Chapter 4 Relations (Q & A)

An ordered pair is a pair of objects where the order matters. It is typically denoted by $(x, y)$, where $x$ is the first element and $y$ is the second element.

Let's analyze the given options:

(A) $\{2, 3\}$ represents a set, where the order of elements does not matter. So, $\{2, 3\}$ is the same as $\{3, 2\}$.

(B) $(3, 2)$ is enclosed in parentheses, indicating that the order of the elements is significant. This is the standard notation for an ordered pair.

(C) $\{a, b\}$ represents a set, where the order of elements does not matter. So, $\{a, b\}$ is the same as $\{b, a\}$.

(D) $[1, 5]$ typically denotes a closed interval on the number line, representing all real numbers between 1 and 5, inclusive. While notation can vary, in the context of ordered pairs, square brackets usually do not represent ordered pairs.

Therefore, the correct option representing an ordered pair is (B).

We are given two ordered pairs that are equal:

For two ordered pairs to be equal, their corresponding elements must be equal.

Equating the first elements:

Equating the second elements:

Solving equation (ii) for $x$:

Subtract 1 from both sides:

Solving equation (iii) for $y$:

Add 2 to both sides:

Thus, the values are $x=2$ and $y=7$.

This corresponds to option (B).

An ordered pair is a pair of objects where the order matters.

It is represented in the form $(x, y)$, where $x$ is the first component (also called the abscissa or the first coordinate) and $y$ is the second component (also called the ordinate or the second coordinate).

In the given ordered pair $(a, b)$:

The element '$a$' comes first in the notation and is therefore the first component.

The element '$b$' comes second in the notation and is therefore the second component.

Thus, the element that represents the first component is $a$.

This corresponds to option (A).

The definition of equality for ordered pairs states that two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if their corresponding components are equal.

This means that the first component of the first ordered pair must be equal to the first component of the second ordered pair, and the second component of the first ordered pair must be equal to the second component of the second ordered pair.

Mathematically, this can be written as:

Let's examine the given options:

(A) $a=c$ or $b=d$: This condition is not sufficient for equality. For example, $(2, 5) = (2, 3)$ is false, but $a=c$ is true.

(B) $a=b$ and $c=d$: This condition relates the components within each ordered pair, not the equality between the ordered pairs themselves.

(C) $a=c$ and $b=d$: This condition exactly matches the definition of equality for ordered pairs.

(D) $a=d$ and $b=c$: This condition implies that the components are swapped, which is generally not the case for equality of ordered pairs unless $a=b$ and $c=d$ as well.

Therefore, the condition that must be met for two ordered pairs $(a, b)$ and $(c, d)$ to be equal is $a=c$ and $b=d$.

This corresponds to option (C).

Let's analyze the Assertion (A) and Reason (R) separately.

Assertion (A): The ordered pairs $(2, 3)$ and $(3, 2)$ are not equal.

For two ordered pairs $(a, b)$ and $(c, d)$ to be equal, their corresponding elements must be equal, i.e., $a=c$ and $b=d$.

In this case, we have $(2, 3)$ and $(3, 2)$.

The first component of the first pair is $2$, and the first component of the second pair is $3$. Since $2 \neq 3$, the first components are not equal.

The second component of the first pair is $3$, and the second component of the second pair is $2$. Since $3 \neq 2$, the second components are not equal.

Because the corresponding components are not equal, the ordered pairs $(2, 3)$ and $(3, 2)$ are not equal.

Therefore, Assertion (A) is true.

Reason (R): In an ordered pair, the order of elements matters.

This statement is the fundamental definition of an ordered pair. The very characteristic that distinguishes an ordered pair from a set is that the sequence in which the elements are listed is important.

For example, in the ordered pair $(2, 3)$, $2$ is the first element and $3$ is the second. In the ordered pair $(3, 2)$, $3$ is the first element and $2$ is the second. Since the order is different, they are considered different ordered pairs.

Therefore, Reason (R) is true.

Now, let's check if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that $(2, 3) \neq (3, 2)$.

Reason (R) explains *why* this is the case: because the order of elements matters in an ordered pair.

The fact that the order matters directly leads to the conclusion that $(2, 3)$ is different from $(3, 2)$, as the positions of $2$ and $3$ are swapped.

Therefore, Reason (R) is indeed the correct explanation for Assertion (A).

Both A and R are true, and R is the correct explanation of A.

This corresponds to option (A).

The Cartesian product of two sets $A$ and $B$, denoted by $A \times B$, is the set of all possible ordered pairs $(a, b)$ where $a$ is an element of $A$ and $b$ is an element of $B$.

Given the sets:

$A = \{1, 2\}$

$B = \{a, b\}$

To find the Cartesian product $A \times B$, we take each element from set $A$ and pair it with each element from set $B$, forming ordered pairs.

For the first element of $A$, which is $1$, we pair it with each element of $B$:

$1$ paired with $a$ gives the ordered pair $(1, a)$.

$1$ paired with $b$ gives the ordered pair $(1, b)$.

For the second element of $A$, which is $2$, we pair it with each element of $B$:

$2$ paired with $a$ gives the ordered pair $(2, a)$.

$2$ paired with $b$ gives the ordered pair $(2, b)$.

The set of all these ordered pairs is the Cartesian product $A \times B$.

$A \times B = \{(1, a), (1, b), (2, a), (2, b)\}$

Comparing this result with the given options, we find that option (C) matches our calculated Cartesian product.

Given:

The number of elements in set A, $\text{n}(A) = 3$.

The number of elements in set B, $\text{n}(B) = 4$.

To Find:

The number of elements in the Cartesian product $A \times B$, i.e., $\text{n}(A \times B)$.

Solution:

The number of elements in the Cartesian product of two finite sets A and B is given by the product of the number of elements in each set.

Substituting the given values:

Therefore, the number of elements in the Cartesian product $A \times B$ is 12.

The correct option is (B).

Given:

Set $A = \{a, b\}$

Set $B = \{c, d\}$

To Find:

The Cartesian product $B \times A$.

Solution:

The Cartesian product $B \times A$ is the set of all ordered pairs $(x, y)$ where $x$ is an element of $B$ and $y$ is an element of $A$.

To find $B \times A$, we pair each element of $B$ with each element of $A$:

First element of B is 'c'. Pair it with elements of A:

$(c, a)$

$(c, b)$

Second element of B is 'd'. Pair it with elements of A:

$(d, a)$

$(d, b)$

So, $B \times A = \{(c, a), (c, b), (d, a), (d, b)\}$.

The correct option is (B).

Given:

Set $A = \{1, 2, 3\}$

Set $B = \emptyset$ (the empty set)

To Find:

The Cartesian product $A \times B$.

Solution:

The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

Since set $B$ is the empty set, it contains no elements. Therefore, it is impossible to form any ordered pair $(x, y)$ where the second element $y$ belongs to $B$.

Hence, the Cartesian product $A \times B$ is also the empty set.

This is because $\text{n}(A \times B) = \text{n}(A) \times \text{n}(B) = 3 \times 0 = 0$. A set with 0 elements is the empty set.

The correct option is (B).

Let's analyze each statement:

Statement (A): $A \times B = B \times A$ is always true.

This statement is INCORRECT.

The Cartesian product is not generally commutative. For $A \times B$ to be equal to $B \times A$, every ordered pair $(a, b)$ in $A \times B$ must also be in $B \times A$. This means if $(a, b) \in A \times B$, then $(b, a) \in B \times A$. This is true. However, for $A \times B = B \times A$ as sets, they must contain exactly the same elements. If $A \neq B$, then $A \times B$ will generally be different from $B \times A$. For example, if $A = \{1, 2\}$ and $B = \{3, 4\}$, then $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$ and $B \times A = \{(3, 1), (3, 2), (4, 1), (4, 2)\}$. Clearly, $A \times B \neq B \times A$. The equality $A \times B = B \times A$ holds only if $A = B$.

Statement (B): $\text{n}(A \times B) = \text{n}(A) \times \text{n}(B)$.

This statement is CORRECT.

This is the definition of the number of elements in the Cartesian product of two finite sets.

Statement (C): If $A = \emptyset$, then $A \times B = \emptyset$.

This statement is CORRECT.

As established in a previous question, if one of the sets in a Cartesian product is empty, the resulting Cartesian product is also empty.

Statement (D): The elements of $A \times B$ are ordered pairs.

This statement is CORRECT.

By definition, the Cartesian product of two sets $A$ and $B$ is the set of all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

Since the question asks for the INCORRECT statement, it is statement (A).

The INCORRECT statement is (A).

Given:

The number of elements in set A, $\text{n}(A) = p$.

The number of elements in set B, $\text{n}(B) = q$.

To Find:

The number of elements in the Cartesian product $A \times B$, i.e., $\text{n}(A \times B)$.

Solution:

The number of elements in the Cartesian product of two finite sets $A$ and $B$ is the product of the number of elements in each set.

This is a fundamental property of Cartesian products.

Substituting the given values $p$ for $\text{n}(A)$ and $q$ for $\text{n}(B)$:

Therefore, $\text{n}(A \times B) = pq$.

The correct option is (B).

Given:

Set $A = \{1, 2\}$

To Find:

The Cartesian product $A \times A \times A$.

Solution:

The Cartesian product $A \times A \times A$ is the set of all ordered triples $(x, y, z)$ where $x \in A$, $y \in A$, and $z \in A$.

Since set $A$ has 2 elements, $\text{n}(A) = 2$.

The number of elements in $A \times A \times A$ is $\text{n}(A \times A \times A) = \text{n}(A) \times \text{n}(A) \times \text{n}(A) = 2 \times 2 \times 2 = 8$.

We need to list all possible combinations of elements from $A$ taken three at a time, in order:

Starting with the first element of A (1):

• (1, 1, 1)
• (1, 1, 2)
• (1, 2, 1)
• (1, 2, 2)

Starting with the second element of A (2):

• (2, 1, 1)
• (2, 1, 2)
• (2, 2, 1)
• (2, 2, 2)

Combining all these ordered triples, we get:

$A \times A \times A = \{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\}$

The correct option is (B).

First, let's determine the elements of set A and set B.

Set A is defined as $A = \{x \in \mathbb{N} : x < 3\}$.

Here, $\mathbb{N}$ represents the set of natural numbers. Natural numbers typically start from 1 (i.e., {1, 2, 3, ...}).

So, the natural numbers less than 3 are 1 and 2.

Therefore, $A = \{1, 2\}$.

Set B is defined as $B = \{y \in \mathbb{Z} : -1 \leq y \leq 1\}$.

Here, $\mathbb{Z}$ represents the set of integers.

The integers $y$ such that $-1 \leq y \leq 1$ are -1, 0, and 1.

Therefore, $B = \{-1, 0, 1\}$.

Now, we need to find the Cartesian product $A \times B$.

This is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

We pair each element of A with each element of B:

For the element 1 from set A:

• $(1, -1)$
• $(1, 0)$
• $(1, 1)$

For the element 2 from set A:

• $(2, -1)$
• $(2, 0)$
• $(2, 1)$

Combining these ordered pairs, we get:

$A \times B = \{(1, -1), (1, 0), (1, 1), (2, -1), (2, 0), (2, 1)\}$

The correct option is (A).

A relation from a set A to a set B is formally defined as any subset of the Cartesian product $A \times B$.

Let's analyze the given options:

(A) A collection of elements from A and B.

This is too general. While a relation involves elements from A and B, this description doesn't specify the structure or the condition that makes it a relation.

(B) A subset of $A \cup B$.

A subset of $A \cup B$ could contain elements that are not ordered pairs, or ordered pairs where both elements are from A, or both from B. A relation is specifically about connections between elements of A and elements of B, represented as ordered pairs from $A \times B$.

(C) A subset of $A \times B$.

This aligns with the formal definition of a relation. The Cartesian product $A \times B$ contains all possible ordered pairs where the first element is from A and the second is from B. A relation is simply a selection of these possible connections.

(D) A function from A to B.

A function is a specific type of relation where each element in the domain (set A) is related to exactly one element in the codomain (set B). While all functions are relations, not all relations are functions.

Therefore, the most accurate and general definition of a relation from set A to set B is a subset of $A \times B$.

The correct option is (C).

Given:

Set $A = \{1, 2, 3\}$

Set $B = \{x, y\}$

To Find:

Which of the given options is a relation from A to B.

Solution:

A relation from set A to set B is a subset of the Cartesian product $A \times B$. The Cartesian product $A \times B$ consists of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

First, let's find the Cartesian product $A \times B$:

$A \times B = \{(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)\}$

Now, let's examine each option to see if it is a subset of $A \times B$.

(A) $\{(1, x), (2, y), (4, x)\}$

This set contains the ordered pair $(4, x)$. Since $4 \notin A$, this ordered pair is not in $A \times B$. Therefore, this is not a relation from A to B.

(B) $\{(x, 1), (y, 2)\}$

This set contains ordered pairs where the first element is from B and the second element is from A. This represents a relation from B to A, not from A to B.

(C) $\{(1, x), (2, y)\}$

Both $(1, x)$ and $(2, y)$ are elements of $A \times B$ because $1 \in A$, $2 \in A$, $x \in B$, and $y \in B$. Since this set is a collection of ordered pairs where the first element is from A and the second is from B, it is a subset of $A \times B$. Therefore, this is a relation from A to B.

(D) $\{(1, 2), (x, y)\}$

This set contains the ordered pair $(1, 2)$. Since $2 \notin B$, this ordered pair is not in $A \times B$. It also contains $(x, y)$, where $x \notin A$. Therefore, this is not a relation from A to B.

The correct option is (C).

Given:

Set $A = \{a, b\}$

To Find:

The number of possible relations from A to A.

Solution:

A relation from set A to set A is a subset of the Cartesian product $A \times A$.

First, let's find the Cartesian product $A \times A$:

$A \times A = \{(a, a), (a, b), (b, a), (b, b)\}$

The number of elements in $A \times A$ is $\text{n}(A \times A) = \text{n}(A) \times \text{n}(A) = 2 \times 2 = 4$.

The number of possible relations from A to A is the number of subsets of $A \times A$. If a set has $n$ elements, the number of its subsets is $2^n$.

In this case, the set $A \times A$ has 4 elements. So, the number of possible relations is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Therefore, there are 16 possible relations that can be defined from A to A.

The correct option is (C).

Given:

A relation $R = \{(1, 2), (2, 3), (3, 4)\}$

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

Set $B = \{1, 2, 3, 4\}$

To Find:

The domain of the relation R.

Solution:

The domain of a relation $R$ from set A to set B is the set of all first elements (or the first components) of the ordered pairs in R.

The relation R is given as: $R = \{(1, 2), (2, 3), (3, 4)\}$.

The ordered pairs in R are:

• $(1, 2)$: The first element is 1.
• $(2, 3)$: The first element is 2.
• $(3, 4)$: The first element is 3.

The set of all first elements from these ordered pairs is $\{1, 2, 3\}$.

Therefore, the domain of R is $\{1, 2, 3\}$.

Note: The domain of a relation is a subset of the first set from which the relation is defined (in this case, set A). While the domain is $\{1, 2, 3\}$, it is indeed a subset of $A = \{1, 2, 3, 4\}$. Option (A) and (D) are the same, but they include an element (4) that is not present as a first element in any ordered pair of R.

The correct option is (C).

Given:

A relation $R = \{(1, 2), (2, 3), (3, 4)\}$

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

Set $B = \{1, 2, 3, 4\}$

To Find:

The range of the relation R.

Solution:

The range of a relation $R$ from set A to set B is the set of all second elements (or the second components) of the ordered pairs in R.

The relation R is given as: $R = \{(1, 2), (2, 3), (3, 4)\}$.

The ordered pairs in R are:

• $(1, 2)$: The second element is 2.
• $(2, 3)$: The second element is 3.
• $(3, 4)$: The second element is 4.

The set of all second elements from these ordered pairs is $\{2, 3, 4\}$.

Therefore, the range of R is $\{2, 3, 4\}$.

Note: The range of a relation is a subset of the second set from which the relation is defined (in this case, set B). While the range is $\{2, 3, 4\}$, it is indeed a subset of $B = \{1, 2, 3, 4\}$.

The correct option is (B).

In the context of relations (and functions) between sets, if a relation $R$ is defined from set A to set B:

• Set A is called the **domain** of the relation. It is the set of all possible first elements of the ordered pairs in the relation.
• Set B is called the **codomain** of the relation. It is the set that contains all the possible second elements of the ordered pairs in the relation.
• The **range** of the relation is the set of all actual second elements of the ordered pairs in the relation, which is a subset of the codomain.

Therefore, if a relation R is defined from set A to set B, then B is the codomain of the relation R.

The correct option is (B).

Let's analyze the Assertion (A) and the Reason (R) separately.

Assertion (A): If $\text{n}(A)=m$ and $\text{n}(B)=n$, the number of relations from A to B is $2^{mn}$.

To determine the number of relations from A to B, we first need to consider the Cartesian product $A \times B$.

The number of elements in $A \times B$ is given by $\text{n}(A \times B) = \text{n}(A) \times \text{n}(B) = m \times n = mn$.

A relation from A to B is defined as any subset of $A \times B$. The number of subsets of a set with $k$ elements is $2^k$.

In this case, the set $A \times B$ has $mn$ elements.

Therefore, the number of possible subsets of $A \times B$, which represents the number of relations from A to B, is $2^{mn}$.

So, Assertion (A) is TRUE.

Reason (R): A relation from A to B is any subset of $A \times B$, and the number of subsets of a set with $k$ elements is $2^k$.

The first part of the reason states that "A relation from A to B is any subset of $A \times B$." This is the correct definition of a relation.

The second part of the reason states that "the number of subsets of a set with $k$ elements is $2^k$." This is also a correct fundamental principle in set theory (the power set theorem).

Thus, Reason (R) is TRUE.

Now, let's determine if Reason (R) correctly explains Assertion (A).

Assertion (A) calculates the number of relations using the definition provided in Reason (R) and the formula for the number of subsets.

The number of elements in $A \times B$ is $mn$ (as derived from $\text{n}(A)=m$ and $\text{n}(B)=n$).

According to Reason (R), the number of subsets of a set with $k$ elements is $2^k$. Applying this to $A \times B$ (where $k = mn$), the number of subsets (relations) is $2^{mn}$.

Therefore, Reason (R) provides the exact logic and formulas used to arrive at the conclusion in Assertion (A).

Both A and R are true, and R is the correct explanation of A.

The correct option is (A).

Given:

Department Sales (S) employees: $\{E1, E2\}$

Department Production (P) employees: $\{M1, M2, M3\}$

Relation R from S to P is defined as $(e, m) \in S \times P$ if employee $e$ from Sales coordinates with employee $m$ from Production.

The relation is given by $R = \{(E1, M1), (E1, M3), (E2, M2)\}$.

To Find:

The domain of the relation R.

Solution:

The domain of a relation R from set S to set P is the set of all first elements of the ordered pairs in R.

The given relation is $R = \{(E1, M1), (E1, M3), (E2, M2)\}$.

Let's identify the first element of each ordered pair in R:

• In $(E1, M1)$, the first element is $E1$.
• In $(E1, M3)$, the first element is $E1$.
• In $(E2, M2)$, the first element is $E2$.

The set of all these first elements, without repetition, is $\{E1, E2\}$.

Therefore, the domain of the relation R is $\{E1, E2\}$.

The correct option is (A).

Given:

Department Sales (S) employees: $\{E1, E2\}$

Department Production (P) employees: $\{M1, M2, M3\}$

Relation R from S to P is defined as $(e, m) \in S \times P$ if employee $e$ from Sales coordinates with employee $m$ from Production.

The relation is given by $R = \{(E1, M1), (E1, M3), (E2, M2)\}$.

To Find:

The range of the relation R.

Solution:

The range of a relation R from set S to set P is the set of all second elements of the ordered pairs in R.

The given relation is $R = \{(E1, M1), (E1, M3), (E2, M2)\}$.

Let's identify the second element of each ordered pair in R:

• In $(E1, M1)$, the second element is $M1$.
• In $(E1, M3)$, the second element is $M3$.
• In $(E2, M2)$, the second element is $M2$.

The set of all these second elements, without repetition, is $\{M1, M3, M2\}$.

This set is equivalent to $\{M1, M2, M3\}$.

Therefore, the range of the relation R is $\{M1, M2, M3\}$.

The correct option is (C).

Let's match each concept with its correct definition:

(i) Ordered Pair

An ordered pair is an ordered collection of two elements, typically written as $(a, b)$, where the order matters.

Matches with (c) An ordered collection of two elements.

(ii) Cartesian Product $A \times B$

The Cartesian product of two sets A and B is the set of all possible ordered pairs $(a, b)$ where the first element $a$ is from set A and the second element $b$ is from set B.

Matches with (d) The set of all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

(iii) Relation from A to B

A relation from set A to set B is a subset of the Cartesian product $A \times B$. It represents a specific set of connections or pairings between elements of A and elements of B, often defined by a particular rule or property.

Matches with (a) A collection of ordered pairs $(a, b)$ where $a \in A$ and $b \in B$, formed according to a specific rule.

(iv) Domain of a Relation

The domain of a relation is the set of all the first elements (or the first components) of the ordered pairs that constitute the relation.

Matches with (b) The set of all first elements of the ordered pairs in a relation.

Therefore, the correct matching is:

(i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

The correct option is (A).

Given:

Set $A = \{1, 2\}$

Set $B = \{3, 4\}$

To Find:

Which of the given options is NOT a relation from A to B.

Solution:

A relation from set A to set B is defined as any subset of the Cartesian product $A \times B$.

First, let's find the Cartesian product $A \times B$:

$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$

Now, we need to check each option to see if it is a subset of $A \times B$. If any option contains an element not present in $A \times B$, or if it's a relation from B to A, then it's not a relation from A to B.

(A) $\{(1, 3), (2, 4)\}$

Both $(1, 3)$ and $(2, 4)$ are elements of $A \times B$. Therefore, this is a subset of $A \times B$ and hence a relation from A to B.

(B) $\{(1, 4), (2, 3)\}$

Both $(1, 4)$ and $(2, 3)$ are elements of $A \times B$. Therefore, this is a subset of $A \times B$ and hence a relation from A to B.

(C) $\{(1, 3), (1, 4), (2, 3), (2, 4)\}$

This set contains all the elements of $A \times B$. Since any set is a subset of itself, this is also a relation from A to B (it's the universal relation).

(D) $\{(3, 1), (4, 2)\}$

The ordered pairs in this set are $(3, 1)$ and $(4, 2)$. For these to be in $A \times B$, the first element must be from A and the second from B. Here, $3 \notin A$ and $4 \notin A$. These ordered pairs represent elements from $B \times A$. Therefore, this set is not a subset of $A \times B$ and is not a relation from A to B.

The option that is NOT a relation from A to B is (D).

Given the equality of two ordered pairs:

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

To solve for x:

Subtract 1 from both sides:

Divide both sides by 2:

Equating the second components:

To solve for y:

Add 3 to both sides:

We need to find the value of $x+y$.

Substitute the values of x and y we found:

The correct option is (B).

Given:

Set $A = \{p, q, r\}$

Set $B = \{x, y, z\}$

Relation $R = \{(p, x), (q, y), (r, z)\}$ is defined from set A to set B.

To Find:

The codomain of the relation R.

Solution:

The codomain of a relation R defined from set A to set B is the set B itself. The codomain is the set that contains all possible second elements of the ordered pairs in the relation, and it is explicitly defined as the target set for the relation.

In this case, the relation R is defined from set A to set B.

Therefore, the codomain of R is set B.

Set B is given as $\{x, y, z\}$.

The range of the relation R is $\{x, y, z\}$, which is a subset of the codomain. In this specific case, the range is equal to the codomain because every element in B is mapped to by an element in A.

However, the question asks for the codomain.

The correct option is (B).

Given:

The number of elements in set A, $\text{n}(A) = 5$.

The number of elements in set B, $\text{n}(B) = 2$.

To Find:

The number of relations from A to B.

Solution:

The number of relations from set A to set B is equal to the number of subsets of the Cartesian product $A \times B$.

First, we find the number of elements in the Cartesian product $A \times B$:

Substitute the given values:

The number of relations from A to B is the number of subsets of $A \times B$. If a set has $k$ elements, it has $2^k$ subsets.

Here, the number of elements in $A \times B$ is 10.

So, the number of relations from A to B is $2^{10}$.

The correct option is (A).

The question asks for the set of all ordered pairs $(a, b)$ such that $a$ belongs to the set $\{1, 2\}$ and $b$ belongs to the set $\{3, 4\}$.

This is precisely the definition of the Cartesian product of the two sets.

Let $A = \{1, 2\}$ and $B = \{3, 4\}$.

The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

Therefore, the set of all ordered pairs $(a, b)$ such that $a \in \{1, 2\}$ and $b \in \{3, 4\}$ is represented by $\{1, 2\} \times \{3, 4\}$.

Let's examine the other options to confirm why they are incorrect:

• (B) $\{(1, 3), (2, 4)\}$: This is a subset of the Cartesian product, and thus a relation, but it's not the set of *all* such ordered pairs.
• (C) $\{ (1, 3), (1, 4), (2, 3), (2, 4) \}$: This is indeed the set of all ordered pairs $(a, b)$ where $a \in \{1, 2\}$ and $b \in \{3, 4\}$. This set is the result of the operation $\{1, 2\} \times \{3, 4\}$. So, while this option lists the elements correctly, option (A) is the notation that represents this set directly from the given conditions. The question asks "Which of the following *is* the set of all ordered pairs...", and (A) is the definition of that set.
• (D) $\{1, 2, 3, 4\}$: This is the union of the two sets, not a set of ordered pairs.

The question is asking for the notation that describes the set of all ordered pairs satisfying the given conditions. That notation is the Cartesian product.

The correct option is (A).

Given:

Set $A = \{a\}$

To Find:

The Cartesian product $A \times A$.

Solution:

The Cartesian product $A \times A$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in A$.

In this case, the only element in set A is 'a'.

So, we need to form ordered pairs where both the first and second elements are 'a'.

The only possible ordered pair is $(a, a)$.

Therefore, $A \times A = \{(a, a)\}$.

Let's look at the options:

• (A) $\{a\}$: This is a set containing the element 'a', not an ordered pair.
• (B) $\{(a)\}$: This is a set containing a set $\{a\}$. This notation is unusual for sets of numbers or symbols in this context and does not represent an ordered pair.
• (C) $\{(a, a)\}$: This is a set containing exactly one element, which is the ordered pair $(a, a)$. This correctly represents $A \times A$.
• (D) $\{a, a\}$: Sets do not contain duplicate elements, so this is equivalent to $\{a\}$.

The correct option is (C).

Given:

Set $A = \{1, 2, 3\}$.

To Find:

The universal relation on A.

Solution:

A universal relation on a set A is a relation from A to A that includes all possible ordered pairs in the Cartesian product $A \times A$. In other words, it is the largest possible relation from A to A.

First, let's find the Cartesian product $A \times A$ for the given set $A = \{1, 2, 3\}$:

$A \times A = \{(a, b) \mid a \in A \text{ and } b \in A\}$

This means we take each element of A and pair it with every element of A:

• Pairs starting with 1: $(1, 1), (1, 2), (1, 3)$
• Pairs starting with 2: $(2, 1), (2, 2), (2, 3)$
• Pairs starting with 3: $(3, 1), (3, 2), (3, 3)$

So, $A \times A = \{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$.

The universal relation on A is this set $A \times A$ itself.

Now let's compare this with the given options:

• (A) $\{(1,1), (2,2), (3,3)\}$: This is the identity relation, not the universal relation.
• (B) $\{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)\}$: This set is exactly the Cartesian product $A \times A$.
• (C) $\emptyset$: This is the empty relation, the smallest possible relation.
• (D) $\{(1,1)\}$: This is a specific relation, not the universal relation.

Therefore, the universal relation on A is the set of all possible ordered pairs in $A \times A$.

The correct option is (B).

First, we need to determine the elements of sets A and B.

Set A is defined as $A = \{x \in \mathbb{Z} : 1 \leq x \leq 2\}$.

Here, $\mathbb{Z}$ represents the set of integers.

The integers $x$ such that $1 \leq x \leq 2$ are 1 and 2.

So, $A = \{1, 2\}$.

The number of elements in A is $\text{n}(A) = 2$.

Set B is defined as $B = \{y \in \mathbb{N} : y^2 = 4\}$.

Here, $\mathbb{N}$ represents the set of natural numbers. Natural numbers are typically considered as positive integers {1, 2, 3, ...}.

We need to find natural numbers $y$ such that $y^2 = 4$.

Taking the square root of both sides, $y = \sqrt{4}$, which gives $y = \pm 2$.

Since $y$ must be a natural number, we take the positive value.

So, $y = 2$.

Therefore, $B = \{2\}$.

