Chapter 1 Sets

Given the equation:

$x^2 + x - 2 = 0$

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We need to find the values of $x$ that satisfy this quadratic equation.

We can solve this equation by factoring the quadratic expression.

We are looking for two numbers that multiply to the constant term $(-2)$ and add up to the coefficient of the $x$ term $(1)$.

These two numbers are $+2$ and $-1$, since $2 \times (-1) = -2$ and $2 + (-1) = 1$.

We can rewrite the middle term ($+x$) using these numbers:

$x^2 + 2x - x - 2 = 0$

Now, group the terms and factor by grouping:

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

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

Factor out the common binomial factor $(x + 2)$:

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

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

So, we have two possibilities:

$x + 2 = 0$ or $x - 1 = 0$

Solving the first equation for $x$:

$x = -2$

Solving the second equation for $x$:

$x = 1$

The roots of the equation are $-2$ and $1$.

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The solution set is the set containing these roots.

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

The given set is defined as $\{x : x \text{ is a positive integer and } x^2 < 40\}$.

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The conditions for an element $x$ to be in the set are:

1. $x$ is a positive integer.

2. The square of $x$, $x^2$, is less than 40.

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Let's check the positive integers one by one and see which ones satisfy the second condition:

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

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

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

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

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

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

For $x = 7$, $x^2 = 7^2 = 49$. Since $49 \not< 40$, $x=7$ is not in the set.

Any positive integer greater than 6 will have a square greater than or equal to $7^2=49$, so none of them will satisfy $x^2 < 40$.

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Therefore, the positive integers $x$ that satisfy the condition $x^2 < 40$ are $1, 2, 3, 4, 5, 6$.

The set in roster form is the list of these elements enclosed in curly braces.

The solution set in roster form is {1, 2, 3, 4, 5, 6}.

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

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We observe a pattern in the elements of the set:

$1 = 1^2$

$4 = 2^2$

$9 = 3^2$

$16 = 4^2$

$25 = 5^2$

and so on.

Each element in the set is the square of a positive integer.

Let $x$ represent an element of the set. Then $x$ can be expressed as the square of some positive integer, say $n$.

So, $x = n^2$, where $n$ is a positive integer ($n \in \{1, 2, 3, ...\}$).

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In set-builder form, we describe the elements of the set by stating the property they satisfy.

The set A can be written as $\{x : x \text{ is the square of a positive integer}\}$.

Alternatively, using symbols:

$A = \{x : x = n^2 \text{ for some positive integer } n\}$

or

$A = \{n^2 : n \text{ is a positive integer}\}$

or

$A = \{n^2 : n \in \mathbb{Z}^+\}$

or

$A = \{n^2 : n \in \mathbb{N}\}$ (if $\mathbb{N}$ denotes the set of positive integers {1, 2, 3, ...})

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Using the first symbolic form for clarity as per previous examples:

The set A in set-builder form is $A = \{x : x = n^2 \text{ for some positive integer } n\}$.

The given set is $A = \left\{ \frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7} \right\}$.

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We observe a pattern in the elements of the set.

Each element is a fraction where the numerator is a positive integer and the denominator is one more than the numerator.

Let's list the elements and see the pattern for the numerator and denominator:

$\frac{1}{2}$: Numerator is 1, Denominator is $1+1=2$. Let $n=1$. The element is $\frac{n}{n+1}$.

$\frac{2}{3}$: Numerator is 2, Denominator is $2+1=3$. Let $n=2$. The element is $\frac{n}{n+1}$.

$\frac{3}{4}$: Numerator is 3, Denominator is $3+1=4$. Let $n=3$. The element is $\frac{n}{n+1}$.

$\frac{4}{5}$: Numerator is 4, Denominator is $4+1=5$. Let $n=4$. The element is $\frac{n}{n+1}$.

$\frac{5}{6}$: Numerator is 5, Denominator is $5+1=6$. Let $n=5$. The element is $\frac{n}{n+1}$.

$\frac{6}{7}$: Numerator is 6, Denominator is $6+1=7$. Let $n=6$. The element is $\frac{n}{n+1}$.

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So, each element in the set is of the form $\frac{n}{n+1}$, where $n$ is a positive integer.

The values of $n$ for the elements in the set are 1, 2, 3, 4, 5, and 6.

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In set-builder form, we describe the elements by the property they satisfy.

The set can be written as the set of all elements of the form $\frac{n}{n+1}$ such that $n$ is a positive integer and $n$ is between 1 and 6, inclusive.

Using mathematical notation, this is:

The set in set-builder form is $\left\{ x : x = \frac{n}{n+1}, \text{where } n \text{ is a positive integer and } 1 \leq n \leq 6 \right\}$.

Alternatively, we can write it as:

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

or

$\left\{ \frac{n}{n+1} : n \in \mathbb{Z}^+, 1 \leq n \leq 6 \right\}$

Let's analyse each set in roster form and match it with the corresponding description in set-builder form.

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

This set consists of the unique letters present in the word "PRINCIPAL".

Matching with the options on the right:

(a) describes positive integer divisors of 18.

(b) describes integer solutions to $x^2 - 9 = 0$.

(c) describes integer solution to $x + 1 = 1$.

(d) describes the letters of the word PRINCIPAL.

Thus, set (i) matches with set (d).

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Set (ii): { 0 }

This set contains only the integer 0.

Matching with the remaining options on the right:

(a) describes positive integer divisors of 18 (0 is not positive).

(b) describes integer solutions to $x^2 - 9 = 0$. Let's solve $x^2 - 9 = 0$:

$x^2 = 9$

$x = \pm \sqrt{9}$

$x = 3$ or $x = -3$. The set is {3, -3}. This does not match {0}.

(c) describes integer solution to $x + 1 = 1$. Let's solve $x + 1 = 1$:

$x = 1 - 1$

$x = 0$. The set is {0}. This matches {0}.

Thus, set (ii) matches with set (c).

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

This set consists of positive integers.

Matching with the remaining options on the right:

(a) describes positive integers which are divisors of 18. Let's find the positive divisors of 18: 1, 2, 3, 6, 9, 18. This matches the set {1, 2, 3, 6, 9, 18}.

(b) describes integer solutions to $x^2 - 9 = 0$, which is {3, -3}. This does not match.

Thus, set (iii) matches with set (a).

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Set (iv): {3, –3}

This set consists of the integers 3 and -3.

Matching with the remaining option on the right:

(b) describes integer solutions to $x^2 - 9 = 0$, which we found to be {3, -3}. This matches the set {3, -3}.

Thus, set (iv) matches with set (b).

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Summary of the matches:

(i) $\leftrightarrow$ (d)

(ii) $\leftrightarrow$ (c)

(iii) $\leftrightarrow$ (a)

(iv) $\leftrightarrow$ (b)

A collection is considered a set if it is well-defined, meaning there is a clear and unambiguous criterion for determining whether an object belongs to the collection or not.

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

This is a set.

Justification: The months of a year are fixed, and the criterion "beginning with the letter J" is specific and unambiguous. The collection is {January, June, July}.

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

This is not a set.

Justification: The term "most talented" is subjective and varies from person to person. There is no clear and objective criterion to determine which ten writers are the most talented.

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

This is not a set.

Justification: The term "best-cricket batsmen" is subjective. The criteria for "best" (e.g., average, strike rate, performance in specific conditions) are not universally agreed upon, making the selection ambiguous.

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

This is a set.

Justification: Assuming "your class" refers to a specific, defined class, and the distinction between a "boy" and not a "boy" is clear, the members of this collection can be precisely identified.

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

This is a set.

Justification: Natural numbers are clearly defined (usually {1, 2, 3, ...}), and the condition "less than 100" is objective. The collection is {1, 2, 3, ..., 99}.

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

This is a set.

Justification: Munshi Prem Chand is a specific writer, and the novels he has written are specific, identifiable works. The collection is well-defined by the author.

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(vii) The collection of all even integers.

This is a set.

Justification: Integers are clearly defined, and the property of being "even" (divisible by 2) is a clear and unambiguous criterion. The collection is {..., -4, -2, 0, 2, 4, ...}.

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(viii) The collection of questions in this Chapter.

This is a set.

Justification: "This Chapter" refers to a specific, defined part of a text, and the questions within it are specific, identifiable items. The collection is well-defined.

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

This is not a set.

Justification: The term "most dangerous" is subjective and depends on how danger is defined (e.g., to humans, in terms of venom, size, aggression, number of fatalities caused per year, etc.). There is no single, universally accepted measure of "most dangerous," making the collection ambiguous.

