2$ are both odd.

Then $f(n_1) = n_1 - 1$ and $f(n_2) = n_2 - 1$.

$f(n_1) = f(n_2) \implies n_1 - 1 = n_2 - 1 \implies n_1 = n_2$.

Case (ii): $n_1$ and $n_2$ are both even.

Then $f(n_1) = n_1 + 1$ and $f(n_2) = n_2 + 1$.

$f(n_1) = f(n_2) \implies n_1 + 1 = n_2 + 1 \implies n_1 = n_2$.

Case (iii): $n_1$ is odd and $n_2$ is even.

Then $f(n_1) = n_1 - 1$. Since $n_1$ is odd ($n_1 \ge 1$), $n_1 - 1$ is an even whole number ($n_1 - 1 \ge 0$).

And $f(n_2) = n_2 + 1$. Since $n_2$ is even ($n_2 \ge 0$), $n_2 + 1$ is an odd whole number ($n_2 + 1 \ge 1$).

An even number cannot be equal to an odd number. Thus, $f(n_1) \neq f(n_2)$. This case contradicts the assumption $f(n_1) = f(n_2)$.

Case (iv): $n_1$ is even and $n_2$ is odd.

Then $f(n_1) = n_1 + 1$ (odd) and $f(n_2) = n_2 - 1$ (even).

Again, an odd number cannot be equal to an even number. Thus, $f(n_1) \neq f(n_2)$. This case also contradicts the assumption $f(n_1) = f(n_2)$.

From the cases considered, we conclude that $f(n_1) = f(n_2)$ implies $n_1 = n_2$.

Therefore, f is one-to-one (injective).

2. Checking for Onto (Surjectivity):

Let $m$ be an arbitrary element in the codomain $W$. We need to find a pre-image $n \in W$ (domain) such that $f(n) = m$.

Case (i): $m$ is an odd whole number ($m \in \{1, 3, 5, ...\}$).

We need $f(n) = m$. If $n$ were odd, $f(n) = n-1$ would be even, which cannot equal $m$. So, $n$ must be even.

Let $n$ be even. Then $f(n) = n + 1$.

We set $f(n) = m \implies n + 1 = m \implies n = m - 1$.

Since $m$ is odd and $m \ge 1$, $n = m - 1$ is an even whole number ($n \ge 0$).

So, for every odd $m$ in the codomain $W$, there exists $n = m - 1 \in W$ (which is even) such that $f(n) = f(m-1) = (m-1) + 1 = m$.

Case (ii): $m$ is an even whole number ($m \in \{0, 2, 4, ...\}$).

We need $f(n) = m$. If $n$ were even, $f(n) = n+1$ would be odd, which cannot equal $m$. So, $n$ must be odd.

Let $n$ be odd. Then $f(n) = n - 1$.

We set $f(n) = m \implies n - 1 = m \implies n = m + 1$.

Since $m$ is even and $m \ge 0$, $n = m + 1$ is an odd whole number ($n \ge 1$).

So, for every even $m$ in the codomain $W$, there exists $n = m + 1 \in W$ (which is odd) such that $f(n) = f(m+1) = (m+1) - 1 = m$.

Since every element $m$ in the codomain $W$ has a pre-image $n$ in the domain $W$, the function f is onto (surjective).

Since $f$ is both one-to-one and onto, $f$ is bijective.

Therefore, f is invertible.

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Finding the Inverse $f^{-1}$:

Let $f(n) = m$. Then $n = f^{-1}(m)$.

From the onto proof:

If $m$ is odd, the pre-image is $n = m - 1$. So, $f^{-1}(m) = m - 1$.

If $m$ is even, the pre-image is $n = m + 1$. So, $f^{-1}(m) = m + 1$.

Thus, the inverse function $f^{-1} : W \to W$ is defined as:

$f^{-1}(m) = \begin{cases} m - 1 & , & \text{if } m \text{ is odd} \\ m + 1 & , & \text{if } m \text{ is even} \end{cases}$

Comparing the definition of $f^{-1}(m)$ with $f(n)$, we see that they have the same rule.

Let's verify by checking if $(f \circ f)(n) = n$ for all $n \in W$.

If $n$ is odd: $f(n) = n - 1$ (which is even).