The number of elements in B is $\text{n}(B) = 1$.

We need to find $\text{n}(A \times B)$, which is the number of elements in the Cartesian product of A and B.

The number of elements in the Cartesian product is given by the product of the number of elements in each set:

Substitute the values of $\text{n}(A)$ and $\text{n}(B)$:

The correct option is (A).

Given:

Set $A = \{1, 2, 3\}$.

A relation R on A is defined by $R = \{(x, y) : x, y \in A, x < y\}$.

We need to find the roster form of R, which means listing all the ordered pairs $(x, y)$ that satisfy the condition $x < y$, where $x$ and $y$ are elements of A.

Let's consider each element of A as 'x' and find corresponding elements of A as 'y' such that $x < y$.

Case 1: If $x = 1$

We need to find $y \in A$ such that $1 < y$. The elements in A greater than 1 are 2 and 3.

So, we get the ordered pairs: $(1, 2)$ and $(1, 3)$.

Case 2: If $x = 2$

We need to find $y \in A$ such that $2 < y$. The only element in A greater than 2 is 3.

So, we get the ordered pair: $(2, 3)$.

Case 3: If $x = 3$

We need to find $y \in A$ such that $3 < y$. There are no elements in A that are greater than 3.

So, there are no ordered pairs with $x = 3$ that satisfy the condition.

Combining all the ordered pairs we found:

$R = \{(1, 2), (1, 3), (2, 3)\}$.

Now, let's compare this with the given options:

• (A) $\{(1, 2), (1, 3), (2, 3)\}$: This matches our derived relation.
• (B) $\{(2, 1), (3, 1), (3, 2)\}$: These pairs satisfy $x > y$, not $x < y$.
• (C) $\{(1, 1), (2, 2), (3, 3)\}$: These pairs satisfy $x = y$, not $x < y$.
• (D) $\{(1, 2), (2, 3), (3, 1)\}$: The pair $(3, 1)$ does not satisfy $3 < 1$.

The correct option is (A).

Given:

Relation $R = \{(1, 2), (1, 3), (2, 4), (3, 4)\}$

Set $A = \{1, 2, 3\}$ (the source set for the domain)

Set $B = \{2, 3, 4\}$ (the source set for the codomain and range)

To Find:

Which element is in the domain of R but not in the range of R.

Solution:

First, let's find the domain of the relation R. The domain is the set of all first elements of the ordered pairs in R.

From $R = \{(1, 2), (1, 3), (2, 4), (3, 4)\}$:

The first elements are 1, 1, 2, 3.

The domain of R is the set of unique first elements: Domain(R) = $\{1, 2, 3\}$.

Note that the domain is a subset of A, which is $\{1, 2, 3\}$. In this case, the domain is equal to A.

Next, let's find the range of the relation R. The range is the set of all second elements of the ordered pairs in R.

From $R = \{(1, 2), (1, 3), (2, 4), (3, 4)\}$:

The second elements are 2, 3, 4, 4.

The range of R is the set of unique second elements: Range(R) = $\{2, 3, 4\}$.

Note that the range is a subset of B, which is $\{2, 3, 4\}$. In this case, the range is equal to B.

Now, we need to find the element that is in the domain but not in the range.

Domain(R) = $\{1, 2, 3\}$

Range(R) = $\{2, 3, 4\}$

We are looking for an element $e$ such that $e \in \text{Domain(R)}$ and $e \notin \text{Range(R)}$.

Let's check each element in the domain:

• Is 1 in the domain? Yes, $1 \in \{1, 2, 3\}$.
• Is 1 in the range? No, $1 \notin \{2, 3, 4\}$.

So, the element 1 is in the domain but not in the range.

Let's check the other domain elements:

• Is 2 in the domain? Yes, $2 \in \{1, 2, 3\}$.
• Is 2 in the range? Yes, $2 \in \{2, 3, 4\}$. So, 2 is in both.
• Is 3 in the domain? Yes, $3 \in \{1, 2, 3\}$.
• Is 3 in the range? Yes, $3 \in \{2, 3, 4\}$. So, 3 is in both.

The element that is in the domain but not in the range is 1.

The correct option is (A).

Given:

The number of elements in the Cartesian product $A \times B$, $\text{n}(A \times B) = 12$.

The number of elements in set A, $\text{n}(A) = 3$.

To Find:

The number of elements in set B, $\text{n}(B)$.

Solution:

We know the relationship between the number of elements in the Cartesian product of two sets and the number of elements in each set:

We are given $\text{n}(A \times B) = 12$ and $\text{n}(A) = 3$. We need to find $\text{n}(B)$.

Substitute the given values into the formula:

To find $\text{n}(B)$, divide both sides of the equation by 3:

So, the number of elements in set B is 4.

The correct option is (A).

Let's analyze each statement about a relation R from set A to set B.

(A) Domain of R is always equal to A.

This statement is FALSE. The domain of a relation is the set of all first elements of the ordered pairs in the relation. It is possible for a relation to not include all elements of A in its domain. For example, if $A = \{1, 2, 3\}$ and $R = \{(1, a), (2, b)\}$, the domain is $\{1, 2\}$, which is a proper subset of A.

(B) Range of R is always equal to B.

This statement is FALSE. The range of a relation is the set of all second elements of the ordered pairs in the relation. It is possible for a relation to not map to all elements of B. For example, if $A = \{1, 2\}$ and $B = \{a, b, c\}$ and $R = \{(1, a), (2, b)\}$, the range is $\{a, b\}$, which is a proper subset of B.

(C) Range of R is a subset of Codomain (B).

This statement is TRUE. By definition, a relation from A to B is a subset of $A \times B$. The second elements of the ordered pairs in $A \times B$ are all the elements of B. Since the range is formed by these second elements, the range must be a subset of B. The codomain of a relation from A to B is precisely B.

(D) Domain of R is a subset of B.

This statement is FALSE. The domain of a relation from A to B consists of elements from set A, not set B. Therefore, the domain is a subset of A, not B.

The only TRUE statement is (C).

We are given the equality of two ordered pairs:

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

We are also given that $p$ is a positive integer. Taking the square root of both sides of the equation:

This gives $p = \pm 2$.

Since $p$ is a positive integer, we choose the positive value:

Equating the second components:

To solve for q, subtract 1 from both sides:

We need to find the value of $p \times q$.

Substitute the values of p and q we found:

The correct option is (A).

Given:

Set $A = \{1, 2\}$.

To Find:

The number of subsets of $A \times A$.

Solution:

First, we need to find the Cartesian product $A \times A$.

$A \times A = \{(x, y) \mid x \in A \text{ and } y \in A\}$

Since $A = \{1, 2\}$, we pair each element of A with every element of A:

• Pairs starting with 1: $(1, 1), (1, 2)$
• Pairs starting with 2: $(2, 1), (2, 2)$

So, $A \times A = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$.

The number of elements in $A \times A$ is $\text{n}(A \times A) = 4$.

The number of subsets of a set with $k$ elements is $2^k$.

In this case, the set is $A \times A$, which has $k = 4$ elements.

Therefore, the number of subsets of $A \times A$ is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Thus, the number of subsets of $A \times A$ is 16.

The correct option is (C).

Given:

Set $A = \{a, b, c\}$.

The empty set, $\emptyset$.

To Find:

The number of relations from A to the empty set $\emptyset$.

Solution:

The number of relations from a set A to a set B is given by $2^{\text{n}(A \times B)}$.

First, let's find the number of elements in the Cartesian product $A \times \emptyset$.

The number of elements in set A is $\text{n}(A) = 3$.

The number of elements in the empty set is $\text{n}(\emptyset) = 0$.

The number of elements in the Cartesian product is:

Substitute the values:

So, the Cartesian product $A \times \emptyset$ is the empty set itself, $A \times \emptyset = \emptyset$.

The number of relations from A to $\emptyset$ is the number of subsets of $A \times \emptyset$ (which is $\emptyset$).

The number of subsets of a set with $k$ elements is $2^k$. Here, $k=0$.

Number of relations = $2^{\text{n}(A \times \emptyset)} = 2^0$.

There is only one subset of the empty set, which is the empty set itself ($\emptyset$). Therefore, there is exactly one relation from A to the empty set (which is the empty relation).

The correct option is (B).

The relation R is defined as $R = \{(x, y) : x \in \mathbb{N}, y \in \mathbb{N}, x+y=5\}$.

Here, $\mathbb{N}$ represents the set of natural numbers, which are positive integers {1, 2, 3, ...}.

We need to find all pairs of natural numbers $(x, y)$ such that their sum is 5.

Let's list the possible pairs:

• If $x = 1$ (which is in $\mathbb{N}$), then $y = 5 - 1 = 4$. Since $4 \in \mathbb{N}$, the pair $(1, 4)$ is in R.
• If $x = 2$ (which is in $\mathbb{N}$), then $y = 5 - 2 = 3$. Since $3 \in \mathbb{N}$, the pair $(2, 3)$ is in R.
• If $x = 3$ (which is in $\mathbb{N}$), then $y = 5 - 3 = 2$. Since $2 \in \mathbb{N}$, the pair $(3, 2)$ is in R.
• If $x = 4$ (which is in $\mathbb{N}$), then $y = 5 - 4 = 1$. Since $1 \in \mathbb{N}$, the pair $(4, 1)$ is in R.
• If $x = 5$ (which is in $\mathbb{N}$), then $y = 5 - 5 = 0$. Since $0 \notin \mathbb{N}$, the pair $(5, 0)$ is not in R.
• If $x > 4$, then $y$ will be less than 1, and hence not a natural number.

So, the relation R in roster form is $R = \{(1, 4), (2, 3), (3, 2), (4, 1)\}$.

The question asks for the range of this relation. The range is the set of all second elements (y-values) in the ordered pairs of R.

The second elements in the ordered pairs of R are 4, 3, 2, and 1.

Therefore, the range of R is $\{4, 3, 2, 1\}$, which can be written in ascending order as $\{1, 2, 3, 4\}$.

Let's check the options:

• (A) $\{1, 2, 3, 4\}$: This matches our calculated range.
• (B) $\{1, 2, 3, 4\}$: This is the same as (A).
• (C) $\{ (1,4), (2,3), (3,2), (4,1) \}$: This is the roster form of the relation R itself, not its range.
• (D) $\{5\}$: This is the sum, not the range.

The correct option is (A) (or (B), as they are identical).

We need to find the range for each described relation.

(i) $R = \{(x, x^2) : x \in \{1, 2, 3\}\}$

Let's find the ordered pairs:

• When $x=1$, $x^2 = 1^2 = 1$. So, $(1, 1)$ is in R.
• When $x=2$, $x^2 = 2^2 = 4$. So, $(2, 4)$ is in R.
• When $x=3$, $x^2 = 3^2 = 9$. So, $(3, 9)$ is in R.

The relation is $R = \{(1, 1), (2, 4), (3, 9)\}$.

The range is the set of second elements: $\{1, 4, 9\}$.

Matches with (c) $\{1, 4, 9\}$.

(ii) $R = \{(x, y) : y = x+1, x \in \{1, 2, 3\}\}$

Let's find the ordered pairs:

• When $x=1$, $y = 1+1 = 2$. So, $(1, 2)$ is in R.
• When $x=2$, $y = 2+1 = 3$. So, $(2, 3)$ is in R.
• When $x=3$, $y = 3+1 = 4$. So, $(3, 4)$ is in R.

The relation is $R = \{(1, 2), (2, 3), (3, 4)\}$.

The range is the set of second elements: $\{2, 3, 4\}$.

Matches with (b) $\{2, 3, 4\}$.

(iii) $R = \{(1, 5), (2, 5), (3, 5)\}$

The ordered pairs are already given.

The range is the set of second elements: $\{5, 5, 5\}$. Since sets do not have duplicate elements, the range is $\{5\}$.

Matches with (a) $\{5\}$.

(iv) $R = \{(x, y) : x \text{ is the capital of } y, y \in \{\text{Bihar, Goa, Assam}\}\}

Here, the second elements ($y$) are given as \{Bihar, Goa, Assam\}. The relation connects each of these states to its capital (which would be the 'x' element).

Let's assume the capitals are Patna for Bihar, Panaji for Goa, and Dispur for Assam.

So, R would be something like $R = \{(\text{Patna, Bihar}), (\text{Panaji, Goa}), (\text{Dispur, Assam})\}$.

The range is the set of second elements, which are the states themselves: \{Bihar, Goa, Assam}.

Matches with (d) $\{\text{Bihar, Goa, Assam}\}$.

Therefore, the correct matching is:

(i)-(c), (ii)-(b), (iii)-(a), (iv)-(d)

The correct option is (A).

Given:

Set $A = \{1, 2, 3\}$.

To Find:

The identity relation on A.

Solution:

An identity relation on a set A is a relation where every element of A is related to itself. In terms of ordered pairs, the identity relation consists of all pairs $(x, x)$ where $x$ is an element of A.

For the given set $A = \{1, 2, 3\}$, the identity relation on A would be:

$I_A = \{(x, y) : x \in A, y \in A, x = y\}$

Listing the pairs:

• When $x=1$, $y$ must be 1. So, $(1, 1)$ is in the relation.
• When $x=2$, $y$ must be 2. So, $(2, 2)$ is in the relation.
• When $x=3$, $y$ must be 3. So, $(3, 3)$ is in the relation.

Therefore, the identity relation on A is $\{(1, 1), (2, 2), (3, 3)\}$.

Let's examine the options:

• (A) $\{(1, 2), (2, 3), (3, 1)\}$: This is a relation, but not the identity relation (it's a cyclic relation).
• (B) $\{(1, 1), (2, 2), (3, 3)\}$: This matches our derived identity relation.
• (C) $A \times A$: This is the universal relation, which contains all possible ordered pairs, not just those where the elements are equal.
• (D) $\emptyset$: This is the empty relation, which contains no ordered pairs.

The correct option is (B).

Let's analyze each statement to determine which one is FALSE.

(A) An ordered pair $(a, b)$ is different from $(b, a)$ if $a \neq b$.

This statement is TRUE. The definition of an ordered pair emphasizes that the order of the elements matters. For example, $(1, 2)$ is different from $(2, 1)$.

(B) The Cartesian product of two non-empty sets A and B is empty if and only if A is empty or B is empty.

This statement is FALSE. The Cartesian product $A \times B$ is empty if and only if A is empty OR B is empty. The statement says "if and only if A is empty OR B is empty". This is correct. The Cartesian product $A \times B$ contains ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. If either A or B is empty, then no such pairs can be formed, making $A \times B$ empty. If both A and B are non-empty, then $A \times B$ is also non-empty.

Re-reading the statement: "The Cartesian product of two non-empty sets A and B is empty if and only if A is empty or B is empty."

Let's break this down:

Part 1: "The Cartesian product of two non-empty sets A and B is empty"

This part is incorrect. If A and B are non-empty, then $A \times B$ is NON-EMPTY.

Part 2: "...if and only if A is empty or B is empty."

This part is correct. $A \times B$ is empty if and only if A is empty or B is empty.

The statement as a whole is saying: (Non-empty A and Non-empty B $\implies$ $A \times B$ is empty) AND ($A \times B$ is empty $\iff$ A is empty or B is empty).

The first implication is false. Therefore, the statement as a whole is FALSE.

(C) A relation is always a function.

This statement is FALSE. A function is a special type of relation where each element in the domain is related to exactly one element in the codomain. There can be relations that are not functions (e.g., a relation where one element in the domain is related to multiple elements in the codomain, or a relation where some elements in the domain are not related to any element in the codomain).

(D) The domain of a relation is a subset of the first set in the Cartesian product.

This statement is TRUE. If a relation is defined from set A to set B, then the relation is a subset of $A \times B$. The domain of this relation consists of the first elements of the ordered pairs in the relation. Since all first elements must come from set A, the domain is always a subset of A.

We have identified two FALSE statements: (B) and (C).

Let's re-examine (B). "The Cartesian product of two non-empty sets A and B is empty if and only if A is empty or B is empty."

The statement is structured as: P $\iff$ Q.

P: "The Cartesian product of two non-empty sets A and B is empty". This premise is false. If A and B are non-empty, $A \times B$ is non-empty.

Q: "A is empty or B is empty". This is true.

The statement claims P $\iff$ Q. Since P is false and Q is true, P $\iff$ Q is false.

Now let's re-examine (C). "A relation is always a function." This is unequivocally false, as a function is a specific *type* of relation.

In questions of this type, we are looking for the most fundamentally false statement. The definition of a relation and a function has a strict hierarchy where functions are a subset of relations. Saying a relation is *always* a function is a direct contradiction of this hierarchy.

Statement (B) is also flawed in its construction, but the core fact it's trying to convey about when a Cartesian product is empty is correct in its second part.

Let's confirm the standard definitions. The number of relations is $2^{mn}$. The number of functions from A to B is $n^m$. Since $2^{mn}$ is generally much larger than $n^m$ (e.g., if m=2, n=2, $2^{4}=16$ relations, $2^2=4$ functions), it's clear that not all relations are functions.

Considering the options, statement (C) is a direct misstatement of fundamental definitions in set theory concerning relations and functions.

The FALSE statement is (C).

Let R be a relation from set A to set B.

The definition of a relation R from set A to set B is that R is a subset of the Cartesian product $A \times B$. This means that every element $(a, b)$ in R must have $a \in A$ and $b \in B$.

The domain of R is the set of all first elements of the ordered pairs in R. Since all first elements come from A, the domain is a subset of A.

The range of R is the set of all second elements of the ordered pairs in R. Since all second elements come from B, the range is a subset of B.

The codomain of R is B itself.

Let's analyze the options:

• (A) A: The range is not necessarily a subset of A. For example, if $A=\{1\}$ and $B=\{2\}$ and $R=\{(1,2)\}$, the range is $\{2\}$, which is not a subset of A.
• (B) B: The range consists of the second elements of the ordered pairs in R. Since every second element must belong to set B, the range is indeed a subset of B.
• (C) $A \times B$: The range is a set of single elements (the second components), not ordered pairs. $A \times B$ is a set of ordered pairs. Thus, the range is not a subset of $A \times B$ in general.
• (D) Domain of R: The range and the domain are generally distinct sets. For example, in $R=\{(1,2)\}$, the domain is $\{1\}$ and the range is $\{2\}$. The range is not a subset of the domain.

Therefore, the range of R is a subset of B.

The correct option is (B).

We are given the equality of two ordered pairs:

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

Equating the second components:

From the first equation, we directly get $x = 3$.

Let's check if this value of x satisfies the second equation:

If $x = 3$, then $x^2 = 3^2 = 9$. This matches the second component of the given ordered pair.

The second equation, $x^2 = 9$, also implies that $x = 3$ or $x = -3$. However, for the equality of ordered pairs $(x, x^2) = (3, 9)$ to hold, the value of $x$ must satisfy *both* component equalities simultaneously.

The first component equality $x = 3$ *uniquely* determines the value of $x$. The second component equality $x^2 = 9$ is consistent with $x=3$ but also allows for $x=-3$. However, the first component fixes $x$ to be 3.

Therefore, the only value of $x$ that satisfies both conditions is $x = 3$.

The correct option is (A).

Given:

Set $A = \{1, 2\}$.

Set $B = \{3, 4\}$.

To Find:

The number of subsets of the Cartesian product $A \times B$.

Solution:

First, we need to find the Cartesian product $A \times B$.

$A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}$

This means we take each element of A and pair it with every element of B:

• Pairs starting with 1 from A: $(1, 3), (1, 4)$
• Pairs starting with 2 from A: $(2, 3), (2, 4)$

So, $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

The number of elements in $A \times B$ is $\text{n}(A \times B) = \text{n}(A) \times \text{n}(B) = 2 \times 2 = 4$.

The number of subsets of a set with $k$ elements is $2^k$.

In this case, the set is $A \times B$, which has $k = 4$ elements.

Therefore, the number of subsets of $A \times B$ is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Thus, the Cartesian product $A \times B$ has 16 subsets. These subsets are the possible relations from A to B.

The correct option is (C).

Given:

Set $A = \{1, 2\}$.

To Find:

The identity relation on set A.

Solution:

The identity relation on a set A is a relation that contains all ordered pairs $(x, x)$ where $x$ is an element of A. In other words, each element is related only to itself.

For the set $A = \{1, 2\}$, the identity relation consists of the ordered pairs where the first element is equal to the second element, and both elements are from A.

• For the element 1 in A, the identity relation includes the pair $(1, 1)$.
• For the element 2 in A, the identity relation includes the pair $(2, 2)$.

Therefore, the identity relation on A is $\{(1, 1), (2, 2)\}$.

Let's examine the given options:

• (A) $\{(1, 1)\}$: This is a subset of the identity relation, but it does not include the pair $(2, 2)$, so it is not the complete identity relation.
• (B) $\{(1, 1), (2, 2)\}$: This matches the definition of the identity relation for the set A.
• (C) $\{(1, 2), (2, 1)\}$: This is a relation, but it is not the identity relation. It represents a pairing where elements are related to other distinct elements.
• (D) $\emptyset$: This is the empty relation, which contains no pairs, and therefore cannot be the identity relation unless the set A itself is empty.

The correct option is (B).

We know that the number of elements in the Cartesian product of two sets A and B is given by the formula:

We are given that $\text{n}(A \times B) = 0$.

Substituting this into the formula:

For the product of two numbers to be zero, at least one of the numbers must be zero.

Therefore, either $\text{n}(A) = 0$ or $\text{n}(B) = 0$ (or both).

Let's analyze the given options:

• (A) $\text{n}(A) = 0$ or $\text{n}(B) = 0$. This statement correctly reflects our conclusion from the equation $0 = \text{n}(A) \times \text{n}(B)$.
• (B) $\text{n}(A) = 0$ and $\text{n}(B) = 0$. This is a possibility, but it's not the only one. For example, if $\text{n}(A) = 0$ and $\text{n}(B) = 5$, then $\text{n}(A \times B) = 0 \times 5 = 0$. So, it's not necessarily true that *both* must be zero.
• (C) $\text{n}(A) > 0$ and $\text{n}(B) > 0$. If both $\text{n}(A)$ and $\text{n}(B)$ are greater than 0, their product $\text{n}(A) \times \text{n}(B)$ would be greater than 0, contradicting $\text{n}(A \times B) = 0$. So, this statement is false.
• (D) $A=B$. This condition is not necessary for $\text{n}(A \times B) = 0$. For example, if $A = \{1\}$ and $B = \emptyset$, then $\text{n}(A \times B) = 0$, but $A \neq B$.

Therefore, the statement that MUST be true is that $\text{n}(A) = 0$ or $\text{n}(B) = 0$.

The correct option is (A).

Given:

Set $A = \{1, 2, 3\}$.

Set $B = \{a, b\}$.

The relation R is defined by $R = \{(x, y) : x \in A, y \in B, x \text{ is odd and } y=a\}$.

To Find:

The domain of the relation R.

Solution:

We need to find the ordered pairs $(x, y)$ that satisfy the given conditions:

1. $x$ must be from set A ($x \in \{1, 2, 3\}$).
2. $y$ must be from set B ($y \in \{a, b\}$).
3. $x$ must be odd.
4. $y$ must be equal to 'a'.

Let's go through the elements of A for the condition "$x$ is odd":

• If $x = 1$: 1 is odd. The condition $y=a$ must also be met. Since $a \in B$, the pair $(1, a)$ satisfies all conditions.
• If $x = 2$: 2 is not odd. So, no pair with $x=2$ will satisfy the condition.
• If $x = 3$: 3 is odd. The condition $y=a$ must also be met. Since $a \in B$, the pair $(3, a)$ satisfies all conditions.

So, the relation R in roster form is $R = \{(1, a), (3, a)\}$.

The domain of a relation is the set of all first elements of the ordered pairs in the relation.

The first elements in R are 1 and 3.

Therefore, the domain of R is $\{1, 3\}$.

The correct option is (A).

Given:

Set $A = \{1, 2\}$.

A relation R on A is defined by $R = \{(x, y) : x \in A, y \in A, x \geq y\}$.

To Find:

The range of the relation R.

Solution:

First, we need to find the ordered pairs $(x, y)$ that satisfy the condition $x \geq y$, where $x$ and $y$ are elements of A.

Let's consider each element of A for $x$ and find the corresponding $y$ from A such that $x \geq y$:

Case 1: If $x = 1$

We need $y \in A$ such that $1 \geq y$. The elements in A that satisfy this are $y=1$.

So, we get the ordered pair: $(1, 1)$.

Case 2: If $x = 2$

We need $y \in A$ such that $2 \geq y$. The elements in A that satisfy this are $y=1$ and $y=2$.

So, we get the ordered pairs: $(2, 1)$ and $(2, 2)$.

Combining all the ordered pairs that satisfy the condition, the relation R in roster form is $R = \{(1, 1), (2, 1), (2, 2)\}$.

The range of a relation is the set of all second elements of the ordered pairs in the relation.

The second elements in the ordered pairs of R are 1, 1, and 2.

The set of unique second elements is $\{1, 2\}$.

Therefore, the range of R is $\{1, 2\}$.

Let's check the options:

• (A) $\{1\}$: This is only one of the second elements.
• (B) $\{2\}$: This is only one of the second elements.
• (C) $\{1, 2\}$: This is the set of all unique second elements in R.
• (D) $\{(1,1), (2,1), (2,2)\}$: This is the roster form of the relation R itself, not its range.

The correct option is (C).

The Cartesian product of three sets A, B, and C, denoted as $A \times B \times C$, is the set of all possible ordered triples $(a, b, c)$ such that $a$ is an element of A, $b$ is an element of B, and $c$ is an element of C.

Let's analyze the given options:

(A) A set of ordered pairs $(a, b)$ where $a \in A$ and $b \in B \times C$.

This describes the Cartesian product $A \times (B \times C)$. While $A \times (B \times C)$ is often considered equivalent to $A \times B \times C$ in terms of elements, this representation is in terms of ordered pairs where the second element is itself an ordered pair, not an ordered triple directly.

(B) A set of ordered triples $(a, b, c)$ where $a \in A, b \in B, c \in C$.

This is the standard and most direct definition of the Cartesian product $A \times B \times C$. It correctly describes the elements as ordered triples with components from each respective set.

(C) A set of ordered pairs $(a, (b, c))$ where $a \in A$ and $(b, c) \in B \times C$.

This describes the Cartesian product $A \times (B \times C)$. This is a valid way to construct the elements, and it is often considered equivalent to $A \times B \times C$ in set theory, due to the associative property of Cartesian products (i.e., $A \times (B \times C) = (A \times B) \times C = A \times B \times C$). The elements are indeed ordered pairs, where the first element is from A and the second element is an ordered pair from $B \times C$.

(D) Both (B) and (C) are equivalent representations.

As discussed above, in set theory, the Cartesian product is associative, meaning $A \times (B \times C)$ is equivalent to $A \times B \times C$. Option (B) gives the direct definition of ordered triples, while option (C) gives a nested representation. Both correctly capture the relationships between elements of A, B, and C and are typically considered equivalent representations of the Cartesian product of three sets.

Given the options, (B) is the most direct and universally accepted definition of $A \times B \times C$. Option (C) is also a valid representation due to the associativity of the Cartesian product. Therefore, the statement that both (B) and (C) are equivalent representations is the most comprehensive and correct answer.

The correct option is (D).

A relation on a set A is defined as any subset of the Cartesian product $A \times A$.

The empty relation on a set A is the relation that contains no ordered pairs.

The empty set, denoted by $\emptyset$, is a subset of every set, including $A \times A$.

Therefore, the empty relation on a set A is the subset $\emptyset$ of $A \times A$.

Let's look at the options:

• (A) A: This is the set itself, not necessarily a subset of ordered pairs from $A \times A$.
• (B) $\emptyset$: This is the empty set, which contains no elements. The empty relation is defined as the empty set.
• (C) $\{(a,a) : a \in A\}$: This represents the identity relation, not the empty relation.
• (D) $A \times A$: This represents the universal relation, which contains all possible ordered pairs.

Thus, the empty relation on a set A is the subset $\emptyset$ of $A \times A$.

The correct option is (B).

Given:

Set $A = \{ \text{Red}, \text{Blue} \}$

Set $B = \{ \text{Car}, \text{Bike} \}$

To Find:

Which of the following is an element of $B \times A$.

Solution:

The Cartesian product $B \times A$ is the set of all ordered pairs $(b, a)$ such that $b \in B$ and $a \in A$.

Let's list the elements of $B \times A$ by taking each element from B and pairing it with each element from A:

• For the element "Car" from B:
• ("Car", "Red")
• ("Car", "Blue")
• For the element "Bike" from B:
• ("Bike", "Red")
• ("Bike", "Blue")

So, $B \times A = \{ (\text{Car}, \text{Red}), (\text{Car}, \text{Blue}), (\text{Bike}, \text{Red}), (\text{Bike}, \text{Blue}) \}$.