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

We need to determine if the given number is an element of set A ($\in$) or not an element of set A ($\notin$).

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(i) 5 . . . A

The number 5 is present in the set A.

So, 5 is an element of A.

The appropriate symbol is $\in$.

5 $\in$ A

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(ii) 8 . . . A

The number 8 is not present in the set A.

So, 8 is not an element of A.

The appropriate symbol is $\notin$.

8 $\notin$ A

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(iii) 0 . . .A

The number 0 is not present in the set A.

So, 0 is not an element of A.

The appropriate symbol is $\notin$.

0 $\notin$ A

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(iv) 4 . . . A

The number 4 is present in the set A.

So, 4 is an element of A.

The appropriate symbol is $\in$.

4 $\in$ A

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(v) 2 . . . A

The number 2 is present in the set A.

So, 2 is an element of A.

The appropriate symbol is $\in$.

2 $\in$ A

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(vi) 10 . . . A

The number 10 is not present in the set A.

So, 10 is not an element of A.

The appropriate symbol is $\notin$.

10 $\notin$ A

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

The integers greater than or equal to -3 and less than 7 are -3, -2, -1, 0, 1, 2, 3, 4, 5, and 6.

So, A = {-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 1, 2, 3, ... The natural numbers less than 6 are 1, 2, 3, 4, and 5.

So, B = {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}

We list the two-digit numbers whose digits add up to 8.

17 ($1+7=8$), 26 ($2+6=8$), 35 ($3+5=8$), 44 ($4+4=8$), 53 ($5+3=8$), 62 ($6+2=8$), 71 ($7+1=8$), 80 ($8+0=8$).

So, C = {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 positive divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Next, identify which of these are prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

The prime divisors are 2, 3, and 5.

So, D = {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 a set, elements are not repeated. The distinct letters are T, R, I, G, O, N, M, E, Y.

So, E = {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.

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

So, F = {B, E, T, R}.

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

The elements are 3, 6, 9, 12. These are multiples of 3.

$3 = 3 \times 1$

$6 = 3 \times 2$

$9 = 3 \times 3$

$12 = 3 \times 4$

The elements are of the form $3n$, where $n$ is a natural number from 1 to 4.

Set-builder form: $\{x : x = 3n, n \in \mathbb{N}, 1 \leq n \leq 4\}$.

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

The elements are 2, 4, 8, 16, 32. These are powers of 2.

$2 = 2^1$

$4 = 2^2$

$8 = 2^3$

$16 = 2^4$

$32 = 2^5$

The elements are of the form $2^n$, where $n$ is a natural number from 1 to 5.

Set-builder form: $\{x : x = 2^n, n \in \mathbb{N}, 1 \leq n \leq 5\}$.

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

The elements are 5, 25, 125, 625. These are powers of 5.

$5 = 5^1$

$25 = 5^2$

$125 = 5^3$

$625 = 5^4$

The elements are of the form $5^n$, where $n$ is a natural number from 1 to 4.

Set-builder form: $\{x : x = 5^n, n \in \mathbb{N}, 1 \leq n \leq 4\}$.

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

The elements are 2, 4, 6, and the pattern continues indefinitely. These are even natural numbers.

An even natural number can be written in the form $2n$, where $n$ is a natural number ($n \in \{1, 2, 3, ...\}$).

Set-builder form: $\{x : x \text{ is an even natural number}\}$ or $\{x : x = 2n, n \in \mathbb{N}\}$.

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

The elements are 1, 4, 9, ..., 100. These are perfect squares.

$1 = 1^2$

$4 = 2^2$

$9 = 3^2$

... and so on, up to $100 = 10^2$.

The elements are of the form $n^2$, where $n$ is a natural number from 1 to 10.

Set-builder form: $\{x : x = n^2, n \in \mathbb{N}, 1 \leq n \leq 10\}$.

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

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

Odd natural numbers are those that are not divisible by 2. These are 1, 3, 5, 7, ...

The set A consists of all odd natural numbers.

A = {1, 3, 5, 7, ...}

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

We need to find the integers $x$ such that $-\frac{1}{2} < x < \frac{9}{2}$.

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

The integers between -0.5 and 4.5 (exclusive) are 0, 1, 2, 3, and 4.

B = {0, 1, 2, 3, 4}

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

We need to find integers $x$ such that the square of $x$ is less than or equal to 4.

Let's test some integers:

$0^2 = 0 \leq 4$

$1^2 = 1 \leq 4$

$(-1)^2 = 1 \leq 4$

$2^2 = 4 \leq 4$

$(-2)^2 = 4 \leq 4$

$3^2 = 9 \not\leq 4$

$(-3)^2 = 9 \not\leq 4$

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

The integers satisfying $x^2 \leq 4$ are -2, -1, 0, 1, and 2.

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.

In a set, we list each unique element only once.

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

D = {L, O, Y, A}

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

We list the months of the year and check their number of days:

January (31), February (28/29), 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 February, April, June, September, and November.

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 English alphabet up to k is A, B, C, D, E, F, G, H, I, J, K.

The letters that precede k are A, B, C, D, E, F, G, H, I, J.

We need to identify the consonants among these letters. Vowels are A, E, I, O, U.

The consonants in the list A, B, C, D, E, F, G, H, I, J are B, C, D, F, G, H, J.

F = {B, C, D, F, G, H, J}

We need to match the sets given in roster form with their equivalent description in set-builder form.

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Let's examine each set in roster form and its properties:

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

The elements are 1, 2, 3, and 6. These are all positive integers.

Let's check if they are divisors of 6. $6 \div 1 = 6$, $6 \div 2 = 3$, $6 \div 3 = 2$, $6 \div 6 = 1$. All elements are divisors of 6.

Let's check the options on the right:

(a) {x : x is a prime number and a divisor of 6}. The prime divisors of 6 are 2 and 3. This set is {2, 3}. Does not match.

(b) {x : x is an odd natural number less than 10}. This set is {1, 3, 5, 7, 9}. Does not match.

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

(d) {x : x is a letter of the word MATHEMATICS}. This set is {M, A, T, H, E, I, C, S}. Does not match.

So, (i) matches with (c).

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(ii) {2, 3}

The elements are 2 and 3. These are positive integers.

Let's check if they are prime numbers. 2 is prime, 3 is prime. Both are prime.

Let's check if they are divisors of 6. $6 \div 2 = 3$, $6 \div 3 = 2$. Both are divisors of 6.

Let's check the remaining options on the right:

(a) {x : x is a prime number and a divisor of 6}. The prime numbers that are divisors of 6 are 2 and 3. This set is {2, 3}. Matches.

(b) {x : x is an odd natural number less than 10}. This set is {1, 3, 5, 7, 9}. Does not match.

(d) {x : x is a letter of the word MATHEMATICS}. This set is {M, A, T, H, E, I, C, S}. Does not match.

So, (ii) matches with (a).

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

The elements are letters.

Let's look at the letters in the word "MATHEMATICS". The letters are M, A, T, H, E, M, A, T, I, C, S.

The unique letters are M, A, T, H, E, I, C, S. This matches the set.

Let's check the remaining options on the right:

(b) {x : x is an odd natural number less than 10}. This set is {1, 3, 5, 7, 9}. Does not match.

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

So, (iii) matches with (d).

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

The elements are 1, 3, 5, 7, and 9. These are all natural numbers.

They are all odd numbers.

They are all less than 10.

Let's check the last remaining option on the right:

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

So, (iv) matches with (b).

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The final matches are:

(i) $\leftrightarrow$ (c)

(ii) $\leftrightarrow$ (a)

(iii) $\leftrightarrow$ (d)

(iv) $\leftrightarrow$ (b)

A set is finite if it is either empty or its elements can be listed by a finite number of counting process. Otherwise, the set is infinite.

N denotes the set of natural numbers, $\mathbb{N} = \{1, 2, 3, ...\}$.

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

We need to find the natural numbers $x$ that satisfy $(x - 1)(x - 2) = 0$.

The solutions to $(x - 1)(x - 2) = 0$ are $x - 1 = 0$ or $x - 2 = 0$.

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

Both 1 and 2 are natural numbers.

So, the set is {1, 2}.

This set has exactly two elements, which is a finite number.

Therefore, this set is finite.

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(ii) {x : x ∈ N and x² = 4}

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

The solutions to $x^2 = 4$ are $x = \sqrt{4}$ or $x = -\sqrt{4}$.