Then $(f \circ f)(n) = f(f(n)) = f(n - 1) = (n - 1) + 1 = n$.

If $n$ is even: $f(n) = n + 1$ (which is odd).

Then $(f \circ f)(n) = f(f(n)) = f(n + 1) = (n + 1) - 1 = n$.

In both cases, $(f \circ f)(n) = n$. This confirms that $f$ is its own inverse.

Therefore, the inverse of $f$ is $f$ itself: $\mathbf{f^{-1} = f}$.

Given:

The function $f : R \to R$ is defined by:

$f(x) = x^2 - 3x + 2$

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To Find:

We need to find the expression for $f(f(x))$.

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Solution:

To find $f(f(x))$, we substitute $f(x)$ into the function $f$.

$f(f(x)) = f(x^2 - 3x + 2)$

Now, we replace $x$ in the expression for $f(x)$ with $(x^2 - 3x + 2)$:

$f(f(x)) = (x^2 - 3x + 2)^2 - 3(x^2 - 3x + 2) + 2$

First, let's expand the term $(x^2 - 3x + 2)^2$:

$(x^2 - 3x + 2)^2 = (x^2 - 3x + 2)(x^2 - 3x + 2)$

$= x^2(x^2 - 3x + 2) - 3x(x^2 - 3x + 2) + 2(x^2 - 3x + 2)$

$= (x^4 - 3x^3 + 2x^2) - (3x^3 - 9x^2 + 6x) + (2x^2 - 6x + 4)$

$= x^4 - 3x^3 + 2x^2 - 3x^3 + 9x^2 - 6x + 2x^2 - 6x + 4$

Combining like terms:

$= x^4 + (-3 - 3)x^3 + (2 + 9 + 2)x^2 + (-6 - 6)x + 4$

$= x^4 - 6x^3 + 13x^2 - 12x + 4$

Next, let's expand the term $-3(x^2 - 3x + 2)$:

$-3(x^2 - 3x + 2) = -3x^2 + 9x - 6$

Now, substitute these expanded forms back into the expression for $f(f(x))$:

$f(f(x)) = (x^4 - 6x^3 + 13x^2 - 12x + 4) + (-3x^2 + 9x - 6) + 2$

Combine like terms again:

$f(f(x)) = x^4 - 6x^3 + (13 - 3)x^2 + (-12 + 9)x + (4 - 6 + 2)$

$f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x + 0$

$f(f(x)) = x^4 - 6x^3 + 10x^2 - 3x$

Therefore, the expression for $f(f(x))$ is $\mathbf{x^4 - 6x^3 + 10x^2 - 3x}$.

Given:

The function $f : R \to \{x \in R : -1 < x < 1\}$ is defined by:

$f(x) = \frac{x}{1 + |x|}$, for all $x \in R$.

The codomain is the open interval $(-1, 1)$.

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To Show:

The function $f$ is one-one (injective) and onto (surjective).

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Proof:

First, let's express the function $f(x)$ in a piecewise form based on the definition of $|x|$:

$|x| = \begin{cases} x & , & x \ge 0 \\ -x & , & x < 0 \end{cases}$

So, $f(x) = \begin{cases} \frac{x}{1 + x} & , & x \ge 0 \\ \frac{x}{1 - x} & , & x < 0 \end{cases}$

1. One-to-One (Injectivity):

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$. We need to show that $x_1 = x_2$.

We consider the following cases:

Case (i): $x_1 \ge 0$ and $x_2 \ge 0$.

Then $f(x_1) = \frac{x_1}{1 + x_1}$ and $f(x_2) = \frac{x_2}{1 + x_2}$.

$f(x_1) = f(x_2) \implies \frac{x_1}{1 + x_1} = \frac{x_2}{1 + x_2}$

$\implies x_1(1 + x_2) = x_2(1 + x_1)$

$\implies x_1 + x_1 x_2 = x_2 + x_1 x_2$

$\implies x_1 = x_2$

Case (ii): $x_1 < 0$ and $x_2 < 0$.

Then $f(x_1) = \frac{x_1}{1 - x_1}$ and $f(x_2) = \frac{x_2}{1 - x_2}$.