Now let's examine the options to see which one is an element of this set:

• (A) $(\text{Red}, \text{Car})$: This is an ordered pair where the first element is from A and the second is from B. This would be an element of $A \times B$, not $B \times A$.
• (B) $(\text{Blue}, \text{Bike})$: This is an ordered pair where the first element is from A and the second is from B. This would be an element of $A \times B$, not $B \times A$.
• (C) $(\text{Car}, \text{Red})$: This is an ordered pair where "Car" is from B and "Red" is from A. This matches the format $(b, a)$ with $b \in B$ and $a \in A$. Thus, $(\text{Car}, \text{Red})$ is an element of $B \times A$.
• (D) $\{(\text{Red}, \text{Car}), (\text{Blue}, \text{Bike})\}$: This is a set of ordered pairs, not a single ordered pair, so it cannot be an element of $B \times A$.

The correct option is (C).

Given that R is a relation from set A to set B, and the domain of R is equal to A.

The domain of a relation R is the set of all first elements of the ordered pairs in R. If the domain of R is equal to A, it means that every element in set A appears as the first element in at least one ordered pair in the relation R.

Let's analyze the implications:

(A) R is a function.

This is not necessarily true. For R to be a function, each element in the domain must be related to *exactly one* element in the codomain. A relation where domain(R) = A could still have one element from A related to multiple elements in B. For example, if $A=\{1,2\}$ and $B=\{a,b\}$, and $R=\{(1,a), (1,b), (2,a)\}$, then domain(R) = A, but R is not a function because 1 is related to both a and b.

(B) Every element in A is related to at least one element in B.

This statement is the direct consequence of the domain of R being equal to A. Since the domain is the set of all first elements of the ordered pairs in R, and the domain is A, it means every element of A must appear as a first element in at least one ordered pair in R. These ordered pairs connect elements from A to elements in B.

(C) Every element in B is related to at least one element in A.

This statement describes a property related to the range of the relation being equal to B, or a surjective property if R were a function. The domain of R being equal to A does not guarantee this. For example, if $A=\{1,2\}$ and $B=\{a,b\}$, and $R=\{(1,a), (2,a)\}$, then domain(R) = A, but the element 'b' in B is not related to any element in A.

(D) The range of R is equal to B.

This is not necessarily true. The range is the set of second elements. Just because every element in A is involved in a relation does not mean every element in B is related to. For example, if $A=\{1,2\}$ and $B=\{a,b\}$, and $R=\{(1,a), (2,a)\}$, then domain(R) = A, but the range is $\{a\}$, which is not equal to B.

Therefore, if the domain of a relation R from A to B is equal to A, it implies that every element in A is related to at least one element in B.

The correct option is (B).

We are given:

The number of elements in set A, $\text{n}(A) = 3$.

The number of elements in the Cartesian product $A \times B$, $\text{n}(A \times B) = 15$.

To find:

The number of elements in set B, $\text{n}(B)$.

Solution:

The number of elements in the Cartesian product of two sets is the product of the number of elements in each set:

Substitute the given values into the formula:

To find $\text{n}(B)$, divide both sides of the equation by 3:

Therefore, the number of elements in set B is 5.

The correct option is (A).

The relation R on the set of integers $\mathbb{Z}$ is defined by $a R b$ if and only if $a$ divides $b$. This means that $b$ must be an integer multiple of $a$. Mathematically, this can be written as $b = k \times a$ for some integer $k$, or $b/a$ must be an integer, provided $a \neq 0$.

We need to check each ordered pair $(a, b)$ to see if $a$ divides $b$ (i.e., if $b/a$ is an integer).

(A) $(3, 5)$

Does 3 divide 5? $5 / 3$ is not an integer. So, $(3, 5)$ does not belong to R.

(B) $(6, 2)$

Does 6 divide 2? $2 / 6 = 1/3$, which is not an integer. So, $(6, 2)$ does not belong to R.

(C) $(-4, 8)$

Does -4 divide 8? $8 / (-4) = -2$. Since -2 is an integer, -4 divides 8. So, $(-4, 8)$ belongs to R.

(D) $(7, 3)$

Does 7 divide 3? $3 / 7$ is not an integer. So, $(7, 3)$ does not belong to R.

Therefore, the ordered pair that belongs to the relation R is $(-4, 8)$.

The correct option is (C).

We are given that R is a relation from set A to set B, and the range of R is equal to B.

The range of a relation R is the set of all second elements of the ordered pairs in R. If the range of R is equal to B, it means that every element in set B appears as the second element in at least one ordered pair in the relation R.

Let's analyze the implications:

(A) Every element in B is related to at least one element in A.

This statement is the direct consequence of the range of R being equal to B. For an element $b \in B$ to be in the range, there must be some $a \in A$ such that $(a, b)$ is in the relation R. This means $b$ is related to $a$. Thus, every element in B is related to at least one element in A.

(B) Every element in A is related to at least one element in B.

This statement describes the condition where the domain of R is equal to A, not the range being equal to B. For example, if $A=\{1,2\}$ and $B=\{a,b\}$, and $R=\{(1,a), (2,a)\}$, then domain(R) = A, but the range is $\{a\}$, which is not equal to B. So, this statement is not implied by range(R) = B.

(C) R is a surjective function.

A function is a relation where each element in the domain maps to exactly one element in the codomain. If R is a relation such that range(R) = B, and additionally, if R were a function (meaning each element of A maps to exactly one element of B), then R would be a surjective function. However, the condition range(R) = B alone does not guarantee that R is a function. A relation can have range(R) = B without being a function (e.g., if one element of A maps to multiple elements of B, and these elements cover all of B).

(D) R is an injective function.

An injective function (or one-to-one function) means that each element in the domain maps to a *unique* element in the codomain. The condition range(R) = B does not imply injectivity. For example, if $A=\{1,2\}$ and $B=\{a,b\}$, and $R=\{(1,a), (2,a)\}$, then range(R) = $\{a\}$, which is not equal to B. If $A=\{1,2\}$ and $B=\{a,b\}$ and $R=\{(1,a), (2,a)\}$, range is $\{a\}$, domain is $\{1,2\}$. If $A=\{1,2\}$ and $B=\{a,b\}$ and $R=\{(1,a), (2,b)\}$, then domain=A, range=B, and it is injective and surjective (a bijection). If $A=\{1,2,3\}$ and $B=\{a,b\}$, and $R=\{(1,a), (2,b), (3,a)\}$, then domain=A, range=B, but it's not injective. So range=B does not imply injectivity.

The direct implication of the range of a relation R being equal to B is that every element in B is "hit" by at least one element from A through the relation.

The correct option is (A).

Given:

Set $A = \{a, b\}$. The number of elements in A is $\text{n}(A) = 2$.

Set $B = \{1, 2\}$. The number of elements in B is $\text{n}(B) = 2$.

To Find:

The total number of relations from B to A.

Solution:

The number of relations from a set X to a set Y is given by $2^{\text{n}(X \times Y)}$.

In this case, we are looking for relations from B to A. So, we need to calculate the number of elements in the Cartesian product $B \times A$.

The number of elements in $B \times A$ is:

Substitute the given values:

The total number of relations from B to A is the number of subsets of $B \times A$. If a set has $k$ elements, it has $2^k$ subsets.

Here, $k = \text{n}(B \times A) = 4$.

So, the total number of relations is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

The calculation involves multiplying the number of elements ($2 \times 2$), not adding them ($2+2$).

So, the correct representation is $2^{2 \times 2} = 2^4 = 16$.

The correct option is (A).

Given:

Set $A = \{1, 2\}$.

Set $B = \{3, 4\}$.

To Find:

The universal relation R from A to B.

Solution:

The universal relation from a set A to a set B is defined as the Cartesian product $A \times B$ itself. It is the relation that contains all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

Let's find the Cartesian product $A \times B$:

$A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}$

Substitute the elements of A and B:

• Taking the first element of A, 1, and pairing it with each element of B: $(1, 3), (1, 4)$.
• Taking the second element of A, 2, and pairing it with each element of B: $(2, 3), (2, 4)$.

So, $A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

The universal relation R from A to B is this set $A \times B$ itself.

Now, let's compare this with the given options:

• (A) $\{(1, 3), (2, 4)\}$: This is a subset of $A \times B$, but not the universal relation.
• (B) $\{(1, 3), (1, 4), (2, 3), (2, 4)\}$: This is exactly the Cartesian product $A \times B$.
• (C) $\{(1, 3), (2, 4), (3, 1), (4, 2)\}$: This set contains pairs where the first element is from B and the second is from A, and also includes pairs not possible from A to B or B to A. This is not related to the universal relation from A to B.
• (D) $\emptyset$: This is the empty relation, not the universal relation.

Therefore, the universal relation R from A to B is $\{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

The correct option is (B).

We are given the equality of two ordered pairs:

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

To solve for a, add 2 to both sides:

Equating the second components:

To solve for b, subtract 3 from both sides:

We need to find the ordered pair $(b, a)$.

Substitute the values of b and a we found:

$(b, a) = (-2, 6)$.

The correct option is (C).

Given:

The set $A = \{1, 2, 3\}$.

The relation R on A is defined by $R = \{(x, y) : x \in A, y \in A, x \leq y\}$.

To Find:

The domain and range of R.

Solution:

First, let's list the ordered pairs $(x, y)$ that satisfy the condition $x \leq y$, where $x, y \in \{1, 2, 3\}$.

If $x = 1$: $1 \leq y$. Possible values for $y$ from A are 1, 2, 3. This gives pairs: $(1, 1), (1, 2), (1, 3)$.

If $x = 2$: $2 \leq y$. Possible values for $y$ from A are 2, 3. This gives pairs: $(2, 2), (2, 3)$.

If $x = 3$: $3 \leq y$. The only possible value for $y$ from A is 3. This gives the pair: $(3, 3)$.

So, the relation R in roster form is $R = \{(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 3)\}$.

Now, let's find the domain of R. The domain is the set of all first elements of the ordered pairs in R.

The first elements are 1, 1, 1, 2, 2, 3.

The unique first elements are $\{1, 2, 3\}$.

So, the Domain of R is $\{1, 2, 3\}$.

Next, let's find the range of R. The range is the set of all second elements of the ordered pairs in R.

The second elements are 1, 2, 3, 2, 3, 3.

The unique second elements are $\{1, 2, 3\}$.

So, the Range of R is $\{1, 2, 3\}$.

Therefore, the domain is $\{1, 2, 3\}$ and the range is $\{1, 2, 3\}$.

The correct option is (A).

We are given that set A has $m$ elements, so $\text{n}(A) = m$.

Set B has $n$ elements, so $\text{n}(B) = n$.

The number of elements in the Cartesian product $A \times B$ is given by the formula:

We are given that $\text{n}(A \times B) = 20$.

So, we have the equation:

We need to find which of the given pairs $(m, n)$ satisfies this equation.

Let's check each option:

(A) $m=6, n=14$: $m \times n = 6 \times 14 = 84$. This is not 20.

(B) $m=4, n=5$: $m \times n = 4 \times 5 = 20$. This matches the given condition.

(C) $m=10, n=10$: $m \times n = 10 \times 10 = 100$. This is not 20.

(D) $m=3, n=7$: $m \times n = 3 \times 7 = 21$. This is not 20.

The only pair of values for $m$ and $n$ that satisfies the condition $m \times n = 20$ is $m=4$ and $n=5$.

The correct option is (B).

An ordered triple is an ordered collection of three elements. The standard notation for an ordered triple is $(x, y, z)$, where the order of the elements matters.

Let's analyze the options:

(A) $\{a, b, c\}$

This represents a set of three elements. In sets, the order of elements does not matter, and duplicate elements are not listed. So, this is not an ordered triple.

(B) $(a, b, c)$

This is the standard notation for an ordered triple. It represents a sequence of three elements where the order is significant.

(C) $((a, b), c)$

This is an ordered pair where the first element is itself an ordered pair $((a, b))$, and the second element is $c$. While it involves three elements in a specific order, it is technically an ordered pair. However, in the context of the Cartesian product $A \times B \times C$, it can be seen as an equivalent representation of an ordered triple. $A \times B \times C$ can be thought of as $A \times (B \times C)$, which would result in elements of the form $(a, (b, c))$.

(D) Both (B) and (C)

Option (B) is the most direct representation of an ordered triple. Option (C) is a nested representation that is equivalent to an ordered triple in set theory due to the associative property of the Cartesian product ($A \times (B \times C) \equiv A \times B \times C$). Therefore, both (B) and (C) represent the concept of an ordered triple, with (B) being the canonical form and (C) being an equivalent structure.

Since the question asks which *represents* an ordered triple, and (B) is the standard form, and (C) is an equivalent representation often used in the context of Cartesian products of three sets, option (D) is the most comprehensive answer.

The correct option is (D).

The relation R is defined from the set of natural numbers ($\mathbb{N}$) to the set of natural numbers ($\mathbb{N}$) by $R = \{(x, y) : y = x^2\}$.

This means that for every natural number $x$, the corresponding element $y$ in the relation is its square, $y = x^2$. The output $y$ must also be a natural number.

Let's consider some examples:

• If $x = 1$, then $y = 1^2 = 1$. Since $1 \in \mathbb{N}$, the pair $(1, 1)$ is in R.
• If $x = 2$, then $y = 2^2 = 4$. Since $4 \in \mathbb{N}$, the pair $(2, 4)$ is in R.
• If $x = 3$, then $y = 3^2 = 9$. Since $9 \in \mathbb{N}$, the pair $(3, 9)$ is in R.
• If $x = 4$, then $y = 4^2 = 16$. Since $16 \in \mathbb{N}$, the pair $(4, 16)$ is in R.

The range of the relation R is the set of all possible second elements ($y$) that result from the definition $y = x^2$, where $x$ is a natural number.

The values of $y$ are $1^2, 2^2, 3^2, 4^2, \ldots$, which are $1, 4, 9, 16, \ldots$.

These numbers are the squares of natural numbers. Numbers that are the square of an integer are called perfect squares. Since the domain is natural numbers, the resulting squares will be natural numbers.

The set of all natural numbers is $\mathbb{N} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \ldots \}$.

The set of perfect squares that are also natural numbers is $\{1, 4, 9, 16, 25, \ldots \}$. This set is a subset of $\mathbb{N}$ because not every natural number is a perfect square (e.g., 2, 3, 5, 6, 7, 8 are not perfect squares).

Let's examine the options:

• (A) The set of all perfect squares. This accurately describes the range, as the outputs are squares of natural numbers, and these are precisely the perfect squares that are natural numbers.
• (B) $\mathbb{N}$. This is the set of all natural numbers. The range is a subset of $\mathbb{N}$, but not all natural numbers are perfect squares. So, the range is not equal to $\mathbb{N}$.
• (C) The set of all natural numbers. This is the same as (B) and is incorrect for the same reason.
• (D) The set of all integers. The range consists only of positive numbers (squares of natural numbers), so it's not the set of all integers.

Therefore, the most accurate description of the range of R is the set of all perfect squares.

The correct option is (A).

Given:

Set $A = \{1, 2\}$.

To Find:

Which of the given options is NOT a subset of $A \times A$.

Solution:

First, let's determine the Cartesian product $A \times A$.

$A \times A = \{(x, y) \mid x \in A \text{ and } y \in A\}$

Since $A = \{1, 2\}$, the elements of $A \times A$ are:

• Pairing 1 from A with elements of A: $(1, 1), (1, 2)$
• Pairing 2 from A with elements of A: $(2, 1), (2, 2)$

So, $A \times A = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$.

Now, let's check which of the given options is NOT a subset of $A \times A$. A set is a subset of another set if all its elements are also elements of the other set.

(A) $\emptyset$

The empty set is a subset of every set. Thus, $\emptyset$ is a subset of $A \times A$. This option is a subset.

(B) $\{(1, 2)\}$

The set $\{(1, 2)\}$ contains only one element, $(1, 2)$. Since $(1, 2)$ is an element of $A \times A$, this set is a subset of $A \times A$. This option is a subset.

(C) $\{(2, 1), (2, 3)\}$

This set contains two elements: $(2, 1)$ and $(2, 3)$.

The element $(2, 1)$ is in $A \times A$ because $2 \in A$ and $1 \in A$.

However, the element $(2, 3)$ is NOT in $A \times A$ because $3 \notin A$. Since at least one element of this set is not in $A \times A$, this set is NOT a subset of $A \times A$.

(D) $\{(1, 1), (2, 2)\}$

Both $(1, 1)$ and $(2, 2)$ are elements of $A \times A$. Thus, this set is a subset of $A \times A$. This option is a subset.

Therefore, the set that is NOT a subset of $A \times A$ is $\{(2, 1), (2, 3)\}$.

The correct option is (C).

We are given the equality of two ordered pairs:

For two ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

Equating the second components:

We are asked to find the possible values of $x$, given that $x$ is an integer.

From the equation $x^2 = 4$, we need to find the integer values of $x$ that satisfy this equation.

Taking the square root of both sides:

This gives two possible values for $x$: $x = 2$ and $x = -2$.

Since both 2 and -2 are integers, both are possible values for $x$.

The value of $y$ is determined to be 5 from the second component, but the question only asks for the possible values of $x$.

Therefore, the possible values of $x$ are 2 and -2.

The correct option is (C).

Given:

Set $A = \{p, q\}$.

The number of elements in set A is $\text{n}(A) = 2$.

To Find:

The total number of distinct relations that can be defined from A to A.

Solution:

The number of relations from a set X to a set Y is given by $2^{\text{n}(X \times Y)}$.

In this case, we are looking for relations from A to A. So, we need to calculate the number of elements in the Cartesian product $A \times A$.

The number of elements in $A \times A$ is:

Substitute the value of $\text{n}(A)$:

The total number of relations from A to A is the number of subsets of $A \times A$. If a set has $k$ elements, it has $2^k$ subsets.

Here, $k = \text{n}(A \times A) = 4$.

So, the total number of relations is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Therefore, there are 16 distinct relations that can be defined from A to A.

The correct option is (C).

Given:

The set $S = \{1, 2, 3, 4\}$.

The relation R is defined as $R = \{(x, y) : x \in S, y \in S, x+y=6\}$.

To Find:

The domain of the relation R.

Solution:

We need to find all ordered pairs $(x, y)$ such that $x$ and $y$ are from the set $\{1, 2, 3, 4\}$ and their sum $x+y=6$.

Let's systematically find these pairs:

• If $x=1$: We need $1+y=6$, so $y=5$. However, $5 \notin S$. Thus, no pair starts with 1.
• If $x=2$: We need $2+y=6$, so $y=4$. Since $4 \in S$, the pair $(2, 4)$ is in R.
• If $x=3$: We need $3+y=6$, so $y=3$. Since $3 \in S$, the pair $(3, 3)$ is in R.
• If $x=4$: We need $4+y=6$, so $y=2$. Since $2 \in S$, the pair $(4, 2)$ is in R.

So, the relation R in roster form is $R = \{(2, 4), (3, 3), (4, 2)\}$.

The domain of a relation is the set of all the first elements of the ordered pairs in the relation.

The first elements in R are 2, 3, and 4.

Therefore, the domain of R is $\{2, 3, 4\}$.

The correct option is (C).

Given:

Set $A = \{a, b, c\}$. The number of elements in A is $\text{n}(A) = 3$.

Set $B = \{p, q\}$. The number of elements in B is $\text{n}(B) = 2$.

To Find:

The number of elements in the Cartesian product $B \times A$, denoted as $\text{n}(B \times A)$.

Solution:

The number of elements in the Cartesian product of two sets is the product of the number of elements in each set:

Substitute the given values:

Therefore, the number of elements in the Cartesian product $B \times A$ is 6.

The correct option is (B).

Let's examine each of the given properties of the Cartesian product.

(A) $A \times (B \cup C) = (A \times B) \cup (A \times C)$

This is the distributive property of the Cartesian product over the union of sets. Let's verify this property. An element $(x, y)$ is in $A \times (B \cup C)$ if $x \in A$ and $y \in (B \cup C)$. This means $x \in A$ and ($y \in B$ or $y \in C$). If $x \in A$ and $y \in B$, then $(x, y) \in A \times B$. If $x \in A$ and $y \in C$, then $(x, y) \in A \times C$. Thus, $(x, y) \in (A \times B) \cup (A \times C)$. Conversely, if $(x, y) \in (A \times B) \cup (A \times C)$, then $(x, y) \in A \times B$ or $(x, y) \in A \times C$. This means ($x \in A$ and $y \in B$) or ($x \in A$ and $y \in C$). In either case, $x \in A$ and ($y \in B$ or $y \in C$), which means $y \in B \cup C$. So, $(x, y) \in A \times (B \cup C)$. This property is TRUE.

(B) $A \times (B \cap C) = (A \times B) \cap (A \times C)$

This is the distributive property of the Cartesian product over the intersection of sets. Let's verify this. An element $(x, y)$ is in $A \times (B \cap C)$ if $x \in A$ and $y \in (B \cap C)$. This means $x \in A$ and ($y \in B$ and $y \in C$). If $x \in A$ and $y \in B$, then $(x, y) \in A \times B$. If $x \in A$ and $y \in C$, then $(x, y) \in A \times C$. Thus, if $x \in A$ and ($y \in B$ and $y \in C$), then $(x, y) \in (A \times B)$ and $(x, y) \in (A \times C)$. This means $(x, y) \in (A \times B) \cap (A \times C)$. Conversely, if $(x, y) \in (A \times B) \cap (A \times C)$, then $(x, y) \in A \times B$ and $(x, y) \in A \times C$. This means ($x \in A$ and $y \in B$) and ($x \in A$ and $y \in C$). This simplifies to $x \in A$ and ($y \in B$ and $y \in C$), which means $x \in A$ and $y \in B \cap C$. So, $(x, y) \in A \times (B \cap C)$. This property is TRUE.

(C) $A \times (B - C) = (A \times B) - (A \times C)$

This property relates the Cartesian product with set difference. Let's verify this. An element $(x, y)$ is in $A \times (B - C)$ if $x \in A$ and $y \in (B - C)$. This means $x \in A$ and ($y \in B$ and $y \notin C$). If $(x, y) \in A \times (B - C)$, then $x \in A$ and $y \in B$ and $y \notin C$. If $x \in A$ and $y \in B$, then $(x, y) \in A \times B$. If $x \in A$ and $y \notin C$, then $(x, y) \notin A \times C$. So, if $(x, y) \in A \times (B - C)$, then $(x, y) \in (A \times B)$ and $(x, y) \notin (A \times C)$. This means $(x, y) \in (A \times B) - (A \times C)$. Conversely, if $(x, y) \in (A \times B) - (A \times C)$, then $(x, y) \in (A \times B)$ and $(x, y) \notin (A \times C)$. This means ($x \in A$ and $y \in B$) and it is NOT the case that ($x \in A$ and $y \in C$). Since $x \in A$ is true, it must be that $y \notin C$. So, $x \in A$ and $y \in B$ and $y \notin C$. This means $x \in A$ and $y \in B - C$. Therefore, $(x, y) \in A \times (B - C)$. This property is TRUE.

(D) All of the above

Since properties (A), (B), and (C) have all been verified to be true, this option is correct.

The correct option is (D).

In an arrow diagram representing a relation R from set A to set B:

• Set A is typically shown on the left, and set B is shown on the right.
• Elements of set A are represented by points or nodes on the left.
• Elements of set B are represented by points or nodes on the right.
• An arrow is drawn from an element $a \in A$ to an element $b \in B$ if and only if the ordered pair $(a, b)$ is in the relation R.

The domain of a relation is the set of all first elements of the ordered pairs in the relation. In an arrow diagram, these are the elements in set A from which at least one arrow originates.

The range of a relation is the set of all second elements of the ordered pairs in the relation. In an arrow diagram, these are the elements in set B that have at least one arrow pointing to them.

The codomain of a relation from A to B is set B itself.

The question states that "The elements in set A that have at least one outgoing arrow" are being considered. These are precisely the elements of A that are involved as the first component in the ordered pairs of the relation.

Therefore, the elements in set A that have at least one outgoing arrow form the domain of the relation.

The correct option is (C).

Given:

Set of students $S = \{Ram, Sita, John\}$.

Set of subjects $Sub = \{Maths, Physics, Chemistry\}$.

Relation $R = \{(Ram, Maths), (Sita, Physics), (Ram, Physics), (John, Chemistry)\}$.

To Find:

Which subject is chosen by more than one student?

Solution:

We need to look at the second elements (subjects) in the ordered pairs of the relation R and see which subject appears more than once.

The ordered pairs in R are:

• $(Ram, Maths)$: Ram chose Maths.
• $(Sita, Physics)$: Sita chose Physics.
• $(Ram, Physics)$: Ram chose Physics.
• $(John, Chemistry)$: John chose Chemistry.

Let's list the subjects chosen by each student and count how many students chose each subject:

• Maths was chosen by Ram. (1 student)
• Physics was chosen by Sita and Ram. (2 students)
• Chemistry was chosen by John. (1 student)

The subject chosen by more than one student is Physics.

The correct option is (B).

Given:

Set of students $S = \{Ram, Sita, John\}$.

Set of subjects $Sub = \{Maths, Physics, Chemistry\}$.

Relation $R = \{(Ram, Maths), (Sita, Physics), (Ram, Physics), (John, Chemistry)\}$.

To Find:

The range of the relation R.

Solution:

The range of a relation is the set of all second elements (subjects in this case) of the ordered pairs in the relation.

The ordered pairs in R are:

• $(Ram, Maths)$: The second element is Maths.
• $(Sita, Physics)$: The second element is Physics.
• $(Ram, Physics)$: The second element is Physics.
• $(John, Chemistry)$: The second element is Chemistry.

The second elements are Maths, Physics, Physics, and Chemistry.

The range is the set of unique second elements:

Range(R) = \{Maths, Physics, Chemistry\}

Now let's look at the options:

• (A) \{Ram, Sita, John\}: This is the set of students (the domain), not the subjects (the range).
• (B) \{Maths, Physics, Chemistry\}: This correctly lists the unique subjects chosen.
• (C) \{Maths, Physics, Chemistry\}: This is identical to (B).
• (D) \{Maths, Physics, Chemistry\}: This is identical to (B) and (C).

All options (B), (C), and (D) correctly state the range. Since they are identical, any of them would be considered correct. However, in a multiple-choice question, if identical correct options are presented, it often implies a convention or a need to pick one of them. Assuming the question intends for a single best answer, and options B, C, and D are identical and correct, we select one of them.

The correct option is (B) (or (C) or (D)).

Given:

Set $A = \{1, 2\}$.

Relation R on A is defined by $R = \{(x, y) : x \in A, y \in A, x+y \text{ is even}\}$.

To Find:

The range of the relation R.

Solution:

First, we need to find the ordered pairs $(x, y)$ from $A \times A$ such that their sum $x+y$ is an even number.

Let's consider all possible pairs from $A \times A$ and check the sum:

• Pair $(1, 1)$: $1 + 1 = 2$. 2 is even. So, $(1, 1) \in R$.
• Pair $(1, 2)$: $1 + 2 = 3$. 3 is odd. So, $(1, 2) \notin R$.
• Pair $(2, 1)$: $2 + 1 = 3$. 3 is odd. So, $(2, 1) \notin R$.
• Pair $(2, 2)$: $2 + 2 = 4$. 4 is even. So, $(2, 2) \in R$.

The relation R in roster form is $R = \{(1, 1), (2, 2)\}$.

The range of a relation is the set of all second elements of the ordered pairs in the relation.

The second elements in R are 1 (from $(1, 1)$) and 2 (from $(2, 2)$).

Therefore, the range of R is $\{1, 2\}$.

Let's check the options:

• (A) $\{1, 2\}$: This matches our calculated range.
• (B) $\{2\}$: This only includes one of the second elements.
• (C) $\{1\}$: This only includes one of the second elements.
• (D) $\{ (1,1), (2,2) \}$: This is the roster form of the relation R itself, not its range.

The correct option is (A).

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.

We are given that $A \times B$ contains the elements $(1, 3), (1, 4), (2, 3), (2, 4)$.

This means that:

• The first elements of these ordered pairs must belong to set A. The first elements are 1, 1, 2, 2. So, set A must contain $\{1, 2\}$.
• The second elements of these ordered pairs must belong to set B. The second elements are 3, 4, 3, 4. So, set B must contain $\{3, 4\}$.

To form all the given pairs, the smallest possible sets for A and B would be $A = \{1, 2\}$ and $B = \{3, 4\}$. Let's check if this works:

$A \times B = \{1, 2\} \times \{3, 4\}$

$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$.

This matches the given elements.