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

We require $x$ to be a natural number ($\in \mathbb{N}$). The number 2 is a natural number, but -2 is not.

So, the only element in the set is 2.

The set is {2}.

This set has exactly one element, which is a finite number.

Therefore, this set is finite.

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(iii) {x : x ∈ N and 2x –1 = 0}

We need to find the natural numbers $x$ that satisfy $2x - 1 = 0$.

Solving for $x$: $2x = 1 \implies x = \frac{1}{2}$.

We require $x$ to be a natural number ($\in \mathbb{N}$). The number $\frac{1}{2}$ is not a natural number.

So, there are no natural numbers that satisfy the given condition.

The set is {}, which is the empty set.

The empty set is considered a finite set.

Therefore, this set is finite.

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(iv) {x : x ∈ N and x is prime}

We need to list the natural numbers that are prime.

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

The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, ...

This sequence of prime numbers continues indefinitely, as proven by Euclid.

The set is {2, 3, 5, 7, 11, ...}.

This set has an infinite number of elements.

Therefore, this set is infinite.

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(v) {x : x ∈ N and x is odd}

We need to list the natural numbers that are odd.

Odd natural numbers are natural numbers not divisible by 2. These are 1, 3, 5, 7, 9, 11, ...

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

This sequence of odd natural numbers continues indefinitely.

This set has an infinite number of elements.

Therefore, this set is infinite.

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

Let's determine the elements of each set:

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Set A = {0}

This set contains the single element 0.

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

We are looking for numbers $x$ that are simultaneously greater than 15 and less than 5.

There is no number that satisfies both conditions.

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

B = { }

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Set C = {x : x – 5 = 0 }

We need to find the value(s) of $x$ that satisfy the equation $x - 5 = 0$.

$x - 5 = 0$

$x = 5$

So, set C contains the single element 5.

C = {5}

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Set D = {x: x² = 25}

We need to find the value(s) of $x$ that satisfy the equation $x^2 = 25$.

$x^2 = 25$

$x = \pm \sqrt{25}$

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

So, set D contains the elements 5 and -5.

D = {5, -5}

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

We need to find the integer and positive roots of the equation $x^2 – 2x – 15 = 0$.

We can solve the quadratic equation by factoring:

$x^2 - 2x - 15 = 0$

We look for two numbers that multiply to -15 and add to -2. These numbers are -5 and 3.

$(x - 5)(x + 3) = 0$

This gives the roots:

$x - 5 = 0$ or $x + 3 = 0$

$x = 5$ or $x = -3$

The roots are 5 and -3. We need the integral positive root.

Both roots are integers. The positive root is 5.

So, set E contains the single element 5.

E = {5}

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Comparing the elements of each set:

A = {0}

B = { }

C = {5}

D = {5, -5}

E = {5}

We can see that Set C and Set E contain exactly the same element, which is 5.

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The pairs of equal sets are C and E.

Reason: Set C contains the solution to the equation $x-5=0$, which is $x=5$. Set E contains the integral positive root of the equation $x^2 - 2x - 15 = 0$. The roots of this equation are $x=5$ and $x=-3$. Among these, the integral positive root is $x=5$. Therefore, both sets C and E contain exactly the same element, 5.

Two sets are equal if they contain exactly the same elements, regardless of the order in which the elements are listed or how many times they are repeated.

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

Let's list the unique elements of set X:

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

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

So, X = {A, L, O, Y}.

Let's list the unique elements of set B:

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

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

So, B = {L, O, Y, A}.

Comparing the elements, we see that both sets contain the same unique letters: A, L, O, Y.

Therefore, sets X and B are equal.

Justification: Set X = {A, L, O, Y} and Set B = {L, O, Y, A}. Both sets contain exactly the same elements. The order in which the elements are listed in a set does not affect its equality with another set.

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

Let's determine the elements of set A:

A = {n : n ∈ Z and n² ≤ 4}.

We need to find all integers $n$ whose square is less than or equal to 4.

• If $n=0$, $n^2 = 0 \leq 4$. So, 0 is in A.
• If $n=1$, $n^2 = 1 \leq 4$. So, 1 is in A.
• If $n=-1$, $n^2 = (-1)^2 = 1 \leq 4$. So, -1 is in A.
• If $n=2$, $n^2 = 4 \leq 4$. So, 2 is in A.
• If $n=-2$, $n^2 = (-2)^2 = 4 \leq 4$. So, -2 is in A.
• If $n=3$, $n^2 = 9 \not\leq 4$.
• If $n=-3$, $n^2 = (-3)^2 = 9 \not\leq 4$.

So, the integers whose square is less than or equal to 4 are -2, -1, 0, 1, and 2.

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

Let's determine the elements of set B:

B = {x : x ∈ R and x² – 3x + 2 = 0}.

We need to find the real roots of the quadratic equation $x^2 – 3x + 2 = 0$.

We can solve the quadratic equation by factoring:

$x^2 – 3x + 2 = 0$

Find two numbers that multiply to +2 and add to -3. These numbers are -1 and -2.

So, we can factor the equation as $(x - 1)(x - 2) = 0$.

This equation is satisfied if $x - 1 = 0$ or $x - 2 = 0$.

Solving for $x$, we get $x = 1$ or $x = 2$.

The roots are 1 and 2. Both are real numbers.

So, B = {1, 2}.

Comparing the elements of Set A and Set B:

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

Set B = {1, 2}

The sets A and B do not contain the same elements (for example, -2 is in A but not in B, and 0 is in A but not in B).

Therefore, sets A and B are not equal.

Justification: Set A contains the elements {-2, -1, 0, 1, 2}, while Set B contains only the elements {1, 2}. Since the elements are not exactly the same, the sets are not equal.

A null set (or empty set) is a set that contains no elements. We need to check if each described set contains any elements based on the given conditions.

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(i) Set of odd natural numbers divisible by 2

Natural numbers are {1, 2, 3, 4, ...}.

Odd natural numbers are {1, 3, 5, 7, ...}.

Numbers divisible by 2 are even numbers {2, 4, 6, 8, ...}.

A number cannot be both odd and divisible by 2. By definition, odd numbers are not divisible by 2.

There are no elements that satisfy both conditions simultaneously.

This is a null set.

---

(ii) Set of even prime numbers

Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves: {2, 3, 5, 7, 11, ...}.

Even numbers are integers divisible by 2: {..., -4, -2, 0, 2, 4, ...}.

We are looking for prime numbers that are also even.

The number 2 is a prime number (only divisible by 1 and 2) and it is also an even number.

The set contains the element {2}.

Since the set contains an element (namely 2), it is not empty.

This is not a null set.

---

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

We are looking for a natural number $x$ such that $x < 5$ and $x > 7$.

If a number $x$ is less than 5, it belongs to the set {1, 2, 3, 4} (among natural numbers).

If a number $x$ is greater than 7, it belongs to the set {8, 9, 10, 11, ...} (among natural numbers).

There is no natural number that is simultaneously less than 5 and greater than 7.

There are no elements that satisfy both conditions.

This is a null set.

---

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

In geometry, parallel lines are lines in a plane that never meet, no matter how far they are extended.

A "point common to two lines" is an intersection point.

By definition, parallel lines have no intersection points.

Therefore, there is no point that is common to any two parallel lines.

There are no elements that satisfy the condition.

This is a null set.

A set is finite if its elements can be counted, resulting in a non-negative integer (the number of elements). Otherwise, the set is infinite.

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(i) The set of months of a year

The months of a year are January, February, March, April, May, June, July, August, September, October, November, and December.

There are exactly 12 months in a year.

The number of elements is 12, which is a finite number.

Therefore, this set is finite.

---

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

This set represents the set of all natural numbers {1, 2, 3, 4, 5, ...}.

The "..." indicates that the list continues indefinitely without end.

The number of natural numbers is not finite.

Therefore, this set is infinite.

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

This set represents the collection of natural numbers starting from 1 and ending at 100.

The elements are 1, 2, 3, ..., 99, 100.

The number of elements in this set is 100, which is a finite number.

Therefore, this set is finite.

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(iv) The set of positive integers greater than 100

Positive integers are {1, 2, 3, ...}.

The positive integers greater than 100 are {101, 102, 103, 104, ...}.

This list starts at 101 and continues indefinitely.

The number of elements in this set is not finite.

Therefore, this set is infinite.

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(v) The set of prime numbers less than 99

Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves.

We need to list prime numbers $p$ such that $p < 99$.

The prime numbers less than 99 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

The set is {2, 3, 5, 7, ..., 97}.

