Chapter 1 Sets

The given equation is $x^2 + x - 2 = 0$.

To find the solution set, we need to find the values of $x$ that satisfy this equation. We can solve this quadratic equation by factoring.

We need to find two numbers that multiply to -2 and add to +1 (the coefficient of the $x$ term).

These two numbers are +2 and -1.

So, we can factor the quadratic equation as:

$(x + 2)(x - 1) = 0$

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

Case 1: $x + 2 = 0$

Subtracting 2 from both sides, we get $x = -2$.

Case 2: $x - 1 = 0$

Adding 1 to both sides, we get $x = 1$.

The solutions to the equation are $x = -2$ and $x = 1$.

The solution set is the set containing these solutions. In roster form, we list the elements within curly braces.

Therefore, the solution set of the equation $x^2 + x - 2 = 0$ in roster form is $\{-2, 1\}$.

We are asked to write the set $S = \{x : x \text{ is a positive integer and } x^2 < 40\}$ in roster form.

The condition requires $x$ to be a positive integer, which means $x$ can be $1, 2, 3, 4, \dots$.

The second condition is that the square of $x$, $x^2$, must be less than 40.

Let's find the squares of the positive integers starting from 1 and check if they are less than 40:

$1^2 = 1$. Since $1 < 40$, $x=1$ is in the set.

$2^2 = 4$. Since $4 < 40$, $x=2$ is in the set.

$3^2 = 9$. Since $9 < 40$, $x=3$ is in the set.

$4^2 = 16$. Since $16 < 40$, $x=4$ is in the set.

$5^2 = 25$. Since $25 < 40$, $x=5$ is in the set.

$6^2 = 36$. Since $36 < 40$, $x=6$ is in the set.

$7^2 = 49$. Since $49$ is not less than $40$ ($49 \not< 40$), $x=7$ is not in the set.

Since the squares of integers increase as the integers increase, any integer greater than 6 will also have a square greater than 40.

Thus, the positive integers $x$ such that $x^2 < 40$ are 1, 2, 3, 4, 5, and 6.

In roster form, we list these elements enclosed in curly braces.

Therefore, the set in roster form is $\{1, 2, 3, 4, 5, 6\}$.

The given set is A = {1, 4, 9, 16, 25, . . . }.

We need to identify the common property shared by all the elements in the set A.

Let's examine the elements:

$1 = 1^2$

$4 = 2^2$

$9 = 3^2$

$16 = 4^2$

$25 = 5^2$

And so on.

We observe that each element in the set A is the square of a natural number (positive integer).

Let $x$ be an arbitrary element of the set A.

Then $x$ can be written in the form $n^2$, where $n$ is a natural number ($n = 1, 2, 3, \dots$).

The set of natural numbers is denoted by $\mathbb{N}$.

Therefore, the set A can be described in set-builder form as the set of all elements $x$ such that $x$ is the square of a natural number $n$.

In set-builder notation, this is written as:

$A = \{x : x = n^2, \text{ where } n \in \mathbb{N}\}$

Alternatively, it can also be written as:

$A = \{x \mid x \text{ is the square of a natural number}\}$

Let the given set be denoted by $A$. So, $A = \left\{ \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7} \right\}$.

We need to find a common property or rule that describes all the elements of the set $A$.

Let's analyze the elements:

The first element is $\frac{1}{2}$. Here, the numerator is 1 and the denominator is $1+1=2$.

The second element is $\frac{2}{3}$. Here, the numerator is 2 and the denominator is $2+1=3$.

The third element is $\frac{3}{4}$. Here, the numerator is 3 and the denominator is $3+1=4$.

The fourth element is $\frac{4}{5}$. Here, the numerator is 4 and the denominator is $4+1=5$.

The fifth element is $\frac{5}{6}$. Here, the numerator is 5 and the denominator is $5+1=6$.

The sixth element is $\frac{6}{7}$. Here, the numerator is 6 and the denominator is $6+1=7$.

We observe a pattern: each element is a fraction where the denominator is one more than the numerator.

Let the numerator be represented by a variable $n$. Then the denominator is $n+1$.

So, each element $x$ in the set $A$ can be expressed in the form $x = \frac{n}{n+1}$.

Now we need to determine the possible values for $n$.

The numerators in the set are 1, 2, 3, 4, 5, and 6.

These are natural numbers (positive integers) starting from 1 and ending at 6.

So, $n$ is a natural number such that $1 \le n \le 6$. We can denote the set of natural numbers by $\mathbb{N}$.

Therefore, the set $A$ can be described in set-builder form as the set of all $x$ such that $x = \frac{n}{n+1}$, where $n$ is a natural number and $1 \le n \le 6$.

In set-builder notation, this can be written as:

$A = \left\{ x : x = \frac{n}{n+1}, \text{ where } n \in \mathbb{N} \text{ and } 1 \le n \le 6 \right\}$

Alternatively, we can write:

$A = \left\{ \frac{n}{n+1} : n \in \{1, 2, 3, 4, 5, 6\} \right\}$

We need to match the sets given in roster form on the left with their corresponding set-builder form descriptions on the right.

(i) {P, R, I, N, C, A, L}

This set contains the distinct letters present in the word PRINCIPAL. Looking at the options on the right:

(d) {x : x is a letter of the word PRINCIPAL} describes exactly this set. The letters in PRINCIPAL are P, R, I, N, C, I, P, A, L. The set of distinct letters is {P, R, I, N, C, A, L}.

So, (i) matches with (d).

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(ii) { 0 }

This set contains only the integer 0. Let's examine the set-builder descriptions:

(a) Describes divisors of 18, which are {1, 2, 3, 6, 9, 18}.

(b) Describes integers satisfying $x^2 - 9 = 0$, which are {3, -3}.

(c) Describes integers satisfying $x + 1 = 1$. Solving this equation: $x = 1 - 1 = 0$. The set is {0}.

(d) Describes letters of a word.

So, (ii) matches with (c).

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(iii) {1, 2, 3, 6, 9, 18}

This set contains positive integers. Let's check the properties described on the right:

(a) { x : x is a positive integer and is a divisor of 18}. The positive divisors of 18 are indeed 1, 2, 3, 6, 9, and 18. This matches the given set.

(b) Describes {3, -3}.

(c) Describes {0}.

(d) Describes letters.

So, (iii) matches with (a).

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(iv) {3, –3}

This set contains the integers 3 and -3. Let's examine the remaining description:

(b) { x : x is an integer and x² – 9 = 0}. We need to solve the equation $x^2 - 9 = 0$.

$x^2 = 9$

$x = \pm\sqrt{9}$

$x = 3$ or $x = -3$.

The set described is {3, -3}, which matches the given set.

So, (iv) matches with (b).

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Summary of Matches:

Roster Form	Set-Builder Form
(i) {P, R, I, N, C, A, L}	(d) {x : x is a letter of the word PRINCIPAL}
(ii) { 0 }	(c) {x : x is an integer and x + 1= 1}
(iii) {1, 2, 3, 6, 9, 18}	(a) { x : x is a positive integer and is a divisor of 18}
(iv) {3, –3}	(b) { x : x is an integer and x2 – 9 = 0}

A collection is considered a set if it is well-defined. This means that for any given object, it must be possible to determine definitively whether or not that object belongs to the collection. There should be no ambiguity.

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(i) The collection of all the months of a year beginning with the letter J.

Justification: The months of the year are fixed. We can clearly and definitively identify the months that begin with the letter J (January, June, July). The criteria for membership are clear and unambiguous.

Conclusion: This collection is well-defined and therefore, it is a set.

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(ii) The collection of ten most talented writers of India.

Justification: The term "most talented" is subjective and varies from person to person. There is no objective standard to measure talent that would allow everyone to agree on the same ten writers. The criteria for membership are ambiguous.

Conclusion: This collection is not well-defined and therefore, it is not a set.

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(iii) A team of eleven best-cricket batsmen of the world.

Justification: Similar to (ii), the term "best" is subjective. The criteria for selecting the "best" batsmen (e.g., highest average, most runs, impact) are not universally agreed upon, leading to different selections by different people.

Conclusion: This collection is not well-defined and therefore, it is not a set.

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(iv) The collection of all boys in your class.

Justification: Assuming "your class" refers to a specific, identifiable group of students, we can definitively list all the students in that class. We can also objectively determine which of these students are boys. The criteria for membership are clear.

Conclusion: This collection is well-defined and therefore, it is a set.

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(v) The collection of all natural numbers less than 100.

Justification: Natural numbers are well-defined (1, 2, 3,...). The condition "less than 100" is a clear, objective criterion. We can definitively list all the natural numbers that satisfy this condition (1, 2, ..., 99).

Conclusion: This collection is well-defined and therefore, it is a set.

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(vi) A collection of novels written by the writer Munshi Prem Chand.

Justification: Munshi Prem Chand is a specific writer. Whether a work is a novel and whether it was written by him are generally verifiable facts. We can, in principle, definitively determine whether any given novel belongs to this collection.

Conclusion: This collection is well-defined and therefore, it is a set.

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(vii) The collection of all even integers.

Justification: Integers are well-defined. An integer is even if it is divisible by 2, which is a clear and objective property. We can definitively determine whether any given integer is even or not.

Conclusion: This collection is well-defined and therefore, it is a set.

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(viii) The collection of questions in this Chapter.

Justification: Assuming "this Chapter" refers to a specific chapter in a particular book, the questions within that chapter are fixed and identifiable. We can definitively list all the questions that belong to it.

Conclusion: This collection is well-defined and therefore, it is a set.

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(ix) A collection of most dangerous animals of the world.

Justification: The term "most dangerous" is subjective and lacks a single, clear definition. Does it mean most venomous, most aggressive, responsible for most human deaths? Different criteria lead to different collections of animals. The criteria for membership are ambiguous.

Conclusion: This collection is not well-defined and therefore, it is not a set.

The given set is A = {1, 2, 3, 4, 5, 6}.

The symbol ∈ means "is an element of" or "belongs to".

The symbol ∉ means "is not an element of" or "does not belong to".

We need to check if the given numbers are elements of set A.

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(i) 5 . . . A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 5 is present in the set A.

Therefore, 5 ∈ A.

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(ii) 8 . . . A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 8 is not present in the set A.

Therefore, 8 ∉ A.

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(iii) 0 . . .A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 0 is not present in the set A.

Therefore, 0 ∉ A.

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(iv) 4 . . . A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 4 is present in the set A.

Therefore, 4 ∈ A.

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(v) 2 . . . A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 2 is present in the set A.

Therefore, 2 ∈ A.

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(vi) 10 . . . A

We look at the elements of set A: {1, 2, 3, 4, 5, 6}. The number 10 is not present in the set A.

Therefore, 10 ∉ A.

(i) A = {x : x is an integer and –3 ≤ x < 7}

The elements of this set are integers $x$ such that $x$ is greater than or equal to -3 and strictly less than 7.

The integers satisfying this condition are -3, -2, -1, 0, 1, 2, 3, 4, 5, and 6.

Therefore, the set A in roster form is {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.

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(ii) B = {x : x is a natural number less than 6}

Natural numbers are positive integers {1, 2, 3, ...}.

The elements of this set are natural numbers $x$ such that $x$ is less than 6.

The natural numbers satisfying this condition are 1, 2, 3, 4, and 5.

Therefore, the set B in roster form is {1, 2, 3, 4, 5}.

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(iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8}

The elements of this set are two-digit natural numbers (from 10 to 99).

Let the two-digit number be $10a + b$, where $a$ is the tens digit ($a \in \{1, 2, ..., 9\}$) and $b$ is the units digit ($b \in \{0, 1, ..., 9\}$).

The condition is that the sum of the digits is 8, i.e., $a + b = 8$.

We find the pairs $(a, b)$ such that $a+b=8$ and $a \ne 0$:

If $a=1$, $b=7 \implies$ Number = 17.

If $a=2$, $b=6 \implies$ Number = 26.

If $a=3$, $b=5 \implies$ Number = 35.

If $a=4$, $b=4 \implies$ Number = 44.

If $a=5$, $b=3 \implies$ Number = 53.

If $a=6$, $b=2 \implies$ Number = 62.

If $a=7$, $b=1 \implies$ Number = 71.

If $a=8$, $b=0 \implies$ Number = 80.

These are all the two-digit natural numbers whose digits sum to 8.

Therefore, the set C in roster form is {17, 26, 35, 44, 53, 62, 71, 80}.

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(iv) D = {x : x is a prime number which is divisor of 60}

First, find the prime factorization of 60.

$\begin{array}{c|cc} 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$60 = 2^2 \times 3 \times 5$.

The prime divisors of 60 are the prime numbers that appear in its prime factorization.

The prime divisors are 2, 3, and 5.

Therefore, the set D in roster form is {2, 3, 5}.

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(v) E = The set of all letters in the word TRIGONOMETRY

The letters in the word TRIGONOMETRY are T, R, I, G, O, N, O, M, E, T, R, Y.

In roster form, we list each distinct letter only once.

The distinct letters are T, R, I, G, O, N, M, E, Y.

Therefore, the set E in roster form is {T, R, I, G, O, N, M, E, Y}.

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(vi) F = The set of all letters in the word BETTER

The letters in the word BETTER are B, E, T, T, E, R.

In roster form, we list each distinct letter only once.

The distinct letters are B, E, T, R.

Therefore, the set F in roster form is {B, E, T, R}.

(i) {3, 6, 9, 12}

We observe that the elements are consecutive multiples of 3, starting from $3 \times 1$ and ending at $3 \times 4$.

$3 = 3 \times 1$

$6 = 3 \times 2$

$9 = 3 \times 3$

$12 = 3 \times 4$

So, each element $x$ can be written as $x = 3n$, where $n$ is a natural number ($\mathbb{N}$) and $1 \le n \le 4$.

Therefore, the set-builder form is $\{x : x = 3n, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}$.

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(ii) {2, 4, 8, 16, 32}

We observe that the elements are powers of 2, starting from $2^1$ and ending at $2^5$.

$2 = 2^1$

$4 = 2^2$

$8 = 2^3$

$16 = 2^4$

$32 = 2^5$

So, each element $x$ can be written as $x = 2^n$, where $n$ is a natural number ($\mathbb{N}$) and $1 \le n \le 5$.

Therefore, the set-builder form is $\{x : x = 2^n, n \in \mathbb{N} \text{ and } 1 \le n \le 5\}$.

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(iii) {5, 25, 125, 625}

We observe that the elements are powers of 5, starting from $5^1$ and ending at $5^4$.

$5 = 5^1$

$25 = 5^2$

$125 = 5^3$

$625 = 5^4$

So, each element $x$ can be written as $x = 5^n$, where $n$ is a natural number ($\mathbb{N}$) and $1 \le n \le 4$.

Therefore, the set-builder form is $\{x : x = 5^n, n \in \mathbb{N} \text{ and } 1 \le n \le 4\}$.

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(iv) {2, 4, 6, . . .}

We observe that the elements are all the positive even integers (or even natural numbers).

Each element $x$ can be written as $x = 2n$, where $n$ is any natural number ($\mathbb{N}$).

Therefore, the set-builder form is $\{x : x = 2n, n \in \mathbb{N}\}$.

Alternatively, it can be written as $\{x : x \text{ is an even natural number}\}$.

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(v) {1, 4, 9, . . . ,100}

We observe that the elements are the squares of the first 10 natural numbers.

$1 = 1^2$

$4 = 2^2$

$9 = 3^2$

...

$100 = 10^2$

So, each element $x$ can be written as $x = n^2$, where $n$ is a natural number ($\mathbb{N}$) and $1 \le n \le 10$.

Therefore, the set-builder form is $\{x : x = n^2, n \in \mathbb{N} \text{ and } 1 \le n \le 10\}$.

(i) A = {x : x is an odd natural number}

Natural numbers are 1, 2, 3, 4, 5, ...

Odd natural numbers are those natural numbers which are not divisible by 2.

The elements are 1, 3, 5, 7, 9, ...

Since the set is infinite, we represent it using ellipsis.

Listing the elements: A = {1, 3, 5, 7, ...}

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(ii) B = {x : x is an integer, $-\frac{1}{2} < x < \frac{9}{2}$}

We are looking for integers $x$ that satisfy the inequality $-\frac{1}{2} < x < \frac{9}{2}$.

Converting the fractions to decimals, we have $-0.5 < x < 4.5$.

The integers strictly greater than -0.5 are 0, 1, 2, 3, 4, ...

The integers strictly less than 4.5 are ..., 2, 3, 4.

The integers satisfying both conditions are 0, 1, 2, 3, 4.

Listing the elements: B = {0, 1, 2, 3, 4}

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(iii) C = {x : x is an integer, x² ≤ 4}

We are looking for integers $x$ whose square is less than or equal to 4.

Let's test some integers:

If $x = 0$, $x^2 = 0^2 = 0$. Since $0 \le 4$, 0 is in the set.

If $x = 1$, $x^2 = 1^2 = 1$. Since $1 \le 4$, 1 is in the set.

If $x = -1$, $x^2 = (-1)^2 = 1$. Since $1 \le 4$, -1 is in the set.

If $x = 2$, $x^2 = 2^2 = 4$. Since $4 \le 4$, 2 is in the set.

If $x = -2$, $x^2 = (-2)^2 = 4$. Since $4 \le 4$, -2 is in the set.

If $x = 3$, $x^2 = 3^2 = 9$. Since $9 \not\le 4$, 3 is not in the set.

If $x = -3$, $x^2 = (-3)^2 = 9$. Since $9 \not\le 4$, -3 is not in the set.

Any integer with absolute value greater than 2 will have a square greater than 4.

The integers satisfying the condition are -2, -1, 0, 1, 2.

Listing the elements: C = {-2, -1, 0, 1, 2}

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(iv) D = {x : x is a letter in the word “LOYAL”}

The letters in the word LOYAL are L, O, Y, A, L.

A set lists each element only once.

The distinct letters are L, O, Y, A.

Listing the elements: D = {L, O, Y, A}

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(v) E = {x : x is a month of a year not having 31 days}

The months of the year are: January (31), February (28 or 29), March (31), April (30), May (31), June (30), July (31), August (31), September (30), October (31), November (30), December (31).

The months not having 31 days are those with 30 days or 28/29 days.

These months are February, April, June, September, November.

Listing the elements: E = {February, April, June, September, November}

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(vi) F = {x : x is a consonant in the English alphabet which precedes k }.

The letters in the English alphabet that precede (come before) 'k' are a, b, c, d, e, f, g, h, i, j.

The vowels in this list are a, e, i.

The consonants in this list are the letters that are not vowels: b, c, d, f, g, h, j.

Listing the elements: F = {b, c, d, f, g, h, j}

We need to match each set given in roster form (left side) with its corresponding description in set-builder form (right side).

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Matching (i) {1, 2, 3, 6}:

Let's analyze the descriptions on the right:

(a) {x : x is a prime number and a divisor of 6}: Divisors of 6 are 1, 2, 3, 6. Prime divisors are 2, 3. Set is {2, 3}. Doesn't match.

(b) {x : x is an odd natural number less than 10}: Odd natural numbers less than 10 are 1, 3, 5, 7, 9. Set is {1, 3, 5, 7, 9}. Doesn't match.

(c) {x : x is natural number and divisor of 6}: Natural number divisors of 6 are 1, 2, 3, 6. Set is {1, 2, 3, 6}. Matches.

(d) {x : x is a letter of the word MATHEMATICS}: Set is {M, A, T, H, E, I, C, S}. Doesn't match.

So, (i) matches with (c).