Now let's check the other options:

• (B) $A = \{1, 3\}, B = \{2, 4\}$: $A \times B = \{(1, 2), (1, 4), (3, 2), (3, 4)\}$. This does not match the given elements.
• (C) $A = \{1, 2, 3, 4\}, B = \emptyset$: If $B$ is empty, $A \times B$ would be empty. This does not match.
• (D) $A = \{1, 4\}, B = \{2, 3\}$: $A \times B = \{(1, 2), (1, 3), (4, 2), (4, 3)\}$. This does not match the given elements.

Thus, the only sets A and B that correctly form the given Cartesian product are $A = \{1, 2\}$ and $B = \{3, 4\}$.

The correct option is (A).

The relation R is defined as $R = \{(x, x^2) : x \text{ is a prime number less than 10}\}$.

The domain of a relation is the set of all first elements of the ordered pairs in the relation.

In this relation, the ordered pairs are of the form $(x, x^2)$, where $x$ is a prime number less than 10.

First, let's identify the prime numbers less than 10:

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers less than 10 are: 2, 3, 5, and 7.

So, the values of $x$ for which the ordered pairs are formed are $x \in \{2, 3, 5, 7\}$.

The domain of R is the set of these first elements, which are the prime numbers less than 10.

Therefore, the domain of R is $\{2, 3, 5, 7\}$.

Let's examine the options:

• (A) $\{x^2 : x \text{ is a prime number less than 10}\}$: This describes the range of R, not the domain.
• (B) $\{2, 3, 5, 7\}$: This is the set of prime numbers less than 10, which are the first elements ($x$) in the ordered pairs, thus representing the domain.
• (C) $\{4, 9, 25, 49\}$: This is the set of squares of the prime numbers less than 10 (i.e., the range of R).
• (D) $\{2, 3, 5, 7, 11, \dots\}$: This represents the set of all prime numbers, not just those less than 10.

The correct option is (B).

We are given the equality of an ordered triple:

For two ordered triples to be equal, their corresponding components must be equal.

This means:

• First component equality: $x = 2$
• Second component equality: $y = 3$
• Third component equality: $z = 5$

Now let's check each option based on these equalities:

(A) $x=2$

This statement is correct, as derived from the first components.

(B) $y=3$

This statement is correct, as derived from the second components.

(C) $z=5$

This statement is correct, as derived from the third components.

(D) $\{x, y, z\} = \{2, 3, 5\}$

Substituting the values of $x, y,$ and $z$, we get $\{2, 3, 5\} = \{2, 3, 5\}$. This statement is correct. Sets are collections of distinct elements, and the order does not matter. Therefore, the set formed by the elements of the ordered triple $(2, 3, 5)$ is indeed $\{2, 3, 5\}$.

Let's re-evaluate the question and options carefully. It asks which of the following is NOT correct.

All statements (A), (B), (C), and (D) appear to be correct based on the initial equality of ordered triples.

There might be a subtle interpretation being tested. Let's consider the structure of an ordered triple vs. a set.

An ordered triple $(x, y, z)$ has order and distinguishes elements by position. A set $\{x, y, z\}$ does not have order and lists distinct elements. The conversion from an ordered triple to a set of its elements is a valid operation.

Let's assume there might be an error in the question or options, or a very specific interpretation is intended.

If we assume the question is testing the fundamental difference between ordered triples and sets, then stating that the *ordered triple* is *equal* to the *set* might be considered incorrect because they are different mathematical objects, even if they contain the same elements. However, usually, equality in such contexts means that the collection of elements is the same, disregarding the order and multiplicity.

Let's review the basics:

An ordered triple $(x, y, z)$ implies order. $(2, 3, 5)$ is different from $(3, 2, 5)$.

A set $\{x, y, z\}$ has no order. $\{2, 3, 5\}$ is the same as $\{3, 2, 5\}$.

The statement $\{x, y, z\} = \{2, 3, 5\}$ is correct in terms of the elements contained. However, saying the ordered triple $(x, y, z)$ *is* the set $\{2, 3, 5\}$ would be incorrect.

The statement is $\{x, y, z\} = \{2, 3, 5\}$. This statement is correct. It's not claiming $(x, y, z) = \{2, 3, 5\}$, but rather the *set formed by* $x, y, z$ is equal to the set $\{2, 3, 5\}$.

Let's consider if there's any other way to interpret the question.

Given $(x, y, z) = (2, 3, 5)$, we have $x=2$, $y=3$, $z=5$. All are correct.

The set of these values is $\{2, 3, 5\}$. This is also correct.

If the question is strictly about notation, then perhaps the statement implying equality of a structure that has order (ordered triple) with a structure that does not have order (set) might be intended as incorrect in a pedantic sense. However, in typical mathematical contexts, this is accepted as correct in terms of element composition.

Let's assume there might be a typo or a standard interpretation issue.

If the question intended to ask which statement is incorrect in a fundamental way:

• A, B, C are direct consequences of ordered triple equality.
• D is about forming a set from the elements of the ordered triple.

It's highly likely that all statements are intended to be correct. However, if one must be chosen as incorrect, it would hinge on a very strict interpretation of "equality" between an ordered structure and an unordered structure.

Given the typical nature of such questions, it's more probable that there's a misunderstanding or an error in the question/options. However, if forced to choose the least "correct" statement based on strict definition: an ordered triple IS NOT a set, even if they comprise the same elements.

Let's re-read the question: "which of the following is NOT correct?"

Let's consider the possibility that the question tests the uniqueness of representation.

The statements $x=2$, $y=3$, $z=5$ are direct consequences of the equality of ordered triples. The statement $\{x, y, z\} = \{2, 3, 5\}$ is also correct, as the set formed by the elements of the ordered triple is indeed the set $\{2, 3, 5\}$.

There's no obvious error in any of the statements. It's possible the question is flawed.

However, let's consider if any statement makes an unnecessary or incorrect assumption.

All statements are derived directly from $(x, y, z) = (2, 3, 5)$.

Let's assume the question is valid and there is a subtle error in one of the options.

If we consider the possibility of multiplicity of elements if the definition of the ordered triple allowed it, then the set notation would lose that information. But ordered triples don't have multiplicity in their definition.

Let's reconsider the equality statement itself: $(x, y, z) = (2, 3, 5)$. This is an equality between two ordered triples.

A: $x=2$. Correct.

B: $y=3$. Correct.

C: $z=5$. Correct.

D: $\{x, y, z\} = \{2, 3, 5\}$. This statement asserts equality between a set formed from the elements of the triple and another set. Since $x=2, y=3, z=5$, the set $\{x, y, z\}$ is $\{2, 3, 5\}$. So, $\{2, 3, 5\} = \{2, 3, 5\}$, which is true.

There seems to be no incorrect statement among the options.

Could the question implicitly assume that $x, y, z$ are distinct or have some other property? No, that's not stated.

Let's consider if the question might be testing a very subtle point about the definition of equality.

In most contexts, statement (D) is considered correct. However, if the question is designed to be tricky, it might be aiming at the difference between an ordered structure and an unordered structure.

If we strictly interpret "correct representation", then saying the ordered triple *is* the set might be seen as not fully correct, even though the elements match.

Given that A, B, and C are direct assignments from the ordered triple equality, and D is about the set formed from those elements, and all are factually correct statements, it is possible the question is flawed.

However, if forced to pick the "least correct" or "most potentially incorrect" statement based on the fundamental difference between ordered structures and sets, it would be (D) because it equates an ordered entity with an unordered one, even though the elements are the same.

Let's assume, for the sake of providing an answer, that the question intends to highlight the difference between an ordered triple and a set.

Therefore, statement (D) could be considered "not correct" in the sense that an ordered triple is fundamentally different in structure from a set, even if they contain the same elements.

If forced to choose, the most likely intended answer for "NOT correct" would be (D), playing on the distinction between ordered triples and sets.

Given:

Set $A = \{1, 2\}$. The number of elements in A is $\text{n}(A) = 2$.

Set $B = \{3\}$. The number of elements in B is $\text{n}(B) = 1$.

To Find:

The total number of relations from A to B.

Solution:

The number of relations from a set A to a set B is given by $2^{\text{n}(A \times B)}$.

First, we need to find the number of elements in the Cartesian product $A \times B$.

The number of elements in $A \times B$ is:

Substitute the given values:

The total number of relations from A to B is the number of subsets of $A \times B$. If a set has $k$ elements, it has $2^k$ subsets.

Here, $k = \text{n}(A \times B) = 2$.

So, the total number of relations is $2^2$.

Calculating $2^2$:

$2^2 = 2 \times 2 = 4$.

Therefore, there are 4 distinct relations that can be defined from A to B.

The correct option is (B).

A relation R from set A to set B is defined as any subset of the Cartesian product $A \times B$.

The empty set, denoted by $\emptyset$, is a subset of every set. Therefore, $\emptyset$ is always a subset of $A \times B$, regardless of whether A and B are empty or non-empty.

When a relation R is the empty set ($\emptyset$), it means that there are no ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$ that satisfy the specific condition defining the relation.

Let's analyze the options:

(A) True

This statement asserts that a relation can be empty. As established, the empty set is always a subset of $A \times B$, so a relation can indeed be empty.

(B) False

This is incorrect, as explained above.

(C) True, only if A or B is empty.

While it is true that if A or B is empty, then $A \times B$ is empty, and the only subset of an empty set is the empty set (meaning the only relation is the empty relation), it is also possible for the empty relation to exist even when A and B are non-empty. For example, if $A=\{1,2\}$ and $B=\{3,4\}$, and we define a relation $R = \{(x,y) : x > y\}$. There are no such pairs in $A \times B$, so the relation R is empty, even though A and B are non-empty.

(D) True, for any non-empty sets A and B as well.

This is also true. As demonstrated in the example above, even if A and B are non-empty, a relation can be empty if no elements satisfy the defining condition.

The question asks which statement is true about whether a relation can be empty. The statement "A relation R from set A to set B can be empty" is fundamentally true.

Options (C) and (D) provide additional conditions. While (D) is true, (A) is the most direct and encompassing answer. The existence of an empty relation is a general property, not restricted to cases where A or B are empty.

The most accurate and fundamental answer is (A) True.

The relation R is defined as $R = \{(x, y) : x, y \in \mathbb{N}, x+y \leq 4\}$.

Here, $\mathbb{N}$ represents the set of natural numbers, which are positive integers $\{1, 2, 3, \dots\}$.

We need to find all pairs of natural numbers $(x, y)$ such that their sum $x+y$ is less than or equal to 4.

Let's systematically find these pairs:

• If $x=1$: We need $1+y \leq 4$, which means $y \leq 3$. Since $y$ must be a natural number, possible values for $y$ are 1, 2, 3. The pairs are $(1, 1), (1, 2), (1, 3)$.
• If $x=2$: We need $2+y \leq 4$, which means $y \leq 2$. Since $y$ must be a natural number, possible values for $y$ are 1, 2. The pairs are $(2, 1), (2, 2)$.
• If $x=3$: We need $3+y \leq 4$, which means $y \leq 1$. Since $y$ must be a natural number, the only possible value for $y$ is 1. The pair is $(3, 1)$.
• If $x=4$: We need $4+y \leq 4$, which means $y \leq 0$. There are no natural numbers $y$ that satisfy $y \leq 0$.
• If $x > 4$, then $x+y$ will always be greater than 4 for any natural number $y$.

So, the relation R in roster form is $R = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}$.

The question asks for the range of R. The range is the set of all second elements ($y$) of the ordered pairs in R.

The second elements are 1, 2, 3, 1, 2, 1.

The unique second elements are $\{1, 2, 3\}$.

Therefore, the range of R is $\{1, 2, 3\}$.

Let's check the options:

• (A) $\{1, 2, 3\}$: This matches our calculated range.
• (B) $\{1, 2, 3\}$: This is identical to (A).
• (C) $\{ (1,1), (1,2), (1,3), (2,1), (2,2), (3,1) \}$: This is the roster form of the relation R itself, not its range.
• (D) $\{1, 2, 3, 4\}$: This includes 4, which is not a second element in any of the pairs in R.

The correct option is (A) (or (B)).

A relation is a set of ordered pairs. The question asks which of the following *pairs* represents a relation on the set of students in a class. This phrasing is a bit loose, as a relation is a *set* of ordered pairs, not a single pair.

However, the examples describe potential *types* of relations or components of relations involving students.

Let's consider each option as a description of a relation involving students:

(A) (Student A, Student B) if they are siblings.

This describes a relation where the domain and codomain are both the set of students in the class. The ordered pair (Student A, Student B) is in the relation if Student A and Student B are siblings. This is a valid type of relation on the set of students.

(B) (Student A, Subject) if Student A studies that subject.

This describes a relation from the set of students to the set of subjects. The ordered pair (Student A, Subject) is in the relation if Student A studies that subject. This is a valid type of relation involving students.

(C) (Student A, Score) if Student A got that score in a test.

This describes a relation from the set of students to the set of scores. The ordered pair (Student A, Score) is in the relation if Student A achieved that score. This is also a valid type of relation involving students.

(D) All of the above are examples of relations.

Since options (A), (B), and (C) all describe valid scenarios for relations involving students (either within the set of students, or from the set of students to another set), they are all examples of relations.

Therefore, the statement that all of the above are examples of relations is correct.

The correct option is (D).

Given:

Set $A = \{1, 2, 3\}$. The number of elements in A is $\text{n}(A) = 3$.

Set $B = \{a, b\}$. The number of elements in B is $\text{n}(B) = 2$.

To Find:

The number of proper subsets of the Cartesian product $A \times B$.

Solution:

First, we need to find the number of elements in the Cartesian product $A \times B$.

The number of elements in $A \times B$ is:

Substitute the given values:

The total number of subsets of a set with $k$ elements is $2^k$.

In this case, the set is $A \times B$, which has $k = 6$ elements.

So, the total number of subsets of $A \times B$ is $2^6$.

Calculating $2^6$:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

A proper subset of a set is any subset of that set, excluding the set itself. So, if a set has $N$ total subsets, it has $N-1$ proper subsets.

Number of proper subsets = (Total number of subsets) - 1

Number of proper subsets = $64 - 1 = 63$.

Therefore, $A \times B$ has 63 proper subsets.

The correct option is (A).

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

We are given $A \times B = \{(1, p), (1, q), (2, p), (2, q)\}$.

To find set A, we need to identify all the first elements of the ordered pairs in $A \times B$.

The first elements in the given set are 1, 1, 2, 2.

The set A consists of the unique first elements.

Therefore, set A is $\{1, 2\}$.

Similarly, to find set B, we would identify all the second elements of the ordered pairs in $A \times B$. The second elements are p, q, p, q. The unique second elements are $\{p, q\}$. So, set B is $\{p, q\}$.

The question asks for set A.

The correct option is (A).

First, let's identify the elements of sets A and B.

Set A is the set of the first five natural numbers. The natural numbers are $\{1, 2, 3, 4, 5, \dots\}$.

So, $A = \{1, 2, 3, 4, 5\}$.

The number of elements in set A is $\text{n}(A) = 5$.

Set B is the set of vowels in the English alphabet. The vowels are A, E, I, O, U.

So, $B = \{A, E, I, O, U\}$.

The number of elements in set B is $\text{n}(B) = 5$.

We need to find $\text{n}(A \times B)$, which is the number of elements in the Cartesian product of A and B.

The number of elements in the Cartesian product is given by the product of the number of elements in each set:

Substitute the values of $\text{n}(A)$ and $\text{n}(B)$:

Therefore, the number of elements in the Cartesian product $A \times B$ is 25.

The correct option is (C).

Given two ordered pairs $(a, b)$ and $(c, d)$ are equal, then $a=c$ and $b=d$.

The given ordered pairs are $(x+y, 3)$ and $(5, x-y)$.

Since the ordered pairs are equal, we can equate their corresponding components:

Now we have a system of two linear equations with two variables:

Equation (i): $x+y = 5$

Equation (ii): $x-y = 3$

We can solve this system using either the substitution method or the elimination method. Let's use the elimination method.

Add Equation (i) and Equation (ii):

This simplifies to:

Divide both sides by 2 to find the value of $x$:

Now, substitute the value of $x$ (which is 4) into Equation (i) to find the value of $y$:

Subtract 4 from both sides to find the value of $y$:

To verify the solution, we can substitute $x=4$ and $y=1$ into Equation (ii):

This matches the given equation (ii), so our values for $x$ and $y$ are correct.

Therefore, the values are:

$x = 4$

$y = 1$

Given two sets $A = \{1, 2\}$ and $B = \{a, b\}$.

The Cartesian product of two sets $A$ and $B$, denoted by $A \times B$, is the set of all ordered pairs $(a, b)$ where $a$ is an element of $A$ and $b$ is an element of $B$.

To find $A \times B$, we take each element from set $A$ and pair it with each element from set $B$:

For the element '1' in set $A$, we pair it with 'a' and 'b' from set $B$:

$(1, a)$

$(1, b)$

For the element '2' in set $A$, we pair it with 'a' and 'b' from set $B$:

$(2, a)$

$(2, b)$

So, the Cartesian product $A \times B$ is:

$A \times B = \{(1, a), (1, b), (2, a), (2, b)\}$

The Cartesian product of two sets $B$ and $A$, denoted by $B \times A$, is the set of all ordered pairs $(b, a)$ where $b$ is an element of $B$ and $a$ is an element of $A$.

To find $B \times A$, we take each element from set $B$ and pair it with each element from set $A$:

For the element 'a' in set $B$, we pair it with '1' and '2' from set $A$:

$(a, 1)$

$(a, 2)$

For the element 'b' in set $B$, we pair it with '1' and '2' from set $A$:

$(b, 1)$

$(b, 2)$

So, the Cartesian product $B \times A$ is:

$B \times A = \{(a, 1), (a, 2), (b, 1), (b, 2)\}$

Given the set $A = \{1, 2, 3\}$.

The Cartesian product $A \times A$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in A$.

We will pair each element of $A$ with every element of $A$:

For the element '1' in $A$:

$(1, 1), (1, 2), (1, 3)$

For the element '2' in $A$:

$(2, 1), (2, 2), (2, 3)$

For the element '3' in $A$:

$(3, 1), (3, 2), (3, 3)$

Therefore, the Cartesian product $A \times A$ is:

$A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$

To find the number of elements in $A \times A$, we can use the formula:

Number of elements in $A \times A = (\text{Number of elements in } A) \times (\text{Number of elements in } A)$

In this case, the number of elements in set $A$ is 3, since $A = \{1, 2, 3\}$.

So, the number of elements in $A \times A$ is:

Thus, there are 9 elements in $A \times A$.

Given the number of elements in set $A$ is $|A| = 5$ and the number of elements in set $B$ is $|B| = 3$.

The number of elements in the Cartesian product of two sets $A$ and $B$, denoted by $|A \times B|$, is given by the product of the number of elements in each set.

The formula for the number of elements in a Cartesian product is:

Substituting the given values:

So, there are 15 elements in $A \times B$.

Similarly, the number of elements in the Cartesian product of sets $B$ and $A$, denoted by $|B \times A|$, is given by the product of the number of elements in each set.

The formula is:

Substituting the given values:

So, there are 15 elements in $B \times A$.

A relation $R$ from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$.

In this case, we are looking for relations from set $A = \{a, b, c\}$ to the empty set $\phi$.

First, let's determine the Cartesian product $A \times \phi$.

The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.

Since the set $\phi$ contains no elements, it is impossible to form any ordered pair $(x, y)$ where $y$ is an element of $\phi$.

Therefore, the Cartesian product of any set with the empty set is always the empty set:

Now, a relation from $A$ to $\phi$ is any subset of $A \times \phi$.

The subsets of the empty set $\phi$ are only the empty set itself.

Therefore, the only possible relation from $A$ to the empty set $\phi$ is the empty set itself.

The set of all possible relations from $A$ to $\phi$ is the power set of $A \times \phi$.

Since $A \times \phi = \phi$, the power set of $\phi$ is $\{\phi\}$.

Thus, there is only one possible relation from $A$ to the empty set $\phi$, which is the empty relation.

The only possible relation from $A$ to $\phi$ is $\phi$.

Given the set $A = \{1, 2, 3, 4, 5, 6\}$.

A relation $R$ on $A$ is defined by $R = \{(x, y) : y = x+1, x, y \in A\}$.

To write the relation $R$ in roster form, we need to find all ordered pairs $(x, y)$ from $A \times A$ such that $y = x+1$.

We will consider each element $x$ in set $A$ and check if $x+1$ is also in set $A$.

If $x=1$, then $y = 1+1 = 2$. Since $2 \in A$, the pair $(1, 2)$ is in $R$.

If $x=2$, then $y = 2+1 = 3$. Since $3 \in A$, the pair $(2, 3)$ is in $R$.

If $x=3$, then $y = 3+1 = 4$. Since $4 \in A$, the pair $(3, 4)$ is in $R$.

If $x=4$, then $y = 4+1 = 5$. Since $5 \in A$, the pair $(4, 5)$ is in $R$.

If $x=5$, then $y = 5+1 = 6$. Since $6 \in A$, the pair $(5, 6)$ is in $R$.

If $x=6$, then $y = 6+1 = 7$. Since $7 \notin A$, the pair $(6, 7)$ is not in $R$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$

The domain of a relation $R$ is the set of all first elements (x-coordinates) of the ordered pairs in $R$.

From the roster form of $R$, the first elements are 1, 2, 3, 4, and 5.

So, the domain of $R$ is:

Domain$(R) = \{1, 2, 3, 4, 5\}$

The range of a relation $R$ is the set of all second elements (y-coordinates) of the ordered pairs in $R$.

From the roster form of $R$, the second elements are 2, 3, 4, 5, and 6.

So, the range of $R$ is:

Range$(R) = \{2, 3, 4, 5, 6\}$

Given the sets $A = \{x, y, z\}$ and $B = \{1, 2\}$.

A relation $R$ from $A$ to $B$ is a subset of the Cartesian product $A \times B$.

First, let's find the Cartesian product $A \times B$:

$A \times B = \{(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)\}$

A relation $R$ from $A$ to $B$ is any combination of these ordered pairs. We can choose any subset of $A \times B$ to define the relation $R$. Let's define one such relation.

Let's choose the following subset for our relation $R$:

$R = \{(x, 1), (y, 2)\}$

This is a valid relation because each ordered pair in $R$ is also an element of $A \times B$.

Now, let's define the domain and range of this relation $R$.

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

In $R = \{(x, 1), (y, 2)\}$, the first elements are $x$ and $y$.

So, the domain of $R$ is:

Domain$(R) = \{x, y\}$

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

In $R = \{(x, 1), (y, 2)\}$, the second elements are $1$ and $2$.

So, the range of $R$ is:

Range$(R) = \{1, 2\}$

Alternative Relation:

Let's consider another possible relation $R'$ from $A$ to $B$:

$R' = \{(y, 1), (z, 1), (x, 2)\}$

For this relation $R'$:

The domain of $R'$ is the set of the first elements: $\{y, z, x\}$.

Domain$(R') = \{x, y, z\}$

The range of $R'$ is the set of the second elements: $\{1, 2\}$.

Range$(R') = \{1, 2\}$

The relation $R$ on the set of integers $\mathbb{Z}$ is defined as $R = \{(x, y) : x - y \text{ is an even integer}\}$.

An integer is considered even if it is divisible by 2. That is, an integer $k$ is even if $k = 2m$ for some integer $m$. Equivalently, an integer $k$ is even if $k \pmod 2 = 0$.

We need to determine if the given ordered pairs belong to the relation $R$ by checking if the difference between the first and second element is an even integer.

1. Is $(2, 4) \in R$?

To check if $(2, 4) \in R$, we need to calculate the difference $x - y$ for $x=2$ and $y=4$ and see if the result is an even integer.

$x - y = 2 - 4 = -2$.

Since $-2 = 2 \times (-1)$, $-2$ is an even integer.

Therefore, $(2, 4) \in R$.

2. Is $(3, 5) \in R$?

To check if $(3, 5) \in R$, we need to calculate the difference $x - y$ for $x=3$ and $y=5$ and see if the result is an even integer.

$x - y = 3 - 5 = -2$.

Since $-2 = 2 \times (-1)$, $-2$ is an even integer.

Therefore, $(3, 5) \in R$.

3. Is $(2, 5) \in R$?

To check if $(2, 5) \in R$, we need to calculate the difference $x - y$ for $x=2$ and $y=5$ and see if the result is an even integer.

$x - y = 2 - 5 = -3$.

An integer is even if it can be written in the form $2k$ for some integer $k$. $-3$ cannot be expressed in this form, as $-3 = 2 \times (-1.5)$, and $-1.5$ is not an integer. Thus, $-3$ is an odd integer.

Therefore, $(2, 5) \notin R$.

Justification Summary:

A pair $(x, y)$ is in the relation $R$ if $x-y$ is an even integer.

For $(2, 4)$: $2 - 4 = -2$. Since $-2$ is an even integer, $(2, 4) \in R$.

For $(3, 5)$: $3 - 5 = -2$. Since $-2$ is an even integer, $(3, 5) \in R$.

For $(2, 5)$: $2 - 5 = -3$. Since $-3$ is an odd integer, $(2, 5) \notin R$.

Given a relation $R$ from a set $A$ to a set $B$, the inverse relation $R^{-1}$ from $B$ to $A$ is defined as $R^{-1} = \{(y, x) : (x, y) \in R\}$.

In this problem, the relation $R$ is on the set $A = \{1, 2, 3\}$. This means that $R$ is a subset of $A \times A$. The inverse relation $R^{-1}$ will be from $A$ to $A$ as well.

The given relation is $R = \{(1, 2), (1, 3), (2, 3), (3, 2)\}$.

To find the inverse relation $R^{-1}$, we simply reverse the order of the elements in each ordered pair of $R$.

For each pair $(x, y)$ in $R$, we create a pair $(y, x)$ for $R^{-1}$.

• For $(1, 2) \in R$, we have $(2, 1) \in R^{-1}$.
• For $(1, 3) \in R$, we have $(3, 1) \in R^{-1}$.
• For $(2, 3) \in R$, we have $(3, 2) \in R^{-1}$.
• For $(3, 2) \in R$, we have $(2, 3) \in R^{-1}$.

Combining all these reversed pairs, we get the inverse relation $R^{-1}$:

$R^{-1} = \{(2, 1), (3, 1), (3, 2), (2, 3)\}$

The identity relation on a set $A$, denoted by $I_A$ or $Id_A$, is a relation on $A$ such that every element of $A$ is related only to itself.

In other words, for a set $A$, the identity relation $I_A$ consists of all ordered pairs $(x, y)$ where $x \in A$, $y \in A$, and $x = y$.

Given the set $A = \{1, 2, 3, 4\}$.

To define the identity relation on $A$, we need to form ordered pairs where the first and second elements are the same and are elements of $A$.

The elements of $A$ are 1, 2, 3, and 4.

So, the ordered pairs where the elements are equal and belong to $A$ are:

• $(1, 1)$
• $(2, 2)$
• $(3, 3)$
• $(4, 4)$

Therefore, the identity relation on the set $A = \{1, 2, 3, 4\}$, written in roster form, is:

$I_A = \{(1, 1), (2, 2), (3, 3), (4, 4)\}$

Given that two ordered pairs are equal, their corresponding components must be equal.

We are given the equality of two ordered pairs:

Equating the first components:

Equating the second components:

Now we solve these two simple equations for $a$ and $b$.

From equation (i):

$a+1 = 3$

Subtract 1 from both sides:

From equation (ii):

$b-2 = 1$

Add 2 to both sides:

So, the values of $a$ and $b$ are:

$a = 2$

$b = 3$

Given the sets $P = \{a, b\}$ and $Q = \{c, d\}$.

The Cartesian product $P \times Q$ is the set of all ordered pairs $(x, y)$ where $x \in P$ and $y \in Q$.

To find the elements of $P \times Q$, we pair each element of $P$ with each element of $Q$:

For the element 'a' in $P$:

• $(a, c)$
• $(a, d)$

For the element 'b' in $P$:

• $(b, c)$
• $(b, d)$

So, the elements of $P \times Q$ are:

$P \times Q = \{(a, c), (a, d), (b, c), (b, d)\}$

The Cartesian product $Q \times P$ is the set of all ordered pairs $(y, x)$ where $y \in Q$ and $x \in P$.

To find the elements of $Q \times P$, we pair each element of $Q$ with each element of $P$:

For the element 'c' in $Q$:

• $(c, a)$
• $(c, b)$

For the element 'd' in $Q$:

• $(d, a)$
• $(d, b)$

So, the elements of $Q \times P$ are:

$Q \times P = \{(c, a), (c, b), (d, a), (d, b)\}$

Now, we need to determine if $P \times Q = Q \times P$.

Two sets are equal if and only if they contain the exact same elements. The order of elements in a set does not matter, but the order of elements within an ordered pair does matter.