The number of elements in this set is 25, which is a finite number.

Therefore, this set is finite.

A set is finite if its elements can be counted, resulting in a non-negative integer (the number of elements). Otherwise, the set is infinite.

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

Lines parallel to the x-axis have the equation $y = c$, where $c$ is a real number. The value of $c$ can be any real number (positive, negative, or zero).

There are infinitely many possible values for $c$ (corresponding to different horizontal lines).

Therefore, there are infinitely many lines parallel to the x-axis.

This set is infinite.

---

(ii) The set of letters in the English alphabet

The English alphabet consists of 26 letters: A, B, C, ..., Z.

The number of elements is 26, which is a finite number.

Therefore, this set is finite.

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

Multiples of 5 can be positive or negative: {..., -10, -5, 0, 5, 10, 15, ...}.

This list continues indefinitely in both positive and negative directions.

The number of multiples of 5 is not finite.

Therefore, this set is infinite.

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(iv) The set of animals living on the earth

While the exact number of animals is very large and constantly changing, it is a concrete number at any given point in time.

It is possible, in principle, to count all the animals, even though it is practically impossible.

The number of animals is a finite number, even if very large.

Therefore, this set is finite.

---

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

A circle is defined by its center $(h, k)$ and radius $r$. The equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$.

If a circle passes through the origin (0, 0), then the point (0, 0) must satisfy the equation:

$(0-h)^2 + (0-k)^2 = r^2$

$h^2 + k^2 = r^2$

There are infinitely many possible values for $h$ and $k$ (real numbers) that can satisfy this equation for a given radius, or infinitely many combinations of $h$, $k$, and $r$. For example, consider circles with their center on the x-axis $(h, 0)$ passing through (0, 0). Then $h^2 + 0^2 = r^2$, so $r = |h|$. The equation is $(x-h)^2 + y^2 = h^2$. For every real number $h$, this represents a circle passing through the origin.

This demonstrates that there are infinitely many such circles.

Therefore, this set is infinite.

Two sets A and B are equal, denoted by A = B, if they have exactly the same elements. The order of elements does not matter, and repetition of elements is not considered (as elements are unique within a set).

---

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

B = { d, c, b, a }

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

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

The two sets contain the same elements. The order is different, but this does not affect equality of sets.

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, and 16.

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

Comparing the elements, we see that 12 is an element of A but not of B, and 18 is an element of B but not of A.

Since the sets do not contain exactly the same elements, they are not equal.

Therefore, A $\neq$ 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 as {2, 4, 6, 8, 10}.

Let's list the elements of set B based on its description:

We need positive even integers that are less than or equal to 10.

Positive integers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Positive even integers: 2, 4, 6, 8, 10, 12, ...

Positive even integers less than or equal to 10: 2, 4, 6, 8, 10.

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

Comparing the elements of A and B, we see they are exactly the same.

Therefore, A = B.

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(iv) A = { x : x is a multiple of 10}

B = { 10, 15, 20, 25, 30, . . . }

Let's list some elements of set A:

Multiples of 10 are numbers obtained by multiplying 10 by an integer. These include positive and negative multiples and zero.

A = {..., -20, -10, 0, 10, 20, 30, 40, ...}

Set B is given in roster form as {10, 15, 20, 25, 30, ...}. The "..." suggests the pattern continues.

Comparing the elements, we see that 15 is an element of B but not a multiple of 10 (and thus not in A). Similarly, 25 is in B but not in A.

Also, 0 is in A but not in the listed elements of B, and it's unlikely to be in the pattern suggested by B.

Since the sets do not contain exactly the same elements, they are not equal.

Therefore, A $\neq$ B.

Two sets are equal if and only if they have 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 {2, 3}.

Let's find the elements of set B by solving the equation $x^2 + 5x + 6 = 0$.

We need to find the roots of the quadratic equation.

$x^2 + 5x + 6 = 0$

We look for two numbers that multiply to 6 and add up to 5. These numbers are 2 and 3.

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

This gives the solutions:

$x + 2 = 0$ or $x + 3 = 0$

$x = -2$ or $x = -3$

So, set B contains the elements -2 and -3.

B = {-2, -3}.

Comparing Set A = {2, 3} and Set B = {-2, -3}, we see that the elements are different.

Therefore, sets A and B are not equal.

Reason: Set A contains the elements 2 and 3. Set B contains the elements -2 and -3. The elements are not the same.

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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}

Let's list the elements of set A:

The letters in the word "FOLLOW" are F, O, L, L, O, W.

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

So, A = {F, O, L, W}.

Let's list the elements of set B:

The letters in the word "WOLF" are W, O, L, F.

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

So, B = {W, O, L, F}.

Comparing the elements, we see that both sets contain the same unique letters: F, O, L, W.

Therefore, sets A and B are equal.

Reason: The elements of A are {F, O, L, W}, and the elements of B are {W, O, L, F}. Since the order of elements in a set does not matter, the sets contain exactly the same elements.

Two sets are considered equal if they contain exactly the same elements. The order of elements in a set does not affect its equality with another set.

---

Let's list the elements of each given set:

Set A = {2, 4, 8, 12}

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

Set C = {4, 8, 12, 14}

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

Set E = {-1, 1}

Set F = {0, a}

Set G = {1, -1}

Set H = {0, 1}

---

Now, let's compare the sets to find pairs with identical elements:

Compare A with others:

A = {2, 4, 8, 12}. None of the other sets contain exactly these elements.

For example, B has {1, 2, 3, 4}, C has {4, 8, 12, 14}, etc. None match A.

---

Compare B with others:

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

Look at D = {3, 1, 4, 2}. Rearranging the elements of D gives {1, 2, 3, 4}.

Sets B and D contain exactly the same elements.

So, B = D.

None of the remaining sets (C, E, F, G, H) contain the elements {1, 2, 3, 4}.

---

Compare C with others:

C = {4, 8, 12, 14}. We already checked A and B. D is equal to B. Check E, F, G, H. None match C.

---

Compare D with others:

D = {3, 1, 4, 2}. We already found D = B. Check E, F, G, H. None match D.

---

Compare E with others:

E = {-1, 1}.

Look at G = {1, -1}. Rearranging the elements of G gives {-1, 1}.

Sets E and G contain exactly the same elements.

So, E = G.

None of the remaining sets (F, H) contain the elements {-1, 1}.

---

Compare F with others:

F = {0, a}. We already checked others. H = {0, 1}. These are not equal as F has 'a' and H has '1', and neither is the other.

---

Compare G with others:

G = {1, -1}. We already found G = E. Check H. Not equal.

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Compare H with others:

H = {0, 1}. We already checked others. Not equal to any other set.

---

The pairs of equal sets are those that contain precisely the same elements.

Based on our comparison:

B = {1, 2, 3, 4} and D = {3, 1, 4, 2} are equal sets.

E = {-1, 1} and G = {1, -1} are equal sets.

---

The equal sets are:

B and D

E and G

The symbol $\subset$ means 'is a subset of'. A set P is a subset of set Q if every element of P is also an element of Q. The symbol $\not\subset$ means 'is not a subset of'.

The given sets are:

$\Phi$ = { }

A = {1, 3}

B = {1, 5, 9}

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

---

(i) $\Phi$ . . . B

We need to check if $\Phi$ is a subset of B.

The empty set $\Phi$ has no elements.

The definition of a subset states that every element of the first set must be in the second set. Since there are no elements in $\Phi$, this condition is vacuously true for any set B.

So, every element of $\Phi$ (none) is in B.

Therefore, $\Phi$ is a subset of B.

The symbol is $\subset$.

$\Phi \subset$ B

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(ii) A . . . B

We need to check if A is a subset of B.

A = {1, 3}, B = {1, 5, 9}

Let's check the elements of A:

Is 1 in B? Yes, $1 \in$ B.

Is 3 in B? No, $3 \notin$ B.

Since there is an element in A (the number 3) that is not in B, A is not a subset of B.

The symbol is $\not\subset$.

A $\not\subset$ B

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(iii) A . . . C

We need to check if A is a subset of C.

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

Let's check the elements of A:

Is 1 in C? Yes, $1 \in$ C.

Is 3 in C? Yes, $3 \in$ C.

Since every element in A is also in C, A is a subset of C.

The symbol is $\subset$.

A $\subset$ C

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(iv) B . . . C

We need to check if B is a subset of C.

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

Let's check the elements of B:

Is 1 in C? Yes, $1 \in$ C.

Is 5 in C? Yes, $5 \in$ C.