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Matching (ii) {2, 3}:

From our analysis above:

(a) {x : x is a prime number and a divisor of 6}: Prime divisors of 6 are 2, 3. Set is {2, 3}. Matches.

So, (ii) matches with (a).

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Matching (iii) {M, A, T, H, E, I, C, S}:

From our analysis above:

(d) {x : x is a letter of the word MATHEMATICS}: The letters in MATHEMATICS are M, A, T, H, E, M, A, T, I, C, S. The set of distinct letters is {M, A, T, H, E, I, C, S}. Matches.

So, (iii) matches with (d).

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Matching (iv) {1, 3, 5, 7, 9}:

From our analysis above:

(b) {x : x is an odd natural number less than 10}: Odd natural numbers less than 10 are 1, 3, 5, 7, 9. Set is {1, 3, 5, 7, 9}. Matches.

So, (iv) matches with (b).

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Summary of Matches:

Roster Form	Set-Builder Form
(i) {1, 2, 3, 6}	(c) {x : x is natural number and divisor of 6}
(ii) {2, 3}	(a) {x : x is a prime number and a divisor of 6}
(iii) {M, A, T, H, E, I, C, S}	(d) {x : x is a letter of the word MATHEMATICS}
(iv) {1, 3, 5, 7, 9}	(b) {x : x is an odd natural number less than 10}

A set is finite if it is empty or consists of a definite number of elements. Otherwise, the set is called infinite.

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(i) {x : x ∈ N and (x – 1) (x –2) = 0}

The given set contains natural numbers $x$ such that $(x - 1)(x - 2) = 0$.

Solving the equation $(x - 1)(x - 2) = 0$, we get $x - 1 = 0$ or $x - 2 = 0$.

This gives $x = 1$ or $x = 2$.

Both 1 and 2 are natural numbers (elements of N).

So, the set is {1, 2}.

This set has a definite number of elements (two elements).

Therefore, the set is finite.

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(ii) {x : x ∈ N and x² = 4}

The given set contains natural numbers $x$ such that $x^2 = 4$.

Solving the equation $x^2 = 4$, we get $x = \pm\sqrt{4}$, so $x = 2$ or $x = -2$.

We require $x$ to be a natural number ($x \in N$).

Out of the solutions {2, -2}, only 2 is a natural number.

So, the set is {2}.

This set has a definite number of elements (one element).

Therefore, the set is finite.

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(iii) {x : x ∈ N and 2x –1 = 0}

The given set contains natural numbers $x$ such that $2x - 1 = 0$.

Solving the equation $2x - 1 = 0$, we get $2x = 1$, so $x = \frac{1}{2}$.

We require $x$ to be a natural number ($x \in N$).

However, $x = \frac{1}{2}$ is not a natural number.

There are no natural numbers that satisfy the given condition.

So, the set is the empty set, { } or $\emptyset$.

The empty set has 0 elements, which is a definite number.

Therefore, the set is finite.

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(iv) {x : x ∈ N and x is prime}

The given set contains all natural numbers $x$ such that $x$ is a prime number.

This set is {2, 3, 5, 7, 11, 13, 17, 19, ...}.

There are infinitely many prime numbers (as proven by Euclid). The list of prime numbers goes on forever.

The set does not have a definite number of elements.

Therefore, the set is infinite.

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(v) {x : x ∈ N and x is odd}

The given set contains all natural numbers $x$ such that $x$ is odd.

This set is {1, 3, 5, 7, 9, 11, ...}.

The list of odd natural numbers continues indefinitely. For any odd number $n$, $n+2$ is also an odd number.

The set does not have a definite number of elements.

Therefore, the set is infinite.

Two sets are said to be equal if they have exactly the same elements.

Let's determine the elements of each set:

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Set A:

A = {0}. This set contains only the element 0.

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Set B:

B = {x : x > 15 and x < 5}.

This set contains numbers $x$ that are simultaneously greater than 15 and less than 5.

There is no number that satisfies both conditions.

Therefore, B is the empty set. B = {} or $B = \emptyset$.

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Set C:

C = {x : x – 5 = 0 }.

We solve the equation $x - 5 = 0$.

$x = 5$.

Therefore, C = {5}.

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Set D:

D = {x: x² = 25}.

We solve the equation $x^2 = 25$.

$x = \pm\sqrt{25}$.

$x = 5$ or $x = -5$.

Therefore, D = {-5, 5}.

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Set E:

E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}.

First, we solve the quadratic equation $x^2 - 2x - 15 = 0$.

We can factor the quadratic: $(x - 5)(x + 3) = 0$.

The roots are $x = 5$ and $x = -3$.

The set E requires the roots to be:

1. Integral (integers): Both 5 and -3 are integers.

2. Positive: Only 5 is positive.

So, the only element satisfying both conditions is 5.

Therefore, E = {5}.

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Comparison of Sets:

A = {0}

B = {}

C = {5}

D = {-5, 5}

E = {5}

Comparing these sets, we see that set C and set E have exactly the same element, which is 5.

Conclusion:

The only pair of equal sets is C = E.

Reason: Both sets C and E contain only one element, which is 5. Set C is defined by the solution to $x-5=0$, which is $x=5$. Set E is defined as the positive integer root of $x^2-2x-15=0$; the roots are 5 and -3, and the only positive integer root is 5. Since both sets contain exactly the same element, they are equal.

Two sets are considered equal if they contain exactly the same elements, regardless of the order in which the elements are listed.

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(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.

First, let's find the set X by listing the distinct letters in the word "ALLOY".

The letters are A, L, L, O, Y.

The set of distinct letters is X = {A, L, O, Y}.

Next, let's find the set B by listing the distinct letters in the word "LOYAL".

The letters are L, O, Y, A, L.

The set of distinct letters is B = {L, O, Y, A}.

Now we compare set X and set B.

X = {A, L, O, Y}

B = {L, O, Y, A}

Since the order of elements does not matter in a set, both sets contain exactly the same elements: A, L, O, and Y.

Justification: Both sets consist of the same collection of letters {A, L, O, Y}.

Therefore, X and B are equal sets (X = B).

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(ii) A = {n : n ∈ Z and n² ≤ 4} and B = {x : x ∈ R and x² – 3x + 2 = 0}.

First, let's find the elements of set A.

A contains integers $n$ (where Z denotes the set of integers) such that $n^2$ is less than or equal to 4.

We test integers:

If $n=0$, $n^2=0$. $0 \le 4$, so 0 ∈ A.

If $n=1$, $n^2=1$. $1 \le 4$, so 1 ∈ A.

If $n=-1$, $n^2=1$. $1 \le 4$, so -1 ∈ A.

If $n=2$, $n^2=4$. $4 \le 4$, so 2 ∈ A.

If $n=-2$, $n^2=4$. $4 \le 4$, so -2 ∈ A.

If $n=3$, $n^2=9$. $9 \not\le 4$.

If $n=-3$, $n^2=9$. $9 \not\le 4$.

Any integer with an absolute value greater than 2 will have a square greater than 4.

So, the integers satisfying $n^2 \le 4$ are -2, -1, 0, 1, 2.

Therefore, A = {-2, -1, 0, 1, 2}.

Next, let's find the elements of set B.

B contains real numbers $x$ (where R denotes the set of real numbers) such that $x^2 - 3x + 2 = 0$.

We solve the quadratic equation $x^2 - 3x + 2 = 0$.

Factoring the quadratic, we get $(x - 1)(x - 2) = 0$.

The solutions are $x = 1$ and $x = 2$.

Both 1 and 2 are real numbers.

Therefore, B = {1, 2}.

Now we compare set A and set B.

A = {-2, -1, 0, 1, 2}

B = {1, 2}

The sets A and B do not contain the same elements. For instance, 0 is an element of A but not an element of B. Also, -1 and -2 are in A but not in B.

Justification: The elements of set A are {-2, -1, 0, 1, 2}, while the elements of set B are {1, 2}. Since the sets do not contain exactly the same elements, they are not equal.

Therefore, A and B are not equal sets (A ≠ B).

The null set (also called the empty set) is the set which contains no elements. It is denoted by $\emptyset$ or { }.

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(i) Set of odd natural numbers divisible by 2

Justification: Natural numbers are {1, 2, 3, 4, ...}. Odd natural numbers are {1, 3, 5, 7, ...}. A number is divisible by 2 if it is an even number. By definition, an odd number is a number that is not divisible by 2. Therefore, there cannot be any odd natural number that is also divisible by 2.

Conclusion: This set contains no elements. Thus, it is an example of the null set.

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(ii) Set of even prime numbers

Justification: Prime numbers are natural numbers greater than 1 whose only divisors are 1 and themselves ({2, 3, 5, 7, ...}). Even numbers are integers divisible by 2 ({..., -2, 0, 2, 4, ...}). We are looking for numbers that are both even and prime. The number 2 is prime (divisors are 1 and 2) and it is also even. Any other even number greater than 2 (like 4, 6, 8, ...) is divisible by 2, and hence has more than two divisors (1, 2, and itself), so it cannot be prime.

Conclusion: This set contains exactly one element, which is 2. The set is {2}. Since it contains an element, it is not an example of the null set.

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(iii) { x : x is a natural number, x < 5 and x > 7 }

Justification: We are looking for a natural number $x$ that satisfies two conditions simultaneously: $x < 5$ and $x > 7$. Natural numbers less than 5 are {1, 2, 3, 4}. Natural numbers greater than 7 are {8, 9, 10, ...}. There is no natural number which is both less than 5 and greater than 7 at the same time.

Conclusion: This set contains no elements. Thus, it is an example of the null set.

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(iv) { y : y is a point common to any two parallel lines}

Justification: Parallel lines are defined as lines in a plane that never intersect, no matter how far they are extended. If they never intersect, they have no points in common.

Conclusion: This set contains no elements, as there is no point $y$ that can be common to two distinct parallel lines. Thus, it is an example of the null set.

A set is called finite if it consists of a definite number of elements (i.e., the counting process of elements can come to an end). Otherwise, the set is called infinite.

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(i) The set of months of a year

Justification: There are exactly 12 months in a year (January, February, ..., December). The number of elements is definite and countable.

Conclusion: This set is finite.

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(ii) {1, 2, 3, . . .}

Justification: This set represents the set of all natural numbers. The ellipsis (...) indicates that the list continues indefinitely. There is no largest natural number, so the number of elements is unlimited.

Conclusion: This set is infinite.

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(iii) {1, 2, 3, . . .99, 100}

Justification: This set represents the natural numbers from 1 to 100, inclusive. There is a specific first element (1) and a specific last element (100). The number of elements is exactly 100, which is a definite number.

Conclusion: This set is finite.

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(iv) The set of positive integers greater than 100

Justification: This set is {101, 102, 103, ...}. It includes all integers starting from 101 and continuing without any upper limit. The number of elements is unlimited.

Conclusion: This set is infinite.

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(v) The set of prime numbers less than 99

Justification: Prime numbers are natural numbers greater than 1 with only two divisors: 1 and themselves. The condition "less than 99" sets an upper bound. The prime numbers less than 99 are {2, 3, 5, 7, ..., 97}. Although we might need to list them to find the exact count, we know that the number of primes below any given number is finite. The elements are definite and countable.

Conclusion: This set is finite.

A set is finite if it contains a definite number of elements or is empty. A set is infinite if the number of elements is unlimited.

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(i) The set of lines which are parallel to the x-axis

Justification: A line parallel to the x-axis has the equation $y = c$, where $c$ is a real constant representing the y-intercept. Since there are infinitely many possible real values for $c$, there are infinitely many distinct lines parallel to the x-axis.

Conclusion: This set is infinite.

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(ii) The set of letters in the English alphabet

Justification: The English alphabet consists of 26 distinct letters (A, B, ..., Z). The number of elements is fixed and countable.

Conclusion: This set is finite.

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(iii) The set of numbers which are multiple of 5

Justification: This set includes all numbers that can be expressed as $5n$, where $n$ is an integer. The set is {..., -15, -10, -5, 0, 5, 10, 15, ...}. This list continues indefinitely in both the positive and negative directions. The number of elements is unlimited.

Conclusion: This set is infinite.

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(iv) The set of animals living on the earth

Justification: Although the total number of animals living on Earth is very large and practically difficult to count exactly, it is a definite, countable number at any specific point in time. The number is not conceptually unlimited like the set of integers or real numbers.

Conclusion: This set is finite.

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(v) The set of circles passing through the origin (0, 0)

Justification: A circle is defined by its center $(h, k)$ and radius $r$. For a circle to pass through the origin $(0, 0)$, its equation $(x-h)^2 + (y-k)^2 = r^2$ must satisfy $h^2 + k^2 = r^2$. We can choose infinitely many points in the plane to serve as the center $(h, k)$ (e.g., (1,0), (2,0), (0,1), (1,1), etc.). For each distinct center (other than the origin itself), there is a unique radius $r = \sqrt{h^2+k^2}$ that defines a circle passing through the origin. Since there are infinitely many possible centers, there are infinitely many such circles.

Conclusion: This set is infinite.

Two sets A and B are equal (A = B) if they have exactly the same elements. The order in which the elements are listed does not matter.

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(i) A = { a, b, c, d }; B = { d, c, b, a }

Set A contains the elements a, b, c, d.

Set B contains the elements d, c, b, a.

Comparing the elements, we see that both sets contain the same four elements: a, b, c, and d.

Therefore, A = B.

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(ii) A = { 4, 8, 12, 16 }; B = { 8, 4, 16, 18}

Set A contains the elements 4, 8, 12, 16.

Set B contains the elements 8, 4, 16, 18.

Let's compare the elements:

4 ∈ A and 4 ∈ B.

8 ∈ A and 8 ∈ B.

12 ∈ A but 12 ∉ B.

16 ∈ A and 16 ∈ B.

18 ∈ B but 18 ∉ A.

Since the elements are not exactly the same (A contains 12 while B contains 18), the sets are not equal.

Therefore, A ≠ B.

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(iii) A = {2, 4, 6, 8, 10}; B = { x : x is positive even integer and x ≤ 10}

Set A is given in roster form: A = {2, 4, 6, 8, 10}.

Set B is given in set-builder form. We need to list its elements.

The condition "x is positive even integer" means x can be 2, 4, 6, 8, 10, 12, ...

The condition "x ≤ 10" means x must be less than or equal to 10.

Combining these conditions, the elements of B are the positive even integers that are less than or equal to 10. These are 2, 4, 6, 8, 10.

So, B = {2, 4, 6, 8, 10}.

Comparing A and B, we find they contain exactly the same elements.

Therefore, A = B.

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(iv) A = { x : x is a multiple of 10}; B = { 10, 15, 20, 25, 30, . . . }

Set A is given in set-builder form. The multiples of 10 are numbers that can be expressed as $10n$, where $n$ is an integer. Assuming $x$ refers to positive multiples (implied by comparison with set B), A = {10, 20, 30, 40, 50, ...}.

Set B is given in roster form: B = {10, 15, 20, 25, 30, ...}. The elements listed are 10, 15, 20, 25, 30. The pattern suggests these are multiples of 5 starting from 10.

Let's compare the elements:

10 ∈ A and 10 ∈ B.

15 ∉ A (15 is not a multiple of 10) but 15 ∈ B.

20 ∈ A and 20 ∈ B.

25 ∉ A (25 is not a multiple of 10) but 25 ∈ B.

Since set B contains elements (like 15, 25) that are not present in set A, the sets are not equal.

Therefore, A ≠ B.

Two sets are equal if they contain exactly the same elements.

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(i) A = {2, 3}, B = {x : x is solution of x² + 5x + 6 = 0}

Set A is given as A = {2, 3}.

To find the elements of set B, we need to solve the quadratic equation $x^2 + 5x + 6 = 0$.

We can factor the quadratic equation:

$x^2 + 2x + 3x + 6 = 0$

$x(x + 2) + 3(x + 2) = 0$

$(x + 2)(x + 3) = 0$

The solutions are $x + 2 = 0$ or $x + 3 = 0$.

This gives $x = -2$ or $x = -3$.

So, the set B = {-2, -3}.

Now, we compare set A = {2, 3} and set B = {-2, -3}.

The elements of set A are 2 and 3.

The elements of set B are -2 and -3.

Reason: The elements in set A (2 and 3) are different from the elements in set B (-2 and -3). Since the sets do not contain exactly the same elements, they are not equal.

Therefore, A and B are not equal sets (A ≠ B).

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(ii) A = { x : x is a letter in the word FOLLOW}, B = { y : y is a letter in the word WOLF}

To find the elements of set A, we list the distinct letters in the word "FOLLOW".

The letters are F, O, L, L, O, W.

The set of distinct letters is A = {F, O, L, W}.

To find the elements of set B, we list the distinct letters in the word "WOLF".

The letters are W, O, L, F.

The set of distinct letters is B = {W, O, L, F}.

Now, we compare set A = {F, O, L, W} and set B = {W, O, L, F}.

Both sets contain the same four letters: F, O, L, and W. The order of elements does not matter in a set.

Reason: Both sets contain exactly the same elements {F, O, L, W}.

Therefore, A and B are equal sets (A = B).

We are given the following sets:

A = { 2, 4, 8, 12}

B = { 1, 2, 3, 4}

C = { 4, 8, 12, 14}

D = { 3, 1, 4, 2}

E = {–1, 1}

F = { 0, a}

G = {1, –1}

H = { 0, 1}

Two sets are equal if they contain exactly the same elements, irrespective of the order in which the elements are listed.

We compare the sets to find pairs that have identical elements:

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Comparing B and D:

B = { 1, 2, 3, 4}

D = { 3, 1, 4, 2}

Both sets contain the elements 1, 2, 3, and 4. The order is different, but the elements are identical.

Thus, B = D.

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Comparing E and G:

E = {–1, 1}

G = {1, –1}

Both sets contain the elements -1 and 1. The order is different, but the elements are identical.

Thus, E = G.

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Let's check other pairs quickly:

A ≠ B, A ≠ C, A ≠ D (different elements)

B ≠ C (different elements)

C ≠ D (different elements)

E ≠ F (unless a = -1, but F also contains 0)

E ≠ H (-1 is in E but not H; 0 is in H but not E)

F ≠ G (different elements, unless a=-1 and 0=1 which is false)

F ≠ H (unless a = 1)

G ≠ H (-1 is in G but not H; 0 is in H but not G)

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Therefore, the pairs of equal sets selected from the given sets are:

B = D

E = G

We are given the following sets:

Φ (the empty set)

A = { 1, 3 }

B = {1, 5, 9}

C = {1, 3, 5, 7, 9}

The symbol ⊂ means "is a subset of". A set X is a subset of a set Y (X ⊂ Y) if every element of X is also an element of Y.

The symbol ⊄ means "is not a subset of". A set X is not a subset of a set Y (X ⊄ Y) if there is at least one element in X that is not in Y.

Recall that the empty set Φ is a subset of every set.

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(i) φ . . . B

The empty set Φ contains no elements. By definition, the empty set is a subset of every set, including B = {1, 5, 9}.

Therefore, φ ⊂ B.

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(ii) A . . . B

We compare A = {1, 3} and B = {1, 5, 9}.

We check if every element of A is also an element of B.

The element 1 is in A and also in B.

The element 3 is in A, but it is not in B.

Since not all elements of A are in B, A is not a subset of B.

Therefore, A ⊄ B.

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(iii) A . . . C

We compare A = {1, 3} and C = {1, 3, 5, 7, 9}.

We check if every element of A is also an element of C.

The element 1 is in A and also in C.

The element 3 is in A and also in C.