Comparing the elements of $P \times Q$ and $Q \times P$:

$P \times Q = \{(a, c), (a, d), (b, c), (b, d)\}$

$Q \times P = \{(c, a), (c, b), (d, a), (d, b)\}$

We can see that the ordered pairs in $P \times Q$ are different from the ordered pairs in $Q \times P$. For example, $(a, c) \in P \times Q$, but $(a, c) \notin Q \times P$ (since the first element must be from $Q$ and the second from $P$ for $Q \times P$). Instead, we have $(c, a) \in Q \times P$.

Therefore, $P \times Q \neq Q \times P$.

A relation from a set $A$ to a set $B$ is defined as a subset of the Cartesian product $A \times B$.

We are given a set $A$ with $|A| = m$ and a set $B$ with $|B| = n$.

The number of elements in the Cartesian product $A \times B$ is given by the product of the number of elements in $A$ and the number of elements in $B$.

So, the Cartesian product $A \times B$ contains $mn$ ordered pairs.

The number of relations that can be defined from $A$ to $B$ is equal to the number of possible subsets of $A \times B$.

The number of subsets of a set with $k$ elements is $2^k$. This is known as the size of the power set.

In this case, the set $A \times B$ has $mn$ elements. Therefore, the number of subsets of $A \times B$ is $2^{mn}$.

Each of these subsets is a relation from $A$ to $B$.

Thus, the total number of relations that can be defined from $A$ to $B$ is $2^{mn}$.

The formula for the number of relations from $A$ to $B$ is:

Number of relations = $2^{|A| \times |B|}$

Substituting the given values:

Number of relations = $2^{m \times n} = 2^{mn}$

Given the set $A = \{1, 2, 3\}$ and set $B = \{2, 4, 6\}$.

The relation $R$ is defined as "is less than" from set $A$ to set $B$. This means that an ordered pair $(x, y)$ is in $R$ if $x \in A$, $y \in B$, and $x < y$.

To write $R$ in roster form, we need to find all such ordered pairs $(x, y)$ where $x$ is from set $A$, $y$ is from set $B$, and $x < y$.

Let's consider each element of set $A$ and compare it with each element of set $B$:

For $x = 1 \in A$:

• Is $1 < 2$? Yes. So, $(1, 2) \in R$.
• Is $1 < 4$? Yes. So, $(1, 4) \in R$.
• Is $1 < 6$? Yes. So, $(1, 6) \in R$.

For $x = 2 \in A$:

• Is $2 < 2$? No.
• Is $2 < 4$? Yes. So, $(2, 4) \in R$.
• Is $2 < 6$? Yes. So, $(2, 6) \in R$.

For $x = 3 \in A$:

• Is $3 < 2$? No.
• Is $3 < 4$? Yes. So, $(3, 4) \in R$.
• Is $3 < 6$? Yes. So, $(3, 6) \in R$.

Combining all the ordered pairs that satisfy the condition $x < y$, we get the relation $R$ in roster form:

$R = \{(1, 2), (1, 4), (1, 6), (2, 4), (2, 6), (3, 4), (3, 6)\}$

Let $R$ be a relation from a set $A$ to a set $B$. This means $R$ is a subset of the Cartesian product $A \times B$, so $R \subseteq A \times B$.

Definition of Domain:

The domain of a relation $R$ from set $A$ to set $B$ is the set of all first elements of the ordered pairs in $R$. It is the set of all elements in $A$ that are related to at least one element in $B$.

Mathematically, the domain of $R$, denoted as Domain$(R)$ or Dom$(R)$, is defined as:

Domain$(R) = \{x \in A \mid \exists y \in B \text{ such that } (x, y) \in R\}$

Definition of Range:

The range of a relation $R$ from set $A$ to set $B$ is the set of all second elements of the ordered pairs in $R$. It is the set of all elements in $B$ that are related to at least one element in $A$.

Mathematically, the range of $R$, denoted as Range$(R)$ or Ran$(R)$, is defined as:

Range$(R) = \{y \in B \mid \exists x \in A \text{ such that } (x, y) \in R\}$

Given the set $A = \{-1, 0, 1\}$.

The Cartesian product $A \times A \times A$ is the set of all ordered triples $(x, y, z)$ where $x \in A$, $y \in A$, and $z \in A$.

Since the set $A$ has 3 elements, the number of elements in $A \times A \times A$ will be $|A| \times |A| \times |A| = 3 \times 3 \times 3 = 27$.

To list all the elements, we can systematically combine each element of $A$ with every possible ordered pair from $A \times A$. Let's first consider $A \times A$:

$A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}$

Now, for each element in $A \times A$, we will append each element of $A$ as the third component:

For the first element of $A$, which is $-1$:

• $(-1, -1, -1)$
• $(-1, -1, 0)$
• $(-1, -1, 1)$
• $(-1, 0, -1)$
• $(-1, 0, 0)$
• $(-1, 0, 1)$
• $(-1, 1, -1)$
• $(-1, 1, 0)$
• $(-1, 1, 1)$

For the second element of $A$, which is $0$:

• $(0, -1, -1)$
• $(0, -1, 0)$
• $(0, -1, 1)$
• $(0, 0, -1)$
• $(0, 0, 0)$
• $(0, 0, 1)$
• $(0, 1, -1)$
• $(0, 1, 0)$
• $(0, 1, 1)$

For the third element of $A$, which is $1$:

• $(1, -1, -1)$
• $(1, -1, 0)$
• $(1, -1, 1)$
• $(1, 0, -1)$
• $(1, 0, 0)$
• $(1, 0, 1)$
• $(1, 1, -1)$
• $(1, 1, 0)$
• $(1, 1, 1)$

So, $A \times A \times A$ is the set containing all these 27 ordered triples:

$A \times A \times A = \{(-1, -1, -1), (-1, -1, 0), (-1, -1, 1), (-1, 0, -1), (-1, 0, 0), (-1, 0, 1), (-1, 1, -1), (-1, 1, 0), (-1, 1, 1), (0, -1, -1), (0, -1, 0), (0, -1, 1), (0, 0, -1), (0, 0, 0), (0, 0, 1), (0, 1, -1), (0, 1, 0), (0, 1, 1), (1, -1, -1), (1, -1, 0), (1, -1, 1), (1, 0, -1), (1, 0, 0), (1, 0, 1), (1, 1, -1), (1, 1, 0), (1, 1, 1)\}$

Given the set $A = \{1, 2, 3, 4\}$.

The relation $R$ on set $A$ is defined by $R = \{(x, y) : x, y \in A, x \text{ divides } y\}$. This means that an ordered pair $(x, y)$ is in $R$ if $x$ is an element of $A$, $y$ is an element of $A$, and $x$ divides $y$ evenly (i.e., $y$ is a multiple of $x$).

To write $R$ in roster form, we need to consider every possible pair $(x, y)$ from $A \times A$ and check if $x$ divides $y$.

Let's check for each element $x$ in $A$:

For $x = 1$:

• Does 1 divide 1? Yes ($1 = 1 \times 1$). So, $(1, 1) \in R$.
• Does 1 divide 2? Yes ($2 = 1 \times 2$). So, $(1, 2) \in R$.
• Does 1 divide 3? Yes ($3 = 1 \times 3$). So, $(1, 3) \in R$.
• Does 1 divide 4? Yes ($4 = 1 \times 4$). So, $(1, 4) \in R$.

For $x = 2$:

• Does 2 divide 1? No.
• Does 2 divide 2? Yes ($2 = 2 \times 1$). So, $(2, 2) \in R$.
• Does 2 divide 3? No.
• Does 2 divide 4? Yes ($4 = 2 \times 2$). So, $(2, 4) \in R$.

For $x = 3$:

• Does 3 divide 1? No.
• Does 3 divide 2? No.
• Does 3 divide 3? Yes ($3 = 3 \times 1$). So, $(3, 3) \in R$.
• Does 3 divide 4? No.

For $x = 4$:

• Does 4 divide 1? No.
• Does 4 divide 2? No.
• Does 4 divide 3? No.
• Does 4 divide 4? Yes ($4 = 4 \times 1$). So, $(4, 4) \in R$.

Combining all the ordered pairs that satisfy the condition, the relation $R$ in roster form is:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)\}$

Let $R$ be a relation from a set $A$ to a set $B$. This means $R$ is a subset of the Cartesian product $A \times B$, i.e., $R \subseteq A \times B$.

The inverse relation of $R$, denoted by $R^{-1}$, is a relation from set $B$ to set $A$. It is formed by reversing the order of the elements in each ordered pair of $R$.

More formally, the inverse relation $R^{-1}$ is defined as:

$R^{-1} = \{(y, x) \mid (x, y) \in R\}$

In simpler terms, if the ordered pair $(x, y)$ is in the relation $R$ (meaning $x$ is related to $y$), then the ordered pair $(y, x)$ is in the inverse relation $R^{-1}$ (meaning $y$ is related to $x$ in the inverse sense).

Given the set $A = \{x, y\}$.

A relation from $A$ to $A$ is any subset of the Cartesian product $A \times A$.

First, let's find the Cartesian product $A \times A$:

$A \times A = \{(x, x), (x, y), (y, x), (y, y)\}$

The number of elements in $A \times A$ is $|A \times A| = |A| \times |A| = 2 \times 2 = 4$.

The number of possible relations from $A$ to $A$ is equal to the number of subsets of $A \times A$. The number of subsets of a set with $k$ elements is $2^k$.

In this case, $k=4$, so the number of relations is $2^4 = 16$.

To list all 16 relations, we need to list all possible subsets of $A \times A$. These are:

1. The empty set (the empty relation):

$\phi$ or $R_1 = \emptyset$

2. Relations with one element:

$R_2 = \{(x, x)\}$

$R_3 = \{(x, y)\}$

$R_4 = \{(y, x)\}$

$R_5 = \{(y, y)\}$

3. Relations with two elements:

$R_6 = \{(x, x), (x, y)\}$

$R_7 = \{(x, x), (y, x)\}$

$R_8 = \{(x, x), (y, y)\}$

$R_9 = \{(x, y), (y, x)\}$

$R_{10} = \{(x, y), (y, y)\}$

$R_{11} = \{(y, x), (y, y)\}$

4. Relations with three elements:

$R_{12} = \{(x, x), (x, y), (y, x)\}$

$R_{13} = \{(x, x), (x, y), (y, y)\}$

$R_{14} = \{(x, x), (y, x), (y, y)\}$

$R_{15} = \{(x, y), (y, x), (y, y)\}$

5. The universal relation (all elements):

$R_{16} = \{(x, x), (x, y), (y, x), (y, y)\}$

To determine the type of relation $R$, we need to check its properties: reflexivity, symmetry, and transitivity.

Given set $A = \{1, 2, 3\}$ and relation $R = \{(1, 1), (2, 2), (3, 3)\}$.

1. Reflexivity:

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

For set $A = \{1, 2, 3\}$, 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:

Since all elements of $A$ are related to themselves in $R$, the relation $R$ is reflexive.

2. Symmetry:

A relation $R$ on a set $A$ is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$.

Let's check the pairs in $R$:

• $(1, 1) \in R$. If $(1, 1) \in R$, then $(1, 1)$ must also be in $R$. This is true.
• $(2, 2) \in R$. If $(2, 2) \in R$, then $(2, 2)$ must also be in $R$. This is true.
• $(3, 3) \in R$. If $(3, 3) \in R$, then $(3, 3)$ must also be in $R$. This is true.

Since for every pair $(a, b) \in R$, the pair $(b, a)$ is also in $R$ (in this case, $a=b$ for all pairs), the relation $R$ is symmetric.

3. Transitivity:

A relation $R$ on a set $A$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Let's check the pairs in $R$. For transitivity, we need to consider cases where the second element of one pair matches the first element of another pair.

The pairs in $R$ are of the form $(a, a)$. So, if we have $(a, b) \in R$ and $(b, c) \in R$, it means $a=b$ and $b=c$. This implies $a=c$. Therefore, $(a, c)$ would be $(a, a)$, which is in $R$.

• Consider $(1, 1) \in R$ and $(1, 1) \in R$. Is $(1, 1) \in R$? Yes.
• Consider $(2, 2) \in R$ and $(2, 2) \in R$. Is $(2, 2) \in R$? Yes.
• Consider $(3, 3) \in R$ and $(3, 3) \in R$. Is $(3, 3) \in R$? Yes.

There are no other combinations of pairs in $R$ where the second element of one matches the first element of another, other than the self-pairs.

Since the condition for transitivity holds for all possible combinations, the relation $R$ is transitive.

Conclusion:

The relation $R = \{(1, 1), (2, 2), (3, 3)\}$ on the set $A = \{1, 2, 3\}$ is reflexive, symmetric, and transitive.

A relation that is reflexive, symmetric, and transitive is called an equivalence relation.

In this specific case, the relation $R$ is also the identity relation on set $A$. The identity relation is always reflexive, symmetric, and transitive.

Therefore, the relation $R$ is an equivalence relation and also the identity relation.

Given the set $A = \{2, 3, 4, 5\}$.

The relation $R$ on $A$ is defined by $R = \{(x, y) : x+y=7\}$, where $x, y \in A$.

To write $R$ in roster form, we need to find all ordered pairs $(x, y)$ from $A \times A$ such that their sum is equal to 7.

Let's consider each element $x$ in set $A$ and find a corresponding $y$ in set $A$ such that $x+y=7$. This means $y = 7-x$.

For $x = 2$:

$y = 7 - 2 = 5$. Since $5 \in A$, the pair $(2, 5)$ is in $R$.

For $x = 3$:

$y = 7 - 3 = 4$. Since $4 \in A$, the pair $(3, 4)$ is in $R$.

For $x = 4$:

$y = 7 - 4 = 3$. Since $3 \in A$, the pair $(4, 3)$ is in $R$.

For $x = 5$:

$y = 7 - 5 = 2$. Since $2 \in A$, the pair $(5, 2)$ is in $R$.

Therefore, the relation $R$ in roster form is:

$R = \{(2, 5), (3, 4), (4, 3), (5, 2)\}$

The range of a relation $R$ is the set of all second elements of the ordered pairs in $R$.

Looking at the roster form of $R = \{(2, 5), (3, 4), (4, 3), (5, 2)\}$:

The second elements of the ordered pairs are 5, 4, 3, and 2.

So, the range of $R$ is:

Range$(R) = \{2, 3, 4, 5\}$

Given that $R$ is a relation from $\mathbb{N}$ to $\mathbb{N}$, where $\mathbb{N}$ represents the set of natural numbers $\{1, 2, 3, \ldots\}$.

The relation is defined by $R = \{(x, y) : y = 2x + 3, x, y \in \mathbb{N}, x \leq 5\}$.

To write $R$ in roster form, we need to find the ordered pairs $(x, y)$ that satisfy the given conditions. The conditions are:

1. $x$ and $y$ must be natural numbers ($x, y \in \mathbb{N}$).
2. $y = 2x + 3$.
3. $x$ must be less than or equal to 5 ($x \leq 5$).

We will iterate through possible values of $x$ starting from 1 up to 5, and for each $x$, we will calculate $y$ using the formula $y = 2x + 3$. We then check if the calculated $y$ is also a natural number.

For $x = 1$:

$y = 2(1) + 3 = 2 + 3 = 5$. Since $5 \in \mathbb{N}$, the pair $(1, 5)$ is in $R$.

For $x = 2$:

$y = 2(2) + 3 = 4 + 3 = 7$. Since $7 \in \mathbb{N}$, the pair $(2, 7)$ is in $R$.

For $x = 3$:

$y = 2(3) + 3 = 6 + 3 = 9$. Since $9 \in \mathbb{N}$, the pair $(3, 9)$ is in $R$.

For $x = 4$:

$y = 2(4) + 3 = 8 + 3 = 11$. Since $11 \in \mathbb{N}$, the pair $(4, 11)$ is in $R$.

For $x = 5$:

$y = 2(5) + 3 = 10 + 3 = 13$. Since $13 \in \mathbb{N}$, the pair $(5, 13)$ is in $R$.

Since the condition is $x \leq 5$, we have considered all possible values for $x$. All calculated $y$ values are natural numbers.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 5), (2, 7), (3, 9), (4, 11), (5, 13)\}$

Given the equality of two ordered pairs:

For these ordered pairs to be equal, their corresponding components must be equal.

Equating the first components:

Equating the second components:

We now have a system of two linear equations with two variables, $a$ and $b$. We can solve this system using the elimination method.

Add Equation (i) and Equation (ii):

This simplifies to:

Divide by 3 to find the value of $a$:

Now, substitute the value of $a$ into Equation (ii) to find the value of $b$:

$a - b = 3$

$\frac{11}{3} - b = 3$

To solve for $b$, subtract $\frac{11}{3}$ from both sides:

To perform the subtraction, find a common denominator:

Multiply both sides by -1 to find $b$:

So, the values of $a$ and $b$ are:

$a = \frac{11}{3}$

$b = \frac{2}{3}$

The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ such that $x$ is an element of set $A$ and $y$ is an element of set $B$.

We are given the Cartesian product $A \times B$ as:

$A \times B = \{(p, q), (p, r), (m, q), (m, r)\}$

To find set $A$, we need to collect all the unique first elements from each ordered pair in $A \times B$.

The first elements are $p$, $p$, $m$, and $m$.

The unique first elements are $p$ and $m$.

Therefore, set $A$ is:

$A = \{p, m\}$

To find set $B$, we need to collect all the unique second elements from each ordered pair in $A \times B$.

The second elements are $q$, $r$, $q$, and $r$.

The unique second elements are $q$ and $r$.

Therefore, set $B$ is:

$B = \{q, r\}$

We can verify this by calculating $A \times B$ using the sets we found:

$A \times B = \{p, m\} \times \{q, r\}$

$A \times B = \{(p, q), (p, r), (m, q), (m, r)\}$

This matches the given Cartesian product, so our sets $A$ and $B$ are correct.

Given the set $A = \{1, 2, 3\}$.

The relation $R$ on $A$ is defined by $R = \{(x, y) : x < y\}$, where $x, y \in A$.

To write $R$ in roster form, we need to find all ordered pairs $(x, y)$ from $A \times A$ such that $x < y$.

Let's check each element $x$ in $A$ against all elements $y$ in $A$:

For $x = 1$:

• Is $1 < 1$? No.
• Is $1 < 2$? Yes. So, $(1, 2) \in R$.
• Is $1 < 3$? Yes. So, $(1, 3) \in R$.

For $x = 2$:

• Is $2 < 1$? No.
• Is $2 < 2$? No.
• Is $2 < 3$? Yes. So, $(2, 3) \in R$.

For $x = 3$:

• Is $3 < 1$? No.
• Is $3 < 2$? No.
• Is $3 < 3$? No.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (1, 3), (2, 3)\}$

Now, we need to find the domain of the inverse relation $R^{-1}$.

First, let's find the inverse relation $R^{-1}$. The inverse relation $R^{-1}$ is obtained by reversing the order of the elements in each ordered pair of $R$.

If $R = \{(1, 2), (1, 3), (2, 3)\}$, then:

• $(1, 2) \in R \implies (2, 1) \in R^{-1}$
• $(1, 3) \in R \implies (3, 1) \in R^{-1}$
• $(2, 3) \in R \implies (3, 2) \in R^{-1}$

So, $R^{-1} = \{(2, 1), (3, 1), (3, 2)\}$.

The domain of a relation is the set of all first elements of its ordered pairs.

Looking at $R^{-1} = \{(2, 1), (3, 1), (3, 2)\}$:

The first elements are 2, 3, and 3.

The unique first elements are 2 and 3.

Therefore, the domain of $R^{-1}$ is:

Domain$(R^{-1}) = \{2, 3\}$

To find the number of possible relations from set $A$ to set $B$, we first need to determine the number of elements in the Cartesian product $A \times B$.

Given:

Set $A = \{a, b, c\}$, so $|A| = 3$.

Set $B = \{d, e\}$, so $|B| = 2$.

The number of elements in the Cartesian product $A \times B$ is given by the formula $|A \times B| = |A| \times |B|$.

This means there are 6 possible ordered pairs in the Cartesian product $A \times B$. Let's list them to be clear:

$A \times B = \{(a, d), (a, e), (b, d), (b, e), (c, d), (c, e)\}$

A relation from $A$ to $B$ is defined as any subset of the Cartesian product $A \times B$.

The number of subsets of a set with $k$ elements is $2^k$. This is also known as the size of the power set.

In this case, the set $A \times B$ has 6 elements. Therefore, the number of possible relations from $A$ to $B$ is $2^6$.

Calculating $2^6$:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

So, there are 64 possible relations from set $A$ to set $B$.

The formula for the number of relations from a set $A$ with $|A|=m$ elements to a set $B$ with $|B|=n$ elements is $2^{mn}$.

In this problem, $m=3$ and $n=2$, so the number of relations is $2^{3 \times 2} = 2^6 = 64$.

Therefore, there are 64 relations possible from $A$ to $B$.

The relation $R$ is defined as $R = \{(x, x^2) : x \text{ is a prime number less than 10}\}$.

First, we need to identify the prime numbers less than 10.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime numbers less than 10 are: 2, 3, 5, and 7.

Now, we use these prime numbers as the first elements ($x$) of our ordered pairs and calculate the corresponding second elements ($x^2$):

For $x = 2$:

$x^2 = 2^2 = 4$. So, the pair is $(2, 4)$.

For $x = 3$:

$x^2 = 3^2 = 9$. So, the pair is $(3, 9)$.

For $x = 5$:

$x^2 = 5^2 = 25$. So, the pair is $(5, 25)$.

For $x = 7$:

$x^2 = 7^2 = 49$. So, the pair is $(7, 49)$.

Therefore, the relation $R$ in roster form is:

$R = \{(2, 4), (3, 9), (5, 25), (7, 49)\}$

The range of a relation is the set of all second elements of the ordered pairs in the relation.

Looking at the roster form of $R = \{(2, 4), (3, 9), (5, 25), (7, 49)\}$:

The second elements are 4, 9, 25, and 49.

So, the range of $R$ is:

Range$(R) = \{4, 9, 25, 49\}$

Definition of Universal Relation:

The universal relation on a set $A$, denoted by $U_A$ or $A \times A$, is a relation in which every element of $A$ is related to every element of $A$. In other words, it is the relation that includes all possible ordered pairs in the Cartesian product $A \times A$.

So, for any non-empty set $A$, the universal relation $U_A$ is defined as:

$U_A = \{(x, y) \mid x \in A \text{ and } y \in A\}$

This is equivalent to saying that the universal relation is simply the set $A \times A$ itself.

Universal Relation on $A = \{1, 2\}$:

Given the set $A = \{1, 2\}$.

The universal relation on $A$ is the Cartesian product $A \times A$.

First, let's find $A \times A$:

$A \times A = \{1, 2\} \times \{1, 2\}$

This involves pairing each element of the first set $A$ with each element of the second set $A$:

• Pairing 1 with 1: $(1, 1)$
• Pairing 1 with 2: $(1, 2)$
• Pairing 2 with 1: $(2, 1)$
• Pairing 2 with 2: $(2, 2)$

Therefore, the universal relation on the set $A = \{1, 2\}$, written in roster form, is:

$U_A = \{(1, 1), (1, 2), (2, 1), (2, 2)\}$

To determine the number of subsets of $A \times B$, we first need to find the number of elements in the Cartesian product $A \times B$.

Given:

The number of elements in set $A$ is $|A| = 3$.

The number of elements in set $B$ is $|B| = 4$.

The number of elements in the Cartesian product $A \times B$ is given by the formula:

Substituting the given values:

So, the Cartesian product $A \times B$ has 12 elements.

The number of subsets of any set with $k$ elements is $2^k$. This is the size of its power set.

In this case, the set $A \times B$ has 12 elements. Therefore, the number of subsets of $A \times B$ is $2^{12}$.

Calculating $2^{12}$:

$2^{12} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 4096$.

Thus, $A \times B$ has 4096 subsets.

The universal relation on a set $A$, denoted by $U_A$ or $A \times A$, is the relation that contains all possible ordered pairs $(x, y)$ where $x \in A$ and $y \in A$. In essence, every element in $A$ is related to every element in $A$.

We are given a relation $R$ on the set of real numbers $\mathbb{R}$ defined by $R = \{(x, y) : x \geq y\}$.

To determine if $R$ is the universal relation on $\mathbb{R}$, we need to check if every ordered pair $(x, y)$ of real numbers satisfies the condition $x \geq y$.

Consider an arbitrary ordered pair of real numbers $(x, y)$. For this pair to be in the universal relation on $\mathbb{R}$, it must be true that $(x, y) \in \mathbb{R} \times \mathbb{R}$. The universal relation would be the set of ALL such pairs.

Now let's examine the condition for $R$: $x \geq y$.

Does every pair of real numbers $(x, y)$ satisfy $x \geq y$? No.

Let's take a counterexample:

Consider the real number $x = 2$ and the real number $y = 5$. Both $x$ and $y$ are in $\mathbb{R}$.

For the pair $(2, 5)$, let's check if it satisfies the condition $x \geq y$:

$2 \geq 5$

This statement is false, as 2 is not greater than or equal to 5. Therefore, the ordered pair $(2, 5)$ is not in the relation $R$.

Since we found at least one ordered pair of real numbers $(2, 5)$ that does not satisfy the condition $x \geq y$, this pair is not in $R$. However, for $R$ to be the universal relation on $\mathbb{R}$, it must include all possible ordered pairs of real numbers.

Thus, the relation $R = \{(x, y) : x \geq y\}$ is not the universal relation on $\mathbb{R}$.

Justification:

The universal relation on $\mathbb{R}$ is $\mathbb{R} \times \mathbb{R}$, which consists of all possible ordered pairs of real numbers. The relation $R = \{(x, y) : x \geq y\}$ only includes pairs where the first element is greater than or equal to the second element. For instance, the pair $(2, 5)$ consists of real numbers, but $2 < 5$, so $(2, 5) \notin R$. Since $R$ does not contain all possible ordered pairs of real numbers, it is not the universal relation on $\mathbb{R}$.

A relation on a set $A$ is a subset of the Cartesian product $A \times A$. To find the number of relations on $A$, we need to determine the number of subsets of $A \times A$.

Given the set $A = \{a, b\}$.

First, let's find the Cartesian product $A \times A$:

$A \times A = \{a, b\} \times \{a, b\}$

This involves pairing each element of the first set $A$ with each element of the second set $A$:

• Pairing 'a' with 'a': $(a, a)$
• Pairing 'a' with 'b': $(a, b)$
• Pairing 'b' with 'a': $(b, a)$
• Pairing 'b' with 'b': $(b, b)$

So, $A \times A = \{(a, a), (a, b), (b, a), (b, b)\}$.

The number of elements in $A \times A$ is $|A \times A| = |A| \times |A| = 2 \times 2 = 4$.

The number of relations on set $A$ is equal to the number of possible subsets of $A \times A$. The number of subsets of a set with $k$ elements is given by $2^k$.

In this case, the set $A \times A$ has 4 elements (where $k=4$).

Therefore, the number of relations that can be defined on $A$ is $2^4$.

Calculating $2^4$:

$2^4 = 2 \times 2 \times 2 \times 2 = 16$.

Thus, there are 16 relations that can be defined on the set $A = \{a, b\}$.

Given the sets $A = \{1, 2\}$ and $B = \{3, 4\}$, and the relation $R = \{(1, 3), (1, 4), (2, 3)\}$.

1. Domain and Range of R:

The domain of a relation $R$ is the set of all first elements of the ordered pairs in $R$.

From $R = \{(1, 3), (1, 4), (2, 3)\}$, the first elements are 1, 1, and 2.

The unique first elements are 1 and 2.

So, the domain of $R$ is:

Domain$(R) = \{1, 2\}$

The range of a relation $R$ is the set of all second elements of the ordered pairs in $R$.

From $R = \{(1, 3), (1, 4), (2, 3)\}$, the second elements are 3, 4, and 3.

The unique second elements are 3 and 4.

So, the range of $R$ is:

Range$(R) = \{3, 4\}$

2. Domain and Range of R^-1:

First, we find the inverse relation $R^{-1}$ by reversing the elements in each ordered pair of $R$.

For $(1, 3) \in R$, we have $(3, 1) \in R^{-1}$.

For $(1, 4) \in R$, we have $(4, 1) \in R^{-1}$.

For $(2, 3) \in R$, we have $(3, 2) \in R^{-1}$.

So, $R^{-1} = \{(3, 1), (4, 1), (3, 2)\}$.

The domain of $R^{-1}$ is the set of all first elements of the ordered pairs in $R^{-1}$.

From $R^{-1} = \{(3, 1), (4, 1), (3, 2)\}$, the first elements are 3, 4, and 3.

The unique first elements are 3 and 4.

So, the domain of $R^{-1}$ is:

Domain$(R^{-1}) = \{3, 4\}$

The range of $R^{-1}$ is the set of all second elements of the ordered pairs in $R^{-1}$.

From $R^{-1} = \{(3, 1), (4, 1), (3, 2)\}$, the second elements are 1, 1, and 2.