Is 9 in C? Yes, $9 \in$ C.

Since every element in B is also in C, B is a subset of C.

The symbol is $\subset$.

B $\subset$ C

Let the given sets be A = {a, e, i, o, u} and B = {a, b, c, d}.

A set P is a subset of set Q (denoted as P $\subset$ Q) if every element of set P is also an element of set Q.

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Is A a subset of B?

To check if A is a subset of B, we need to verify if every element of A is also an element of B.

The elements of A are a, e, i, o, u.

The elements of B are a, b, c, d.

Let's check each element of A:

Is 'a' in B? Yes, a $\in$ B.

Is 'e' in B? No, e $\notin$ B.

Since the element 'e' is in A but not in B, A is not a subset of B.

A $\not\subset$ B

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Is B a subset of A?

To check if B is a subset of A, we need to verify if every element of B is also an element of A.

The elements of B are a, b, c, d.

The elements of A are a, e, i, o, u.

Let's check each element of B:

Is 'a' in A? Yes, a $\in$ A.

Is 'b' in A? No, b $\notin$ A.

Since the element 'b' is in B but not in A, B is not a subset of A.

B $\not\subset$ A

---

Reasons:

A is not a subset of B because the element 'e' (and also 'i', 'o', 'u') is in A but not in B.

B is not a subset of A because the element 'b' (and also 'c', 'd') is in B but not in A.

We are given three sets A, B, and C.

We are given two conditions:

1. $A \in B$

2. $B \subset C$

We need to determine if these conditions imply that $A \subset C$.

---

Let's understand the given conditions:

$A \in B$ means that the set A is an element of the set B.

$B \subset C$ means that set B is a subset of the set C. This implies that every element of B is also an element of C.

---

We want to check if $A \subset C$, which means if every element of A is also an element of C.

Let's consider a counterexample.

Let set A = {1}.

For $A \in B$, the set B must contain A as one of its elements. So, {1} must be an element of B.

Let B = {{1}, 2, 3}. Here, A = {1} is an element of B.

For $B \subset C$, every element of B must be an element of C.

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

So, C must contain {1}, 2, and 3. It can contain other elements as well.

Let C = {{1}, 2, 3, 4, 5}. Here, every element of B ({1}, 2, 3) is in C.

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Now, let's check if $A \subset C$ using our example sets:

A = {1}

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

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

The only element in A is 1.

Is 1 an element of C? The elements listed in C are {1}, 2, 3, 4, 5. The number 1 itself is not listed as an element; the set {1} is an element of C.

Since the element 1 from set A is not found as an element directly within set C, A is not a subset of C.

In our example, $A \in B$ and $B \subset C$ are true, but $A \subset C$ is false.

---

Therefore, it is not true that if $A \in B$ and $B \subset C$, then $A \subset C$.

Counterexample:

Let A = {1}

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

Let C = {{1}, 2, 3, 4}

Here, $A \in B$ because the set {1} is an element of B.

Also, $B \subset C$ because every element of B (which are {1}, 2, and 3) is an element of C.

However, $A \not\subset C$ because the element 1 of A is not an element of C (C contains the set {1}, but not the number 1 itself as a distinct element).

The symbol $\subset$ means 'is a subset of', and $\not\subset$ means 'is not a subset of'. Set A is a subset of Set B if every element in A is also in B.

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

The elements in the first set are 2, 3, and 4. All these elements are also present in the second set {1, 2, 3, 4, 5}.

{ 2, 3, 4 } $\subset$ { 1, 2, 3, 4,5 }

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

The elements in the first set are a, b, and c. The element 'a' is in the first set but not in the second set {b, c, d}.

{ a, b, c } $\not\subset$ { 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}

Every student who is in Class XI of your school is also a student of your school.

{x : x is a student of Class XI of your school} $\subset$ {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}

The first set contains all circles in the plane, regardless of radius. The second set contains only circles with a radius of exactly 1 unit. A circle with a radius of 2 units is in the first set but not in the second set.

{x : x is a circle in the plane} $\not\subset$ {x : x is a circle in the same plane with radius 1 unit}

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(v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane}

A triangle is a three-sided polygon. A rectangle is a four-sided polygon. No triangle is a rectangle. The sets contain different types of geometric shapes.

{x : x is a triangle in a plane} $\not\subset$ {x : x is a rectangle in the plane}

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(vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane}

An equilateral triangle is a type of triangle. Every equilateral triangle is a triangle.

{x : x is an equilateral triangle in a plane} $\subset$ {x : x is a triangle in the same plane}

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(vii) {x : x is an even natural number} . . . {x : x is an integer}

The set of even natural numbers is {2, 4, 6, 8, ...}. The set of integers is {..., -3, -2, -1, 0, 1, 2, 3, ...}. Every even natural number (2, 4, 6, ...) is also an integer.

{x : x is an even natural number} $\subset$ {x : x is an integer}

We will examine each statement based on the definitions of subset ($\subset$) and element of ($\in$).

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

$a \in S$ means that 'a' is an element belonging to the set S.

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(i) { a, b } ⊄ { b, c, a }

Let Set 1 = {a, b} and Set 2 = {b, c, a} = {a, b, c}.

We check if Set 1 is a subset of Set 2. The elements of Set 1 are 'a' and 'b'. Both 'a' and 'b' are present in Set 2 {a, b, c}.

So, { a, b } $\subset$ { b, c, a } is True.

The given statement says { a, b } ⊄ { b, c, a }, which means {a, b} is NOT a subset of {b, c, a}.

This statement is False.

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

Let Set 1 = {a, e}.

Let Set 2 = {x : x is a vowel in the English alphabet}. The vowels in the English alphabet are a, e, i, o, u. So, Set 2 = {a, e, i, o, u}.

We check if Set 1 is a subset of Set 2. The elements of Set 1 are 'a' and 'e'. Both 'a' and 'e' are present in Set 2 {a, e, i, o, u}.

So, { a, e } $\subset$ { a, e, i, o, u } is True.

The given statement is { a, e } ⊂ { x : x is a vowel in the English alphabet}.

This statement is True.

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(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }

Let Set 1 = {1, 2, 3} and Set 2 = {1, 3, 5}.

We check if Set 1 is a subset of Set 2. The elements of Set 1 are 1, 2, and 3.

Is 1 in Set 2? Yes, $1 \in \{1, 3, 5\}$.

Is 2 in Set 2? No, $2 \notin \{1, 3, 5\}$.

Since the element 2 is in Set 1 but not in Set 2, Set 1 is not a subset of Set 2.

So, { 1, 2, 3 } $\not\subset$ { 1, 3, 5 } is True.

The given statement is { 1, 2, 3 } ⊂ { 1, 3, 5 }.

This statement is False.

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(iv) { a } ⊂ { a, b, c }

Let Set 1 = {a} and Set 2 = {a, b, c}.

We check if Set 1 is a subset of Set 2. The only element in Set 1 is 'a'. The element 'a' is present in Set 2 {a, b, c}.

Since every element in Set 1 is also in Set 2, { a } $\subset$ { a, b, c } is True.

The given statement is { a } ⊂ { a, b, c }.

This statement is True.

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(v) { a } ∈ { a, b, c }

Here, we are checking if the set {a} is an element of the set {a, b, c}.

The elements of the set {a, b, c} are 'a', 'b', and 'c'.

The set {a} is a set containing the element 'a', which is different from the element 'a' itself.

Looking at the elements of {a, b, c}, we see 'a', 'b', and 'c', but we do not see the set {a} listed as one of its elements.

So, { a } $\notin$ { a, b, c } is True.

The given statement is { a } ∈ { a, b, c }.

This 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}

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

Natural numbers less than 6 are {1, 2, 3, 4, 5}. The even ones are 2 and 4. So, Set 1 = {2, 4}.

Let Set 2 = { x : x is a natural number which divides 36}.

The natural number divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. So, Set 2 = {1, 2, 3, 4, 6, 9, 12, 18, 36}.

We check if Set 1 is a subset of Set 2. The elements of Set 1 are 2 and 4.

Is 2 in Set 2? Yes, $2 \in \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$.

Is 4 in Set 2? Yes, $4 \in \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$.

Since every element in Set 1 is also in Set 2, { x : x is an even natural number less than 6} $\subset$ { x : x is a natural number which divides 36} is True.

The given statement is { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}.

This statement is True.

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

We will examine each statement.