Since all elements of A are present in C, A is a subset of C.

Therefore, A ⊂ C.

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(iv) B . . . C

We compare B = {1, 5, 9} and C = {1, 3, 5, 7, 9}.

We check if every element of B is also an element of C.

The element 1 is in B and also in C.

The element 5 is in B and also in C.

The element 9 is in B and also in C.

Since all elements of B are present in C, B is a subset of C.

Therefore, B ⊂ C.

We are given the sets:

A = { a, e, i, o, u }

B = { a, b, c, d }

A set X is a subset of a set Y (denoted $X \subset Y$) if every element of X is also an element of Y.

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Is A a subset of B? No.

Why?

To check if A is a subset of B, we must verify if every element of A is present in B.

Let's check the elements of A:

The element 'a' is in A, and 'a' is also in B.

The element 'e' is in A, but 'e' is not in B.

Since we found at least one element ('e') in A that is not present in B, the condition for A being a subset of B is not met.

Therefore, A is not a subset of B (A ⊄ B) because not all elements of A are elements of B. For example, e ∈ A but e ∉ B.

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Is B a subset of A? No.

Why?

To check if B is a subset of A, we must verify if every element of B is present in A.

Let's check the elements of B:

The element 'a' is in B, and 'a' is also in A.

The element 'b' is in B, but 'b' is not in A.

Since we found at least one element ('b') in B that is not present in A, the condition for B being a subset of A is not met.

Therefore, B is not a subset of A (B ⊄ A) because not all elements of B are elements of A. For example, b ∈ B but b ∉ A.

The statement "If A ∈ B and B ⊂ C, then A ⊂ C" is not true in general.

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Reasoning:

The condition $A \in B$ means that the set A itself is considered as a single element within the set B.

The condition $B \subset C$ means that every element of B is also an element of C. Since A is an element of B ($A \in B$), it follows from $B \subset C$ that A must also be an element of C ($A \in C$).

However, the conclusion $A \subset C$ requires that every element *of* A must also be an element of C. Knowing that A *itself* is an element of C ($A \in C$) does not guarantee that the *elements within* A are also elements of C.

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Counterexample:

Let's define the sets A, B, and C as follows:

Let A = {1}

Let B = {{1}, 2}

Let C = {{1}, 2, 3}

Now, let's check the given conditions:

1. Is A ∈ B?

A = {1}. The elements of B are {1} and 2. Yes, the set {1} is an element of B.

So, A ∈ B is true.

2. Is B ⊂ C?

The elements of B are {1} and 2.

The elements of C are {1}, 2, and 3.

Is every element of B also an element of C? Yes, {1} is in C, and 2 is in C.

So, B ⊂ C is true.

Now, let's check the conclusion:

Is A ⊂ C?

A = {1}.

C = {{1}, 2, 3}.

For A to be a subset of C, every element of A must also be an element of C.

The only element of A is 1.

Is 1 an element of C? No. The elements of C are the set {1}, the number 2, and the number 3. The number 1 is not listed as an element of C.

So, A ⊂ C is false.

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Conclusion:

We have found an example where A ∈ B and B ⊂ C are both true, but A ⊂ C is false. Therefore, the original statement is not necessarily true.

The symbol ⊂ denotes "is a subset of", meaning every element of the first set is also an element of the second set.

The symbol ⊄ denotes "is not a subset of", meaning there is at least one element in the first set that is not in the second set.

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(i) { 2, 3, 4 } . . . { 1, 2, 3, 4, 5 }

Reasoning: All elements of the first set {2, 3, 4}, namely 2, 3, and 4, are also present in the second set {1, 2, 3, 4, 5}.

Conclusion: { 2, 3, 4 } ⊂ { 1, 2, 3, 4, 5 }

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(ii) { a, b, c } . . . { b, c, d }

Reasoning: The element 'a' is in the first set {a, b, c} but is not present in the second set {b, c, d}. Since not all elements of the first set are in the second set, it is not a subset.

Conclusion: { a, b, c } ⊄ { b, c, d }

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(iii) {x : x is a student of Class XI of your school}. . .{x : x student of your school}

Reasoning: Every student who is in Class XI of a particular school is necessarily a student of that school. Therefore, every element of the first set is also an element of the second set.

Conclusion: {x : x is a student of Class XI of your school} ⊂ {x : x student of your school}

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(iv) {x : x is a circle in the plane} . . .{x : x is a circle in the same plane with radius 1 unit}

Reasoning: The first set includes all circles in the plane, regardless of radius. The second set includes only circles with a specific radius (1 unit). A circle with radius 2 units is an element of the first set but not an element of the second

(i) { a, b } ⊄ { b, c, a }

Reasoning: A set X is a subset of Y ($X \subset Y$) if every element of X is also in Y. Here, the first set is {a, b} and the second set is {b, c, a}. The element 'a' from the first set is in the second set. The element 'b' from the first set is also in the second set. Since all elements of {a, b} are in {b, c, a}, {a, b} is a subset of {b, c, a}, i.e., {a, b} ⊂ {b, c, a}. The statement claims it is *not* a subset (⊄).

Conclusion: The statement is False.

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(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet}

Reasoning: The second set is the set of vowels in the English alphabet, which is {a, e, i, o, u}. The first set is {a, e}. We need to check if every element of {a, e} is in {a, e, i, o, u}. Both 'a' and 'e' are elements of the set of vowels. Thus, {a, e} is a subset of the set of vowels.

Conclusion: The statement is True.

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(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }

Reasoning: We check if every element of the first set {1, 2, 3} is in the second set {1, 3, 5}. The element 1 is in the second set. The element 3 is in the second set. However, the element 2 is in the first set but not in the second set. Therefore, {1, 2, 3} is not a subset of {1, 3, 5}.

Conclusion: The statement is False.

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(iv) { a } ⊂ { a, b, c }

Reasoning: The first set is {a}. The second set is {a, b, c}. We check if every element of {a} is in {a, b, c}. The only element in the first set is 'a', which is also present in the second set. Therefore, {a} is a subset of {a, b, c}.

Conclusion: The statement is True.

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(v) { a } ∈ { a, b, c }

Reasoning: The symbol ∈ means "is an element of". The statement asks if the set {a} is an element of the set {a, b, c}. The elements of the set {a, b, c} are the items listed inside: 'a', 'b', and 'c'. The item {a} (which is a set containing 'a') is not listed as one of these elements. While 'a' is an element, the set {a} is not.

Conclusion: The statement is False.

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(vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}

Reasoning: First, let's list the elements of both sets.

Set 1 = { x : x is an even natural number less than 6} = {2, 4}.

Set 2 = { x : x is a natural number which divides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}.

Now we check if Set 1 is a subset of Set 2. Is every element of {2, 4} also an element of {1, 2, 3, 4, 6, 9, 12, 18, 36}?

The element 2 from Set 1 is present in Set 2.

The element 4 from Set 1 is present in Set 2.

Since all elements of Set 1 are in Set 2, Set 1 is a subset of Set 2.

Conclusion: The statement is True.

The given set is A = { 1, 2, { 3, 4 }, 5 }.

The elements of A are: 1, 2, the set {3, 4}, and 5.

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(i) {3, 4} ⊂ A

This statement is incorrect.

Why: For {3, 4} to be a subset of A, every element of {3, 4} must be an element of A. The elements of {3, 4} are 3 and 4. However, 3 is not an element of A, and 4 is not an element of A. The elements of A are 1, 2, {3, 4}, and 5. Note that {3, 4} itself is an element, but 3 and 4 individually are not.

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(ii) {3, 4} ∈ A

This statement is correct.

Why: The symbol ∈ denotes "is an element of". Looking at the elements of A, we see that the set {3, 4} is indeed listed as one of the elements of A.

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(iii) {{3, 4}} ⊂ A

This statement is correct.

Why: For {{3, 4}} to be a subset of A, every element of {{3, 4}} must be an element of A. The only element of {{3, 4}} is the set {3, 4}. As established in (ii), {3, 4} is an element of A. Since all elements of {{3, 4}} are in A, it is a subset of A.

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(iv) 1 ∈ A

This statement is correct.

Why: The number 1 is listed as one of the elements of A.

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(v) 1 ⊂ A

This statement is incorrect.

Why: The symbol ⊂ denotes "is a subset of". The subset relation holds between two sets. The number 1 is not a set, it is an element. Therefore, it cannot be a subset of A. (The correct statement relating the element 1 and the set A using the subset symbol would be {1} ⊂ A).

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(vi) {1, 2, 5} ⊂ A

This statement is correct.

Why: For {1, 2, 5} to be a subset of A, every element of {1, 2, 5} must be an element of A. The elements of {1, 2, 5} are 1, 2, and 5. Looking at A = { 1, 2, { 3, 4 }, 5 }, we see that 1, 2, and 5 are all elements of A. Therefore, {1, 2, 5} is a subset of A.

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(vii) {1, 2, 5} ∈ A

This statement is incorrect.

Why: The symbol ∈ denotes "is an element of". We need to check if the set {1, 2, 5} is listed as one of the elements of A = { 1, 2, { 3, 4 }, 5 }. It is not. The elements are 1, 2, {3, 4}, and 5. The set {1, 2, 5} is not one of these.

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(viii) {1, 2, 3} ⊂ A

This statement is incorrect.

Why: For {1, 2, 3} to be a subset of A, every element of {1, 2, 3} must be an element of A. The elements are 1, 2, and 3. While 1 and 2 are elements of A, 3 is not an element of A (only the set {3, 4} is an element). Since not all elements of {1, 2, 3} are in A, it is not a subset of A.

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(ix) φ ∈ A

This statement is incorrect.

Why: The symbol ∈ denotes "is an element of". We need to check if the empty set φ is listed as one of the elements of A = { 1, 2, { 3, 4 }, 5 }. It is not.

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(x) φ ⊂ A

This statement is correct.

Why: The empty set φ is a subset of every set by definition.

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(xi) {φ} ⊂ A

This statement is incorrect.

Why: For {φ} to be a subset of A, every element of {φ} must be an element of A. The only element of {φ} is φ. As established in (ix), φ is not an element of A. Therefore, {φ} is not a subset of A.

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Summary of Incorrect Statements: (i), (v), (vii), (viii), (ix), (xi).

Recall that for a set with $n$ elements, the total number of subsets is $2^n$. The subsets include the empty set (φ) and the set itself.

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(i) {a}

The given set is {a}.

Number of elements $n=1$.

Number of subsets = $2^1 = 2$.

The subsets are:

1. The empty set: φ

2. The set itself: {a}

So, the subsets of {a} are φ, {a}.

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(ii) {a, b}

The given set is {a, b}.

Number of elements $n=2$.

Number of subsets = $2^2 = 4$.

The subsets are:

1. The empty set: φ

2. Subsets with one element: {a}, {b}

3. Subsets with two elements (the set itself): {a, b}

So, the subsets of {a, b} are φ, {a}, {b}, {a, b}.

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(iii) {1, 2, 3}

The given set is {1, 2, 3}.

Number of elements $n=3$.

Number of subsets = $2^3 = 8$.

The subsets are:

1. The empty set: φ

2. Subsets with one element: {1}, {2}, {3}

3. Subsets with two elements: {1, 2}, {1, 3}, {2, 3}

4. Subsets with three elements (the set itself): {1, 2, 3}

So, the subsets of {1, 2, 3} are φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

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(iv) φ

The given set is the empty set φ.

Number of elements $n=0$.

Number of subsets = $2^0 = 1$.

The only subset of the empty set is the empty set itself.

So, the subset of φ is φ.

Given:

The set A is the empty set, $A = \phi$.

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To Find:

The number of elements in the power set of A, denoted as $P(A)$.

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Solution:

The power set $P(A)$ is defined as the set of all subsets of A.

The number of elements in a set A is denoted by $n(A)$.

Since $A = \phi$ (the empty set), it contains no elements.

Therefore, the number of elements in A is $n(A) = 0$.

The number of elements in the power set $P(A)$ is given by the formula $2^{n(A)}$, where $n(A)$ is the number of elements in set A.

Substituting the value of $n(A)$:

Number of elements in $P(A) = 2^{n(A)} = 2^0$.

We know that any non-zero number raised to the power of 0 is 1.

So, $2^0 = 1$.

Thus, the power set $P(A)$ has 1 element.

Specifically, the only subset of the empty set $\phi$ is the empty set $\phi$ itself. Therefore, $P(A) = P(\phi) = \{\phi\}$. The set $\{\phi\}$ contains one element, which is $\phi$.

Interval notation uses parentheses `(` `)` for endpoints that are not included (corresponding to $<$ or $>$) and square brackets `[` `]` for endpoints that are included (corresponding to $≤$ or $≥$). The notation lists the lower bound followed by the upper bound, separated by a comma.

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(i) {x : x ∈ R, – 4 < x ≤ 6}

The set includes all real numbers $x$ such that $x$ is strictly greater than -4 and less than or equal to 6.

The lower bound -4 is not included (use parenthesis `(`).

The upper bound 6 is included (use square bracket `]`).

The interval notation is (-4, 6].

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(ii) {x : x ∈ R, – 12 < x < –10}

The set includes all real numbers $x$ such that $x$ is strictly greater than -12 and strictly less than -10.

The lower bound -12 is not included (use parenthesis `(`).

The upper bound -10 is not included (use parenthesis `(`).

The interval notation is (-12, -10).

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(iii) {x : x ∈ R, 0 ≤ x < 7}

The set includes all real numbers $x$ such that $x$ is greater than or equal to 0 and strictly less than 7.

The lower bound 0 is included (use square bracket `[`).

The upper bound 7 is not included (use parenthesis `(`).

The interval notation is [0, 7).

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(iv) {x : x ∈ R, 3 ≤ x ≤ 4}

The set includes all real numbers $x$ such that $x$ is greater than or equal to 3 and less than or equal to 4.

The lower bound 3 is included (use square bracket `[`).

The upper bound 4 is included (use square bracket `]`).

The interval notation is [3, 4].

Set-builder form describes a set by stating the properties its elements must satisfy. For intervals of real numbers, this involves specifying the range using inequalities.

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(i) (– 3, 0)

This interval represents all real numbers $x$ that are strictly greater than -3 and strictly less than 0.

The condition is $-3 < x < 0$.

The set-builder form is $\{x : x \in \mathbb{R}, -3 < x < 0\}$.

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(ii) [6 , 12]

This interval represents all real numbers $x$ that are greater than or equal to 6 and less than or equal to 12.

The condition is $6 \le x \le 12$.

The set-builder form is $\{x : x \in \mathbb{R}, 6 \le x \le 12\}$.

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(iii) (6, 12]

This interval represents all real numbers $x$ that are strictly greater than 6 and less than or equal to 12.

The condition is $6 < x \le 12$.

The set-builder form is $\{x : x \in \mathbb{R}, 6 < x \le 12\}$.

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(iv) [–23, 5)

This interval represents all real numbers $x$ that are greater than or equal to -23 and strictly less than 5.

The condition is $-23 \le x < 5$.

The set-builder form is $\{x : x \in \mathbb{R}, -23 \le x < 5\}$.

A universal set for a given set must contain all the elements of that given set. We aim to propose a relevant universal set for the context.

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(i) The set of right triangles.

Right triangles are a specific type of triangle. Therefore, a suitable universal set would be one that encompasses all triangles.

Proposed Universal Set(s):

1. The set of all triangles. (This is the most immediate and specific superset).

2. The set of all polygons. (Since triangles are polygons).

3. The set of all two-dimensional geometric shapes. (A broader category).

Any of these could serve as a universal set, with the set of all triangles often being the most contextually appropriate.

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(ii) The set of isosceles triangles.

Isosceles triangles are also a specific type of triangle.

Proposed Universal Set(s):

Similar to the case of right triangles, suitable universal sets include:

1. The set of all triangles.

2. The set of all polygons.

3. The set of all two-dimensional geometric shapes.

Again, the set of all triangles is generally the most relevant universal set for the set of isosceles triangles.

Given Sets:

A = {1, 3, 5}

B = {2, 4, 6}

C = {0, 2, 4, 6, 8}

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Concept:

A set U is considered a universal set for sets A, B, and C if U contains all the elements present in A, all the elements present in B, and all the elements present in C. In other words, A must be a subset of U ($A \subset U$), B must be a subset of U ($B \subset U$), and C must be a subset of U ($C \subset U$).

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Analysis of Options:

First, let's find the union of A, B, and C to identify all the elements that must be present in a universal set for them:

$A \cup B \cup C = \{1, 3, 5\} \cup \{2, 4, 6\} \cup \{0, 2, 4, 6, 8\}$

$A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$

Any proposed universal set must contain all these elements: 0, 1, 2, 3, 4, 5, 6, 8.

Now, let's examine each option:

(i) {0, 1, 2, 3, 4, 5, 6}

This set contains all elements of A ({1, 3, 5}) and B ({2, 4, 6}).

However, it does not contain the element 8, which is present in set C.

Since $8 \in C$ but $8 \notin \{0, 1, 2, 3, 4, 5, 6\}$, this set cannot be the universal set for A, B, and C.

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(ii) φ

The empty set (φ) contains no elements. It cannot contain the elements of A, B, or C.

Therefore, φ cannot be the universal set.

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(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Let's check if this set contains all elements from $A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$.

Yes, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} contains 0, 1, 2, 3, 4, 5, 6, and 8.

Since it contains all elements of A, all elements of B, and all elements of C, this set can be considered a universal set for A, B, and C.

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(iv) {1, 2, 3, 4, 5, 6, 7, 8}

This set contains all elements of A ({1, 3, 5}) and B ({2, 4, 6}).

It also contains 8 from set C.

However, it does not contain the element 0, which is present in set C.

Since $0 \in C$ but $0 \notin \{1, 2, 3, 4, 5, 6, 7, 8\}$, this set cannot be the universal set for A, B, and C.

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Conclusion:

Only the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (Option iii) contains all the elements of sets A, B, and C, and therefore may be considered as a universal set for them.

Given:

Set A = { 2, 4, 6, 8 }

Set B = { 6, 8, 10, 12 }

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To Find:

The union of set A and set B, denoted as A ∪ B.

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Solution:

The union of two sets A and B, $A \cup B$, is the set containing all elements that are in A, or in B, or in both. We list all unique elements from both sets.

The elements of A are 2, 4, 6, 8.

The elements of B are 6, 8, 10, 12.

To find the union $A \cup B$, we combine all elements from A and B, making sure to list each element only once:

Start with the elements of A: {2, 4, 6, 8}.

Add the elements from B that are not already listed: 10, 12. (Note: 6 and 8 are already included from set A).

Combining these gives the set {2, 4, 6, 8, 10, 12}.

Therefore, $A \cup B = \{2, 4, 6, 8, 10, 12\}$.

The union of A and B is {2, 4, 6, 8, 10, 12}.

Given:

Set A = { a, e, i, o, u }

Set B = { a, i, u }

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To Show:

That the union of set A and set B is equal to set A, i.e., $A \cup B = A$.

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Proof:

The union of two sets, $A \cup B$, is the set containing all elements that are in set A, or in set B, or in both, with each element listed only once.

We start by listing the elements of A:

{ a, e, i, o, u }

Next, we consider the elements of B: { a, i, u }.

We add any elements from B that are not already in our list.

The element 'a' from B is already in the list.

The element 'i' from B is already in the list.

The element 'u' from B is already in the list.

Since all elements of B are already present in A, adding the elements of B to A does not introduce any new elements.

Therefore, the union $A \cup B$ is { a, e, i, o, u }.

We are given that A = { a, e, i, o, u }.