The unique second elements are 1 and 2.

So, the range of $R^{-1}$ is:

Range$(R^{-1}) = \{1, 2\}$

Observation:

Notice that Domain$(R)$ = Range$(R^{-1})$ and Range$(R)$ = Domain$(R^{-1})$. This is a general property of inverse relations.

Given the set $A = \{1, 2, 3\}$ and the relation $R = \{(1, 1), (2, 2)\}$ on $A$.

1. Find $A \times A$:

The Cartesian product $A \times A$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in A$.

$A \times A = \{1, 2, 3\} \times \{1, 2, 3\}$

This involves pairing each element of $A$ with every element of $A$:

• For 1: $(1, 1), (1, 2), (1, 3)$
• For 2: $(2, 1), (2, 2), (2, 3)$
• For 3: $(3, 1), (3, 2), (3, 3)$

So, $A \times A$ in roster form is:

$A \times A = \{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)\}$

2. Is $R$ a subset of $A \times A$?

A set $R$ is a subset of a set $S$ if every element in $R$ is also an element in $S$. In this case, $R$ is a relation on $A$, meaning $R$ is a subset of $A \times A$. Let's verify.

The relation $R = \{(1, 1), (2, 2)\}$.

We check if each element of $R$ is present in $A \times A$:

• $(1, 1)$ is in $A \times A$.
• $(2, 2)$ is in $A \times A$.

Since all elements of $R$ are also elements of $A \times A$, $R$ is indeed a subset of $A \times A$.

Yes, $R$ is a subset of $A \times A$.

3. What type of relation is $R$?

To determine the type of relation, we check for reflexivity, symmetry, and transitivity.

Reflexivity:

A relation $R$ on $A$ is reflexive if $(a, a) \in R$ for all $a \in A$. For $A = \{1, 2, 3\}$, we need $(1, 1), (2, 2), (3, 3)$ to be in $R$.

In $R = \{(1, 1), (2, 2)\}$, we have $(1, 1) \in R$ and $(2, 2) \in R$. However, $(3, 3) \notin R$.

Therefore, $R$ is not reflexive.

Symmetry:

A relation $R$ on $A$ is symmetric if whenever $(a, b) \in R$, then $(b, a) \in R$.

Let's check the pairs in $R$:

• $(1, 1) \in R$. If $(1, 1) \in R$, then $(1, 1)$ must be in $R$. This is true.
• $(2, 2) \in R$. If $(2, 2) \in R$, then $(2, 2)$ must be in $R$. This is true.

Since there are no pairs of the form $(a, b)$ where $a \neq b$ in $R$, the condition for symmetry is satisfied trivially.

Therefore, $R$ is symmetric.

Transitivity:

A relation $R$ on $A$ is transitive if whenever $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Let's check the pairs in $R$:

• We have $(1, 1) \in R$ and $(1, 1) \in R$. If $(1, 1) \in R$ and $(1, 1) \in R$, then $(1, 1)$ must be in $R$. This is true.
• We have $(2, 2) \in R$ and $(2, 2) \in R$. If $(2, 2) \in R$ and $(2, 2) \in R$, then $(2, 2)$ must be in $R$. This is true.

There are no other combinations where the second element of one pair matches the first element of another pair in $R$. For example, we don't have $(1, 1) \in R$ and $(1, 2) \in R$ simultaneously.

Therefore, $R$ is transitive.

Conclusion on the type of relation:

The relation $R = \{(1, 1), (2, 2)\}$ on $A = \{1, 2, 3\}$ is symmetric and transitive, but it is not reflexive because $(3, 3) \notin R$. It is not an equivalence relation because it is not reflexive.

We know that the number of elements in the Cartesian product of two sets $A$ and $B$ is given by the product of the number of elements in each set.

The formula is: $|A \times B| = |A| \times |B|$.

We are given:

The number of elements in $A \times B$ is 12, so $|A \times B| = 12$.

The number of elements in set $A$ is 3, so $|A| = 3$.

We need to find the number of elements in set $B$, denoted by $|B|$.

Substitute the given values into the formula:

To find $|B|$, we need to divide both sides of the equation by 3:

Therefore, the number of elements in set $B$ is 4.

Given the set $A = \{1, 2, 3, 4, 5\}$.

The relation $R$ on $A$ is defined by $R = \{(x, y) : x+y \text{ is an odd number}\}$, where $x, y \in A$.

For the sum of two integers to be odd, one integer must be even and the other must be odd. This is because:

• Odd + Odd = Even
• Even + Even = Even
• Odd + Even = Odd
• Even + Odd = Odd

In our set $A = \{1, 2, 3, 4, 5\}$:

The odd numbers are $\{1, 3, 5\}$.

The even numbers are $\{2, 4\}$.

Now, we need to find pairs $(x, y)$ from $A \times A$ where one element is odd and the other is even.

Case 1: $x$ is odd, $y$ is even.

• If $x=1$ (odd):
• $y=2$ (even): $1+2=3$ (odd). Pair: $(1, 2)$.
• $y=4$ (even): $1+4=5$ (odd). Pair: $(1, 4)$.
• If $x=3$ (odd):
• $y=2$ (even): $3+2=5$ (odd). Pair: $(3, 2)$.
• $y=4$ (even): $3+4=7$ (odd). Pair: $(3, 4)$.
• If $x=5$ (odd):
• $y=2$ (even): $5+2=7$ (odd). Pair: $(5, 2)$.
• $y=4$ (even): $5+4=9$ (odd). Pair: $(5, 4)$.

Case 2: $x$ is even, $y$ is odd.

• If $x=2$ (even):
• $y=1$ (odd): $2+1=3$ (odd). Pair: $(2, 1)$.
• $y=3$ (odd): $2+3=5$ (odd). Pair: $(2, 3)$.
• $y=5$ (odd): $2+5=7$ (odd). Pair: $(2, 5)$.
• If $x=4$ (even):
• $y=1$ (odd): $4+1=5$ (odd). Pair: $(4, 1)$.
• $y=3$ (odd): $4+3=7$ (odd). Pair: $(4, 3)$.
• $y=5$ (odd): $4+5=9$ (odd). Pair: $(4, 5)$.

Combining all the pairs where the sum $x+y$ is odd, the relation $R$ in roster form is:

$R = \{(1, 2), (1, 4), (3, 2), (3, 4), (5, 2), (5, 4), (2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5)\}$

Given the set $A = \{2, 4, 6\}$ and set $B = \{1, 3, 5\}$.

The relation $R$ from $A$ to $B$ is defined as $R = \{(x, y) : x < y\}$, where $x \in A$ and $y \in B$.

To write $R$ in roster form, we need to find all ordered pairs $(x, y)$ such that $x$ is from set $A$, $y$ is from set $B$, and $x$ is strictly less than $y$.

Let's consider each element $x$ from set $A$ and check its relation with each element $y$ from set $B$ based on the condition $x < y$:

For $x = 2 \in A$:

• Is $2 < 1$? No.
• Is $2 < 3$? Yes. So, $(2, 3) \in R$.
• Is $2 < 5$? Yes. So, $(2, 5) \in R$.

For $x = 4 \in A$:

• Is $4 < 1$? No.
• Is $4 < 3$? No.
• Is $4 < 5$? Yes. So, $(4, 5) \in R$.

For $x = 6 \in A$:

• Is $6 < 1$? No.
• Is $6 < 3$? No.
• Is $6 < 5$? No.

Combining all the ordered pairs that satisfy the condition $x < y$, the relation $R$ in roster form is:

$R = \{(2, 3), (2, 5), (4, 5)\}$

The inverse relation $R^{-1}$ is obtained by reversing the order of the elements in each ordered pair of the relation $R$.

Given the relation $R = \{(a, 1), (b, 2), (c, 1)\}$.

To find $R^{-1}$, we reverse each pair:

• For $(a, 1) \in R$, the corresponding pair in $R^{-1}$ is $(1, a)$.
• For $(b, 2) \in R$, the corresponding pair in $R^{-1}$ is $(2, b)$.
• For $(c, 1) \in R$, the corresponding pair in $R^{-1}$ is $(1, c)$.

So, the inverse relation $R^{-1}$ is:

$R^{-1} = \{(1, a), (2, b), (1, c)\}$

The domain of a relation is the set of all first elements of the ordered pairs in that relation.

Looking at the inverse relation $R^{-1} = \{(1, a), (2, b), (1, c)\}$:

The first elements are 1, 2, and 1.

The unique first elements are 1 and 2.

Therefore, the domain of $R^{-1}$ is:

Domain$(R^{-1}) = \{1, 2\}$

A relation $R$ on a set $A$ is a subset of the Cartesian product $A \times A$. The empty relation on a set $A$ is the relation that contains no ordered pairs. It is the subset of $A \times A$ that is equal to the empty set, $\phi$.

Given the set $A = \{1, 2, 3, 4\}$.

The Cartesian product $A \times A$ would contain all possible ordered pairs of elements from $A$. However, for the empty relation, we do not include any of these pairs.

The empty relation is simply the empty set.

Therefore, the empty relation $R$ on the set $A = \{1, 2, 3, 4\}$, written in roster form, is:

$R = \emptyset$

or

$R = \{\}$

Two ordered pairs $(p, q)$ and $(r, s)$ are considered equal if and only if their corresponding components are equal.

This means that:

1. The first component of the first ordered pair must be equal to the first component of the second ordered pair.
2. The second component of the first ordered pair must be equal to the second component of the second ordered pair.

Mathematically, the condition for equality of ordered pairs is:

Here, '$ \iff $' means "if and only if".

In summary, for two ordered pairs to be equal, their first elements must be equal, and their second elements must be equal.

Given the sets $A = \{1, 2\}$ and $B = \{p, q\}$.

1. Find $A \times B$:

The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

To find the elements of $A \times B$, we pair each element of $A$ with each element of $B$:

For the element '1' in $A$:

• Pairing with 'p': $(1, p)$
• Pairing with 'q': $(1, q)$

For the element '2' in $A$:

• Pairing with 'p': $(2, p)$
• Pairing with 'q': $(2, q)$

So, the Cartesian product $A \times B$ in roster form is:

$A \times B = \{(1, p), (1, q), (2, p), (2, q)\}$

2. Graphical Representation of $A \times B$:

To represent $A \times B$ graphically, we can use a coordinate plane. We typically place the elements of the first set ($A$) on the horizontal axis (x-axis) and the elements of the second set ($B$) on the vertical axis (y-axis).

Let the horizontal axis represent the elements of $A$: $\{1, 2\}$.

Let the vertical axis represent the elements of $B$: $\{p, q\}$.

We then plot each ordered pair $(x, y)$ from $A \times B$ as a point on this plane.

The points to be plotted are:

• $(1, p)$
• $(1, q)$
• $(2, p)$
• $(2, q)$

Here's a conceptual representation of the graph:

``` q . (1, q) . (2, q) | | p . (1, p) . (2, p) +----------+---------- 1 2 (Elements of A) ```

The graphical representation consists of four distinct points, each representing an ordered pair in the Cartesian product $A \times B$. The points form a rectangular grid on the plane.

To find the number of relations from set $A$ to set $B$, we first need to determine the number of elements in the Cartesian product $A \times B$.

Given:

The number of elements in set $A$ is $|A| = 3$.

The number of elements in set $B$ is $|B| = 2$.

The number of elements in the Cartesian product $A \times B$ is given by the formula:

Substituting the given values:

So, the Cartesian product $A \times B$ has 6 elements.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$.

The number of subsets of a set with $k$ elements is $2^k$.

In this case, the set $A \times B$ has 6 elements (where $k=6$).

Therefore, the number of relations that can be defined from $A$ to $B$ is $2^6$.

Calculating $2^6$:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.

Thus, there are 64 relations that can be defined from set $A$ to set $B$.

The relation $R$ is defined on the set of natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$.

The definition of the relation is $R = \{(x, y) : x+y \leq 5, x, y \in \mathbb{N}\}$.

We need to find all ordered pairs $(x, y)$ where both $x$ and $y$ are natural numbers, and their sum $x+y$ is less than or equal to 5.

Let's systematically find these pairs:

If $x = 1$:

• $1+y \leq 5 \implies y \leq 4$. Since $y \in \mathbb{N}$, possible values for $y$ are 1, 2, 3, 4.
• Pairs: $(1, 1), (1, 2), (1, 3), (1, 4)$.

If $x = 2$:

• $2+y \leq 5 \implies y \leq 3$. Since $y \in \mathbb{N}$, possible values for $y$ are 1, 2, 3.
• Pairs: $(2, 1), (2, 2), (2, 3)$.

If $x = 3$:

• $3+y \leq 5 \implies y \leq 2$. Since $y \in \mathbb{N}$, possible values for $y$ are 1, 2.
• Pairs: $(3, 1), (3, 2)$.

If $x = 4$:

• $4+y \leq 5 \implies y \leq 1$. Since $y \in \mathbb{N}$, the only possible value for $y$ is 1.
• Pair: $(4, 1)$.

If $x = 5$:

• $5+y \leq 5 \implies y \leq 0$. Since $y$ must be a natural number ($y \in \mathbb{N}$), there are no possible values for $y$ that satisfy $y \leq 0$.

We stop here because for any $x > 4$, the condition $x+y \leq 5$ with $y \in \mathbb{N}$ (which implies $y \geq 1$) would not be met ($x+1 \geq 5+1=6$).

Combining all the pairs found:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)\}$

Let $R$ be a relation from set $A$ to set $B$. This means $R$ is a subset of $A \times B$.

1. Domain and Range of R:

The domain of R, denoted as Domain$(R)$, is the set of all first elements of the ordered pairs in $R$. It is the set of all elements $x$ in $A$ such that there exists at least one element $y$ in $B$ for which $(x, y) \in R$.

Mathematically: Domain$(R) = \{x \in A \mid \exists y \in B \text{ such that } (x, y) \in R\}$

The range of R, denoted as Range$(R)$, is the set of all second elements of the ordered pairs in $R$. It is the set of all elements $y$ in $B$ such that there exists at least one element $x$ in $A$ for which $(x, y) \in R$.

Mathematically: Range$(R) = \{y \in B \mid \exists x \in A \text{ such that } (x, y) \in R\}$

2. Domain and Range of R^-1:

The inverse relation $R^{-1}$ is formed by reversing the elements in each ordered pair of $R$. If $R$ is a relation from $A$ to $B$, then $R^{-1}$ is a relation from $B$ to $A$. The definition is:

$R^{-1} = \{(y, x) \mid (x, y) \in R\}$

The domain of R^-1, denoted as Domain$(R^{-1})$, is the set of all first elements of the ordered pairs in $R^{-1}$. These first elements come from set $B$.

Mathematically: Domain$(R^{-1}) = \{y \in B \mid \exists x \in A \text{ such that } (y, x) \in R^{-1}\}$

By the definition of $R^{-1}$, if $(y, x) \in R^{-1}$, then $(x, y) \in R$. Thus, the first elements of $R^{-1}$ are the second elements of $R$.

Therefore: Domain$(R^{-1}) = \{y \in B \mid \exists x \in A \text{ such that } (x, y) \in R\} =$ Range$(R)$

The range of R^-1, denoted as Range$(R^{-1})$, is the set of all second elements of the ordered pairs in $R^{-1}$. These second elements come from set $A$.

Mathematically: Range$(R^{-1}) = \{x \in A \mid \exists y \in B \text{ such that } (y, x) \in R^{-1}\}$

By the definition of $R^{-1}$, if $(y, x) \in R^{-1}$, then $(x, y) \in R$. Thus, the second elements of $R^{-1}$ are the first elements of $R$.

Therefore: Range$(R^{-1}) = \{x \in A \mid \exists y \in B \text{ such that } (x, y) \in R\} =$ Domain$(R)$

3. How are they related?

The domain and range of a relation and its inverse are related in the following way:

• The domain of the inverse relation ($R^{-1}$) is equal to the range of the original relation ($R$).
• The range of the inverse relation ($R^{-1}$) is equal to the domain of the original relation ($R$).

In summary:

Domain$(R^{-1})$ = Range$(R)$

Range$(R^{-1})$ = Domain$(R)$

Let $A$ and $B$ be two non-empty sets.

A relation from set $A$ to set $B$ is formally defined as any subset of the Cartesian product of $A$ and $B$, which is denoted by $A \times B$.

The Cartesian product $A \times B$ is the set of all possible ordered pairs $(x, y)$, where $x$ is an element of set $A$ and $y$ is an element of set $B$. Mathematically, $A \times B = \{(x, y) \mid x \in A \text{ and } y \in B\}$.

Therefore, a relation $R$ from $A$ to $B$ is a set of ordered pairs, $R \subseteq A \times B$.

If an ordered pair $(x, y)$ is an element of a relation $R$ (i.e., $(x, y) \in R$), it means that $x$ is related to $y$ according to the definition of the relation $R$. The specific nature of this relation (e.g., "is less than," "is equal to," "divides") is determined by the rule used to form the relation.

In simpler terms, a relation establishes a connection or a rule between elements of set $A$ and elements of set $B$. This connection is represented by a collection of ordered pairs that satisfy the rule.

The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where the first element $x$ comes from set $A$ and the second element $y$ comes from set $B$.

We are given the Cartesian product $A \times B$ as:

$A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$

To find set $A$, we need to identify all the unique first elements present in the ordered pairs of $A \times B$.

The first elements in the given ordered pairs are $a$, $a$, $b$, and $b$.

The unique first elements are $a$ and $b$.

Therefore, set $A$ is:

$A = \{a, b\}$

To find set $B$, we need to identify all the unique second elements present in the ordered pairs of $A \times B$.

The second elements in the given ordered pairs are $x$, $y$, $x$, and $y$.

The unique second elements are $x$ and $y$.

Therefore, set $B$ is:

$B = \{x, y\}$

We can verify this by calculating $A \times B$ using the sets we found:

$A \times B = \{a, b\} \times \{x, y\}$

$A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$

This matches the given Cartesian product.

Given the set $A = \{1, 2, 3, 4\}$.

The relation $R$ on $A$ is defined by $R = \{(x, y) : x \geq y\}$, where $x, y \in A$. This means that an ordered pair $(x, y)$ is in $R$ if $x$ is an element of $A$, $y$ is an element of $A$, and $x$ is greater than or equal to $y$.

To write $R$ in roster form, we need to consider every possible pair $(x, y)$ from $A \times A$ and check if the condition $x \geq y$ is satisfied.

Let's check for each element $x$ in $A$:

For $x = 1$:

• Is $1 \geq 1$? Yes. So, $(1, 1) \in R$.
• Is $1 \geq 2$? No.
• Is $1 \geq 3$? No.
• Is $1 \geq 4$? No.

For $x = 2$:

• Is $2 \geq 1$? Yes. So, $(2, 1) \in R$.
• Is $2 \geq 2$? Yes. So, $(2, 2) \in R$.
• Is $2 \geq 3$? No.
• Is $2 \geq 4$? No.

For $x = 3$:

• Is $3 \geq 1$? Yes. So, $(3, 1) \in R$.
• Is $3 \geq 2$? Yes. So, $(3, 2) \in R$.
• Is $3 \geq 3$? Yes. So, $(3, 3) \in R$.
• Is $3 \geq 4$? No.

For $x = 4$:

• Is $4 \geq 1$? Yes. So, $(4, 1) \in R$.
• Is $4 \geq 2$? Yes. So, $(4, 2) \in R$.
• Is $4 \geq 3$? Yes. So, $(4, 3) \in R$.
• Is $4 \geq 4$? Yes. So, $(4, 4) \in R$.

Combining all the ordered pairs that satisfy the condition $x \geq y$, the relation $R$ in roster form is:

$R = \{(1, 1), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (4, 4)\}$

Two ordered pairs are equal if and only if their corresponding components are equal.

Given the equality of the ordered pairs:

Equating the first components:

Equating the second components:

Now, we solve each equation for its respective variable.

From equation (i), $x^2 = 9$:

To solve for $x$, we take the square root of both sides:

$x = \pm\sqrt{9}$

$x = \pm 3$

So, the possible values for $x$ are $3$ and $-3$.

From equation (ii), $y+1 = 5$:

To solve for $y$, we subtract 1 from both sides:

So, the value for $y$ is $4$.

Therefore, the possible values are:

Possible values for $x$: 3 and -3

Value for $y$: 4

The universal relation from a set $A$ to a set $B$ is the Cartesian product $A \times B$ itself. It is the relation that includes every possible ordered pair $(x, y)$ where $x \in A$ and $y \in B$.

Given the sets:

$A = \{1, 2, 3\}$

$B = \{4, 5, 6\}$

To define the universal relation from $A$ to $B$, we need to find the Cartesian product $A \times B$. This involves pairing each element of set $A$ with each element of set $B$.

For the element $1 \in A$:

• $(1, 4)$
• $(1, 5)$
• $(1, 6)$

For the element $2 \in A$:

• $(2, 4)$
• $(2, 5)$
• $(2, 6)$

For the element $3 \in A$:

• $(3, 4)$
• $(3, 5)$
• $(3, 6)$

The universal relation from $A$ to $B$, denoted as $A \times B$, is the set of all these ordered pairs:

Universal Relation = $\{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$

The number of elements in the Cartesian product of sets is the product of the number of elements in each individual set.

We are given:

The number of elements in set $A$ is $|A| = m$.

The number of elements in set $B$ is $|B| = n$.

We want to find the number of elements in the Cartesian product $A \times B \times A$.

The number of elements in $A \times B \times A$ is calculated as:

Substituting the given values:

This simplifies to:

Therefore, the number of elements in $A \times B \times A$ is $m^2n$.

The relation $R$ is defined on the set of integers $\mathbb{Z}$ by $R = \{(x, y) : x \cdot y \text{ is even}\}$.

An integer is even if it is divisible by 2. The product of two integers $x$ and $y$ ($x \cdot y$) is even if at least one of the integers ($x$ or $y$) is even.

Let's check each given pair:

1. Is $(1, 3) \in R$?

We need to check if the product of the components, $1 \cdot 3$, is an even number.

$1 \times 3 = 3$.

The number 3 is not divisible by 2, so it is an odd number.

Therefore, $(1, 3) \notin R$.

2. Is $(2, 4) \in R$?

We need to check if the product of the components, $2 \cdot 4$, is an even number.

$2 \times 4 = 8$.

The number 8 is divisible by 2 ($8 = 2 \times 4$), so it is an even number.

Therefore, $(2, 4) \in R$.

3. Is $(1, 2) \in R$?

We need to check if the product of the components, $1 \cdot 2$, is an even number.

$1 \times 2 = 2$.

The number 2 is divisible by 2 ($2 = 2 \times 1$), so it is an even number.

Therefore, $(1, 2) \in R$.

Justification Summary:

For $(1, 3)$: $1 \times 3 = 3$, which is odd. Thus, $(1, 3) \notin R$.

For $(2, 4)$: $2 \times 4 = 8$, which is even. Thus, $(2, 4) \in R$.

For $(1, 2)$: $1 \times 2 = 2$, which is even. Thus, $(1, 2) \in R$.

Given:

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

Set $B = \{2, 4, 6, 8\}$

Relation $R$ from $A$ to $B$ is defined as $R = \{(x, y) : y \text{ is a multiple of } x, x \in A, y \in B\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Codomain of $R$.

4. Range of $R$.

5. Inverse relation $R^{-1}$.

6. Domain of $R^{-1}$.

7. Range of $R^{-1}$.

Solution:

We need to find pairs $(x, y)$ such that $x \in A$, $y \in B$, and $y$ is a multiple of $x$. Let's check each element of $A$:

For $x=1$: Multiples of 1 in $B$ are 2, 4, 6, 8. So, the pairs are (1, 2), (1, 4), (1, 6), (1, 8).

For $x=2$: Multiples of 2 in $B$ are 2, 4, 6, 8. So, the pairs are (2, 2), (2, 4), (2, 6), (2, 8).

For $x=3$: Multiples of 3 in $B$ are 6. So, the pair is (3, 6).

For $x=4$: Multiples of 4 in $B$ are 4, 8. So, the pairs are (4, 4), (4, 8).

1. Relation $R$ in roster form:

$R = \{(1, 2), (1, 4), (1, 6), (1, 8), (2, 2), (2, 4), (2, 6), (2, 8), (3, 6), (4, 4), (4, 8)\}$

2. Domain of $R$:

The domain of a relation is the set of all first elements of the ordered pairs.

Domain$(R) = \{1, 2, 3, 4\}$

3. Codomain of $R$:

The codomain is the set to which the second elements of the ordered pairs belong.

Codomain$(R) = B = \{2, 4, 6, 8\}$

4. Range of $R$:

The range of a relation is the set of all second elements of the ordered pairs.

Range$(R) = \{2, 4, 6, 8\}$

5. Inverse relation $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$.

$R^{-1} = \{(2, 1), (4, 1), (6, 1), (8, 1), (2, 2), (4, 2), (6, 2), (8, 2), (6, 3), (4, 4), (8, 4)\}$

6. Domain of $R^{-1}$:

The domain of $R^{-1}$ is the set of all first elements of the ordered pairs in $R^{-1}$. This is equivalent to the range of $R$.

Domain$(R^{-1}) = \{1, 2, 3, 4\}$

7. Range of $R^{-1}$:

The range of $R^{-1}$ is the set of all second elements of the ordered pairs in $R^{-1}$. This is equivalent to the domain of $R$.

Range$(R^{-1}) = \{2, 4, 6, 8\}$

Given:

Set $A = \{x \in \mathbb{N} : x \leq 5\}$

Set $B = \{x \in \mathbb{Z} : -1 \leq x \leq 2\}$

To Find:

1. Sets $A$ and $B$ in roster form.

2. Cartesian product $A \times B$.

3. The number of relations that can be defined from $A$ to $B$.

Solution:

1. Writing $A$ and $B$ in roster form:

The set $A$ consists of natural numbers less than or equal to 5. The natural numbers are $1, 2, 3, \dots$.

So, $A = \{1, 2, 3, 4, 5\}$

The set $B$ consists of integers between -1 and 2, inclusive.

So, $B = \{-1, 0, 1, 2\}$

2. Finding the Cartesian product $A \times B$:

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

The number of elements in $A$ is $|A| = 5$.

The number of elements in $B$ is $|B| = 4$.

The number of elements in $A \times B$ is $|A \times B| = |A| \times |B| = 5 \times 4 = 20$.

$A \times B = \{(1, -1), (1, 0), (1, 1), (1, 2),$

$(2, -1), (2, 0), (2, 1), (2, 2),$

$(3, -1), (3, 0), (3, 1), (3, 2),$

$(4, -1), (4, 0), (4, 1), (4, 2),$

$(5, -1), (5, 0), (5, 1), (5, 2)\}$

3. Number of relations that can be defined from $A$ to $B$:

A relation from set $A$ to set $B$ is a subset of the Cartesian product $A \times B$.

The total number of subsets of a set with $n$ elements is $2^n$.

In this case, the set is $A \times B$, and it has $|A \times B| = 20$ elements.

Therefore, the number of possible relations from $A$ to $B$ is $2^{|A \times B|}$.

Number of relations = $2^{20}$

Calculating $2^{20}$:

$2^{10} = 1024$

$2^{20} = (2^{10})^2 = 1024^2 = 1048576$

So, there can be $2^{20}$ or $1,048,576$ relations defined from $A$ to $B$.

Given:

The relation $R$ on the set of integers $\mathbb{Z}$ is defined by $R = \{(x, y) : x, y \in \mathbb{Z}, x^2 + y^2 = 25\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

4. Inverse relation $R^{-1}$.

Solution:

We need to find integer pairs $(x, y)$ such that $x^2 + y^2 = 25$. We can list the possible integer squares that sum up to 25:

• $0^2 + 5^2 = 0 + 25 = 25$
• $1^2 + (\sqrt{24})^2 \neq 25$ (not integers)
• $2^2 + (\sqrt{21})^2 \neq 25$ (not integers)
• $3^2 + 4^2 = 9 + 16 = 25$
• $4^2 + 3^2 = 16 + 9 = 25$
• $5^2 + 0^2 = 25 + 0 = 25$

Considering both positive and negative integer values for $x$ and $y$:

If $x=0$, $y^2 = 25 \implies y = \pm 5$. Pairs: $(0, 5), (0, -5)$.

If $x=3$, $y^2 = 25 - 3^2 = 25 - 9 = 16 \implies y = \pm 4$. Pairs: $(3, 4), (3, -4)$.

If $x=4$, $y^2 = 25 - 4^2 = 25 - 16 = 9 \implies y = \pm 3$. Pairs: $(4, 3), (4, -3)$.

If $x=5$, $y^2 = 25 - 5^2 = 25 - 25 = 0 \implies y = 0$. Pair: $(5, 0)$.

Now consider the negative values for $x$:

If $x=-3$, $y^2 = 25 - (-3)^2 = 25 - 9 = 16 \implies y = \pm 4$. Pairs: $(-3, 4), (-3, -4)$.

If $x=-4$, $y^2 = 25 - (-4)^2 = 25 - 16 = 9 \implies y = \pm 3$. Pairs: $(-4, 3), (-4, -3)$.

If $x=-5$, $y^2 = 25 - (-5)^2 = 25 - 25 = 0 \implies y = 0$. Pair: $(-5, 0)$.

1. Relation $R$ in roster form:

$R = \{(0, 5), (0, -5), (3, 4), (3, -4), (4, 3), (4, -3), (5, 0), (-3, 4), (-3, -4), (-4, 3), (-4, -3), (-5, 0)\}$

2. Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{0, 3, 4, 5, -3, -4, -5\}$

3. Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{5, -5, 4, -4, 3, -3, 0\}$

4. Inverse relation $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$. Since the defining equation $x^2 + y^2 = 25$ is symmetric with respect to $x$ and $y$, the relation $R$ is its own inverse.

$R^{-1} = \{(5, 0), (-5, 0), (4, 3), (-4, 3), (3, 4), (-3, 4), (0, 5), (4, -3), (-4, -3), (3, -4), (-3, -4), (0, -5)\}$

We can observe that $R^{-1}$ contains the same set of ordered pairs as $R$. Therefore, $R^{-1} = R$.

Domain of $R^{-1}$ = Range of $R$ = $\{0, 3, 4, 5, -3, -4, -5\}$

Range of $R^{-1}$ = Domain of $R$ = $\{0, 3, 4, 5, -3, -4, -5\}$

Given:

Set $A = \{1, 2, 3\}$

Set $B = \{3, 4, 5\}$

Relation $R_1$ from $A$ to $B$ defined by $R_1 = \{(x, y) : x+y = 5\}$.

Relation $R_2$ from $A$ to $B$ defined by $R_2 = \{(x, y) : x < y\}$.

To Find:

1. $R_1$ in roster form.

2. $R_2$ in roster form.

3. $R_1 \cup R_2$.

4. $R_1 \cap R_2$.

5. $R_1 \setminus R_2$.

6. $R_2 \setminus R_1$.

Solution:

1. Writing $R_1$ in roster form:

We need to find pairs $(x, y)$ from $A \times B$ such that $x+y=5$.

For $x=1 \in A$: $1+y=5 \implies y=4$. Since $4 \in B$, $(1, 4) \in R_1$.

For $x=2 \in A$: $2+y=5 \implies y=3$. Since $3 \in B$, $(2, 3) \in R_1$.

For $x=3 \in A$: $3+y=5 \implies y=2$. Since $2 \notin B$, there is no pair with $x=3$.

$R_1 = \{(1, 4), (2, 3)\}$

2. Writing $R_2$ in roster form:

We need to find pairs $(x, y)$ from $A \times B$ such that $x < y$.

For $x=1 \in A$: $1 < y$. So, for $y=3, 4, 5$, we have $(1, 3), (1, 4), (1, 5)$.

For $x=2 \in A$: $2 < y$. So, for $y=3, 4, 5$, we have $(2, 3), (2, 4), (2, 5)$.

For $x=3 \in A$: $3 < y$. So, for $y=4, 5$, we have $(3, 4), (3, 5)$.

$R_2 = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5)\}$

3. Finding $R_1 \cup R_2$:

$R_1 \cup R_2$ is the set of all ordered pairs that are in $R_1$ or in $R_2$ or in both.

$R_1 \cup R_2 = \{(1, 4), (2, 3)\} \cup \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5)\}$

$R_1 \cup R_2 = \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5)\}$

4. Finding $R_1 \cap R_2$:

$R_1 \cap R_2$ is the set of all ordered pairs that are common to both $R_1$ and $R_2$.

$R_1 \cap R_2 = \{(1, 4), (2, 3)\} \cap \{(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5)\}$

$R_1 \cap R_2 = \{(1, 4), (2, 3)\}$

5. Finding $R_1 \setminus R_2$:

$R_1 \setminus R_2$ is the set of all ordered pairs that are in $R_1$ but not in $R_2$.

$R_1 \setminus R_2 = \{(x, y) : (x, y) \in R_1 \text{ and } (x, y) \notin R_2\}$

Checking elements of $R_1$: $(1, 4)$ is in $R_2$. $(2, 3)$ is in $R_2$.

Since all elements of $R_1$ are also in $R_2$, $R_1 \setminus R_2$ is an empty set.

$R_1 \setminus R_2 = \emptyset$

6. Finding $R_2 \setminus R_1$:

$R_2 \setminus R_1$ is the set of all ordered pairs that are in $R_2$ but not in $R_1$.

$R_2 \setminus R_1 = \{(x, y) : (x, y) \in R_2 \text{ and } (x, y) \notin R_1\}$

Elements in $R_2$ but not in $R_1$ are:

$(1, 3), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)$

$R_2 \setminus R_1 = \{(1, 3), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)\}$

To Prove:

For any two finite sets $A$ and $B$, the cardinality of their Cartesian product is equal to the product of their cardinalities: $|A \times B| = |A| \times |B|$.

Proof:

Let $A$ be a finite set with cardinality $|A| = m$, and let $B$ be a finite set with cardinality $|B| = n$. This means that the elements of $A$ can be listed as $A = \{a_1, a_2, \dots, a_m\}$ and the elements of $B$ can be listed as $B = \{b_1, b_2, \dots, b_n\}$.

The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in B$.

To form an element of $A \times B$, we need to choose one element from set $A$ and one element from set $B$.

We can construct the elements of $A \times B$ by systematically pairing each element of $A$ with each element of $B$.

For the first element $a_1 \in A$, we can pair it with all $n$ elements of $B$: $(a_1, b_1), (a_1, b_2), \dots, (a_1, b_n)$. This gives us $n$ ordered pairs.

For the second element $a_2 \in A$, we can pair it with all $n$ elements of $B$: $(a_2, b_1), (a_2, b_2), \dots, (a_2, b_n)$. This again gives us $n$ ordered pairs.

We continue this process for all $m$ elements of $A$. For each of the $m$ elements in $A$, there are $n$ possible elements in $B$ to form an ordered pair.

Using the multiplication principle of counting, the total number of ordered pairs $(x, y)$ that can be formed is the product of the number of choices for $x$ and the number of choices for $y$.

Number of choices for $x$ (from set $A$) = $|A| = m$.

Number of choices for $y$ (from set $B$) = $|B| = n$.

Therefore, the total number of ordered pairs in $A \times B$ is $m \times n$.

This means $|A \times B| = |A| \times |B|$.

This proves the statement.

Illustration with an Example:

Let $A = \{1, 2\}$ and $B = \{a, b, c\}$.

First, let's find the cardinalities of $A$ and $B$.

$|A| = 2$ (The elements are 1 and 2).

$|B| = 3$ (The elements are a, b, and c).

According to the statement we need to prove, $|A \times B|$ should be equal to $|A| \times |B|$.

$|A| \times |B| = 2 \times 3 = 6$.

Now, let's find the Cartesian product $A \times B$ in roster form to verify this.

$A \times B = \{(x, y) : x \in A, y \in B\}$

We pair each element of $A$ with each element of $B$:

For $x=1$: $(1, a), (1, b), (1, c)$.

For $x=2$: $(2, a), (2, b), (2, c)$.

So, $A \times B = \{(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)\}$.

Now, let's find the cardinality of $A \times B$. By counting the elements in the roster form:

$|A \times B| = 6$.

We see that $|A \times B| = 6$ and $|A| \times |B| = 6$.

Thus, $|A \times B| = |A| \times |B|$ is illustrated with this example.

Given:

The relation $R$ on the set of real numbers $\mathbb{R}$ is defined by $R = \{(x, y) : y = |x|\}$.

To Do:

1. Draw the graph of the relation $R$.

2. Find the domain of $R$.

3. Find the range of $R$.

Solution:

1. Drawing the graph of the relation $R$:

The relation is given by the equation $y = |x|$. The absolute value function $|x|$ is defined as:

$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

This means:

• If $x \geq 0$, then $y = x$. This is a straight line passing through the origin with a slope of 1 in the first quadrant.
• If $x < 0$, then $y = -x$. This is a straight line passing through the origin with a slope of -1 in the second quadrant.

To draw the graph, we can consider a few points:

• If $x=0$, $y = |0| = 0$. Point: (0, 0).
• If $x=1$, $y = |1| = 1$. Point: (1, 1).
• If $x=2$, $y = |2| = 2$. Point: (2, 2).
• If $x=-1$, $y = |-1| = 1$. Point: (-1, 1).
• If $x=-2$, $y = |-2| = 2$. Point: (-2, 2).

The graph of $y = |x|$ is a V-shaped graph with its vertex at the origin (0, 0). The right arm of the V is the line $y=x$ for $x \ge 0$, and the left arm of the V is the line $y=-x$ for $x < 0$.

[Here, a graphical representation of y = |x| would be displayed. It would show a V-shape with the vertex at the origin, extending into the first and second quadrants.]

2. Finding the domain of $R$:

The domain of a relation is the set of all possible input values (x-values) for which the relation is defined.

In this case, the relation $y = |x|$ is defined for all real numbers $x$. We can input any real number into the absolute value function, and we will get a real number as the output.

Therefore, the domain of $R$ is all real numbers.

Domain$(R) = \mathbb{R}$

3. Finding the range of $R$:

The range of a relation is the set of all possible output values (y-values) that the relation can produce.

The absolute value of any real number $x$, denoted as $|x|$, is always non-negative. That is, $|x| \geq 0$ for all $x \in \mathbb{R}$.

The smallest possible value for $|x|$ is 0 (when $x=0$). As $|x|$ increases, $y$ also increases without any upper bound.

Therefore, the range of $R$ is the set of all non-negative real numbers.

Range$(R) = \{y \in \mathbb{R} : y \geq 0\}$ or $[0, \infty)$

Given:

Set $A = \{1, 2, 3\}$

Set $B = \{a, b\}$

Relation $R$ from $A$ to $B$ is given by $R = \{(1, a), (2, b), (3, a)\}$.

To Find:

1. The inverse relation $R^{-1}$.

2. Arrow diagram representation of $R$.

3. Arrow diagram representation of $R^{-1}$.

Solution:

1. Finding the inverse relation $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$. The domain of $R^{-1}$ will be the range of $R$, and the range of $R^{-1}$ will be the domain of $R$.

Given $R = \{(1, a), (2, b), (3, a)\}$.

To find $R^{-1}$, we swap the elements in each pair:

• From $(1, a)$, we get $(a, 1)$.
• From $(2, b)$, we get $(b, 2)$.
• From $(3, a)$, we get $(a, 3)$.

Therefore, $R^{-1} = \{(a, 1), (b, 2), (a, 3)\}$.

2. Arrow diagram representation of $R$:

To represent $R$ using an arrow diagram, we draw two ovals, one for set $A$ and one for set $B$. We list the elements of each set within their respective ovals. Then, for each ordered pair $(x, y) \in R$, we draw an arrow from the element $x$ in set $A$ to the element $y$ in set $B$.

For $R = \{(1, a), (2, b), (3, a)\}$:

• Draw an oval for $A$ with elements {1, 2, 3}.
• Draw an oval for $B$ with elements {a, b}.
• Draw an arrow from 1 to a.
• Draw an arrow from 2 to b.
• Draw an arrow from 3 to a.

[Here, an arrow diagram for R would be shown. It would have two sets of nodes, A = {1, 2, 3} and B = {a, b}. Arrows would go from 1 to a, 2 to b, and 3 to a.]

3. Arrow diagram representation of $R^{-1}$:

To represent $R^{-1}$ using an arrow diagram, we will consider the domain of $R^{-1}$ (which is the range of $R$) and the range of $R^{-1}$ (which is the domain of $R$).

Domain of $R^{-1}$ is the set of second elements of $R$: $\{a, b\}$. This corresponds to set $B$.

Range of $R^{-1}$ is the set of first elements of $R$: $\{1, 2, 3\}$. This corresponds to set $A$.

For $R^{-1} = \{(a, 1), (b, 2), (a, 3)\}$:

• Draw an oval for the domain of $R^{-1}$ (set $B$) with elements {a, b}.
• Draw an oval for the range of $R^{-1}$ (set $A$) with elements {1, 2, 3}.
• Draw an arrow from a to 1.
• Draw an arrow from b to 2.
• Draw an arrow from a to 3.

[Here, an arrow diagram for R⁻¹ would be shown. It would have two sets of nodes, B = {a, b} and A = {1, 2, 3}. Arrows would go from a to 1, b to 2, and a to 3. Note that there are two arrows originating from 'a' in this diagram.]

Given:

Set $A$ is the set of the first five natural numbers.

Relation $R$ on set $A$ is defined by $R = \{(x, y) : x+y < 7\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

4. Domain of $R^{-1}$.

5. Range of $R^{-1}$.

Solution:

1. Writing $A$ in roster form:

The first five natural numbers are 1, 2, 3, 4, 5.

So, $A = \{1, 2, 3, 4, 5\}$.

2. Writing $R$ in roster form:

We need to find pairs $(x, y)$ such that $x \in A$, $y \in A$, and $x+y < 7$. Let's systematically check pairs:

• If $x=1$:
• $1+y < 7 \implies y < 6$. Since $y \in A$, possible values for $y$ are 1, 2, 3, 4, 5.
• Pairs: $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5)$
• If $x=2$:
• $2+y < 7 \implies y < 5$. Since $y \in A$, possible values for $y$ are 1, 2, 3, 4.
• Pairs: $(2, 1), (2, 2), (2, 3), (2, 4)$
• If $x=3$:
• $3+y < 7 \implies y < 4$. Since $y \in A$, possible values for $y$ are 1, 2, 3.
• Pairs: $(3, 1), (3, 2), (3, 3)$
• If $x=4$:
• $4+y < 7 \implies y < 3$. Since $y \in A$, possible values for $y$ are 1, 2.
• Pairs: $(4, 1), (4, 2)$
• If $x=5$:
• $5+y < 7 \implies y < 2$. Since $y \in A$, the only possible value for $y$ is 1.
• Pair: $(5, 1)$

So, the relation $R$ in roster form is:

$R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)\}$

3. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2, 3, 4, 5\}$

4. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{1, 2, 3, 4, 5\}$

5. Finding $R^{-1}$ and its domain and range:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$.

$R^{-1} = \{(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (1, 2), (2, 2), (3, 2), (4, 2), (1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (1, 5)\}$

The domain of $R^{-1}$ is the set of all first elements of the ordered pairs in $R^{-1}$. This is the same as the range of $R$.

Domain$(R^{-1}) = \{1, 2, 3, 4, 5\}$

The range of $R^{-1}$ is the set of all second elements of the ordered pairs in $R^{-1}$. This is the same as the domain of $R$.

Range$(R^{-1}) = \{1, 2, 3, 4, 5\}$

Given:

Set $A$ has 4 elements, so $|A| = 4$.

Set $B$ has 3 elements, so $|B| = 3$.

Set $C$ is a subset of $A \times B$, and $|C| = 6$.

To Determine:

Is $C$ necessarily a relation from $A$ to $B$?

Answer:

Yes, $C$ is necessarily a relation from $A$ to $B$.

Justification:

A relation from a set $A$ to a set $B$ is defined as any subset of the Cartesian product $A \times B$. The Cartesian product $A \times B$ is the set of all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

We are given that $A$ has 4 elements and $B$ has 3 elements. The total number of elements in the Cartesian product $A \times B$ is given by the formula $|A \times B| = |A| \times |B|$.

$|A \times B| = 4 \times 3 = 12$.

So, $A \times B$ contains 12 possible ordered pairs.

We are also given that $C$ is a subset of $A \times B$, and $C$ has 6 elements. By the definition of a relation, any subset of the Cartesian product $A \times B$ is a relation from $A$ to $B$. Since $C$ is a subset of $A \times B$, it automatically fulfills the definition of a relation from $A$ to $B$. The number of elements in $C$ (which is 6) is less than or equal to the total number of elements in $A \times B$ (which is 12), so it is possible for such a subset to exist.

Therefore, any subset of $A \times B$, regardless of its size (as long as it's not larger than $|A \times B|$), is considered a relation from $A$ to $B$. In this case, $C$ is a subset of $A \times B$ and has 6 elements, so it is indeed a relation from $A$ to $B$.

Example:

Let $A = \{1, 2, 3, 4\}$ and $B = \{a, b, c\}$.

Then $|A| = 4$ and $|B| = 3$.

$|A \times B| = 4 \times 3 = 12$.

Let's define a subset $C$ of $A \times B$ with 6 elements:

$C = \{(1, a), (1, b), (2, c), (3, a), (3, b), (4, c)\}$

Here, $C$ is a subset of $A \times B$, and it has 6 elements. By definition, $C$ is a relation from $A$ to $B$.

Given:

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

Set $B = \{p, q, r\}$

Relation $R$ from $A$ to $B$ is $R = \{(1, p), (1, q), (2, q), (3, r), (4, p)\}$.

To Find:

1. Domain of $R$.

2. Range of $R$.

3. Whether $R^{-1}$ is a relation from $B$ to $A$.

4. $R^{-1}$ in roster form.

5. Domain of $R^{-1}$.

6. Range of $R^{-1}$.

Solution:

1. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2, 3, 4\}$

2. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{p, q, r\}$

3. Is $R^{-1}$ a relation from $B$ to $A$?

A relation from set $B$ to set $A$ is defined as any subset of the Cartesian product $B \times A$. The inverse relation $R^{-1}$ is formed by reversing the order of elements in each pair of $R$. If $R$ is a relation from $A$ to $B$, then $R^{-1}$ will be a collection of ordered pairs where the first element comes from the set of second elements of $R$ (which is a subset of $B$) and the second element comes from the set of first elements of $R$ (which is a subset of $A$).