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(i) {3, 4} ⊂ A

This statement claims that the set {3, 4} is a subset of A. For this to be true, every element of the set {3, 4} must be an element of A. The elements of the set {3, 4} are 3 and 4. Are 3 and 4 elements of A? No, A contains the set {3, 4} as a single element, but not 3 and 4 individually.

This statement is Incorrect. Reason: The elements 3 and 4 are not elements of set A.

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(ii) {3, 4} ∈ A

This statement claims that the set {3, 4} is an element of A. Looking at the elements listed in A, we see 1, 2, {3, 4}, and 5. The set {3, 4} is indeed one of the elements of A.

This statement is Correct.

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(iii) {{3, 4}} ⊂ A

This statement claims that the set containing the set {3, 4} as its only element is a subset of A. For this to be true, the only element of {{3, 4}}, which is {3, 4}, must be an element of A. From (ii), we know that {3, 4} is an element of A.

This statement is Correct.

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(iv) 1 ∈ A

This statement claims that the number 1 is an element of A. Looking at the elements listed in A, we see 1, 2, {3, 4}, and 5. The number 1 is indeed one of the elements of A.

This statement is Correct.

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(v) 1 ⊂ A

This statement claims that the number 1 is a subset of A. The subset relation is defined between two sets. The number 1 is an element, not a set (in this context). An element cannot be a subset unless it is itself a set and meets the subset definition.

This statement is Incorrect. Reason: The number 1 is an element, not a set. The subset relation is between sets.

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(vi) {1, 2, 5} ⊂ A

This statement claims that the set {1, 2, 5} is a subset of A. For this to be true, every element of {1, 2, 5} must be an element of A. The elements are 1, 2, and 5. Looking at the elements of A {1, 2, {3, 4}, 5}, we see that 1, 2, and 5 are all elements of A.

This statement is Correct.

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(vii) {1, 2, 5} ∈ A

This statement claims that the set {1, 2, 5} is an element of A. Looking at the elements listed in A, we see 1, 2, {3, 4}, and 5. The set {1, 2, 5} is not listed as one of the elements of A.

This statement is Incorrect. Reason: The set {1, 2, 5} is not one of the listed elements in set A.

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(viii) {1, 2, 3} ⊂ A

This statement claims that the set {1, 2, 3} is a subset of A. For this to be true, every element of {1, 2, 3} must be an element of A. The elements are 1, 2, and 3. Looking at the elements of A {1, 2, {3, 4}, 5}, we see that 1 and 2 are elements of A, but 3 is not an element of A (the set {3, 4} is an element, but 3 itself is not).

This statement is Incorrect. Reason: The element 3 is in the set {1, 2, 3} but not in set A.

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(ix) $\phi$ ∈ A

This statement claims that the empty set $\phi$ is an element of A. Looking at the elements listed in A, which are 1, 2, {3, 4}, and 5, the empty set $\phi$ is not listed as one of the elements of A.

This statement is Incorrect. Reason: The empty set $\phi$ is not one of the listed elements in set A.

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(x) $\phi$ ⊂ A

This statement claims that the empty set $\phi$ is a subset of A. By definition, the empty set is a subset of every set.

This statement is Correct.

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(xi) {$\phi$} ⊂ A

This statement claims that the set containing the empty set as its only element is a subset of A. For this to be true, the only element of {$\phi$}, which is $\phi$, must be an element of A. From (ix), we know that $\phi$ is not an element of A.

This statement is Incorrect. Reason: The element $\phi$ of the set {$\phi$} is not an element of set A.

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The incorrect statements are (i), (v), (vii), (viii), (ix), and (xi).

A subset of a set S is a set containing only elements from S. The empty set ($\phi$) is a subset of every set, and every set is a subset of itself.

If a set has $n$ elements, the number of subsets is $2^n$.

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(i) {a}

Let the set be S = {a}. The number of elements is $n=1$. The number of subsets is $2^1 = 2$.

The subsets are the empty set and the set itself.

The subsets are: $\phi$, {a}.

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(ii) {a, b}

Let the set be S = {a, b}. The number of elements is $n=2$. The number of subsets is $2^2 = 4$.

The subsets are the empty set, sets with one element, and the set itself.

Subsets with 0 elements: $\phi$

Subsets with 1 element: {a}, {b}

Subsets with 2 elements: {a, b}

The subsets are: $\phi$, {a}, {b}, {a, b}.

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(iii) {1, 2, 3}

Let the set be S = {1, 2, 3}. The number of elements is $n=3$. The number of subsets is $2^3 = 8$.

Subsets with 0 elements: $\phi$

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

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

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

The subsets are: $\phi$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

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(iv) $\phi$

Let the set be S = $\phi$ = { }. The number of elements is $n=0$. The number of subsets is $2^0 = 1$.

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

The subset is: $\phi$.

We need to express the given sets in interval notation. For real numbers, the standard interval notations are:

• (a, b) = {x : a < x < b} (Open interval)
• [a, b] = {x : a ≤ x ≤ b} (Closed interval)
• [a, b) = {x : a ≤ x < b} (Half-open or half-closed interval)
• (a, b] = {x : a < x ≤ b} (Half-open or half-closed interval)
• ($-\infty$, b) = {x : x < b}
• ($-\infty$, b] = {x : x ≤ b}
• (a, $\infty$) = {x : x > a}
• [a, $\infty$) = {x : x ≥ a}
• ($-\infty$, $\infty$) = R

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(i) {x : x ∈ R, – 4 < x ≤ 6}

The variable $x$ is a real number. The condition is that $x$ is strictly greater than -4 and less than or equal to 6.

This corresponds to a half-open interval where the left endpoint -4 is excluded (uses parenthesis) and the right endpoint 6 is included (uses square bracket).

The interval is (-4, 6].

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(ii) {x : x ∈ R, – 12 < x < –10}

The variable $x$ is a real number. The condition is that $x$ is strictly greater than -12 and strictly less than -10.

This corresponds to an open interval where both endpoints -12 and -10 are excluded (uses parentheses).

The interval is (-12, -10).

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(iii) {x : x ∈ R, 0 ≤ x < 7}

The variable $x$ is a real number. The condition is that $x$ is greater than or equal to 0 and strictly less than 7.

This corresponds to a half-open interval where the left endpoint 0 is included (uses square bracket) and the right endpoint 7 is excluded (uses parenthesis).

The interval is [0, 7).

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(iv) {x : x ∈ R, 3 ≤ x ≤ 4}

The variable $x$ is a real number. The condition is that $x$ is greater than or equal to 3 and less than or equal to 4.

This corresponds to a closed interval where both endpoints 3 and 4 are included (uses square brackets).

The interval is [3, 4].

We need to express the given intervals in set-builder form. The general format for a set-builder form for intervals of real numbers is {x : x ∈ R, condition on x}.

Recall the meaning of interval notations:

• (a, b) means $a < x < b$
• [a, b] means $a \leq x \leq b$
• (a, b] means $a < x \leq b$
• [a, b) means $a \leq x < b$

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(i) (– 3, 0)

This is an open interval from -3 to 0. It includes all real numbers $x$ that are strictly greater than -3 and strictly less than 0.

The set-builder form is {x : x ∈ R, – 3 < x < 0}.

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(ii) [6 , 12]

This is a closed interval from 6 to 12. It includes all real numbers $x$ that are greater than or equal to 6 and less than or equal to 12.

The set-builder form is {x : x ∈ R, 6 ≤ x ≤ 12}.

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(iii) (6, 12]

This is a half-open/half-closed interval from 6 to 12. It includes all real numbers $x$ that are strictly greater than 6 and less than or equal to 12.

The set-builder form is {x : x ∈ R, 6 < x ≤ 12}.

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(iv) [–23, 5)

This is a half-closed/half-open interval from -23 to 5. It includes all real numbers $x$ that are greater than or equal to -23 and strictly less than 5.

The set-builder form is {x : x ∈ R, –23 ≤ x < 5}.

A universal set is a super set of all the sets under consideration.

For both (i) and (ii), the sets described are collections of triangles with specific properties.

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(i) The set of right triangles.

This set contains all triangles that have one angle equal to $90^\circ$. All such triangles are a part of the larger collection of all triangles.

A suitable universal set would be the set of all triangles.

More broadly, the set of all polygons in a plane, or the set of all two-dimensional geometric figures in a plane could also be considered universal sets.

We propose: The set of all triangles in a plane.

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(ii) The set of isosceles triangles.

This set contains all triangles that have at least two sides of equal length. All such triangles are also a part of the larger collection of all triangles.

A suitable universal set would be the set of all triangles.