Comparing the result for $A \cup B$ with the set A, we see that:

$A \cup B = \{ a, e, i, o, u \}$

$A = \{ a, e, i, o, u \}$

The two sets are identical.

Hence, we have shown that $A \cup B = A$.

(This is a general property: If B is a subset of A ($B \subset A$), then $A \cup B = A$. In this case, B = {a, i, u} is indeed a subset of A = {a, e, i, o, u}).

Given:

Set X = {Ram, Geeta, Akbar} (Class XI students in the hockey team)

Set Y = {Geeta, David, Ashok} (Class XI students in the football team)

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To Find:

1. The union of sets X and Y, $X \cup Y$.

2. The interpretation of the set $X \cup Y$.

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Solution:

1. Finding $X \cup Y$:

The union $X \cup Y$ consists of all elements that are in X, or in Y, or in both. We list all unique elements from both sets.

Elements from X: Ram, Geeta, Akbar.

Elements from Y: Geeta, David, Ashok.

Combining these and listing each unique element once, we get:

$X \cup Y = \{\text{Ram, Geeta, Akbar, David, Ashok}\}$

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2. Interpretation of $X \cup Y$:

Set X represents the students in the hockey team.

Set Y represents the students in the football team.

The union $X \cup Y$ contains all students who belong to either set X or set Y or both.

Therefore, $X \cup Y = \{\text{Ram, Geeta, Akbar, David, Ashok}\}$ is the set of students from Class XI who are either in the school hockey team or in the school football team or in both teams.

Given:

From Example 12, the sets are:

Set A = { 2, 4, 6, 8 }

Set B = { 6, 8, 10, 12 }

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To Find:

The intersection of set A and set B, denoted as A ∩ B.

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Solution:

The intersection of two sets A and B, $A \cap B$, is the set containing all elements that are common to both A and B.

We need to identify the elements that are present in Set A AND also present in Set B.

Let's compare the elements:

Element 2: Is it in B? No.

Element 4: Is it in B? No.

Element 6: Is it in B? Yes.

Element 8: Is it in B? Yes.

Element 10 (from B): Is it in A? No.

Element 12 (from B): Is it in A? No.

The elements that are common to both sets are 6 and 8.

Therefore, the intersection $A \cap B$ is the set containing these common elements.

$A \cap B = \{6, 8\}$

The intersection of A and B is {6, 8}.

Given:

From Example 14, the sets are:

Set X = {Ram, Geeta, Akbar} (Class XI students in the hockey team)

Set Y = {Geeta, David, Ashok} (Class XI students in the football team)

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To Find:

The intersection of sets X and Y, denoted as X ∩ Y.

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Solution:

The intersection of two sets X and Y, $X \cap Y$, is the set containing all elements that are common to both X and Y.

We need to identify the students who are present in both the hockey team list (Set X) AND the football team list (Set Y).

Comparing the elements of X and Y:

X = {Ram, Geeta, Akbar}

Y = {Geeta, David, Ashok}

We look for names that appear in both lists.

Ram is only in X.

Geeta is in both X and Y.

Akbar is only in X.

David is only in Y.

Ashok is only in Y.

The only element common to both sets is Geeta.

Therefore, the intersection $X \cap Y$ is the set containing only Geeta.

$X \cap Y = \{\text{Geeta}\}$

The intersection represents the set of students from Class XI who are in both the hockey team and the football team. In this case, only Geeta is in both teams.

The intersection of X and Y is {Geeta}.

Given:

Set A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Set B = { 2, 3, 5, 7 }

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To Find:

1. The intersection $A \cap B$.

2. Show that $A \cap B = B$.

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Solution:

1. Finding $A \cap B$:

The intersection $A \cap B$ is the set of all elements that are common to both set A and set B.

We compare the elements of B with the elements of A:

The element 2 is in B. Is 2 in A? Yes.

The element 3 is in B. Is 3 in A? Yes.

The element 5 is in B. Is 5 in A? Yes.

The element 7 is in B. Is 7 in A? Yes.

The elements common to both A and B are 2, 3, 5, and 7.

Therefore, $A \cap B = \{2, 3, 5, 7\}$.

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2. Showing $A \cap B = B$:

We found that $A \cap B = \{2, 3, 5, 7\}$.

We are given that $B = \{ 2, 3, 5, 7 \}$.

Comparing the set $A \cap B$ with the set B, we see that they contain exactly the same elements.

Hence, $A \cap B = B$.

(This result occurs because every element of B is also an element of A, meaning B is a subset of A ($B \subset A$). In general, if $B \subset A$, then $A \cap B = B$.)

Given:

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

Set B = { 2, 4, 6, 8 }

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To Find:

1. The set difference $A - B$.

2. The set difference $B - A$.

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Solution:

1. Finding $A - B$:

The set difference $A - B$ consists of all elements that are in set A but are not in set B.

We start with the elements of A: {1, 2, 3, 4, 5, 6}.

We remove the elements from A that are also found in B: {2, 4, 6}.

The elements remaining in A after removing the common elements are {1, 3, 5}.

Therefore, $A - B = \{1, 3, 5\}$.

The set difference $A - B$ is {1, 3, 5}.

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2. Finding $B - A$:

The set difference $B - A$ consists of all elements that are in set B but are not in set A.

We start with the elements of B: {2, 4, 6, 8}.

We remove the elements from B that are also found in A: {2, 4, 6}.

The element remaining in B after removing the common elements is {8}.

Therefore, $B - A = \{8\}$.

The set difference $B - A$ is {8}.

Note that in general, $A - B \neq B - A$.

Given:

Set V = { a, e, i, o, u }

Set B = { a, i, k, u }

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To Find:

1. The set difference $V - B$.

2. The set difference $B - V$.

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Solution:

1. Finding $V - B$:

The set difference $V - B$ consists of all elements that are in set V but are not in set B.

The elements of V are {a, e, i, o, u}.

The elements of B are {a, i, k, u}.

We identify the elements of V that are also in B: {a, i, u}.

We remove these common elements from V.

The remaining elements in V are {e, o}.

Therefore, $V - B = \{e, o\}$.

The set difference $V - B$ is {e, o}.

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2. Finding $B - V$:

The set difference $B - V$ consists of all elements that are in set B but are not in set V.

The elements of B are {a, i, k, u}.

The elements of V are {a, e, i, o, u}.

We identify the elements of B that are also in V: {a, i, u}.

We remove these common elements from B.

The remaining element in B is {k}.

Therefore, $B - V = \{k\}$.

The set difference $B - V$ is {k}.

The union of two sets is the set containing all elements that are in either set (or both), with duplicates removed.

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(i) X = {1, 3, 5}, Y = {1, 2, 3}

To find $X \cup Y$, we list all unique elements from both sets:

Elements in X are 1, 3, 5.

Elements in Y are 1, 2, 3.

Combining them and removing duplicates gives {1, 2, 3, 5}.

Therefore, $X \cup Y = \{$1, 2, 3, 5$\}$.

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(ii) A = { a, e, i, o, u}, B = {a, b, c}

To find $A \cup B$, we list all unique elements from both sets:

Elements in A are a, e, i, o, u.

Elements in B are a, b, c.

Combining them and removing duplicates gives {a, b, c, e, i, o, u}.

Therefore, $A \cup B = \{$a, b, c, e, i, o, u$\}$.

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(iii) A = {x : x is a natural number and multiple of 3}, B = {x : x is a natural number less than 6}

First, list the elements of each set:

A = {3, 6, 9, 12, 15, ...}

B = {1, 2, 3, 4, 5}

To find $A \cup B$, we combine these elements:

Start with elements of B: {1, 2, 3, 4, 5}.

Add elements from A not already present: {6, 9, 12, 15, ...}.

So, $A \cup B = \{1, 2, 3, 4, 5, 6, 9, 12, 15, ...\}$.

This can be described as the set containing 1, 2, 4, 5 and all natural numbers which are multiples of 3.

Therefore, $A \cup B = \{$1, 2, 3, 4, 5, 6, 9, 12, ...$\}$ or $\{x : x \in \mathbb{N} \text{ and } (x < 6 \text{ or } x \text{ is a multiple of } 3)\}$.

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(iv) A = {x : x is a natural number and 1 < x ≤ 6 }, B = {x : x is a natural number and 6 < x < 10 }

First, list the elements of each set:

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

B = {7, 8, 9}

To find $A \cup B$, we combine these elements:

{2, 3, 4, 5, 6, 7, 8, 9}.

This set contains all natural numbers $x$ such that $1 < x < 10$.

Therefore, $A \cup B = \{$2, 3, 4, 5, 6, 7, 8, 9$\}$ or $\{x : x \in \mathbb{N} \text{ and } 1 < x < 10\}$.

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(v) A = {1, 2, 3}, B = φ

To find $A \cup B$, we combine the elements of A and B.

A = {1, 2, 3}

B = φ (empty set, contains no elements).

The union is simply the elements of A.

Therefore, $A \cup B = \{$1, 2, 3$\}$. (Note: $A \cup \phi = A$ for any set A).

Given:

Set A = { a, b }

Set B = { a, b, c }

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Part 1: Is A ⊂ B ?

A set X is a subset of a set Y ($X \subset Y$) if every element of X is also an element of Y.

We need to check if every element of A = {a, b} is also an element of B = {a, b, c}.

The element 'a' is in A, and 'a' is also in B.

The element 'b' is in A, and 'b' is also in B.

Since all elements of A are present in B, A is a subset of B.

Therefore, Yes, A ⊂ B.

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Part 2: What is A ∪ B ?

The union of two sets, $A \cup B$, is the set containing all elements that are in A, or in B, or in both, listing each unique element only once.

Elements of A are {a, b}.

Elements of B are {a, b, c}.

Combining all unique elements from A and B gives {a, b, c}.

Therefore, $A \cup B = \{a, b, c\}$.

We notice that $A \cup B$ is the same as set B.

The union $A \cup B$ is {a, b, c}.

Given:

A and B are two sets such that A is a subset of B, denoted as $A \subset B$.

---

To Find:

The union of A and B, $A \cup B$.

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Solution:

The condition $A \subset B$ means that every element which is in set A is also present in set B.

The union $A \cup B$ is the set containing all elements that belong to set A, or to set B, or to both.

Since every element of A is already contained within B, when we form the union $A \cup B$, we are essentially combining the elements of B with elements that are already in B (because they come from A).

Therefore, adding the elements of A to B does not introduce any new elements that were not already in B.

The resulting union set will contain exactly the same elements as set B.

Thus, if $A \subset B$, then $A \cup B = B$.

Example: Let A = {1, 2} and B = {1, 2, 3}. Here, A ⊂ B.

$A \cup B = \{1, 2\} \cup \{1, 2, 3\} = \{1, 2, 3\}$.

We see that $A \cup B = B$.

Given Sets:

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

C = {5, 6, 7, 8}

D = {7, 8, 9, 10}

The union ($ \cup $) of sets includes all the elements which are in either set, or in both, listed without repetition.

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(i) A ∪ B

We combine the elements of A and B:

$A \cup B = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\}$

$A \cup B = \{$1, 2, 3, 4, 5, 6$\}$

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(ii) A ∪ C

We combine the elements of A and C:

$A \cup C = \{1, 2, 3, 4\} \cup \{5, 6, 7, 8\}$

$A \cup C = \{$1, 2, 3, 4, 5, 6, 7, 8$\}$

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(iii) B ∪ C

We combine the elements of B and C:

$B \cup C = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\}$

$B \cup C = \{$3, 4, 5, 6, 7, 8$\}$

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(iv) B ∪ D

We combine the elements of B and D:

$B \cup D = \{3, 4, 5, 6\} \cup \{7, 8, 9, 10\}$

$B \cup D = \{$3, 4, 5, 6, 7, 8, 9, 10$\}$

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(v) A ∪ B ∪ C

We combine the elements of A, B, and C:

$A \cup B \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\}$

Taking all unique elements: {1, 2, 3, 4, 5, 6, 7, 8}.

$A \cup B \cup C = \{$1, 2, 3, 4, 5, 6, 7, 8$\}$

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(vi) A ∪ B ∪ D

We combine the elements of A, B, and D:

$A \cup B \cup D = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} \cup \{7, 8, 9, 10\}$

Taking all unique elements: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

$A \cup B \cup D = \{$1, 2, 3, 4, 5, 6, 7, 8, 9, 10$\}$

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(vii) B ∪ C ∪ D

We combine the elements of B, C, and D:

$B \cup C \cup D = \{3, 4, 5, 6\} \cup \{5, 6, 7, 8\} \cup \{7, 8, 9, 10\}$

Taking all unique elements: {3, 4, 5, 6, 7, 8, 9, 10}.

$B \cup C \cup D = \{$3, 4, 5, 6, 7, 8, 9, 10$\}$

The intersection ($ \cap $) of two sets is the set containing all elements that are common to both sets.

We refer to the pairs of sets defined in Question 1:

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(i) X = {1, 3, 5}, Y = {1, 2, 3}

We look for elements present in both X and Y.

1 is in both X and Y.

3 is in both X and Y.

2 is only in Y. 5 is only in X.

The common elements are 1 and 3.

Therefore, $X \cap Y = \{$1, 3$\}$.

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(ii) A = { a, e, i, o, u}, B = {a, b, c}

We look for elements present in both A and B.

'a' is in both A and B.

'e', 'i', 'o', 'u' are only in A.

'b', 'c' are only in B.

The only common element is 'a'.

Therefore, $A \cap B = \{$a$\}$.

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(iii) A = {x : x is a natural number and multiple of 3}, B = {x : x is a natural number less than 6}

First, list the elements (or first few elements) of each set:

A = {3, 6, 9, 12, 15, ...}

B = {1, 2, 3, 4, 5}

We look for elements present in both A and B.

Comparing the elements, the only number present in both sets is 3.

Therefore, $A \cap B = \{$3$\}$.

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(iv) A = {x : x is a natural number and 1 < x ≤ 6 }, B = {x : x is a natural number and 6 < x < 10 }

First, list the elements of each set:

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

B = {7, 8, 9}

We look for elements present in both A and B.

There are no elements common to both sets A and B.

Therefore, the intersection is the empty set.

$A \cap B = \{$$\}$ or φ.

---

(v) A = {1, 2, 3}, B = φ

We look for elements present in both A and B.

A = {1, 2, 3}

B = φ (the empty set)

Since B has no elements, there cannot be any elements common to both A and B.

Therefore, the intersection is the empty set.

$A \cap B = $ φ.

Given Sets:

A = { 3, 5, 7, 9, 11 }

B = { 7, 9, 11, 13 }

C = { 11, 13, 15 }

D = { 15, 17 }

The intersection ($ \cap $) of sets includes only the elements common to all the sets involved.

The union ($ \cup $) of sets includes all elements present in any of the sets involved, without repetition.

---

(i) A ∩ B

We find the elements common to A and B.

A = { 3, 5, 7, 9, 11 }

B = { 7, 9, 11, 13 }

The common elements are 7, 9, 11.

$A \cap B = \{$7, 9, 11$\}$.

---

(ii) B ∩ C

We find the elements common to B and C.

B = { 7, 9, 11, 13 }

C = { 11, 13, 15 }

The common elements are 11, 13.

$B \cap C = \{$11, 13$\}$.

---

(iii) A ∩ C ∩ D

We find the elements common to A, C, and D.

A = { 3, 5, 7, 9, 11 }

C = { 11, 13, 15 }

D = { 15, 17 }

First find $A \cap C = \{11\}$.

Then find $(A \cap C) \cap D = \{11\} \cap \{15, 17\}$.

There are no elements common to all three sets.

$A \cap C \cap D = \{$$\}$ or φ.

---

(iv) A ∩ C

We find the elements common to A and C.

A = { 3, 5, 7, 9, 11 }

C = { 11, 13, 15 }

The only common element is 11.

$A \cap C = \{$11$\}$.

---

(v) B ∩ D

We find the elements common to B and D.

B = { 7, 9, 11, 13 }

D = { 15, 17 }

There are no common elements.

$B \cap D = \{$$\}$ or φ.

---

(vi) A ∩ (B ∪ C)

First, find the union $B \cup C$.

$B \cup C = \{ 7, 9, 11, 13 \} \cup \{ 11, 13, 15 \} = \{ 7, 9, 11, 13, 15 \}$.

Now, find the intersection of A with this result.

$A \cap (B \cup C) = \{ 3, 5, \mathbf{7}, \mathbf{9}, \mathbf{11} \} \cap \{ \mathbf{7}, \mathbf{9}, \mathbf{11}, 13, 15 \}$.

The common elements are 7, 9, 11.

$A \cap (B \cup C) = \{$7, 9, 11$\}$.

---

(vii) A ∩ D

We find the elements common to A and D.

A = { 3, 5, 7, 9, 11 }

D = { 15, 17 }

There are no common elements.

$A \cap D = \{$$\}$ or φ.

---

(viii) A ∩ (B ∪ D)

First, find the union $B \cup D$.

$B \cup D = \{ 7, 9, 11, 13 \} \cup \{ 15, 17 \} = \{ 7, 9, 11, 13, 15, 17 \}$.

Now, find the intersection of A with this result.

$A \cap (B \cup D) = \{ 3, 5, \mathbf{7}, \mathbf{9}, \mathbf{11} \} \cap \{ \mathbf{7}, \mathbf{9}, \mathbf{11}, 13, 15, 17 \}$.

The common elements are 7, 9, 11.

$A \cap (B \cup D) = \{$7, 9, 11$\}$.

---

(ix) ( A ∩ B ) ∩ ( B ∪ C )

First, find $A \cap B$. From part (i), $A \cap B = \{ 7, 9, 11 \}$.

Next, find $B \cup C$. From part (vi), $B \cup C = \{ 7, 9, 11, 13, 15 \}$.

Now, find the intersection of these two results.

$( A \cap B ) \cap ( B \cup C ) = \{ \mathbf{7}, \mathbf{9}, \mathbf{11} \} \cap \{ \mathbf{7}, \mathbf{9}, \mathbf{11}, 13, 15 \}$.

The common elements are 7, 9, 11.

$( A \cap B ) \cap ( B ∪ C ) = \{$7, 9, 11$\}$.

---

(x) ( A ∪ D) ∩ ( B ∪ C)

First, find $A \cup D$.

$A \cup D = \{ 3, 5, 7, 9, 11 \} \cup \{ 15, 17 \} = \{ 3, 5, 7, 9, 11, 15, 17 \}$.

Next, find $B \cup C$. From part (vi), $B \cup C = \{ 7, 9, 11, 13, 15 \}$.

Now, find the intersection of these two results.

$( A \cup D) \cap ( B \cup C) = \{ 3, 5, \mathbf{7}, \mathbf{9}, \mathbf{11}, \mathbf{15}, 17 \} \cap \{ \mathbf{7}, \mathbf{9}, \mathbf{11}, 13, \mathbf{15} \}$.

The common elements are 7, 9, 11, 15.

$( A \cup D) \cap ( B \cup C) = \{$7, 9, 11, 15$\}$.

Given Sets:

A = {x : x is a natural number} = {1, 2, 3, 4, ...}

B = {x : x is an even natural number} = {2, 4, 6, 8, ...}

C = {x : x is an odd natural number} = {1, 3, 5, 7, ...}

D = {x : x is a prime number} = {2, 3, 5, 7, 11, 13, ...}

The intersection ($ \cap $) of sets includes only the elements common to both sets involved.

---

(i) A ∩ B

We find the elements common to A (all natural numbers) and B (even natural numbers).