Since all the first elements of the pairs in $R$ are from $A$, and all the second elements are from $B$, when we reverse them for $R^{-1}$, the first elements will be from $B$ and the second elements will be from $A$. Therefore, $R^{-1}$ is a relation from $B$ to $A$. The set of all pairs in $R^{-1}$ will be a subset of $B \times A$.

So, yes, $R^{-1}$ is a relation from $B$ to $A$.

4. Finding $R^{-1}$ in roster form:

To find $R^{-1}$, we reverse each ordered pair in $R$:

• From $(1, p)$, we get $(p, 1)$.
• From $(1, q)$, we get $(q, 1)$.
• From $(2, q)$, we get $(q, 2)$.
• From $(3, r)$, we get $(r, 3)$.
• From $(4, p)$, we get $(p, 4)$.

$R^{-1} = \{(p, 1), (q, 1), (q, 2), (r, 3), (p, 4)\}$

5. Finding the Domain of $R^{-1}$:

The domain of $R^{-1}$ is the set of all first elements of the ordered pairs in $R^{-1}$. This is equivalent to the range of $R$.

Domain$(R^{-1}) = \{p, q, r\}$

6. Finding the Range of $R^{-1}$:

The range of $R^{-1}$ is the set of all second elements of the ordered pairs in $R^{-1}$. This is equivalent to the domain of $R$.

Range$(R^{-1}) = \{1, 2, 3, 4\}$

Given:

Set $A = \{1, 2, 3, 4, 5\}$.

Relation $R$ on $A$ is defined by $R = \{(x, y) : y = x^2 - 2x + 3, x, y \in A\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

Solution:

1. Writing $R$ in roster form:

We need to find pairs $(x, y)$ such that $x \in A$, $y \in A$, and $y = x^2 - 2x + 3$. We will substitute each value of $x$ from set $A$ into the equation and check if the resulting $y$ is also in set $A$.

• For $x=1$:
• $y = (1)^2 - 2(1) + 3 = 1 - 2 + 3 = 2$.
• Since $y=2 \in A$, the pair $(1, 2)$ is in $R$.
• For $x=2$:
• $y = (2)^2 - 2(2) + 3 = 4 - 4 + 3 = 3$.
• Since $y=3 \in A$, the pair $(2, 3)$ is in $R$.
• For $x=3$:
• $y = (3)^2 - 2(3) + 3 = 9 - 6 + 3 = 6$.
• Since $y=6 \notin A$, the pair $(3, 6)$ is not in $R$.
• For $x=4$:
• $y = (4)^2 - 2(4) + 3 = 16 - 8 + 3 = 11$.
• Since $y=11 \notin A$, the pair $(4, 11)$ is not in $R$.
• For $x=5$:
• $y = (5)^2 - 2(5) + 3 = 25 - 10 + 3 = 18$.
• Since $y=18 \notin A$, the pair $(5, 18)$ is not in $R$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (2, 3)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{2, 3\}$

Given:

The relation $R$ on the set of real numbers $\mathbb{R}$ is defined by $R = \{(x, y) : x^2 + y^2 = 1\}$.

To Do:

1. Draw the graph of the relation $R$.

2. Find the domain of $R$.

3. Find the range of $R$.

Solution:

1. Drawing the graph of the relation $R$:

The equation $x^2 + y^2 = 1$ represents a circle centered at the origin $(0, 0)$ with a radius of 1. This is because the general equation of a circle centered at $(h, k)$ with radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. In this case, $h=0$, $k=0$, and $r=1$.

To confirm this, we can solve for $y$: $y^2 = 1 - x^2$ $y = \pm \sqrt{1 - x^2}$

This shows that for a given value of $x$ (within the domain), there can be zero, one, or two corresponding values of $y$.

[Here, a graphical representation of the unit circle would be displayed. It would show a circle centered at (0,0) with a radius of 1, passing through the points (1,0), (-1,0), (0,1), and (0,-1).]

2. Finding the Domain of $R$:

The domain of the relation is the set of all possible $x$-values for which the equation $x^2 + y^2 = 1$ has a real solution for $y$.

From $y^2 = 1 - x^2$, for $y$ to be a real number, $y^2$ must be non-negative. Thus, $1 - x^2 \geq 0$.

$1 \geq x^2$

Taking the square root of both sides:

$\sqrt{1} \geq \sqrt{x^2}$

$1 \geq |x|$

This inequality means $-1 \leq x \leq 1$.

Therefore, the domain of $R$ is all real numbers between -1 and 1, inclusive.

Domain$(R) = \{x \in \mathbb{R} : -1 \leq x \leq 1\}$ or $[-1, 1]$.

3. Finding the Range of $R$:

The range of the relation is the set of all possible $y$-values for which the equation $x^2 + y^2 = 1$ has a real solution for $x$.

From $x^2 = 1 - y^2$, for $x$ to be a real number, $x^2$ must be non-negative. Thus, $1 - y^2 \geq 0$.

$1 \geq y^2$

Taking the square root of both sides:

$1 \geq |y|$

This inequality means $-1 \leq y \leq 1$.

Therefore, the range of $R$ is all real numbers between -1 and 1, inclusive.

Range$(R) = \{y \in \mathbb{R} : -1 \leq y \leq 1\}$ or $[-1, 1]$.

Given:

Set $A = \{x : x^2 - 4x + 3 = 0\}$

Set $B = \{x : x \in \mathbb{N}, x < 5\}$

To Find:

1. Sets $A$ and $B$ in roster form.

2. Cartesian product $A \times B$.

3. All possible relations from $A$ to $B$ that contain the ordered pair $(3, 2)$.

Solution:

1. Writing $A$ and $B$ in roster form:

For set $A$, we need to solve the quadratic equation $x^2 - 4x + 3 = 0$.

Factoring the equation:

$(x - 1)(x - 3) = 0$

This gives us $x - 1 = 0$ or $x - 3 = 0$.

So, $x = 1$ or $x = 3$.

Therefore, $A = \{1, 3\}$.

For set $B$, we need natural numbers $x$ such that $x < 5$. The natural numbers are $\{1, 2, 3, 4, 5, \dots\}$.

The natural numbers less than 5 are 1, 2, 3, 4.

Therefore, $B = \{1, 2, 3, 4\}$.

2. Finding the Cartesian product $A \times B$:

$A = \{1, 3\}$ and $B = \{1, 2, 3, 4\}$.

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

$A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2), (3, 3), (3, 4)\}$

The number of elements in $A \times B$ is $|A| \times |B| = 2 \times 4 = 8$.

3. Listing all possible relations from $A$ to $B$ that contain the ordered pair $(3, 2)$:

A relation from $A$ to $B$ is a subset of $A \times B$. We are looking for all subsets of $A \times B$ that must include the specific ordered pair $(3, 2)$.

The set $A \times B$ has 8 elements. The total number of possible relations from $A$ to $B$ is $2^{|A \times B|} = 2^8 = 256$.

We need to find subsets of $A \times B$ that contain $(3, 2)$. This means that $(3, 2)$ must be one of the elements in our relations.

Let $S = A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2), (3, 3), (3, 4)\}$.

We are looking for subsets of $S$ that contain $(3, 2)$. This means that for each of the other 7 elements in $S \setminus \{(3, 2)\}$, we can either include it in the relation or not include it.

The set $S \setminus \{(3, 2)\}$ has $8 - 1 = 7$ elements: $\{(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 3), (3, 4)\}$.

For each of these 7 elements, there are 2 choices (include or don't include). So, the number of such relations is $2^7$.

$2^7 = 128$.

Listing all 128 relations is impractical. However, we can describe them. Any relation $R$ that contains $(3, 2)$ will be of the form:

$R = \{(3, 2)\} \cup T$, where $T$ is any subset of $A \times B \setminus \{(3, 2)\}$.

Examples of such relations:

• The smallest such relation is just $\{(3, 2)\}$ itself.
• Another relation could be $\{(3, 2), (1, 1)\}$.
• Another relation could be $\{(3, 2), (1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 3), (3, 4)\}$ (which is $A \times B$ itself).
• Another relation could be $\{(3, 2), (1, 4), (3, 1)\}$.

In general, any subset of $A \times B$ that has $(3, 2)$ as one of its elements is a valid relation satisfying the condition.

Given:

The set $S = \{1, 2, 3, 4, 5, 6\}$.

The relation $R$ on $S$ is defined by $R = \{(x, y) : y \text{ is the successor of } x\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

4. Inverse relation $R^{-1}$.

Solution:

1. Writing $R$ in roster form:

The condition "$y$ is the successor of $x$" means that $y = x + 1$. We need to find pairs $(x, y)$ such that both $x$ and $y$ are elements of the set $S = \{1, 2, 3, 4, 5, 6\}$, and $y = x + 1$.

• If $x=1$, then $y = 1+1 = 2$. Since $2 \in S$, the pair $(1, 2)$ is in $R$.
• If $x=2$, then $y = 2+1 = 3$. Since $3 \in S$, the pair $(2, 3)$ is in $R$.
• If $x=3$, then $y = 3+1 = 4$. Since $4 \in S$, the pair $(3, 4)$ is in $R$.
• If $x=4$, then $y = 4+1 = 5$. Since $5 \in S$, the pair $(4, 5)$ is in $R$.
• If $x=5$, then $y = 5+1 = 6$. Since $6 \in S$, the pair $(5, 6)$ is in $R$.
• If $x=6$, then $y = 6+1 = 7$. Since $7 \notin S$, the pair $(6, 7)$ is not in $R$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2, 3, 4, 5\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{2, 3, 4, 5, 6\}$

4. Finding the Inverse Relation $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$. The domain of $R^{-1}$ is the range of $R$, and the range of $R^{-1}$ is the domain of $R$.

To find $R^{-1}$, we swap the elements in each pair:

• From $(1, 2)$, we get $(2, 1)$.
• From $(2, 3)$, we get $(3, 2)$.
• From $(3, 4)$, we get $(4, 3)$.
• From $(4, 5)$, we get $(5, 4)$.
• From $(5, 6)$, we get $(6, 5)$.

Therefore, $R^{-1} = \{(2, 1), (3, 2), (4, 3), (5, 4), (6, 5)\}$

The domain of $R^{-1}$ is the range of $R$: Domain$(R^{-1}) = \{2, 3, 4, 5, 6\}$.

The range of $R^{-1}$ is the domain of $R$: Range$(R^{-1}) = \{1, 2, 3, 4, 5\}$.

Given:

Set $A = \{a, b, c\}$

Set $B = \{1, 2, 3, 4\}$

Relation $R_1$ from $A$ to $B$ is $R_1 = \{(a, 1), (b, 2), (c, 3)\}$.

Relation $R_2$ from $A$ to $B$ is $R_2 = \{(a, 2), (b, 3), (c, 4)\}$.

To Find:

1. $R_1 \cup R_2$.

2. $R_1 \cap R_2$.

3. Whether $R_1 \cup R_2$ is a relation from $A$ to $B$ and justify.

4. Whether $R_1 \cap R_2$ is a relation from $A$ to $B$ and justify.

Solution:

1. Finding $R_1 \cup R_2$:

$R_1 \cup R_2$ is the union of the two relations, which means it contains all the ordered pairs that are in $R_1$ or in $R_2$ (or both).

$R_1 \cup R_2 = \{(a, 1), (b, 2), (c, 3)\} \cup \{(a, 2), (b, 3), (c, 4)\}$

$R_1 \cup R_2 = \{(a, 1), (a, 2), (b, 2), (b, 3), (c, 3), (c, 4)\}$

2. Finding $R_1 \cap R_2$:

$R_1 \cap R_2$ is the intersection of the two relations, which means it contains only the ordered pairs that are common to both $R_1$ and $R_2$.

$R_1 \cap R_2 = \{(a, 1), (b, 2), (c, 3)\} \cap \{(a, 2), (b, 3), (c, 4)\}$

By comparing the elements of $R_1$ and $R_2$, we can see that there are no common ordered pairs.

$R_1 \cap R_2 = \emptyset$ (the empty set).

3. Are $R_1 \cup R_2$ and $R_1 \cap R_2$ relations from $A$ to $B$? Justify.

A relation from set $A$ to set $B$ is defined as any subset of the Cartesian product $A \times B$.

First, let's determine the Cartesian product $A \times B$.

$A \times B = \{(a, 1), (a, 2), (a, 3), (a, 4), (b, 1), (b, 2), (b, 3), (b, 4), (c, 1), (c, 2), (c, 3), (c, 4)\}$

Justification for $R_1 \cup R_2$:

$R_1 \cup R_2 = \{(a, 1), (a, 2), (b, 2), (b, 3), (c, 3), (c, 4)\}$.

Every ordered pair in $R_1 \cup R_2$ has its first element from set $A$ (a, b, or c) and its second element from set $B$ (1, 2, 3, or 4).

All the elements of $R_1 \cup R_2$ are also present in $A \times B$. Therefore, $R_1 \cup R_2$ is a subset of $A \times B$. By definition, this means $R_1 \cup R_2$ is a relation from $A$ to $B$.

Justification for $R_1 \cap R_2$:

$R_1 \cap R_2 = \emptyset$.

The empty set $\emptyset$ is a subset of every set, including $A \times B$. Therefore, $\emptyset$ is a relation from $A$ to $B$. It is a valid relation, often called the empty relation.

Given:

Set $A$ has 5 elements, so $|A| = 5$.

To Find:

1. The maximum number of relations that can be defined on $A$.

2. The relationship between the domain of a relation $R$ on $A$ and set $A$.

3. The relationship between the range of a relation $R$ on $A$ and set $A$.

Solution:

1. Maximum number of relations that can be defined on $A$:

A relation on set $A$ is a subset of the Cartesian product $A \times A$.

First, we find the cardinality of $A \times A$.

$|A \times A| = |A| \times |A| = 5 \times 5 = 25$.

The total number of possible relations on $A$ is the total number of subsets of $A \times A$. The number of subsets of a set with $n$ elements is $2^n$.

So, the maximum number of relations that can be defined on $A$ is $2^{|A \times A|} = 2^{25}$.

$2^{25} = 33,554,432$.

Thus, there are $2^{25}$ possible relations on set $A$.

2. Relationship between the domain of $R$ and set $A$:

Let $R$ be a relation on set $A$. This means $R$ is a subset of $A \times A$.

The domain of $R$, denoted as Domain$(R)$, is the set of all first elements of the ordered pairs in $R$. Since all ordered pairs in $R$ are of the form $(x, y)$ where $x \in A$ and $y \in A$, the first element $x$ must belong to set $A$. Therefore, every element in the domain of $R$ must be an element of $A$. This means that the domain of $R$ is a subset of $A$.

Relationship: $\text{Domain}(R) \subseteq A$.

3. Relationship between the range of $R$ and set $A$:

The range of $R$, denoted as Range$(R)$, is the set of all second elements of the ordered pairs in $R$. Since all ordered pairs in $R$ are of the form $(x, y)$ where $x \in A$ and $y \in A$, the second element $y$ must belong to set $A$. Therefore, every element in the range of $R$ must be an element of $A$. This means that the range of $R$ is a subset of $A$.

Relationship: $\text{Range}(R) \subseteq A$.

Given:

The relation $R$ on the set of natural numbers $\mathbb{N}$ is defined by $R = \{(x, y) : x+2y = 10, x, y \in \mathbb{N}\}$.

Note: Natural numbers $\mathbb{N} = \{1, 2, 3, 4, \dots\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

4. Check if $(2, 4) \in R$, $(6, 2) \in R$, and $(8, 1) \in R$.

Solution:

1. Writing $R$ in roster form:

We need to find pairs $(x, y)$ such that $x \in \mathbb{N}$, $y \in \mathbb{N}$, and $x+2y = 10$. We can rewrite the equation as $x = 10 - 2y$. Since $x$ must be a natural number, $x > 0$. Also, since $y$ must be a natural number, $y > 0$.

Let's test values for $y \in \mathbb{N}$ and find the corresponding $x$:

• If $y=1$:
• $x = 10 - 2(1) = 10 - 2 = 8$.
• Since $x=8 \in \mathbb{N}$ and $y=1 \in \mathbb{N}$, the pair $(8, 1)$ is in $R$.
• If $y=2$:
• $x = 10 - 2(2) = 10 - 4 = 6$.
• Since $x=6 \in \mathbb{N}$ and $y=2 \in \mathbb{N}$, the pair $(6, 2)$ is in $R$.
• If $y=3$:
• $x = 10 - 2(3) = 10 - 6 = 4$.
• Since $x=4 \in \mathbb{N}$ and $y=3 \in \mathbb{N}$, the pair $(4, 3)$ is in $R$.
• If $y=4$:
• $x = 10 - 2(4) = 10 - 8 = 2$.
• Since $x=2 \in \mathbb{N}$ and $y=4 \in \mathbb{N}$, the pair $(2, 4)$ is in $R$.
• If $y=5$:
• $x = 10 - 2(5) = 10 - 10 = 0$.
• Since $x=0 \notin \mathbb{N}$ (0 is not a natural number), the pair $(0, 5)$ is not in $R$.
• If $y > 5$, then $2y > 10$, which means $10 - 2y < 0$, so $x$ would be negative and not a natural number.

Therefore, the relation $R$ in roster form is:

$R = \{(8, 1), (6, 2), (4, 3), (2, 4)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{8, 6, 4, 2\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{1, 2, 3, 4\}$

4. Checking membership in $R$:

Is $(2, 4) \in R$?

We check if $2+2(4) = 10$.

$2 + 8 = 10$.

Since the equation holds true and both 2 and 4 are natural numbers, yes, $(2, 4) \in R$.

Is $(6, 2) \in R$?

We check if $6+2(2) = 10$.

$6 + 4 = 10$.

Since the equation holds true and both 6 and 2 are natural numbers, yes, $(6, 2) \in R$.

Is $(8, 1) \in R$?

We check if $8+2(1) = 10$.

$8 + 2 = 10$.

Since the equation holds true and both 8 and 1 are natural numbers, yes, $(8, 1) \in R$.

Given:

Set $A = \{1, 2\}$

Set $B = \{3, 4\}$

Relation $R$ from $A$ to $B$ is defined by $R = \{(x, y) : x \text{ is a factor of } y\}$.

To Find:

1. Relation $R$ in roster form.

2. Inverse relation $R^{-1}$.

3. Proof that $(R^{-1})^{-1} = R$.

Solution:

1. Writing $R$ in roster form:

We need to find pairs $(x, y)$ such that $x \in A$, $y \in B$, and $x$ is a factor of $y$. Let's check all possible pairs from $A \times B$.

• For $x=1 \in A$:
• Is 1 a factor of $y=3 \in B$? Yes, $3 = 1 \times 3$. So, $(1, 3) \in R$.
• Is 1 a factor of $y=4 \in B$? Yes, $4 = 1 \times 4$. So, $(1, 4) \in R$.
• For $x=2 \in A$:
• Is 2 a factor of $y=3 \in B$? No, 3 is not divisible by 2.
• Is 2 a factor of $y=4 \in B$? Yes, $4 = 2 \times 2$. So, $(2, 4) \in R$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 3), (1, 4), (2, 4)\}$

2. Finding $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$. The domain of $R^{-1}$ will be the range of $R$, and the range of $R^{-1}$ will be the domain of $R$. The relation $R^{-1}$ will be from set $B$ to set $A$.

To find $R^{-1}$, we swap the elements in each pair of $R$:

• From $(1, 3)$, we get $(3, 1)$.
• From $(1, 4)$, we get $(4, 1)$.
• From $(2, 4)$, we get $(4, 2)$.

Therefore, $R^{-1} = \{(3, 1), (4, 1), (4, 2)\}$.

3. Showing that $(R^{-1})^{-1} = R$:

To find $(R^{-1})^{-1}$, we need to find the inverse of the relation $R^{-1}$. This means we swap the elements in each ordered pair of $R^{-1}$. The domain of $(R^{-1})^{-1}$ will be the range of $R^{-1}$, and the range of $(R^{-1})^{-1}$ will be the domain of $R^{-1}$. The relation $(R^{-1})^{-1}$ will be from set $A$ to set $B$ (since $R^{-1}$ is from $B$ to $A$).

Let's take $R^{-1} = \{(3, 1), (4, 1), (4, 2)\}$ and find its inverse:

• From $(3, 1)$, we get $(1, 3)$.
• From $(4, 1)$, we get $(1, 4)$.
• From $(4, 2)$, we get $(2, 4)$.

So, $(R^{-1})^{-1} = \{(1, 3), (1, 4), (2, 4)\}$.

Now, let's compare this with the original relation $R$:

$R = \{(1, 3), (1, 4), (2, 4)\}$

We can see that $(R^{-1})^{-1}$ contains the exact same ordered pairs as $R$. Therefore, we have shown that $(R^{-1})^{-1} = R$.

Given:

Set $S = \{1, 2, 3, 4, 5, 6\}$.

Relation $R$ on $S$ is defined by $R = \{(x, y) : x \text{ is relatively prime to } y, x, y \in S, x \neq y\}$.

Two numbers are relatively prime if their greatest common divisor (GCD) is 1.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

Solution:

1. Writing $R$ in roster form:

We need to find pairs $(x, y)$ from $S \times S$ such that $x \neq y$ and $\text{gcd}(x, y) = 1$. We will consider each element of $S$ as $x$ and check for possible $y$ values (where $y \in S$ and $y \neq x$).

• For $x=1$:
• $\text{gcd}(1, y) = 1$ for any $y$. Since $y \in S$ and $y \neq 1$, possible values for $y$ are 2, 3, 4, 5, 6.
• Pairs: $(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)$
• For $x=2$:
• We need $y \in S$, $y \neq 2$, and $\text{gcd}(2, y) = 1$.
• $y=1$: $\text{gcd}(2, 1) = 1$. Pair: $(2, 1)$.
• $y=3$: $\text{gcd}(2, 3) = 1$. Pair: $(2, 3)$.
• $y=4$: $\text{gcd}(2, 4) = 2 \neq 1$. Not included.
• $y=5$: $\text{gcd}(2, 5) = 1$. Pair: $(2, 5)$.
• $y=6$: $\text{gcd}(2, 6) = 2 \neq 1$. Not included.
• For $x=3$:
• We need $y \in S$, $y \neq 3$, and $\text{gcd}(3, y) = 1$.
• $y=1$: $\text{gcd}(3, 1) = 1$. Pair: $(3, 1)$.
• $y=2$: $\text{gcd}(3, 2) = 1$. Pair: $(3, 2)$.
• $y=4$: $\text{gcd}(3, 4) = 1$. Pair: $(3, 4)$.
• $y=5$: $\text{gcd}(3, 5) = 1$. Pair: $(3, 5)$.
• $y=6$: $\text{gcd}(3, 6) = 3 \neq 1$. Not included.
• For $x=4$:
• We need $y \in S$, $y \neq 4$, and $\text{gcd}(4, y) = 1$.
• $y=1$: $\text{gcd}(4, 1) = 1$. Pair: $(4, 1)$.
• $y=2$: $\text{gcd}(4, 2) = 2 \neq 1$. Not included.
• $y=3$: $\text{gcd}(4, 3) = 1$. Pair: $(4, 3)$.
• $y=5$: $\text{gcd}(4, 5) = 1$. Pair: $(4, 5)$.
• $y=6$: $\text{gcd}(4, 6) = 2 \neq 1$. Not included.
• For $x=5$:
• We need $y \in S$, $y \neq 5$, and $\text{gcd}(5, y) = 1$.
• $y=1$: $\text{gcd}(5, 1) = 1$. Pair: $(5, 1)$.
• $y=2$: $\text{gcd}(5, 2) = 1$. Pair: $(5, 2)$.
• $y=3$: $\text{gcd}(5, 3) = 1$. Pair: $(5, 3)$.
• $y=4$: $\text{gcd}(5, 4) = 1$. Pair: $(5, 4)$.
• $y=6$: $\text{gcd}(5, 6) = 1$. Pair: $(5, 6)$.
• For $x=6$:
• We need $y \in S$, $y \neq 6$, and $\text{gcd}(6, y) = 1$.
• $y=1$: $\text{gcd}(6, 1) = 1$. Pair: $(6, 1)$.
• $y=2$: $\text{gcd}(6, 2) = 2 \neq 1$. Not included.
• $y=3$: $\text{gcd}(6, 3) = 3 \neq 1$. Not included.
• $y=4$: $\text{gcd}(6, 4) = 2 \neq 1$. Not included.
• $y=5$: $\text{gcd}(6, 5) = 1$. Pair: $(6, 5)$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 5)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2, 3, 4, 5, 6\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{1, 2, 3, 4, 5, 6\}$

Given:

Set $A = \{a, b, c\}$.

A relation $R$ on $A$ must satisfy:

• $R$ is not the identity relation on $A$.
• $R$ is not the universal relation on $A$.
• $R$ contains at least 4 ordered pairs.

To Find:

1. A relation $R$ satisfying the given conditions in roster form.

2. The inverse relation $R^{-1}$ in roster form.

3. Domain and Range of $R$.

4. Domain and Range of $R^{-1}$.

Solution:

Understanding the conditions:

• The identity relation on $A$, denoted as $I_A$, is $\{(a, a), (b, b), (c, c)\}$.
• The universal relation on $A$, denoted as $A \times A$, is $\{(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)\}$. The number of elements in $A \times A$ is $|A|^2 = 3^2 = 9$.
• $R$ must have at least 4 ordered pairs.

We need to construct a relation $R$ that is a subset of $A \times A$, has at least 4 elements, and is not equal to $I_A$ or $A \times A$.

Let's choose a relation $R$ with 4 ordered pairs that meets the criteria. We can pick some elements from $A \times A$ but exclude the universal relation and ensure it's not just the identity relation.

Let's select the following ordered pairs for $R$:

$R = \{(a, a), (b, b), (a, b), (c, a)\}$

• This relation has 4 ordered pairs, satisfying the "at least 4" condition.
• It is not the identity relation because it contains $(a, b)$ and $(c, a)$, and it does not contain $(c, c)$.
• It is not the universal relation because it does not contain all 9 possible pairs (e.g., $(b, a)$, $(b, c)$, $(c, c)$ are missing).

So, this choice of $R$ satisfies all conditions.

1. Relation $R$ in roster form:

$R = \{(a, a), (b, b), (a, b), (c, a)\}$

2. Inverse relation $R^{-1}$ in roster form:

To find $R^{-1}$, we swap the elements in each ordered pair of $R$. The relation $R^{-1}$ will be from $A$ to $A$.

• From $(a, a)$, we get $(a, a)$.
• From $(b, b)$, we get $(b, b)$.
• From $(a, b)$, we get $(b, a)$.
• From $(c, a)$, we get $(a, c)$.

$R^{-1} = \{(a, a), (b, b), (b, a), (a, c)\}$

3. Domain and Range of $R$:

Domain$(R)$ is the set of all first elements in $R$: $\{a, b, c\}$.

Range$(R)$ is the set of all second elements in $R$: $\{a, b\}$.

4. Domain and Range of $R^{-1}$:

Domain$(R^{-1})$ is the set of all first elements in $R^{-1}$: $\{a, b\}$.

Range$(R^{-1})$ is the set of all second elements in $R^{-1}$: $\{a, b, c\}$.

Notice that Domain$(R) = \{a, b, c\}$ and Range$(R^{-1}) = \{a, b, c\}$.

Notice that Range$(R) = \{a, b\}$ and Domain$(R^{-1}) = \{a, b\}$.

Given:

Set $A = \{x \in \mathbb{N} : 1 \leq x \leq 4\}$, so $A = \{1, 2, 3, 4\}$.

Set $B = \{y \in \mathbb{N} : 4 \leq y \leq 7\}$, so $B = \{4, 5, 6, 7\}$.

Relation $R$ from $A$ to $B$ is defined by $R = \{(x, y) : x+y \text{ is an even number}\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

4. Inverse relation $R^{-1}$.

5. Domain of $R^{-1}$.

6. Range of $R^{-1}$.

Solution:

1. Writing $R$ in roster form:

The sum $x+y$ is even if either both $x$ and $y$ are even, or both $x$ and $y$ are odd.

From set $A = \{1, 2, 3, 4\}$:

• Odd numbers in $A$: $\{1, 3\}$
• Even numbers in $A$: $\{2, 4\}$

From set $B = \{4, 5, 6, 7\}$:

• Odd numbers in $B$: $\{5, 7\}$
• Even numbers in $B$: $\{4, 6\}$

Now we find pairs $(x, y)$ where $x \in A, y \in B$ and $x+y$ is even:

• Case 1: $x$ is odd and $y$ is odd.
• $x \in \{1, 3\}$ and $y \in \{5, 7\}$.
• Pairs: $(1, 5), (1, 7), (3, 5), (3, 7)$
• Case 2: $x$ is even and $y$ is even.
• $x \in \{2, 4\}$ and $y \in \{4, 6\}$.
• Pairs: $(2, 4), (2, 6), (4, 4), (4, 6)$

Combining these, the relation $R$ in roster form is:

$R = \{(1, 5), (1, 7), (3, 5), (3, 7), (2, 4), (2, 6), (4, 4), (4, 6)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 3, 2, 4\}$

Rearranging in ascending order: Domain$(R) = \{1, 2, 3, 4\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{5, 7, 4, 6\}$

Rearranging in ascending order: Range$(R) = \{4, 5, 6, 7\}$

4. Finding $R^{-1}$:

The inverse relation $R^{-1}$ is obtained by interchanging the elements in each ordered pair of $R$. The domain of $R^{-1}$ will be the range of $R$, and the range of $R^{-1}$ will be the domain of $R$. The relation $R^{-1}$ will be from set $B$ to set $A$.

To find $R^{-1}$, we swap the elements in each pair of $R$:

• From $(1, 5)$, we get $(5, 1)$.
• From $(1, 7)$, we get $(7, 1)$.
• From $(3, 5)$, we get $(5, 3)$.
• From $(3, 7)$, we get $(7, 3)$.
• From $(2, 4)$, we get $(4, 2)$.
• From $(2, 6)$, we get $(6, 2)$.
• From $(4, 4)$, we get $(4, 4)$.
• From $(4, 6)$, we get $(6, 4)$.

$R^{-1} = \{(5, 1), (7, 1), (5, 3), (7, 3), (4, 2), (6, 2), (4, 4), (6, 4)\}$

5. Finding the Domain of $R^{-1}$:

The domain of $R^{-1}$ is the set of all first elements of the ordered pairs in $R^{-1}$. This is equivalent to the range of $R$.

Domain$(R^{-1}) = \{5, 7, 4, 6\}$

Rearranging in ascending order: Domain$(R^{-1}) = \{4, 5, 6, 7\}$

6. Finding the Range of $R^{-1}$:

The range of $R^{-1}$ is the set of all second elements of the ordered pairs in $R^{-1}$. This is equivalent to the domain of $R$.

Range$(R^{-1}) = \{1, 3, 2, 4\}$

Rearranging in ascending order: Range$(R^{-1}) = \{1, 2, 3, 4\}$

Given:

Set $A = \{1, 2, 3\}$

Set $B = \{4, 5, 6\}$

Relation $R_1$ from $A$ to $B$ is $R_1 = \{(1, 4), (2, 5)\}$.

Relation $R_2$ from $A$ to $B$ is $R_2 = \{(1, 4), (3, 6)\}$.

To Find:

1. $R_1 \cup R_2$.

2. $R_1 \cap R_2$.

3. Domain and Range of $R_1 \cup R_2$.

4. Domain and Range of $R_1 \cap R_2$.

Solution:

1. Finding $R_1 \cup R_2$:

$R_1 \cup R_2$ is the set of all ordered pairs that are in $R_1$ or in $R_2$ (or both).

$R_1 \cup R_2 = \{(1, 4), (2, 5)\} \cup \{(1, 4), (3, 6)\}$

$R_1 \cup R_2 = \{(1, 4), (2, 5), (3, 6)\}$

2. Finding $R_1 \cap R_2$:

$R_1 \cap R_2$ is the set of all ordered pairs that are common to both $R_1$ and $R_2$.

$R_1 \cap R_2 = \{(1, 4), (2, 5)\} \cap \{(1, 4), (3, 6)\}$

$R_1 \cap R_2 = \{(1, 4)\}$

3. Domain and Range of $R_1 \cup R_2$:

For $R_1 \cup R_2 = \{(1, 4), (2, 5), (3, 6)\}$:

Domain$(R_1 \cup R_2) = \{1, 2, 3\}$

Range$(R_1 \cup R_2) = \{4, 5, 6\}$

4. Domain and Range of $R_1 \cap R_2$:

For $R_1 \cap R_2 = \{(1, 4)\}$:

Domain$(R_1 \cap R_2) = \{1\}$

Range$(R_1 \cap R_2) = \{4\}$

Given:

The set $S = \{1, 2, 3, 4, 5\}$.

The relation $R$ on $S$ is defined by $R = \{(x, y) : |x-y| = 1\}$.

To Find:

1. Relation $R$ in roster form.

2. Domain of $R$.

3. Range of $R$.

Solution:

1. Writing $R$ in roster form:

The condition $|x-y|=1$ means that $x-y = 1$ or $x-y = -1$. This is equivalent to saying that $y = x-1$ or $y = x+1$. In other words, $x$ and $y$ must be consecutive integers.

We need to find pairs $(x, y)$ such that $x, y \in S = \{1, 2, 3, 4, 5\}$ and $|x-y| = 1$. We will check each element of $S$ for $x$ and find corresponding $y$ values in $S$.

• For $x=1$:
• $|1-y|=1$. This means $1-y=1$ or $1-y=-1$.
• If $1-y=1$, then $y=0$. Since $0 \notin S$, $(1, 0)$ is not in $R$.
• If $1-y=-1$, then $y=2$. Since $2 \in S$, $(1, 2)$ is in $R$.
• For $x=2$:
• $|2-y|=1$. This means $2-y=1$ or $2-y=-1$.
• If $2-y=1$, then $y=1$. Since $1 \in S$, $(2, 1)$ is in $R$.
• If $2-y=-1$, then $y=3$. Since $3 \in S$, $(2, 3)$ is in $R$.
• For $x=3$:
• $|3-y|=1$. This means $3-y=1$ or $3-y=-1$.
• If $3-y=1$, then $y=2$. Since $2 \in S$, $(3, 2)$ is in $R$.
• If $3-y=-1$, then $y=4$. Since $4 \in S$, $(3, 4)$ is in $R$.
• For $x=4$:
• $|4-y|=1$. This means $4-y=1$ or $4-y=-1$.
• If $4-y=1$, then $y=3$. Since $3 \in S$, $(4, 3)$ is in $R$.
• If $4-y=-1$, then $y=5$. Since $5 \in S$, $(4, 5)$ is in $R$.
• For $x=5$:
• $|5-y|=1$. This means $5-y=1$ or $5-y=-1$.
• If $5-y=1$, then $y=4$. Since $4 \in S$, $(5, 4)$ is in $R$.
• If $5-y=-1$, then $y=6$. Since $6 \notin S$, $(5, 6)$ is not in $R$.

Therefore, the relation $R$ in roster form is:

$R = \{(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4)\}$

2. Finding the Domain of $R$:

The domain of $R$ is the set of all first elements of the ordered pairs in $R$.

Domain$(R) = \{1, 2, 3, 4, 5\}$

3. Finding the Range of $R$:

The range of $R$ is the set of all second elements of the ordered pairs in $R$.

Range$(R) = \{2, 1, 3, 4, 5\}$

Rearranging in ascending order: Range$(R) = \{1, 2, 3, 4, 5\}$

Given:

Let $|A| = m$ and $|B| = n$, where $m, n \geq 1$.

The number of relations from $A$ to $B$ is 64.

$|A| \neq |B|$, which means $m \neq n$.

To Find:

The possible values of $m$ and $n$.

Solution:

The number of relations from set $A$ to set $B$ is $2^{|A \times B|}$.

We know that $|A \times B| = |A| \times |B| = m \times n$.

So, the number of relations from $A$ to $B$ is $2^{mn}$.

We are given that the number of relations is 64. Therefore:

$2^{mn} = 64$

We need to express 64 as a power of 2:

$64 = 2^6$

Equating the exponents, we get:

$mn = 6$

We are also given that $m \neq n$ and $m, n \geq 1$. We need to find pairs of positive integers $(m, n)$ whose product is 6, with the condition that $m \neq n$.

The pairs of factors of 6 are:

• 1 and 6
• 2 and 3

Let's check these pairs against the condition $m \neq n$:

• If $\{m, n\} = \{1, 6\}$:
• Case 1: $m=1$, $n=6$. Here $m \neq n$ is satisfied.
• Case 2: $m=6$, $n=1$. Here $m \neq n$ is satisfied.
• If $\{m, n\} = \{2, 3\}$:
• Case 3: $m=2$, $n=3$. Here $m \neq n$ is satisfied.
• Case 4: $m=3$, $n=2$. Here $m \neq n$ is satisfied.

Therefore, the possible pairs of values for $(m, n)$ are:

$(1, 6), (6, 1), (2, 3), (3, 2)$.

The possible values for $m$ and $n$ are:

$\{m, n\} = \{1, 6\}$ or $\{m, n\} = \{2, 3\}$.

Given:

The relation $R$ on the set of real numbers $\mathbb{R}$ is defined by $R = \{(x, y) : y = x^2\}$.

To Do:

1. Draw the graph of the relation $R$.

2. Find the domain of $R$.

3. Find the range of $R$.

4. Check if $(2, 4) \in R$.

5. Check if $(-2, -4) \in R$.

Solution:

1. Drawing the graph of the relation $R$:

The relation is given by the equation $y = x^2$. This is the equation of a parabola that opens upwards, with its vertex at the origin $(0, 0)$.

To draw the graph, we can consider a few points:

• If $x=0$, $y = 0^2 = 0$. Point: $(0, 0)$.
• If $x=1$, $y = 1^2 = 1$. Point: $(1, 1)$.
• If $x=2$, $y = 2^2 = 4$. Point: $(2, 4)$.
• If $x=-1$, $y = (-1)^2 = 1$. Point: $(-1, 1)$.
• If $x=-2$, $y = (-2)^2 = 4$. Point: $(-2, 4)$.

The graph is symmetric about the y-axis because $x^2 = (-x)^2$.

[Here, a graphical representation of the parabola y = x² would be displayed. It would show a U-shaped curve with the vertex at the origin, opening upwards.]

2. Finding the Domain of $R$:

The domain of a relation is the set of all possible input values (x-values) for which the relation is defined.

In this case, the relation $y = x^2$ is defined for all real numbers $x$. We can substitute any real number for $x$, and we will get a real number as the output for $y$.

Therefore, the domain of $R$ is all real numbers.

Domain$(R) = \mathbb{R}$

3. Finding the Range of $R$:

The range of a relation is the set of all possible output values (y-values) that the relation can produce.

Since $y = x^2$, and the square of any real number is always non-negative ($x^2 \geq 0$), the output $y$ will always be greater than or equal to zero. The smallest value $y$ can take is 0 (when $x=0$), and $y$ can increase without bound as $|x|$ increases.

Therefore, the range of $R$ is the set of all non-negative real numbers.

Range$(R) = \{y \in \mathbb{R} : y \geq 0\}$ or $[0, \infty)$

4. Checking if $(2, 4) \in R$:

To check if $(2, 4)$ is in $R$, we substitute $x=2$ and $y=4$ into the relation's equation $y = x^2$.

Is $4 = 2^2$? Yes, $4 = 4$.

Therefore, $(2, 4) \in R$.

5. Checking if $(-2, -4) \in R$:

To check if $(-2, -4)$ is in $R$, we substitute $x=-2$ and $y=-4$ into the relation's equation $y = x^2$.

Is $-4 = (-2)^2$? No, $(-2)^2 = 4$. So, $-4 \neq 4$.

Therefore, $(-2, -4) \notin R$.