As in part (i), broader sets like the set of all polygons or geometric figures could also serve as universal sets.

We propose: The set of all triangles in a plane.

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In both cases, the most immediate and relevant universal set is the set of all triangles in the plane where these specific types of triangles exist.

A universal set U for a collection of sets is a set such that every set in the collection is a subset of U. In other words, the universal set must contain all the elements of the sets under consideration.

The given sets are: A = {1, 3, 5}, B = {2, 4, 6}, and C = {0, 2, 4, 6, 8}.

For a set U to be a universal set for A, B, and C, all elements of A, B, and C must be present in U.

The combined elements of A, B, and C are the elements in the union of these sets: $A \cup B \cup C$.

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

So, a universal set for A, B, and C must contain at least the elements {0, 1, 2, 3, 4, 5, 6, 8}.

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Let's check each option:

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

This set contains the elements {0, 1, 2, 3, 4, 5, 6}. Does it contain all elements from A, B, and C? No, the element 8 from set C is not present in this set.

So, this set cannot be a universal set for A, B, and C.

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(ii) $\phi$

The empty set $\phi$ contains no elements. It does not contain any of the elements from A, B, or C.

So, this set cannot be a universal set for A, B, and C.

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

This set contains the elements {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let's check if all elements from A, B, and C are present:

Elements of A: {1, 3, 5} are all in this set.

Elements of B: {2, 4, 6} are all in this set.

Elements of C: {0, 2, 4, 6, 8} are all in this set.

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

This set may be considered a universal set.

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

This set contains the elements {1, 2, 3, 4, 5, 6, 7, 8}. Let's check if all elements from A, B, and C are present:

Elements of A: {1, 3, 5} are all in this set.

Elements of B: {2, 4, 6} are all in this set.

Elements of C: {0, 2, 4, 6, 8}. The element 0 from set C is not present in this set.

So, this set cannot be a universal set for A, B, and C.

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Only option (iii) contains all the elements from sets A, B, and C.

Given sets are:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

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The union of two sets A and B, denoted by $A \cup B$, is the set of all elements which are in A or in B or in both A and B.

In set-builder form, $A \cup B = \{x : x \in A \text{ or } x \in B\}$.

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To find $A \cup B$, we list all the elements that are in set A and all the elements that are in set B, without repeating any elements.

Elements in A are 2, 4, 6, 8.

Elements in B are 6, 8, 10, 12.

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

2, 4, 6, 8 (from set A)

10, 12 (elements from set B that are not in set A, 6 and 8 are already listed)

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Thus, $A \cup B$ consists of the elements 2, 4, 6, 8, 10, and 12.

A ∪ B = {2, 4, 6, 8, 10, 12}.

Given sets are:

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

B = {a, i, u}

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We need to find the union of sets A and B, denoted by $A \cup B$.

$A \cup B = \{x : x \in A \text{ or } x \in B\}$

To find $A \cup B$, we combine the elements of A and B, listing each unique element once.

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

Elements in B are a, i, u.

Combining the unique elements from both sets:

The elements from A are a, e, i, o, u.

The elements from B that are not already in A are none, since a, i, and u are all already in A.

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

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Comparing the result with set A:

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

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

We can see that $A \cup B$ has the same elements as set A.

Thus, $A \cup B = A$.

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Reason:

Observe that every element of set B (a, i, u) is also an element of set A. This means that B is a subset of A, i.e., $B \subset A$.

When set B is a subset of set A, the union of A and B ($A \cup B$) will contain all elements that are in A or B. Since all elements of B are already in A, the union will simply be the set of all elements in A. This is a general property of sets: if $B \subset A$, then $A \cup B = A$.

Given sets representing students in different sports teams from Class XI:

X = {Ram, Geeta, Akbar} (Students in the school hockey team)

Y = {Geeta, David, Ashok} (Students in the school football team)

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We need to find the union of sets X and Y, denoted by $X \cup Y$.

$X \cup Y = \{z : z \in X \text{ or } z \in Y\}$

To find $X \cup Y$, we combine the elements of X and Y, listing each unique element once.

Elements in X are Ram, Geeta, Akbar.

Elements in Y are Geeta, David, Ashok.

Combining the unique elements from both sets:

From X: Ram, Geeta, Akbar

From Y: Geeta is already listed. Add David and Ashok.

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

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Interpretation of the set $X \cup Y$:

The union $X \cup Y$ contains all students who are in set X or in set Y or in both.

Set X consists of students in the hockey team.

Set Y consists of students in the football team.

Therefore, $X \cup Y$ is the set of all students from Class XI who are in the school hockey team or in the school football team (or in both).

In simpler terms, $X \cup Y$ is the set of students who participate in at least one of the two sports, hockey or football.

The sets from Example 12 are:

A = {2, 4, 6, 8}

B = {6, 8, 10, 12}

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

In set-builder form, $A \cap B = \{x : x \in A \text{ and } x \in B\}$.

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To find $A \cap B$, we look for the elements that are present in both set A and set B.

Elements in A are 2, 4, 6, 8.

Elements in B are 6, 8, 10, 12.

The elements that appear in both lists are 6 and 8.

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Thus, $A \cap B$ consists of the elements 6 and 8.

A ∩ B = {6, 8}.

The sets from Example 14 are:

X = {Ram, Geeta, Akbar} (Students in the school hockey team)

Y = {Geeta, David, Ashok} (Students in the school football team)

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We need to find the intersection of sets X and Y, denoted by $X \cap Y$.

$X \cap Y = \{z : z \in X \text{ and } z \in Y\}$

To find $X \cap Y$, we look for the elements that are present in both set X and set Y.

Elements in X are Ram, Geeta, Akbar.

Elements in Y are Geeta, David, Ashok.

The element that appears in both lists is Geeta.

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Thus, $X \cap Y$ consists of the element Geeta.

X ∩ Y = {Geeta}.

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Interpretation:

$X \cap Y$ represents the set of students who are in both the school hockey team and the school football team. In this case, Geeta is the only student who is a member of both teams.

Given sets are:

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

B = {2, 3, 5, 7}

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We need to find the intersection of sets A and B, denoted by $A \cap B$.

$A \cap B = \{x : x \in A \text{ and } x \in B\}$

To find $A \cap B$, we look for the elements that are present in both set A and set B.

Elements in A are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Elements in B are 2, 3, 5, 7.

The elements that appear in both lists are 2, 3, 5, and 7.

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Thus, $A \cap B$ consists of the elements 2, 3, 5, and 7.

A ∩ B = {2, 3, 5, 7}.

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Now we need to show that $A \cap B = B$.

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

The given set B is {2, 3, 5, 7}.

Comparing the elements of $A \cap B$ and B, we see they are exactly the same.

Therefore, $A \cap B = B$.

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Reason:

Observe that every element of set B (2, 3, 5, 7) is also an element of set A. This means that B is a subset of A, i.e., $B \subset A$.

When set B is a subset of set A, the intersection of A and B ($A \cap B$) will contain all elements that are common to both A and B. Since all elements of B are already in A, the elements common to both sets will be precisely the elements of B. This is a general property of sets: if $B \subset A$, then $A \cap B = B$.

Given sets are:

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

B = {2, 4, 6, 8}

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The difference between two sets A and B, denoted by $A - B$, is the set of elements which are in A but not in B.

In set-builder form, $A - B = \{x : x \in A \text{ and } x \notin B\}$.

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To find $A - B$, we start with the elements of A and remove any elements that are also present in B.

Elements in A: 1, 2, 3, 4, 5, 6.

Elements in B: 2, 4, 6, 8.

Identify elements common to both sets (this is $A \cap B$): {2, 4, 6}.

Now, take elements from A and remove those in the intersection:

From {1, 2, 3, 4, 5, 6}, remove {2, 4, 6}.

The remaining elements are 1, 3, 5.

A – B = {1, 3, 5}.

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The difference between two sets B and A, denoted by $B - A$, is the set of elements which are in B but not in A.

In set-builder form, $B - A = \{x : x \in B \text{ and } x \notin A\}$.

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To find $B - A$, we start with the elements of B and remove any elements that are also present in A.

Elements in B: 2, 4, 6, 8.

Elements in A: 1, 2, 3, 4, 5, 6.

Identify elements common to both sets (this is $B \cap A = A \cap B$): {2, 4, 6}.

Now, take elements from B and remove those in the intersection:

From {2, 4, 6, 8}, remove {2, 4, 6}.

The remaining element is 8.

B – A = {8}.