All even natural numbers are natural numbers. So, the elements common to both sets are exactly the even natural numbers.

$A \cap B = B = \{$x : x is an even natural number$\}$ = {2, 4, 6, 8, ...}.

---

(ii) A ∩ C

We find the elements common to A (all natural numbers) and C (odd natural numbers).

All odd natural numbers are natural numbers. So, the elements common to both sets are exactly the odd natural numbers.

$A \cap C = C = \{$x : x is an odd natural number$\}$ = {1, 3, 5, 7, ...}.

---

(iii) A ∩ D

We find the elements common to A (all natural numbers) and D (prime numbers).

All prime numbers are natural numbers. So, the elements common to both sets are exactly the prime numbers.

$A \cap D = D = \{$x : x is a prime number$\}$ = {2, 3, 5, 7, 11, ...}.

---

(iv) B ∩ C

We find the elements common to B (even natural numbers) and C (odd natural numbers).

A natural number cannot be both even and odd. There are no common elements.

$B \cap C = \{$$\}$ or φ.

---

(v) B ∩ D

We find the elements common to B (even natural numbers) and D (prime numbers).

We need to find numbers that are both even and prime. The only even prime number is 2.

$B \cap D = \{$2$\}$.

---

(vi) C ∩ D

We find the elements common to C (odd natural numbers) and D (prime numbers).

We need to find numbers that are both odd and prime. These are all prime numbers except for 2.

$C \cap D = \{3, 5, 7, 11, 13, ...\} = \{$x : x is an odd prime number$\}$.

Two sets are called disjoint if their intersection is the empty set (φ), meaning they have no elements in common.

---

(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6 }

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

Let Set B = {x : x is a natural number and 4 ≤ x ≤ 6 }. Listing the elements of Set B, we get B = {4, 5, 6}.

To check if they are disjoint, we find their intersection, $A \cap B$.

$A \cap B = \{1, 2, 3, \mathbf{4}\} \cap \{\mathbf{4}, 5, 6\}$

The common element is 4.

$A \cap B = \{4\}$.

Since the intersection is not the empty set ($A \cap B \neq \phi$), the sets are not disjoint.

---

(ii) { a, e, i, o, u } and { c, d, e, f }

Let Set A = { a, e, i, o, u }.

Let Set B = { c, d, e, f }.

To check if they are disjoint, we find their intersection, $A \cap B$.

$A \cap B = \{ a, \mathbf{e}, i, o, u \} \cap \{ c, d, \mathbf{e}, f \}$

The common element is 'e'.

$A \cap B = \{e\}$.

Since the intersection is not the empty set ($A \cap B \neq \phi$), the sets are not disjoint.

---

(iii) {x : x is an even integer } and {x : x is an odd integer}

Let Set A = {x : x is an even integer} = {..., -4, -2, 0, 2, 4, ...}.

Let Set B = {x : x is an odd integer} = {..., -3, -1, 1, 3, 5, ...}.

To check if they are disjoint, we find their intersection, $A \cap B$.

An integer can be either even or odd, but it cannot be both simultaneously. Therefore, there are no elements common to both Set A and Set B.

$A \cap B = \phi$.

Since the intersection is the empty set, the sets are disjoint.

---

Conclusion:

The only pair of sets that are disjoint is (iii) {x : x is an even integer } and {x : x is an odd integer}.

Given Sets:

A = {3, 6, 9, 12, 15, 18, 21}

B = {4, 8, 12, 16, 20}

C = {2, 4, 6, 8, 10, 12, 14, 16}

D = {5, 10, 15, 20}

The set difference $X - Y$ contains all elements that are in set X but not in set Y.

---

(i) A – B

Elements in A but not in B. Remove elements of B from A.

Common element is 12.

$A - B = \{3, 6, 9, \cancel{12}, 15, 18, 21\} - \{4, 8, 12, 16, 20\} = \{$3, 6, 9, 15, 18, 21$\}$

---

(ii) A – C

Elements in A but not in C. Remove elements of C from A.

Common elements are 6, 12.

$A - C = \{3, \cancel{6}, 9, \cancel{12}, 15, 18, 21\} - \{2, 4, 6, 8, 10, 12, 14, 16\} = \{$3, 9, 15, 18, 21$\}$

---

(iii) A – D

Elements in A but not in D. Remove elements of D from A.

Common element is 15.

$A - D = \{3, 6, 9, 12, \cancel{15}, 18, 21\} - \{5, 10, 15, 20\} = \{$3, 6, 9, 12, 18, 21$\}$

---

(iv) B – A

Elements in B but not in A. Remove elements of A from B.

Common element is 12.

$B - A = \{4, 8, \cancel{12}, 16, 20\} - \{3, 6, 9, 12, 15, 18, 21\} = \{$4, 8, 16, 20$\}$

---

(v) C – A

Elements in C but not in A. Remove elements of A from C.

Common elements are 6, 12.

$C - A = \{2, 4, \cancel{6}, 8, 10, \cancel{12}, 14, 16\} - \{3, 6, 9, 12, 15, 18, 21\} = \{$2, 4, 8, 10, 14, 16$\}$

---

(vi) D – A

Elements in D but not in A. Remove elements of A from D.

Common element is 15.

$D - A = \{5, 10, \cancel{15}, 20\} - \{3, 6, 9, 12, 15, 18, 21\} = \{$5, 10, 20$\}$

---

(vii) B – C

Elements in B but not in C. Remove elements of C from B.

Common elements are 4, 8, 12, 16.

$B - C = \{\cancel{4}, \cancel{8}, \cancel{12}, \cancel{16}, 20\} - \{2, 4, 6, 8, 10, 12, 14, 16\} = \{$20$\}$

---

(viii) B – D

Elements in B but not in D. Remove elements of D from B.

Common element is 20.

$B - D = \{4, 8, 12, 16, \cancel{20}\} - \{5, 10, 15, 20\} = \{$4, 8, 12, 16$\}$

---

(ix) C – B

Elements in C but not in B. Remove elements of B from C.

Common elements are 4, 8, 12, 16.

$C - B = \{2, \cancel{4}, 6, \cancel{8}, 10, \cancel{12}, 14, \cancel{16}\} - \{4, 8, 12, 16, 20\} = \{$2, 6, 10, 14$\}$

---

(x) D – B

Elements in D but not in B. Remove elements of B from D.

Common element is 20.

$D - B = \{5, 10, 15, \cancel{20}\} - \{4, 8, 12, 16, 20\} = \{$5, 10, 15$\}$

---

(xi) C – D

Elements in C but not in D. Remove elements of D from C.

Common element is 10.

$C - D = \{2, 4, 6, 8, \cancel{10}, 12, 14, 16\} - \{5, 10, 15, 20\} = \{$2, 4, 6, 8, 12, 14, 16$\}$

---

(xii) D – C

Elements in D but not in C. Remove elements of C from D.

Common element is 10.

$D - C = \{5, \cancel{10}, 15, 20\} - \{2, 4, 6, 8, 10, 12, 14, 16\} = \{$5, 15, 20$\}$

Given Sets:

X = { a, b, c, d }

Y = { f, b, d, g }

---

(i) X – Y

The set difference $X - Y$ contains all elements that are in set X but are not in set Y.

Elements in X: { a, b, c, d }

Elements in Y: { f, b, d, g }

Elements common to both X and Y are { b, d }.

Remove the common elements from X:

$X - Y = \{ a, \cancel{b}, c, \cancel{d} \} - \{ f, b, d, g \} = \{ a, c \}$.

Therefore, $X - Y = \{$a, c$\}$.

---

(ii) Y – X

The set difference $Y - X$ contains all elements that are in set Y but are not in set X.

Elements in Y: { f, b, d, g }

Elements in X: { a, b, c, d }

Elements common to both Y and X are { b, d }.

Remove the common elements from Y:

$Y - X = \{ f, \cancel{b}, \cancel{d}, g \} - \{ a, b, c, d \} = \{ f, g \}$.

Therefore, $Y - X = \{$f, g$\}$.

---

(iii) X ∩ Y

The intersection $X \cap Y$ contains all elements that are common to both set X and set Y.

Elements in X: { a, b, c, d }

Elements in Y: { f, b, d, g }

The elements found in both sets are b and d.

Therefore, $X \cap Y = \{$b, d$\}$.

Given:

R = The set of real numbers ($\mathbb{R}$)

Q = The set of rational numbers ($\mathbb{Q}$)

---

To Find:

The set difference $R - Q$.

---

Solution:

The set of real numbers ($\mathbb{R}$) is composed of two disjoint types of numbers: rational numbers and irrational numbers.

Rational numbers ($\mathbb{Q}$) are numbers that can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples include 5, -3, $\frac{1}{2}$, 0.75.

Irrational numbers are real numbers that cannot be expressed as a simple fraction. Their decimal representations are non-terminating and non-repeating. Examples include $\sqrt{2}$, $\pi$, $e$. Let's denote the set of irrational numbers as $T$ or $\mathbb{I}$.

The set of real numbers is the union of the set of rational numbers and the set of irrational numbers:

$\mathbb{R} = \mathbb{Q} \cup T$

Also, the set of rational numbers and the set of irrational numbers are disjoint, meaning they have no elements in common:

$\mathbb{Q} \cap T = \phi$

The set difference $R - Q$ ($\mathbb{R} - \mathbb{Q}$) consists of all elements that are in the set of real numbers ($\mathbb{R}$) but are not in the set of rational numbers ($\mathbb{Q}$).

If we take all real numbers and remove the rational numbers, what remains are the irrational numbers.

$R - Q = (\mathbb{Q} \cup T) - \mathbb{Q} = T$

Therefore, $R - Q$ is the set of irrational numbers.

Two sets are defined as disjoint if they have no elements in common, meaning their intersection is the empty set ($\phi$).

---

(i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.

Let A = { 2, 3, 4, 5 } and B = { 3, 6 }.

The intersection $A \cap B$ is the set of common elements.

$A \cap B = \{ 2, \mathbf{3}, 4, 5 \} \cap \{ \mathbf{3}, 6 \} = \{ 3 \}$.

Justification: The intersection of the two sets is {3}, which is not the empty set ($\phi$).

Conclusion: The statement is False.

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(ii) { a, e, i, o, u } and { a, b, c, d }are disjoint sets.

Let A = { a, e, i, o, u } and B = { a, b, c, d }.

The intersection $A \cap B$ is the set of common elements.

$A \cap B = \{ \mathbf{a}, e, i, o, u \} \cap \{ \mathbf{a}, b, c, d \} = \{ a \}$.

Justification: The intersection of the two sets is {a}, which is not the empty set ($\phi$).

Conclusion: The statement is False.

---

(iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.

Let A = { 2, 6, 10, 14 } and B = { 3, 7, 11, 15 }.

The intersection $A \cap B$ is the set of common elements.

Comparing the elements, there are no elements common to both sets.

$A \cap B = \phi$.

Justification: The intersection of the two sets is the empty set ($\phi$).

Conclusion: The statement is True.

---

(iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.

Let A = { 2, 6, 10 } and B = { 3, 7, 11 }.

The intersection $A \cap B$ is the set of common elements.

Comparing the elements, there are no elements common to both sets.

$A \cap B = \phi$.

Justification: The intersection of the two sets is the empty set ($\phi$).

Conclusion: The statement is True.

Given:

The universal set, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

The set A = {1, 3, 5, 7, 9}

---

To Find:

The complement of set A, denoted as A' (or A^c).

---

Solution:

The complement of a set A (with respect to a universal set U) is the set of all elements in U that are not in A. It is calculated as $A' = U - A$.

We need to find the elements that are in U but not in A.

Elements in U: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Elements in A: {1, 3, 5, 7, 9}

We remove the elements of A from U:

The elements remaining in U after removing {1, 3, 5, 7, 9} are {2, 4, 6, 8, 10}.

Therefore, the complement of A is A' = {2, 4, 6, 8, 10}.

A' = {2, 4, 6, 8, 10}.

Given:

The universal set, U = {x : x is a student of Class XI of a coeducational school}.

The set A = {x : x is a girl in Class XI}.

---

To Find:

The complement of set A, denoted as A'.

---

Solution:

The complement of a set A with respect to the universal set U, denoted as A', is the set of all elements in U that are not elements of A. Mathematically, $A' = U - A$.

In this context:

$A' = \{ \text{All students of Class XI} \} - \{ \text{All girls in Class XI} \}$.

Since the school is coeducational, the students in Class XI consist of either boys or girls.

When we remove the set of all girls (A) from the set of all students (U), the remaining elements are the students who are not girls.

In a coeducational Class XI, the students who are not girls must be boys.

Therefore, A' is the set of all boys in Class XI.

A' = {x : x is a boy in Class XI}.

Given:

Universal Set, U = {1, 2, 3, 4, 5, 6}

Set A = {2, 3}

Set B = {3, 4, 5}

---

To Find and Show:

1. Find A'

2. Find B'

3. Find A' ∩ B'

4. Find A ∪ B

5. Show that ( A ∪ B )' = A'∩ B'

---

Solution:

1. Finding A':

A' is the set of elements in U but not in A.

A' = U - A = {1, 2, 3, 4, 5, 6} - {2, 3}

A' = {1, 4, 5, 6}.

So, A' = {1, 4, 5, 6}.

---

2. Finding B':

B' is the set of elements in U but not in B.

B' = U - B = {1, 2, 3, 4, 5, 6} - {3, 4, 5}

B' = {1, 2, 6}.

So, B' = {1, 2, 6}.

---

3. Finding A' ∩ B':

A' ∩ B' is the set of elements common to both A' and B'.

A' = {1, 4, 5, 6}

B' = {1, 2, 6}

The common elements are 1 and 6.

So, A' ∩ B' = {1, 6}.

---

4. Finding A ∪ B:

A ∪ B is the set of all elements in A or B or both.

$A \cup B = \{2, 3\} \cup \{3, 4, 5\}$

$A \cup B = \{2, 3, 4, 5\}$.

So, A ∪ B = {2, 3, 4, 5}.

---

5. Showing that ( A ∪ B )' = A'∩ B':

First, find ( A ∪ B )'.

This is the complement of the set A ∪ B = {2, 3, 4, 5} with respect to U.

$( A \cup B )' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6\} - \{2, 3, 4, 5\}$

$( A \cup B )' = \{1, 6\}$.

Now, compare this result with A' ∩ B' found in step 3.

We found $A' ∩ B' = \{1, 6\}$.

Since $( A \cup B )' = \{1, 6\}$ and $A' ∩ B' = \{1, 6\}$, the two sets are equal.

Hence, it is shown that ( A ∪ B )' = A'∩ B'.

Given Sets:

Universal Set, U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }

A = { 1, 2, 3, 4 }

B = { 2, 4, 6, 8 }

C = { 3, 4, 5, 6 }

The complement of a set X, denoted X', is the set of elements in U that are not in X ($X' = U - X$).

---

(i) A'

A' is the set of elements in U but not in A.

$A' = U - A = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 1, 2, 3, 4 \}$

$A' = \{$5, 6, 7, 8, 9$\}$

---

(ii) B'

B' is the set of elements in U but not in B.

$B' = U - B = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 2, 4, 6, 8 \}$

$B' = \{$1, 3, 5, 7, 9$\}$

---

(iii) (A ∪ C)'

First, find the union $A \cup C$.

$A \cup C = \{ 1, 2, 3, 4 \} \cup \{ 3, 4, 5, 6 \} = \{ 1, 2, 3, 4, 5, 6 \}$

Now, find the complement of this union with respect to U.

$(A \cup C)' = U - (A \cup C) = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 1, 2, 3, 4, 5, 6 \}$

$(A \cup C)' = \{$7, 8, 9$\}$

---

(iv) (A ∪ B)'

First, find the union $A \cup B$.

$A \cup B = \{ 1, 2, 3, 4 \} \cup \{ 2, 4, 6, 8 \} = \{ 1, 2, 3, 4, 6, 8 \}$

Now, find the complement of this union with respect to U.

$(A \cup B)' = U - (A \cup B) = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 1, 2, 3, 4, 6, 8 \}$

$(A \cup B)' = \{$5, 7, 9$\}$

---

(v) (A')'

This is the complement of A'. The complement of the complement of a set is the set itself.

From part (i), $A' = \{ 5, 6, 7, 8, 9 \}$.

$(A')' = U - A' = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 5, 6, 7, 8, 9 \}$

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

This is equal to the original set A.

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

---

(vi) (B – C)'

First, find the set difference $B - C$. This contains elements in B but not in C.

$B - C = \{ 2, 4, 6, 8 \} - \{ 3, 4, 5, 6 \}$

Common elements are 4 and 6. Remove these from B.

$B - C = \{ 2, 8 \}$.

Now, find the complement of this result with respect to U.

$(B - C)' = U - (B - C) = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \} - \{ 2, 8 \}$

$(B - C)' = \{$1, 3, 4, 5, 6, 7, 9$\}$

Given:

Universal Set, U = { a, b, c, d, e, f, g, h }

Sets:

A = {a, b, c}

B = {d, e, f, g}

C = {a, c, e, g}

D = {f, g, h, a}

The complement of a set X (denoted X') is the set of elements in the universal set U that are not in X ($X' = U - X$).

---

(i) Complement of A = {a, b, c}

A' = U - A = { a, b, c, d, e, f, g, h } - { a, b, c }

Remove a, b, c from U.

A' = { d, e, f, g, h }

---

(ii) Complement of B = {d, e, f, g}

B' = U - B = { a, b, c, d, e, f, g, h } - { d, e, f, g }

Remove d, e, f, g from U.

B' = { a, b, c, h }

---

(iii) Complement of C = {a, c, e, g}

C' = U - C = { a, b, c, d, e, f, g, h } - { a, c, e, g }

Remove a, c, e, g from U.

C' = { b, d, f, h }

---

(iv) Complement of D = { f, g, h, a}

D' = U - D = { a, b, c, d, e, f, g, h } - { f, g, h, a }

Remove f, g, h, a from U.

D' = { b, c, d, e }

The universal set is U = $\mathbb{N}$ = {1, 2, 3, 4, ...}, the set of natural numbers.

The complement of a set A, denoted A', is the set $U - A$.

---

(i) Let A = {x : x is an even natural number}

A = {2, 4, 6, 8, ...}

A' = U - A = {All natural numbers} - {All even natural numbers}

A' = {All odd natural numbers}

A' = {x : x is an odd natural number}

---

(ii) Let A = {x : x is an odd natural number}

A = {1, 3, 5, 7, ...}

A' = U - A = {All natural numbers} - {All odd natural numbers}

A' = {All even natural numbers}

A' = {x : x is an even natural number}

---

(iii) Let A = {x : x is a positive multiple of 3}

A = {3, 6, 9, 12, ...}

A' = U - A = {All natural numbers} - {Multiples of 3}

A' = {Natural numbers that are not multiples of 3}

A' = {x : x ∈ $\mathbb{N}$ and x is not a multiple of 3}

---

(iv) Let A = {x : x is a prime number}

A = {2, 3, 5, 7, 11, ...}

A' = U - A = {All natural numbers} - {Prime numbers}

A' = {Natural numbers that are not prime}

Natural numbers are either 1, prime, or composite. So, the complement includes 1 and all composite numbers.

A' = {x : x = 1 or x is a composite number} or {1, 4, 6, 8, 9, ...}

---

(v) Let A = {x : x is a natural number divisible by 3 and 5}

A number divisible by both 3 and 5 must be a multiple of their LCM, which is 15.