$f(x_1) = f(x_2) \implies \frac{x_1}{1 - x_1} = \frac{x_2}{1 - x_2}$

$\implies x_1(1 - x_2) = x_2(1 - x_1)$

$\implies x_1 - x_1 x_2 = x_2 - x_1 x_2$

$\implies x_1 = x_2$

Case (iii): $x_1 \ge 0$ and $x_2 < 0$.

Then $f(x_1) = \frac{x_1}{1 + x_1}$. Since $x_1 \ge 0$, $1+x_1 > 0$. Thus, $f(x_1) \ge 0$.

And $f(x_2) = \frac{x_2}{1 - x_2}$. Since $x_2 < 0$, the numerator $x_2$ is negative. The denominator $1 - x_2 = 1 + (-x_2)$ is positive because $-x_2 > 0$. Thus, $f(x_2) < 0$.

Since $f(x_1) \ge 0$ and $f(x_2) < 0$, it is impossible that $f(x_1) = f(x_2)$ unless $f(x_1) = f(x_2) = 0$.

$f(x_1) = 0 \implies \frac{x_1}{1+x_1} = 0 \implies x_1 = 0$.

$f(x_2) = 0 \implies \frac{x_2}{1-x_2} = 0 \implies x_2 = 0$.

However, this case assumes $x_2 < 0$. So, $f(x_1)$ cannot be equal to $f(x_2)$ in this case.

Case (iv): $x_1 < 0$ and $x_2 \ge 0$.

This is symmetric to Case (iii), and we again find that $f(x_1) < 0$ and $f(x_2) \ge 0$. So, $f(x_1) = f(x_2)$ is impossible unless $x_1 = x_2 = 0$, which contradicts $x_1 < 0$.

From all cases, we see that $f(x_1) = f(x_2)$ implies $x_1 = x_2$.

Therefore, f is one-to-one (injective).

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2. Onto (Surjectivity):

Let $y$ be an arbitrary element in the codomain $(-1, 1)$. We need to find an $x \in R$ (domain) such that $f(x) = y$.

We consider two cases for $y$:

Case (i): $0 \le y < 1$.

We look for an $x \ge 0$ such that $f(x) = y$.

$f(x) = \frac{x}{1+x} = y$

$\implies x = y(1+x)$

$\implies x = y + yx$

$\implies x - yx = y$

$\implies x(1 - y) = y$

$\implies x = \frac{y}{1 - y}$

Since $0 \le y < 1$, we have $y \ge 0$ and $1 - y > 0$. Therefore, $x = \frac{y}{1 - y} \ge 0$.

Thus, for any $y$ in $[0, 1)$, there exists $x = \frac{y}{1 - y} \in R$ with $x \ge 0$ such that $f(x) = y$.

Case (ii): $-1 < y < 0$.

We look for an $x < 0$ such that $f(x) = y$.

$f(x) = \frac{x}{1 - x} = y$

$\implies x = y(1 - x)$

$\implies x = y - yx$

$\implies x + yx = y$

$\implies x(1 + y) = y$

$\implies x = \frac{y}{1 + y}$

Since $-1 < y < 0$, we have $y < 0$ and $1 + y > 0$. Therefore, $x = \frac{y}{1 + y} < 0$.

Thus, for any $y$ in $(-1, 0)$, there exists $x = \frac{y}{1 + y} \in R$ with $x < 0$ such that $f(x) = y$.

Combining both cases, for any $y \in (-1, 1)$, there exists an $x \in R$ such that $f(x) = y$.

Therefore, f is onto (surjective).

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Conclusion:

Since the function $f$ is both one-to-one and onto, it is a bijective function from $R$ to $(-1, 1)$.

Given:

The function $f : R \to R$ is defined by $f(x) = x^3$.

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To Show:

The function $f$ is injective (one-to-one).

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Proof:

To show that $f$ is injective, we need to demonstrate that for any $x_1, x_2 \in R$, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Let $x_1, x_2 \in R$ such that $f(x_1) = f(x_2)$.

By the definition of $f$, we have:

$x_1^3 = x_2^3$

Subtract $x_2^3$ from both sides:

$x_1^3 - x_2^3 = 0$

We can factor the difference of cubes:

$(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0$

This equation implies that either $(x_1 - x_2) = 0$ or $(x_1^2 + x_1x_2 + x_2^2) = 0$.