A = {15, 30, 45, ...} = {x : x is a multiple of 15}

A' = U - A = {All natural numbers} - {Multiples of 15}

A' = {Natural numbers that are not multiples of 15}

A' = {x : x ∈ $\mathbb{N}$ and x is not a multiple of 15}

---

(vi) Let A = {x : x is a perfect square}

A = {1, 4, 9, 16, ...}

A' = U - A = {All natural numbers} - {Perfect squares}

A' = {Natural numbers that are not perfect squares}

A' = {x : x ∈ $\mathbb{N}$ and x is not a perfect square}

---

(vii) Let A = {x : x is a perfect cube}

A = {1, 8, 27, 64, ...}

A' = U - A = {All natural numbers} - {Perfect cubes}

A' = {Natural numbers that are not perfect cubes}

A' = {x : x ∈ $\mathbb{N}$ and x is not a perfect cube}

---

(viii) Let A = {x : x + 5 = 8}

Solving $x + 5 = 8$, we get $x = 3$. So, A = {3}.

A' = U - A = $\mathbb{N}$ - {3}

A' = {All natural numbers except 3}

A' = {x : x ∈ $\mathbb{N}$ and x ≠ 3}

---

(ix) Let A = {x : 2x + 5 = 9}

Solving $2x + 5 = 9$, we get $2x = 4$, so $x = 2$. So, A = {2}.

A' = U - A = $\mathbb{N}$ - {2}

A' = {All natural numbers except 2}

A' = {x : x ∈ $\mathbb{N}$ and x ≠ 2}

---

(x) Let A = {x : x ≥ 7}

Since U = $\mathbb{N}$, A = {7, 8, 9, 10, ...}

A' = U - A = $\mathbb{N}$ - {7, 8, 9, ...}

A' = {Natural numbers less than 7}

A' = {x : x ∈ $\mathbb{N}$ and x < 7} = {1, 2, 3, 4, 5, 6}

---

(xi) Let A = {x : x ∈ N and 2x + 1 > 10}

Solving the inequality: $2x + 1 > 10 \implies 2x > 9 \implies x > 4.5$.

Since x must be a natural number, A = {5, 6, 7, 8, ...}

A' = U - A = $\mathbb{N}$ - {5, 6, 7, 8, ...}

A' = {Natural numbers less than 5}

A' = {x : x ∈ $\mathbb{N}$ and x ≤ 4} = {1, 2, 3, 4}

Given Sets:

Universal Set, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}

B = {2, 3, 5, 7}

---

Calculations for Verification:

First, let's find the complements A' and B'.

A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9}

B' = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7} = {1, 4, 6, 8, 9}

---

(i) Verification of (A ∪ B)' = A' ∩ B' (De Morgan's First Law)

Calculate the Left Hand Side (LHS): (A ∪ B)'

Find A ∪ B: $A \cup B = \{2, 4, 6, 8\} \cup \{2, 3, 5, 7\} = \{2, 3, 4, 5, 6, 7, 8\}$

Find the complement of A ∪ B:

$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 4, 5, 6, 7, 8\}$

$(A \cup B)' = \{1, 9\}$

Calculate the Right Hand Side (RHS): A' ∩ B'

We have A' = {1, 3, 5, 7, 9} and B' = {1, 4, 6, 8, 9}.

Find the intersection A' ∩ B':

$A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\}$

The common elements are 1 and 9.

$A' \cap B' = \{1, 9\}$

Compare LHS and RHS:

LHS = $(A \cup B)' = \{1, 9\}$

RHS = $A' \cap B' = \{1, 9\}$

Since LHS = RHS, the statement (A ∪ B)' = A' ∩ B' is verified.

---

(ii) Verification of (A ∩ B)' = A' ∪ B' (De Morgan's Second Law)

Calculate the Left Hand Side (LHS): (A ∩ B)'

Find A ∩ B:

$A \cap B = \{2, 4, 6, 8\} \cap \{2, 3, 5, 7\}$

The only common element is 2.

$A \cap B = \{2\}$

Find the complement of A ∩ B:

$(A \cap B)' = U - (A \cap B) = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2\}$

$(A \cap B)' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Calculate the Right Hand Side (RHS): A' ∪ B'

We have A' = {1, 3, 5, 7, 9} and B' = {1, 4, 6, 8, 9}.

Find the union A' ∪ B':

$A' \cup B' = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\}$

Combining all unique elements:

$A' \cup B' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Compare LHS and RHS:

LHS = $(A \cap B)' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

RHS = $A' \cup B' = \{1, 3, 4, 5, 6, 7, 8, 9\}$

Since LHS = RHS, the statement (A ∩ B)' = A' ∪ B' is verified.

In the following descriptions, U represents the universal set (usually drawn as a rectangle), and A and B represent two subsets within U (usually drawn as overlapping circles).

---

(i) (A ∪ B)'

Description:

1. Identify the region representing $A \cup B$. This includes all areas inside circle A, inside circle B, and the overlapping region.

2. The complement $(A \cup B)'$ represents everything in the universal set U that is outside the region $A \cup B$.

Venn Diagram Shading: Shade the area within the rectangle U but outside both circles A and B.

---

(ii) A' ∩ B'

Description:

1. Identify the region representing A' (everything outside circle A).

2. Identify the region representing B' (everything outside circle B).

3. The intersection $A' \cap B'$ represents the area that is common to both A' and B', meaning the area that is outside A and outside B.

Venn Diagram Shading: Shade the area within the rectangle U but outside both circles A and B. (This diagram is identical to (i), illustrating De Morgan's Law).

---

(iii) (A ∩ B)'

Description:

1. Identify the region representing $A \cap B$. This is the overlapping area of circles A and B.

2. The complement $(A \cap B)'$ represents everything in the universal set U that is outside the region $A \cap B$.

Venn Diagram Shading: Shade all areas within the rectangle U except for the overlapping region of circles A and B.

---

(iv) A' ∪ B'

Description:

1. Identify the region representing A' (everything outside circle A).

2. Identify the region representing B' (everything outside circle B).

3. The union $A' \cup B'$ represents the area that is in A' or in B' or in both. This includes everything outside A, plus everything outside B.

Venn Diagram Shading: Shade all areas within the rectangle U except for the overlapping region of circles A and B. (This diagram is identical to (iii), illustrating De Morgan's Law).

Given:

The universal set, U = {x : x is a triangle in a plane}.

Set A = {x : x is a triangle with at least one angle different from 60°}.

---

To Find:

The complement of set A, denoted as A'.

---

Solution:

The complement A' consists of all elements in the universal set U that are not in set A.

$A' = U - A$.

So, A' is the set of all triangles in the plane that do not have at least one angle different from 60°.

If a triangle does not have "at least one angle different from 60°", it means that all of its angles must be equal to 60°.

We know that the sum of angles in any triangle is 180°. If all three angles are 60°, then $60° + 60° + 60° = 180°$.

A triangle where all three angles are equal to 60° is defined as an equilateral triangle.

Therefore, the set A' contains all triangles whose angles are all 60°, which is the set of all equilateral triangles.

A' = {x : x is an equilateral triangle}.

These statements relate to the properties of set operations involving complements, the universal set (U), and the empty set (φ).

---

(i) A ∪ A' = . . .

Reasoning: A is a set. A' is the set of all elements in the universal set U that are not in A. The union $A \cup A'$ combines all elements that are in A with all elements that are not in A. Together, this constitutes all the elements in the universal set U.

Answer: A ∪ A' = U

---

(ii) φ' ∩ A = . . .

Reasoning: First, consider φ'. This is the complement of the empty set. The complement of the empty set is the set of all elements in U that are not in φ. Since φ has no elements, its complement is the entire universal set, U. So, φ' = U.

Now the expression becomes U ∩ A. This is the intersection of the universal set U and set A. The intersection contains elements common to both U and A. Since A is always a subset of U, all elements of A are within U. Therefore, the elements common to both are exactly the elements of A.

Answer: φ' ∩ A = A

---

(iii) A ∩ A' = . . .

Reasoning: This is the intersection of set A and its complement A'. A' contains elements that are specifically *not* in A. The intersection requires elements to be in both sets simultaneously. By definition, no element can be both in A and not in A.

Therefore, the intersection is empty.

Answer: A ∩ A' = φ

---

(iv) U' ∩ A = . . .

Reasoning: First, consider U'. This is the complement of the universal set. The complement of U is the set of all elements in U that are not in U. There are no such elements. So, the complement of the universal set is the empty set, φ. Thus, U' = φ.

Now the expression becomes φ ∩ A. This is the intersection of the empty set φ and set A. Since the empty set has no elements, there can be no elements common to both φ and A.

Therefore, the intersection is empty.

Answer: U' ∩ A = φ

Given:

Number of elements in the union of X and Y, $n(X \cup Y) = 50$.

Number of elements in set X, $n(X) = 28$.

Number of elements in set Y, $n(Y) = 32$.

---

To Find:

Number of elements in the intersection of X and Y, $n(X \cap Y)$.

---

Formula Used:

The Principle of Inclusion-Exclusion for two sets states:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

---

Solution:

We can rearrange the formula to solve for $n(X \cap Y)$:

$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$

Now, substitute the given values into the formula:

$n(X \cap Y) = 28 + 32 - 50$

Calculate the sum:

$n(X \cap Y) = 60 - 50$

Calculate the difference:

$n(X \cap Y) = 10$

Therefore, the number of elements in the intersection of X and Y is 10.

The set $X \cap Y$ has 10 elements.

Given:

Let M be the set of teachers who teach Mathematics.

Let P be the set of teachers who teach Physics.

The total number of teachers who teach Mathematics or Physics is $n(M \cup P) = 20$.

The number of teachers who teach Mathematics is $n(M) = 12$.

The number of teachers who teach both Mathematics and Physics is $n(M \cap P) = 4$.

---

To Find:

The number of teachers who teach Physics, $n(P)$.

---

Formula Used:

The Principle of Inclusion-Exclusion for two sets states:

$n(M \cup P) = n(M) + n(P) - n(M \cap P)$

---

Solution:

We can substitute the given values into the formula:

$20 = 12 + n(P) - 4$

Simplify the right side:

$20 = (12 - 4) + n(P)$

$20 = 8 + n(P)$

Now, solve for $n(P)$ by subtracting 8 from both sides:

$n(P) = 20 - 8$

$n(P) = 12$

Therefore, the number of teachers who teach Physics is 12.

12 teachers teach Physics.

Given:

Let C be the set of students who like to play cricket.

Let F be the set of students who like to play football.

Total number of students in the class = 35.

Since each student likes to play at least one of the two games, the total number of students is the number of students in the union of the two sets: $n(C \cup F) = 35$.

Number of students who like to play cricket, $n(C) = 24$.

Number of students who like to play football, $n(F) = 16$.

---

To Find:

The number of students who like to play both cricket and football, which is the number of students in the intersection of the two sets: $n(C \cap F)$.

---

Formula Used:

The Principle of Inclusion-Exclusion for two sets:

$n(C \cup F) = n(C) + n(F) - n(C \cap F)$

---

Solution:

We can rearrange the formula to solve for $n(C \cap F)$:

$n(C \cap F) = n(C) + n(F) - n(C \cup F)$

Substitute the given values into the formula:

$n(C \cap F) = 24 + 16 - 35$

Calculate the sum:

$n(C \cap F) = 40 - 35$

Calculate the difference:

$n(C \cap F) = 5$

Therefore, the number of students who like to play both cricket and football is 5.

5 students like to play both cricket and football.

Given:

Let U be the set of all students surveyed.

Let A be the set of students taking apple juice.

Let O be the set of students taking orange juice.

Total number of students surveyed, $n(U) = 400$.

Number of students taking apple juice, $n(A) = 100$.

Number of students taking orange juice, $n(O) = 150$.

Number of students taking both apple and orange juice, $n(A \cap O) = 75$.

---

To Find:

The number of students who were taking neither apple juice nor orange juice. This is represented by $n(A' \cap O')$, which, by De Morgan's laws, is equal to $n((A \cup O)')$.

---

Formulas Used:

1. Principle of Inclusion-Exclusion for two sets:

$n(A \cup O) = n(A) + n(O) - n(A \cap O)$

2. Complement Rule:

$n((A \cup O)') = n(U) - n(A \cup O)$

---

Solution:

First, we find the number of students taking at least one of the juices (apple or orange), which is $n(A \cup O)$.

Using the Principle of Inclusion-Exclusion:

$n(A \cup O) = n(A) + n(O) - n(A \cap O)$

Substitute the given values:

$n(A \cup O) = 100 + 150 - 75$

$n(A \cup O) = 250 - 75$

$n(A \cup O) = 175$

So, 175 students were taking at least one of the juices.

Now, we find the number of students taking neither apple juice nor orange juice, which is the complement of $A \cup O$.

Using the Complement Rule:

$n((A \cup O)') = n(U) - n(A \cup O)$

Substitute the values:

$n((A \cup O)') = 400 - 175$

$n((A \cup O)') = 225$

Therefore, the number of students taking neither apple juice nor orange juice is 225.

225 students were taking neither apple juice nor orange juice.

Given:

Let U be the set of all individuals with the skin disorder, so $n(U) = 200$. (Although $n(U)$ isn't directly needed for the calculations asked).

Let $C_1$ be the set of individuals exposed to chemical C₁.

Let $C_2$ be the set of individuals exposed to chemical C₂.

Number of individuals exposed to C₁, $n(C_1) = 120$.

Number of individuals exposed to C₂, $n(C_2) = 50$.

Number of individuals exposed to both C₁ and C₂, $n(C_1 \cap C_2) = 30$.

---

To Find:

(i) Number of individuals exposed to Chemical C₁ but not chemical C₂, which is $n(C_1 - C_2)$.

(ii) Number of individuals exposed to Chemical C₂ but not chemical C₁, which is $n(C_2 - C_1)$.

(iii) Number of individuals exposed to Chemical C₁ or chemical C₂, which is $n(C_1 \cup C_2)$.

---

Formulas Used:

1. $n(C_1 - C_2) = n(C_1) - n(C_1 \cap C_2)$

2. $n(C_2 - C_1) = n(C_2) - n(C_1 \cap C_2)$

3. $n(C_1 \cup C_2) = n(C_1) + n(C_2) - n(C_1 \cap C_2)$

---

Solution:

(i) Number of individuals exposed to Chemical C₁ but not chemical C₂:

This corresponds to the number of elements in $C_1$ only.

$n(C_1 - C_2) = n(C_1) - n(C_1 \cap C_2)$

Substitute the given values:

$n(C_1 - C_2) = 120 - 30$

$n(C_1 - C_2) = 90$

So, 90 individuals were exposed to chemical C₁ but not chemical C₂.

---

(ii) Number of individuals exposed to Chemical C₂ but not chemical C₁:

This corresponds to the number of elements in $C_2$ only.

$n(C_2 - C_1) = n(C_2) - n(C_1 \cap C_2)$

Substitute the given values:

$n(C_2 - C_1) = 50 - 30$

$n(C_2 - C_1) = 20$

So, 20 individuals were exposed to chemical C₂ but not chemical C₁.

---

(iii) Number of individuals exposed to Chemical C₁ or chemical C₂:

This corresponds to the number of elements in the union $C_1 \cup C_2$.

Using the Principle of Inclusion-Exclusion:

$n(C_1 \cup C_2) = n(C_1) + n(C_2) - n(C_1 \cap C_2)$

Substitute the given values:

$n(C_1 \cup C_2) = 120 + 50 - 30$

$n(C_1 \cup C_2) = 170 - 30$

$n(C_1 \cup C_2) = 140$

Alternatively, we can sum the results from (i), (ii) and the number exposed to both:

$n(C_1 \cup C_2) = n(C_1 - C_2) + n(C_2 - C_1) + n(C_1 \cap C_2) = 90 + 20 + 30 = 140$.

So, 140 individuals were exposed to chemical C₁ or chemical C₂.

Given:

$n(X) = 17$

$n(Y) = 23$

$n(X \cup Y) = 38$

---

To Find:

$n(X \cap Y)$

---

Solution:

We know the formula relating the number of elements in the union and intersection of two sets:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

We can rearrange this formula to find $n(X \cap Y)$:

$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$

Now, substitute the given values into the formula:

$n(X \cap Y) = 17 + 23 - 38$

$n(X \cap Y) = 40 - 38$

$n(X \cap Y) = 2$

Thus, the number of elements in the intersection of sets X and Y is 2.

Given:

$n(X \cup Y) = 18$

$n(X) = 8$

$n(Y) = 15$

---

To Find:

$n(X \cap Y)$

---

Solution:

We use the formula for the number of elements in the union of two sets:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

We want to find $n(X \cap Y)$, so we rearrange the formula:

$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$

Now, substitute the given values:

$n(X \cap Y) = 8 + 15 - 18$

$n(X \cap Y) = 23 - 18$

$n(X \cap Y) = 5$

Therefore, the number of elements in the intersection of sets X and Y is 5.

Given:

Let H be the set of people who can speak Hindi.

Let E be the set of people who can speak English.

Total number of people in the group (who speak Hindi or English or both), $n(H \cup E) = 400$.

Number of people who can speak Hindi, $n(H) = 250$.

Number of people who can speak English, $n(E) = 200$.

---

To Find:

The number of people who can speak both Hindi and English, $n(H \cap E)$.

---

Solution:

We use the formula relating the number of elements in the union and intersection of two sets:

$n(H \cup E) = n(H) + n(E) - n(H \cap E)$

We need to find $n(H \cap E)$, so we rearrange the formula:

$n(H \cap E) = n(H) + n(E) - n(H \cup E)$

Substitute the given values into the formula:

$n(H \cap E) = 250 + 200 - 400$

$n(H \cap E) = 450 - 400$

$n(H \cap E) = 50$

Therefore, 50 people can speak both Hindi and English.

Given:

$n(S) = 21$

$n(T) = 32$

$n(S \cap T) = 11$

---

To Find:

$n(S \cup T)$

---

Solution:

We use the formula for the number of elements in the union of two sets:

$n(S \cup T) = n(S) + n(T) - n(S \cap T)$

Substitute the given values into the formula:

$n(S \cup T) = 21 + 32 - 11$

$n(S \cup T) = 53 - 11$

$n(S \cup T) = 42$

Therefore, the set S ∪ T has 42 elements.

Given:

$n(X) = 40$

$n(X \cup Y) = 60$

$n(X \cap Y) = 10$

---

To Find:

$n(Y)$

---

Solution:

We use the formula for the number of elements in the union of two sets:

$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$

We need to find $n(Y)$. Let's substitute the given values into the formula:

$60 = 40 + n(Y) - 10$

Simplify the right side:

$60 = 30 + n(Y)$

To find $n(Y)$, subtract 30 from both sides:

$n(Y) = 60 - 30$

$n(Y) = 30$

Therefore, the set Y has 30 elements.

Given:

Let C be the set of people who like coffee.

Let T be the set of people who like tea.

Total number of people in the group (each person likes at least one drink), $n(C \cup T) = 70$.

Number of people who like coffee, $n(C) = 37$.

Number of people who like tea, $n(T) = 52$.

---

To Find:

The number of people who like both coffee and tea, $n(C \cap T)$.

---

Solution:

We use the formula for the number of elements in the union of two sets:

$n(C \cup T) = n(C) + n(T) - n(C \cap T)$

We need to find $n(C \cap T)$, so we rearrange the formula:

$n(C \cap T) = n(C) + n(T) - n(C \cup T)$

Substitute the given values into the formula:

$n(C \cap T) = 37 + 52 - 70$

$n(C \cap T) = 89 - 70$

$n(C \cap T) = 19$

Therefore, 19 people like both coffee and tea.