Case 1: $x_1 - x_2 = 0$

This directly gives $x_1 = x_2$.

Case 2: $x_1^2 + x_1x_2 + x_2^2 = 0$

We can analyze the term $x_1^2 + x_1x_2 + x_2^2$. Let's try to complete the square with respect to $x_1$:

$x_1^2 + x_1x_2 + \left(\frac{x_2}{2}\right)^2 - \left(\frac{x_2}{2}\right)^2 + x_2^2 = 0$

$\left(x_1 + \frac{x_2}{2}\right)^2 + x_2^2 - \frac{x_2^2}{4} = 0$

$\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3}{4}x_2^2 = 0$

Since $x_1$ and $x_2$ are real numbers, $\left(x_1 + \frac{x_2}{2}\right)^2 \ge 0$ and $\frac{3}{4}x_2^2 \ge 0$.

The sum of two non-negative terms can be zero only if both terms are individually zero.

So, we must have $\frac{3}{4}x_2^2 = 0$, which implies $x_2 = 0$.

And we must have $\left(x_1 + \frac{x_2}{2}\right)^2 = 0$. Substituting $x_2 = 0$, we get $(x_1 + 0)^2 = 0$, which implies $x_1^2 = 0$, so $x_1 = 0$.

Therefore, the condition $x_1^2 + x_1x_2 + x_2^2 = 0$ holds only when $x_1 = 0$ and $x_2 = 0$. In this case, we also have $x_1 = x_2$.

Combining both cases, the condition $f(x_1) = f(x_2)$ always implies $x_1 = x_2$ for any real numbers $x_1, x_2$.

Hence, the function $f(x) = x^3$ is injective (one-to-one).

Given:

We need to find two functions $f : N \to Z$ and $g : Z \to Z$ such that their composition $g \circ f$ is injective, but $g$ itself is not injective.

Here, $N = \{1, 2, 3, ...\}$ is the set of natural numbers and $Z = \{..., -2, -1, 0, 1, 2, ...\}$ is the set of integers.

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Solution:

Let's define the functions $f$ and $g$ as suggested by the hint:

Define $f : N \to Z$ by $f(x) = x$.

Define $g : Z \to Z$ by $g(x) = |x|$.

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Step 1: Check if g is injective.

A function $g$ is injective if for every $x_1, x_2$ in its domain $Z$, $g(x_1) = g(x_2)$ implies $x_1 = x_2$.

Consider $x_1 = 1$ and $x_2 = -1$. Both $1$ and $-1$ are in the domain $Z$.

$g(x_1) = g(1) = |1| = 1$

$g(x_2) = g(-1) = |-1| = 1$

Here, $g(1) = g(-1)$, but $1 \neq -1$.

Therefore, the function $\mathbf{g(x) = |x|}$ is not injective.

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Step 2: Find the composition $g \circ f$ and check if it is injective.

The composition function $g \circ f : N \to Z$ is defined as $(g \circ f)(x) = g(f(x))$.

The domain of $g \circ f$ is the domain of $f$, which is $N$.

Substitute $f(x) = x$ into the expression for $g$:

$(g \circ f)(x) = g(f(x)) = g(x) = |x|$.

However, the domain for $g \circ f$ is $N$. Since $x \in N$, $x$ is always a positive integer ($x \ge 1$).

For any $x \in N$, $|x| = x$.

So, $(g \circ f)(x) = x$ for all $x \in N$.

Now, we check if $g \circ f$ is injective. Let $x_1, x_2 \in N$ such that $(g \circ f)(x_1) = (g \circ f)(x_2)$.

$(g \circ f)(x_1) = x_1$

$(g \circ f)(x_2) = x_2$

So, if $(g \circ f)(x_1) = (g \circ f)(x_2)$, then $x_1 = x_2$.

This satisfies the condition for injectivity.

Therefore, the function $\mathbf{(g \circ f)(x) = x}$ for $x \in N$ is injective.

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Conclusion:

We have found functions $f(x) = x$ ($f: N \to Z$) and $g(x) = |x|$ ($g: Z \to Z$) such that $g \circ f$ is injective, but $g$ is not injective. These functions serve as the required example.