Given:

Let C be the set of people who like cricket.

Let T be the set of people who like tennis.

Total number of people in the group, $n(C \cup T) = 65$.

Number of people who like cricket, $n(C) = 40$.

Number of people who like both cricket and tennis, $n(C \cap T) = 10$.

---

To Find:

Number of people who like tennis only (and not cricket).

Number of people who like tennis ($n(T)$).

---

Solution:

We use the formula for the number of elements in the union of two sets:

$n(C \cup T) = n(C) + n(T) - n(C \cap T)$

We need to find $n(T)$. Let's substitute the given values into the formula:

$65 = 40 + n(T) - 10$

Simplify the right side:

$65 = 30 + n(T)$

To find $n(T)$, subtract 30 from both sides:

$n(T) = 65 - 30$

$n(T) = 35$

So, the number of people who like tennis is 35.

Now, we need to find the number of people who like tennis only (and not cricket).

This is given by the number of people who like tennis minus the number of people who like both cricket and tennis:

Number of people who like tennis only = $n(T) - n(C \cap T)$

Substitute the values we found and were given:

Number of people who like tennis only = $35 - 10$

Number of people who like tennis only = $25$

Therefore, 35 people like tennis, and 25 people like tennis only.

Given:

Let F be the set of people who speak French.

Let S be the set of people who speak Spanish.

Number of people who speak French, $n(F) = 50$.

Number of people who speak Spanish, $n(S) = 20$.

Number of people who speak both French and Spanish, $n(F \cap S) = 10$.

---

To Find:

The number of people who speak at least one of the two languages, $n(F \cup S)$.

---

Solution:

The number of people who speak at least one of the two languages is given by the union of the two sets, $n(F \cup S)$.

We use the formula for the number of elements in the union of two sets:

$n(F \cup S) = n(F) + n(S) - n(F \cap S)$

Substitute the given values into the formula:

$n(F \cup S) = 50 + 20 - 10$

$n(F \cup S) = 70 - 10$

$n(F \cup S) = 60$

Therefore, 60 people speak at least one of the two languages.

Given:

The word "CATARACT".

The word "TRACT".

---

To Show:

That the set of letters in "CATARACT" is equal to the set of letters in "TRACT".

---

Solution:

Let A be the set of letters needed to spell the word "CATARACT".

The letters in "CATARACT" are C, A, T, A, R, A, C, T.

When we write a set, we list the unique elements. The unique letters are C, A, T, and R.

So, the set A is:

$A = \{C, A, T, R\}$

Let B be the set of letters needed to spell the word "TRACT".

The letters in "TRACT" are T, R, A, C, T.

The unique letters are T, R, A, and C.

So, the set B is:

$B = \{T, R, A, C\}$

Now, we compare the elements of set A and set B.

Set A contains the elements {C, A, T, R}.

Set B contains the elements {T, R, A, C}.

Recall that the order of elements in a set does not matter.

We can see that both sets contain exactly the same elements: C, A, T, and R.

Since every element in A is in B, and every element in B is in A, the two sets are equal.

Thus, $A = B$.

This shows that the set of letters needed to spell “CATARACT” and the set of letters needed to spell “TRACT” are indeed equal.

Given:

The set A = $\{-1, 0, 1\}$.

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To Find:

All the subsets of set A.

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Solution:

A subset is a set containing some or all of the elements of another set.

For a set with $n$ elements, the total number of subsets is $2^n$.

In this case, the set A = $\{-1, 0, 1\}$ has $n = 3$ elements.

So, the total number of subsets is $2^3 = 8$.

We list all possible subsets:

1. The empty set (which is a subset of every set): $\emptyset$ or \{\}

2. Subsets containing one element:

$\{-1\}$

$\{0\}$

$\{1\}$

3. Subsets containing two elements:

$\{-1, 0\}$

$\{-1, 1\}$

$\{0, 1\}$

4. The subset containing all elements (the set itself):

$\{-1, 0, 1\}$

Listing all the subsets together:

$\emptyset$, $\{-1\}$, $\{0\}$, $\{1\}$, $\{-1, 0\}$, $\{-1, 1\}$, $\{0, 1\}$, $\{-1, 0, 1\}$.

Given:

A and B are two sets such that $A \cup B = A \cap B$.

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To Show:

$A = B$

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Proof:

To show that $A = B$, we need to prove that $A \subseteq B$ and $B \subseteq A$.

Part 1: Show $A \subseteq B$.

Let $x$ be an arbitrary element such that $x \in A$.

If $x \in A$, then $x$ must also be an element of the union of A and B.

$x \in A \implies x \in A \cup B$

We are given that $A \cup B = A \cap B$.

Therefore, since $x \in A \cup B$, it must also be true that $x \in A \cap B$.

$x \in A \cup B \implies x \in A \cap B$

By the definition of intersection, if an element is in the intersection of two sets, it must be in both sets.

$x \in A \cap B \implies x \in A \text{ and } x \in B$

Since we started with $x \in A$ and concluded that $x \in B$, this shows that every element of A is also an element of B.

Thus, $A \subseteq B$.

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Part 2: Show $B \subseteq A$.

Let $y$ be an arbitrary element such that $y \in B$.

If $y \in B$, then $y$ must also be an element of the union of A and B.

$y \in B \implies y \in A \cup B$

We are given that $A \cup B = A \cap B$.

Therefore, since $y \in A \cup B$, it must also be true that $y \in A \cap B$.

$y \in A \cup B \implies y \in A \cap B$

By the definition of intersection, if an element is in the intersection of two sets, it must be in both sets.

$y \in A \cap B \implies y \in A \text{ and } y \in B$

Since we started with $y \in B$ and concluded that $y \in A$, this shows that every element of B is also an element of A.

Thus, $B \subseteq A$.

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Conclusion:

From Part 1, we have shown that $A \subseteq B$.

From Part 2, we have shown that $B \subseteq A$.

By the definition of set equality, if $A \subseteq B$ and $B \subseteq A$, then $A = B$.

Hence, $A \cup B = A \cap B$ implies $A = B$.

Given:

Two sets A and B.

$P(S)$ denotes the power set of a set S (the set of all subsets of S).

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To Show:

$P(A \cap B) = P(A) \cap P(B)$

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Proof:

To show that two sets are equal, we must show that each is a subset of the other.

Part 1: Show $P(A \cap B) \subseteq P(A) \cap P(B)$.

Let X be an arbitrary element of $P(A \cap B)$.

$X \in P(A \cap B)$

By the definition of the power set, this means X is a subset of $A \cap B$.

$X \subseteq A \cap B$

If X is a subset of $A \cap B$, then every element in X must be in both A and B.

This implies that X is a subset of A, and X is also a subset of B.

$X \subseteq A$ and $X \subseteq B$

By the definition of the power set, if X is a subset of a set, then X is an element of the power set of that set.

$X \subseteq A \implies X \in P(A)$

$X \subseteq B \implies X \in P(B)$

Since X is in both $P(A)$ and $P(B)$, by the definition of intersection, X is in their intersection.

$X \in P(A) \text{ and } X \in P(B) \implies X \in P(A) \cap P(B)$

Since we started with an arbitrary element X in $P(A \cap B)$ and showed it is also in $P(A) \cap P(B)$, we have proved that $P(A \cap B) \subseteq P(A) \cap P(B)$.

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Part 2: Show $P(A) \cap P(B) \subseteq P(A \cap B)$.

Let Y be an arbitrary element of $P(A) \cap P(B)$.

$Y \in P(A) \cap P(B)$

By the definition of intersection, this means Y is an element of both $P(A)$ and $P(B)$.

$Y \in P(A)$ and $Y \in P(B)$

By the definition of the power set, if Y is an element of the power set of a set, then Y is a subset of that set.

$Y \in P(A) \implies Y \subseteq A$

$Y \in P(B) \implies Y \subseteq B$

If Y is a subset of A and Y is a subset of B, it means every element in Y is in A, and every element in Y is also in B.

Therefore, every element in Y must be in both A and B, which means every element in Y is in $A \cap B$.

This implies that Y is a subset of $A \cap B$.

$Y \subseteq A \text{ and } Y \subseteq B \implies Y \subseteq A \cap B$

By the definition of the power set, if Y is a subset of $A \cap B$, then Y is an element of $P(A \cap B)$.

$Y \subseteq A \cap B \implies Y \in P(A \cap B)$

Since we started with an arbitrary element Y in $P(A) \cap P(B)$ and showed it is also in $P(A \cap B)$, we have proved that $P(A) \cap P(B) \subseteq P(A \cap B)$.

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Conclusion:

From Part 1, we have $P(A \cap B) \subseteq P(A) \cap P(B)$.

From Part 2, we have $P(A) \cap P(B) \subseteq P(A \cap B)$.

Since both subset conditions are met, by the definition of set equality, we conclude that:

$P(A \cap B) = P(A) \cap P(B)$

Given:

Let U be the set of all consumers surveyed.

Let A be the set of consumers who like product A.

Let B be the set of consumers who like product B.

Total number of consumers surveyed, $n(U) = 1000$.

Number of consumers who like product A, $n(A) = 720$.

Number of consumers who like product B, $n(B) = 450$.

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To Find:

The least number of consumers who liked both products A and B, i.e., the minimum value of $n(A \cap B)$.

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Solution:

We know the formula for the number of elements in the union of two sets:

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

We want to find $n(A \cap B)$, so we rearrange the formula:

$n(A \cap B) = n(A) + n(B) - n(A \cup B)$

To find the least number of people who liked both products ($n(A \cap B)_{\text{min}}$), we need the greatest possible value for $n(A \cup B)$.

The set $A \cup B$ represents the consumers who like product A or product B (or both). The total number of consumers surveyed is 1000.

The maximum possible value for the number of consumers who liked at least one product is the total number of consumers surveyed.

Thus, the greatest possible value for $n(A \cup B)$ is $n(U) = 1000$.

$n(A \cup B)_{\text{max}} = 1000$

Now, substitute the given values and the maximum value of $n(A \cup B)$ into the rearranged formula to find the minimum value of $n(A \cap B)$:

$n(A \cap B)_{\text{min}} = n(A) + n(B) - n(A \cup B)_{\text{max}}$

$n(A \cap B)_{\text{min}} = 720 + 450 - 1000$

$n(A \cap B)_{\text{min}} = 1170 - 1000$

$n(A \cap B)_{\text{min}} = 170$

Therefore, the least number of consumers that must have liked both products is 170.

Given:

Total number of car owners investigated, $n(U) = 500$.

Let A be the set of car owners who owned car A, $n(A) = 400$.

Let B be the set of car owners who owned car B, $n(B) = 200$.

Number of car owners who owned both A and B cars, $n(A \cap B) = 50$.

---

To Check:

Whether the given data is correct.

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Solution:

We can use the formula for the number of elements in the union of two sets to find the number of people who own car A or car B (or both):

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Substitute the given values into the formula:

$n(A \cup B) = 400 + 200 - 50$

$n(A \cup B) = 600 - 50$

$n(A \cup B) = 550$

The number of people who own at least one of the cars A or B is $n(A \cup B) = 550$.

The total number of car owners investigated is $n(U) = 500$.

The number of people who own at least one car ($n(A \cup B)$) cannot be more than the total number of people investigated ($n(U)$).

We found that $n(A \cup B) = 550$, which is greater than $n(U) = 500$.

Since $n(A \cup B) > n(U)$, the given data is contradictory and therefore incorrect.

Given:

Let F be the set of men who received medals in Football.

Let B be the set of men who received medals in Basketball.

Let C be the set of men who received medals in Cricket.

Number of medals in Football, $n(F) = 38$.

Number of medals in Basketball, $n(B) = 15$.

Number of medals in Cricket, $n(C) = 20$.

Total number of men who received medals (at least one), $n(F \cup B \cup C) = 58$.

Number of men who got medals in all three sports, $n(F \cap B \cap C) = 3$.

---

To Find:

The number of men who received medals in exactly two of the three sports.

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Solution:

We use the formula for the number of elements in the union of three sets:

$n(F \cup B \cup C) = n(F) + n(B) + n(C) - [n(F \cap B) + n(B \cap C) + n(C \cap F)] + n(F \cap B \cap C)$

Substitute the given values into the formula:

$58 = 38 + 15 + 20 - [n(F \cap B) + n(B \cap C) + n(C \cap F)] + 3$

$58 = 73 - [n(F \cap B) + n(B \cap C) + n(C \cap F)] + 3$

$58 = 76 - [n(F \cap B) + n(B \cap C) + n(C \cap F)]

Rearrange the equation to find the sum of the pairwise intersections:

$n(F \cap B) + n(B \cap C) + n(C \cap F) = 76 - 58$

$n(F \cap B) + n(B \cap C) + n(C \cap F) = 18$

The number of men who received medals in exactly two sports is given by the sum of the pairwise intersections minus three times the number of men who received medals in all three sports (since those in the intersection of all three are counted in each pairwise intersection).

Number of men who received medals in exactly two sports = $[n(F \cap B) + n(B \cap C) + n(C \cap F)] - 3 \times n(F \cap B \cap C)$

Substitute the values we found:

Number of men who received medals in exactly two sports = $18 - 3 \times 3$

Number of men who received medals in exactly two sports = $18 - 9$

Number of men who received medals in exactly two sports = $9$

Therefore, 9 men received medals in exactly two of the three sports.

Given:

Set A defined by the equation $x^2 - 8x + 12 = 0$.

Set B = $\{2, 4, 6\}$.

Set C = $\{2, 4, 6, 8, ...\}$ (the set of positive even integers).

Set D = $\{6\}$.

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To Determine:

Which of the given sets are subsets of one another.

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Solution:

First, let's find the elements of set A by solving the quadratic equation $x^2 - 8x + 12 = 0$.

We can factor the equation:

$(x - 2)(x - 6) = 0$

The solutions are $x - 2 = 0$ or $x - 6 = 0$.

This gives $x = 2$ or $x = 6$.

So, set A is:

$A = \{2, 6\}$

Now we have the explicit elements of all sets:

$A = \{2, 6\}$

$B = \{2, 4, 6\}$

$C = \{2, 4, 6, 8, 10, ...\}$

$D = \{6\}$

We now check the subset relationships between these sets. A set X is a subset of a set Y ($X \subseteq Y$) if every element of X is also an element of Y.

1. Comparing A and B:

Elements of A are 2 and 6. Both 2 and 6 are present in B = $\{2, 4, 6\}$.

So, $A \subseteq B$.

Elements of B are 2, 4, and 6. The element 4 is in B but not in A. So, $B \not\subseteq A$.

2. Comparing A and C:

Elements of A are 2 and 6. Both 2 and 6 are positive even integers, so they are present in C = $\{2, 4, 6, 8, ...\}$.

So, $A \subseteq C$.

Elements of C include 8, 10, etc., which are not in A. So, $C \not\subseteq A$.

3. Comparing A and D:

Elements of A are 2 and 6. The element 2 is in A but not in D = $\{6\}$. So, $A \not\subseteq D$.

Element of D is 6. The element 6 is present in A = $\{2, 6\}$.

So, $D \subseteq A$.

4. Comparing B and C:

Elements of B are 2, 4, and 6. These are all positive even integers, so they are present in C = $\{2, 4, 6, 8, ...\}$.

So, $B \subseteq C$.

Elements of C include 8, 10, etc., which are not in B. So, $C \not\subseteq B$.

5. Comparing B and D:

Elements of B are 2, 4, and 6. The elements 2 and 4 are in B but not in D = $\{6\}$. So, $B \not\subseteq D$.

Element of D is 6. The element 6 is present in B = $\{2, 4, 6\}$.

So, $D \subseteq B$.

6. Comparing C and D:

Elements of C include 2, 4, etc., which are not in D = $\{6\}$. So, $C \not\subseteq D$.

Element of D is 6. The element 6 is a positive even integer, so it is present in C = $\{2, 4, 6, 8, ...\}$.

So, $D \subseteq C$.

Summarizing the subset relationships:

$A \subseteq B$

$A \subseteq C$

$B \subseteq C$

$D \subseteq A$

$D \subseteq B$

$D \subseteq C$

No other subset relationships exist between these sets.

Statement (i): If $x \in A$ and $A \in B$, then $x \in B$.

Truth Value: False.

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Reason / Counterexample:

Let $A = \{1\}$.

Let $x = 1$. Then $x \in A$.

Let $B = \{A, 2, 3\} = \{\{1\}, 2, 3\}$. Then $A \in B$ (since the set $\{1\}$ is an element of B).

Now, check if $x \in B$. Is $1 \in \{\{1\}, 2, 3\}$? No, the number 1 is not an element of B. The set $\{1\}$ is an element of B, but $1 \ne \{1\}$.

So, the statement is false.

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Statement (ii): If $A \subset B$ and $B \in C$, then $A \in C$.

Truth Value: False.

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Reason / Counterexample:

Let $A = \{1\}$.

Let $B = \{1, 2\}$. Then $A \subset B$ (since every element of A is in B, and B has an element 2 which is not in A).

Let $C = \{B, 3\} = \{\{1, 2\}, 3\}$. Then $B \in C$ (since the set $\{1, 2\}$ is an element of C).

Now, check if $A \in C$. Is $\{1\} \in \{\{1, 2\}, 3\}$? No, the set $\{1\}$ is not an element of C. The set $\{1, 2\}$ is an element of C, but $\{1\} \ne \{1, 2\}$.

So, the statement is false.

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Statement (iii): If $A \subset B$ and $B \subset C$, then $A \subset C$.

Truth Value: True.

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Proof:

Given that $A \subset B$ and $B \subset C$.

By the definition of a proper subset, $A \subset B$ implies $A \subseteq B$ and $A \ne B$.

Also, $B \subset C$ implies $B \subseteq C$ and $B \ne C$.

To show $A \subset C$, we need to prove two things:

1. $A \subseteq C$

2. $A \ne C$

Let's prove $A \subseteq C$. Assume $x \in A$. Since $A \subseteq B$, it follows that $x \in B$. Since $B \subseteq C$, it follows that $x \in C$. Thus, if $x \in A$, then $x \in C$. This proves $A \subseteq C$.

Now, let's prove $A \ne C$. We are given that $B \subset C$, which means $B \ne C$ and $B \subseteq C$. Since $A \subset B$, there exists at least one element $b \in B$ such that $b \notin A$. Since $B \subseteq C$, this element $b$ is also in $C$. So, we have $b \in C$ and $b \notin A$. This shows that there is an element in C that is not in A, which means $A$ cannot be equal to C.

Alternatively, suppose $A=C$. Since $A \subseteq B$, this implies $C \subseteq B$. We were given $B \subseteq C$. If $C \subseteq B$ and $B \subseteq C$, then $B=C$. But we were given $B \subset C$, which means $B \ne C$. This is a contradiction. Therefore, the assumption $A=C$ must be false, so $A \ne C$.

Since $A \subseteq C$ and $A \ne C$, by the definition of a proper subset, we have $A \subset C$.

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Statement (iv): If $A \not\subseteq B$ and $B \not\subseteq C$, then $A \not\subseteq C$.

Truth Value: False.

---

Reason / Counterexample:

Let $A = \{1\}$.

Let $B = \{2\}$.

Let $C = \{1, 3\}$.

Check the premises:

$A \not\subseteq B$: The element $1 \in A$, but $1 \notin B$. So $A \not\subseteq B$ is true.

$B \not\subseteq C$: The element $2 \in B$, but $2 \notin C$. So $B \not\subseteq C$ is true.

Check the conclusion:

$A \not\subseteq C$: The element $1 \in A$, and $1 \in C$. Since the only element in A is also in C, $A \subseteq C$. Thus, $A \not\subseteq C$ is false.

Since the premises are true but the conclusion is false, the statement is false.

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Statement (v): If $x \in A$ and $A \not\subseteq B$, then $x \in B$.

Truth Value: False.

---

Reason / Counterexample:

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

Let $B = \{1, 3\}$.

Check the premise $A \not\subseteq B$: The element $2 \in A$, but $2 \notin B$. So $A \not\subseteq B$ is true.

Let $x = 2$. Then $x \in A$ (since $2 \in \{1, 2\}$).

So, the premise ($x \in A$ and $A \not\subseteq B$) is true for $x=2$.

Check the conclusion $x \in B$: Is $2 \in \{1, 3\}$? No, $2 \notin B$. The conclusion is false for $x=2$.

Since we found a value of $x$ for which the premise is true but the conclusion is false, the statement is false.

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Statement (vi): If $A \subset B$ and $x \notin B$, then $x \notin A$.

Truth Value: True.

---

Proof:

Given that $A \subset B$ and $x \notin B$.

The condition $A \subset B$ implies that $A \subseteq B$ (A is a subset of B).

By the definition of a subset, if $A \subseteq B$, then every element in A is also an element in B.

This can be written as: if $y \in A$, then $y \in B$ for all $y$.

We are given that $x \notin B$, which means x is not an element of B.

We want to show that $x \notin A$.

Assume for contradiction that $x \in A$.

If $x \in A$, and we know that for any element $y$, if $y \in A$ then $y \in B$ (because $A \subseteq B$), then it must follow that $x \in B$.

So, $x \in A \implies x \in B$.

However, we were given that $x \notin B$. This contradicts our conclusion that $x \in B$.

Therefore, our initial assumption that $x \in A$ must be false.

Hence, $x \notin A$.

This proves that if $A \subset B$ and $x \notin B$, then $x \notin A$.

Given:

A, B, and C are sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$.

---

To Show:

$B = C$

---

Proof:

To show that $B = C$, we must prove that $B \subseteq C$ and $C \subseteq B$.

Part 1: Show $B \subseteq C$.

Let $x$ be an arbitrary element such that $x \in B$.

Since $x \in B$, it implies that $x$ is in the union of A and B.

$x \in B \implies x \in A \cup B$

We are given that $A \cup B = A \cup C$. Therefore, if $x \in A \cup B$, then $x \in A \cup C$.

$x \in A \cup B \implies x \in A \cup C$

By the definition of union, $x \in A \cup C$ means either $x \in A$ or $x \in C$.

So, we have two cases:

Case 1: $x \in C$. In this case, we have already shown that if $x \in B$, then $x \in C$.

Case 2: $x \notin C$. If $x \in A \cup C$ and $x \notin C$, it must be the case that $x \in A$. So, if $x \in B$ and $x \notin C$, then $x \in A$.

If $x \in B$ and $x \in A$, then by the definition of intersection, $x \in A \cap B$.

$x \in B \text{ and } x \in A \implies x \in A \cap B$

We are given that $A \cap B = A \cap C$. Therefore, if $x \in A \cap B$, then $x \in A \cap C$.

$x \in A \cap B \implies x \in A \cap C$

By the definition of intersection, $x \in A \cap C$ means $x \in A$ and $x \in C$.

This implies that $x \in C$.

So, in both cases ($x \in C$ or $x \notin C$), starting with $x \in B$ leads to $x \in C$.

Thus, we have shown that if $x \in B$, then $x \in C$. This proves $B \subseteq C$.

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Part 2: Show $C \subseteq B$.

Let $y$ be an arbitrary element such that $y \in C$.

Since $y \in C$, it implies that $y$ is in the union of A and C.

$y \in C \implies y \in A \cup C$

We are given that $A \cup C = A \cup B$. Therefore, if $y \in A \cup C$, then $y \in A \cup B$.

$y \in A \cup C \implies y \in A \cup B$

By the definition of union, $y \in A \cup B$ means either $y \in A$ or $y \in B$.

So, we have two cases:

Case 1: $y \in B$. In this case, we have already shown that if $y \in C$, then $y \in B$.

Case 2: $y \notin B$. If $y \in A \cup B$ and $y \notin B$, it must be the case that $y \in A$. So, if $y \in C$ and $y \notin B$, then $y \in A$.

If $y \in C$ and $y \in A$, then by the definition of intersection, $y \in A \cap C$.

$y \in C \text{ and } y \in A \implies y \in A \cap C$

We are given that $A \cap C = A \cap B$. Therefore, if $y \in A \cap C$, then $y \in A \cap B$.

$y \in A \cap C \implies y \in A \cap B$

By the definition of intersection, $y \in A \cap B$ means $y \in A$ and $y \in B$.

This implies that $y \in B$.

So, in both cases ($y \in B$ or $y \notin B$), starting with $y \in C$ leads to $y \in B$.

Thus, we have shown that if $y \in C$, then $y \in B$. This proves $C \subseteq B$.

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Conclusion:

From Part 1, we have shown that $B \subseteq C$.

From Part 2, we have shown that $C \subseteq B$.

By the definition of set equality, if $B \subseteq C$ and $C \subseteq B$, then $B = C$.

Hence, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.

Given:

Sets A and B.

We are asked to show that the following four conditions are equivalent:

(i) $A \subseteq B$ (Interpretation assuming strict subset '$\subset$' in the question was intended as non-strict '$\subseteq$' for equivalence with the other conditions)

(ii) $A - B = \phi$

(iii) $A \cup B = B$

(iv) $A \cap B = A$

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To Show:

That conditions (i), (ii), (iii), and (iv) are equivalent.

---

Proof:

To show that the four conditions are equivalent, we will prove the following cycle of implications: (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (i).

Proof of (i) $\implies$ (ii): ($A \subseteq B \implies A - B = \phi$)

Assume $A \subseteq B$.

The set $A - B$ contains elements that are in A but not in B.

By the definition of a subset, if $A \subseteq B$, then every element of A is also an element of B.

This means there is no element that is in A but not in B.

Therefore, the set $A - B$ has no elements, which means $A - B = \phi$.

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Proof of (ii) $\implies$ (iii): ($A - B = \phi \implies A \cup B = B$)

Assume $A - B = \phi$.

We want to show $A \cup B = B$. To prove set equality, we must show $A \cup B \subseteq B$ and $B \subseteq A \cup B$.

The inclusion $B \subseteq A \cup B$ is always true by the definition of union (any element in B is in the union of A and B).

Now, let $x$ be an arbitrary element such that $x \in A \cup B$.

By the definition of union, this means $x \in A$ or $x \in B$.

If $x \in B$, the condition $x \in B$ is satisfied.

If $x \in A$, since we assumed $A - B = \phi$, there are no elements in A that are not in B. This implies that if $x \in A$, then $x$ must also be in B.

So, in either case ($x \in A$ or $x \in B$), we conclude that $x \in B$.

Thus, if $x \in A \cup B$, then $x \in B$, which means $A \cup B \subseteq B$.

Since $A \cup B \subseteq B$ and $B \subseteq A \cup B$, we have $A \cup B = B$.

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Proof of (iii) $\implies$ (iv): ($A \cup B = B \implies A \cap B = A$)

Assume $A \cup B = B$.

We want to show $A \cap B = A$. To prove set equality, we must show $A \cap B \subseteq A$ and $A \subseteq A \cap B$.

The inclusion $A \cap B \subseteq A$ is always true by the definition of intersection (any element in the intersection of A and B must be in A).

Now, let $x$ be an arbitrary element such that $x \in A$.

By the definition of union, if $x \in A$, then $x \in A \cup B$.

Given that $A \cup B = B$, it follows that $x \in B$.

So, we have $x \in A$ and $x \in B$.

By the definition of intersection, if $x \in A$ and $x \in B$, then $x \in A \cap B$.

Thus, if $x \in A$, then $x \in A \cap B$, which means $A \subseteq A \cap B$.

Since $A \cap B \subseteq A$ and $A \subseteq A \cap B$, we have $A \cap B = A$.

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Proof of (iv) $\implies$ (i): ($A \cap B = A \implies A \subseteq B$)

Assume $A \cap B = A$.

We want to show $A \subseteq B$. Let $x$ be an arbitrary element such that $x \in A$.

Since $A = A \cap B$, if $x \in A$, it must also be true that $x \in A \cap B$.

By the definition of intersection, if $x \in A \cap B$, then $x \in A$ and $x \in B$.

In particular, this means $x \in B$.

Thus, if $x \in A$, then $x \in B$.

This is the definition of $A \subseteq B$.

Hence, $A \subseteq B$.

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Conclusion:

We have shown that condition (i) implies condition (ii), condition (ii) implies condition (iii), condition (iii) implies condition (iv), and condition (iv) implies condition (i).

This chain of implications proves that all four conditions are equivalent.

That is, $A \subseteq B \iff A - B = \phi \iff A \cup B = B \iff A \cap B = A$.

Given:

Sets A, B, and C.

The condition $A \subset B$.

(Note: The symbol $\subset$ is sometimes used to denote a proper subset ($A \subseteq B$ and $A \neq B$) and sometimes to denote any subset ($A \subseteq B$). Based on common textbook exercises involving set identities, the latter interpretation ($A \subseteq B$) is usually intended when proving inclusions like $C-B \subseteq C-A$. We will provide the proof assuming $A \subseteq B$. If $A \subset B$ strictly is intended, the statement might be false, as shown in thought process, but the inclusion $C-B \subseteq C-A$ always holds when $A \subseteq B$).

We assume the given condition is $A \subseteq B$.

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To Show:

$C - B \subseteq C - A$.

(Assuming the symbol $\subset$ in the conclusion also means $\subseteq$).

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Proof:

We need to show that every element in the set $C - B$ is also an element in the set $C - A$.

Let $x$ be an arbitrary element such that $x \in C - B$.

By the definition of set difference, $x \in C - B$ means that $x \in C$ and $x \notin B$.

We are given the condition that $A \subseteq B$.

By the definition of a subset, if $A \subseteq B$, then every element that is in set A is also in set B.

In other words, there is no element that is in A but not in B.

This is logically equivalent to stating that if an element is not in B, then it cannot be in A.

Since we know that $x \notin B$ (from $x \in C - B$), and given $A \subseteq B$, it must necessarily follow that $x \notin A$.

So, from $x \in C - B$, we have concluded that $x \in C$ and $x \notin A$.

By the definition of set difference, having $x \in C$ and $x \notin A$ means that $x \in C - A$.

Thus, we have shown that if an element $x$ is in $C - B$, then $x$ is also in $C - A$.

Therefore, by the definition of a subset, $C - B \subseteq C - A$.

If the question intended the strict subset $A \subset B$ and $C-B \subset C-A$:
While $A \subset B$ implies $C-B \subseteq C-A$, it does not necessarily imply $C-B \neq C-A$. For example, if $A=\{1\}$, $B=\{1,2\}$, and $C=\{3,4\}$, then $A \subset B$ is true, but $C-B = \{3,4\}$ and $C-A = \{3,4\}$, so $C-B = C-A$, and thus $C-B \not\subset C-A$. However, based on standard set theory exercises, the problem likely assumes $A \subseteq B \implies C-B \subseteq C-A$.

Given:

Two sets A and B such that $P(A) = P(B)$.

$P(S)$ denotes the power set of a set S, which is the set of all subsets of S.

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To Show:

$A = B$

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Proof:

To show that $A = B$, we must prove that $A \subseteq B$ and $B \subseteq A$.

Part 1: Show $A \subseteq B$.

Let $x$ be an arbitrary element of set A.

$x \in A$

Consider the singleton set containing only the element $x$, which is $\{x\}$.

Since $x \in A$, the set $\{x\}$ is a subset of A.

$\{x\} \subseteq A$

By the definition of the power set, if a set is a subset of A, then it is an element of $P(A)$.

Since $\{x\} \subseteq A$, it follows that $\{x\} \in P(A)$.

We are given that $P(A) = P(B)$. Therefore, if $\{x\} \in P(A)$, then $\{x\}$ must also be an element of $P(B)$.

$\{x\} \in P(A) \implies \{x\} \in P(B)$

By the definition of the power set, if a set is an element of $P(B)$, then it is a subset of B.

Since $\{x\} \in P(B)$, it follows that $\{x\} \subseteq B$.

By the definition of a subset, if the set $\{x\}$ is a subset of B, then the element $x$ must be in B.

$\{x\} \subseteq B \implies x \in B$

Thus, we have shown that if $x \in A$, then $x \in B$.

This proves that $A$ is a subset of $B$, i.e., $A \subseteq B$.

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Part 2: Show $B \subseteq A$.

Let $y$ be an arbitrary element of set B.

$y \in B$

Consider the singleton set containing only the element $y$, which is $\{y\}$.

Since $y \in B$, the set $\{y\}$ is a subset of B.

$\{y\} \subseteq B$

By the definition of the power set, if a set is a subset of B, then it is an element of $P(B)$.

Since $\{y\} \subseteq B$, it follows that $\{y\} \in P(B)$.

We are given that $P(B) = P(A)$. Therefore, if $\{y\} \in P(B)$, then $\{y\}$ must also be an element of $P(A)$.

$\{y\} \in P(B) \implies \{y\} \in P(A)$

By the definition of the power set, if a set is an element of $P(A)$, then it is a subset of A.

Since $\{y\} \in P(A)$, it follows that $\{y\} \subseteq A$.

By the definition of a subset, if the set $\{y\}$ is a subset of A, then the element $y$ must be in A.

$\{y\} \subseteq A \implies y \in A$

Thus, we have shown that if $y \in B$, then $y \in A$.

This proves that $B$ is a subset of $A$, i.e., $B \subseteq A$.

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Conclusion:

From Part 1, we have shown that $A \subseteq B$.

From Part 2, we have shown that $B \subseteq A$.

By the definition of set equality, if $A$ is a subset of $B$ and $B$ is a subset of $A$, then $A$ and $B$ are equal.

Therefore, $A = B$.

Statement: For any sets A and B, $P(A) \cup P(B) = P(A \cup B)$.

Truth Value: False.

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Justification (using a Counterexample):

Let's consider specific sets A and B.

Let $A = \{1\}$.

The power set of A is $P(A) = \{\emptyset, \{1\}\}$.

Let $B = \{2\}$.

The power set of B is $P(B) = \{\emptyset, \{2\}\}$.

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Now, let's find the union of the power sets, $P(A) \cup P(B)$.

$P(A) \cup P(B) = \{\emptyset, \{1\}\} \cup \{\emptyset, \{2\}\}$

$P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\}$

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Next, let's find the union of sets A and B, $A \cup B$.

$A \cup B = \{1\} \cup \{2\}$

$A \cup B = \{1, 2\}$

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Finally, let's find the power set of $A \cup B$, which is $P(A \cup B)$.

$P(A \cup B) = P(\{1, 2\})$

$P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$

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Comparing $P(A) \cup P(B)$ and $P(A \cup B)$:

$P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\}$

$P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}$

We can see that the set $\{1, 2\}$ is an element of $P(A \cup B)$, but it is not an element of $P(A) \cup P(B)$ because $\{1, 2\}$ is not a subset of A (since $2 \notin A$) and it is not a subset of B (since $1 \notin B$).

Since we found an element that is in $P(A \cup B)$ but not in $P(A) \cup P(B)$, the two sets are not equal.

Therefore, the statement $P(A) \cup P(B) = P(A \cup B)$ is false.

Given:

Sets A and B.

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To Show:

1. $A = (A \cap B) \cup (A - B)$

2. $A \cup (B - A) = A \cup B$

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Proof of Identity 1: $A = (A \cap B) \cup (A - B)$

To prove that the two sets are equal, we will show that each set is a subset of the other.

Part 1.1: Show $A \subseteq (A \cap B) \cup (A - B)$.

Let $x$ be an arbitrary element such that $x \in A$.

According to the law of excluded middle, an element is either in a set or not in that set. So, either $x \in B$ or $x \notin B$.

Case 1: $x \in B$. If $x \in A$ and $x \in B$, then by the definition of intersection, $x \in A \cap B$. If $x \in A \cap B$, then by the definition of union, $x \in (A \cap B) \cup (A - B)$.

Case 2: $x \notin B$. If $x \in A$ and $x \notin B$, then by the definition of set difference, $x \in A - B$. If $x \in A - B$, then by the definition of union, $x \in (A \cap B) \cup (A - B)$.

In both cases, if $x \in A$, then $x \in (A \cap B) \cup (A - B)$.

Thus, $A \subseteq (A \cap B) \cup (A - B)$.

Part 1.2: Show $(A \cap B) \cup (A - B) \subseteq A$.

Let $y$ be an arbitrary element such that $y \in (A \cap B) \cup (A - B)$.

By the definition of union, this means $y \in (A \cap B)$ or $y \in (A - B)$.

Case 1: $y \in A \cap B$. By the definition of intersection, $y \in A \cap B$ means $y \in A$ and $y \in B$. This directly implies that $y \in A$.

Case 2: $y \in A - B$. By the definition of set difference, $y \in A - B$ means $y \in A$ and $y \notin B$. This directly implies that $y \in A$.

In both cases, if $y \in (A \cap B) \cup (A - B)$, then $y \in A$.

Thus, $(A \cap B) \cup (A - B) \subseteq A$.

Since we have shown that $A \subseteq (A \cap B) \cup (A - B)$ and $(A \cap B) \cup (A - B) \subseteq A$, by the definition of set equality, we conclude that $A = (A \cap B) \cup (A - B)$.

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Proof of Identity 2: $A \cup (B - A) = A \cup B$

To prove that the two sets are equal, we will show that each set is a subset of the other.

Part 2.1: Show $A \cup (B - A) \subseteq A \cup B$.

Let $x$ be an arbitrary element such that $x \in A \cup (B - A)$.

By the definition of union, this means $x \in A$ or $x \in (B - A)$.

Case 1: $x \in A$. By the definition of union, if $x \in A$, then $x \in A \cup B$.

Case 2: $x \in B - A$. By the definition of set difference, $x \in B - A$ means $x \in B$ and $x \notin A$. If $x \in B$, then by the definition of union, $x \in A \cup B$.

In both cases, if $x \in A \cup (B - A)$, then $x \in A \cup B$.

Thus, $A \cup (B - A) \subseteq A \cup B$.

Part 2.2: Show $A \cup B \subseteq A \cup (B - A)$.

Let $y$ be an arbitrary element such that $y \in A \cup B$.

By the definition of union, this means $y \in A$ or $y \in B$.

Case 1: $y \in A$. By the definition of union, if $y \in A$, then $y \in A \cup (B - A)$.

Case 2: $y \in B$. If $y \in B$, we consider two sub-cases: either $y \in A$ or $y \notin A$.

Subcase 2a: $y \in A$. This falls under Case 1, and we concluded $y \in A \cup (B - A)$.

Subcase 2b: $y \notin A$. If $y \in B$ and $y \notin A$, then by the definition of set difference, $y \in B - A$. If $y \in B - A$, then by the definition of union, $y \in A \cup (B - A)$.

In all possible scenarios (if $y \in A$ or if $y \in B$), we have concluded that $y \in A \cup (B - A)$.

Thus, $A \cup B \subseteq A \cup (B - A)$.

Since we have shown that $A \cup (B - A) \subseteq A \cup B$ and $A \cup B \subseteq A \cup (B - A)$, by the definition of set equality, we conclude that $A \cup (B - A) = A \cup B$.