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Note that in general, $A - B \neq B - A$. This example illustrates this point.

Given sets are:

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

B = {a, i, k, u}

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We need to find the difference between sets V and B, denoted by $V - B$.

$V - B = \{x : x \in V \text{ and } x \notin B\}$.

To find $V - B$, we start with the elements of V and remove any elements that are also present in B.

Elements in V: a, e, i, o, u.

Elements in B: a, i, k, u.

Identify elements common to both sets (this is $V \cap B$): {a, i, u}.

Now, take elements from V and remove those in the intersection:

From {a, e, i, o, u}, remove {a, i, u}.

The remaining elements are e, o.

V – B = {e, o}.

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We need to find the difference between sets B and V, denoted by $B - V$.

$B - V = \{x : x \in B \text{ and } x \notin V\}$.

To find $B - V$, we start with the elements of B and remove any elements that are also present in V.

Elements in B: a, i, k, u.

Elements in V: a, e, i, o, u.

Identify elements common to both sets (this is $B \cap V = V \cap B$): {a, i, u}.

Now, take elements from B and remove those in the intersection:

From {a, i, k, u}, remove {a, i, u}.

The remaining element is k.

B – V = {k}.

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As seen in Example 18, $V - B \neq B - V$.

The union of two sets A and B, denoted by $A \cup B$, is the set of all elements that are in A or in B or in both.

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

$X \cup Y$ contains all elements from X and Y, without repetition.

Elements in X: 1, 3, 5

Elements in Y: 1, 2, 3

Combining unique elements: 1, 3, 5, 2.

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

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

$A \cup B$ contains all elements from A and B, without repetition.

Elements in A: a, e, i, o, u

Elements in B: a, b, c

Combining unique elements: a, e, i, o, u, b, c.

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}

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

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

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

$A \cup B$ contains all elements from A and B, without repetition.

Elements from A: 3, 6, 9, 12, ...

Elements from B: 1, 2, 3, 4, 5. (Note that 3 is already in A).

Combining unique elements: 1, 2, 3, 4, 5, 6, 9, 12, ...

A $\cup$ B = {1, 2, 3, 4, 5, 6, 9, 12, ...} or {x : x ∈ {1, 2, 3, 4, 5} or x is a multiple of 3 greater than or equal to 3}.

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

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

B = {x : x is a natural number and 6 < x < 10 }

B = {7, 8, 9}

$A \cup B$ contains all elements from A and B, without repetition.

Elements in A: 2, 3, 4, 5, 6

Elements in B: 7, 8, 9

Combining unique elements: 2, 3, 4, 5, 6, 7, 8, 9.

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

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(v) A = {1, 2, 3}, B = $\phi$

$A \cup B$ contains all elements from A and B, without repetition.

Elements in A: 1, 2, 3

Elements in B: The empty set has no elements.

Combining unique elements: 1, 2, 3.

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

Note that $A \cup \phi = A$, which is a general property of the union operation.

Given sets are:

A = {a, b}

B = {a, b, c}

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First, we check if A is a subset of B. The symbol $\subset$ means 'is a subset of'.

A set A is a subset of set B if every element of A is also an element of B.

The elements of set A are 'a' and 'b'.

Let's check if these elements are in set B:

Is 'a' in B? Yes, a $\in$ B.

Is 'b' in B? Yes, b $\in$ B.

Since every element of A is also an element of B, A is a subset of B.

Yes, A ⊂ B is True.

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Next, we find the union of A and B, denoted by $A \cup B$.

$A \cup B$ is the set of all elements which are in A or in B or in both.

To find $A \cup B$, we list all the elements present in either set, without repeating any.

Elements in A: a, b

Elements in B: a, b, c

Combining the unique elements from both sets: a, b, c.

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

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Note that since A $\subset$ B, the union $A \cup B$ is equal to B. This is a general property: if A is a subset of B, then $A \cup B = B$.

We are given that A and B are two sets such that A $\subset$ B. This means that every element of set A is also an element of set B.

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We need to find the union of A and B, denoted by $A \cup B$.

By definition, $A \cup B = \{x : x \in A \text{ or } x \in B\}$.

Let $x$ be an element of $A \cup B$. By the definition of union, $x$ is either in A or in B (or both).

Case 1: $x \in A$.

Since we are given that A $\subset$ B, if $x \in A$, it implies that $x \in B$ as well.

Case 2: $x \in B$.

In this case, $x$ is already in B.

In both cases, if $x \in A \cup B$, then $x \in B$. This shows that $A \cup B \subset B$.

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Now let's consider an element $y \in B$.

If $y \in B$, then by the definition of union, $y \in A \cup B$ (since the union includes all elements in B).

This shows that $B \subset A \cup B$.

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Since we have shown that $A \cup B \subset B$ and $B \subset A \cup B$, by the definition of set equality, we can conclude that $A \cup B = B$.

---

Alternatively, we can think about combining the elements. Since every element in A is already in B (because A $\subset$ B), when we combine the elements of A and B to form the union, we are essentially just taking all the elements of B (as all elements of A are already included in B).

---

For example, if A = {1, 2} and B = {1, 2, 3}, then A $\subset$ B. $A \cup B = \{1, 2\} \cup \{1, 2, 3\} = \{1, 2, 3\} = B$.

---

If A $\subset$ B, then A ∪ B = B.

The given sets are:

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

C = {5, 6, 7, 8}

D = {7, 8, 9, 10}

---

The union of sets contains all unique elements from the sets being combined.

---

(i) A ∪ B

A ∪ B contains elements that are in A or in B.

Elements of A: 1, 2, 3, 4

Elements of B: 3, 4, 5, 6

Combining unique elements: 1, 2, 3, 4, 5, 6.

A ∪ B = {1, 2, 3, 4, 5, 6}.

---

(ii) A ∪ C

A ∪ C contains elements that are in A or in C.

Elements of A: 1, 2, 3, 4

Elements of C: 5, 6, 7, 8

Combining unique elements: 1, 2, 3, 4, 5, 6, 7, 8.

A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.

---

(iii) B ∪ C

B ∪ C contains elements that are in B or in C.

Elements of B: 3, 4, 5, 6

Elements of C: 5, 6, 7, 8

Combining unique elements: 3, 4, 5, 6, 7, 8.

B ∪ C = {3, 4, 5, 6, 7, 8}.

---

(iv) B ∪ D

B ∪ D contains elements that are in B or in D.

Elements of B: 3, 4, 5, 6

Elements of D: 7, 8, 9, 10

Combining unique elements: 3, 4, 5, 6, 7, 8, 9, 10.

B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.

---

(v) A ∪ B ∪ C

A ∪ B ∪ C contains elements that are in A or in B or in C.

Elements of A: 1, 2, 3, 4

Elements of B: 3, 4, 5, 6

Elements of C: 5, 6, 7, 8

Combining unique elements from all three sets: 1, 2, 3, 4, 5, 6, 7, 8.

A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.

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(vi) A ∪ B ∪ D

A ∪ B ∪ D contains elements that are in A or in B or in D.

Elements of A: 1, 2, 3, 4

Elements of B: 3, 4, 5, 6

Elements of D: 7, 8, 9, 10

Combining unique elements from all three sets: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

---

(vii) B ∪ C ∪ D

B ∪ C ∪ D contains elements that are in B or in C or in D.

Elements of B: 3, 4, 5, 6

Elements of C: 5, 6, 7, 8

Elements of D: 7, 8, 9, 10

Combining unique elements from all three sets: 3, 4, 5, 6, 7, 8, 9, 10.

B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.

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

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

$X \cap Y$ contains elements that are in both X and Y.

Elements in X: 1, 3, 5

Elements in Y: 1, 2, 3

Common elements: 1, 3.

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

---

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

$A \cap B$ contains elements that are in both A and B.

Elements in A: a, e, i, o, u

Elements in B: a, b, c

Common elements: a.

A $\cap$ B = {a}.

---

(iii) A = {x : x is a natural number and multiple of 3}

A = {3, 6, 9, 12, 15, 18, ...}

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

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

$A \cap B$ contains elements that are in both A and B.

Look for elements that are both multiples of 3 and natural numbers less than 6.

From B = {1, 2, 3, 4, 5}, which ones are multiples of 3? Only 3.

A $\cap$ B = {3}.

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

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

B = {x : x is a natural number and 6 < x < 10 }

B = {7, 8, 9}

$A \cap B$ contains elements that are in both A and B.

Elements in A: 2, 3, 4, 5, 6

Elements in B: 7, 8, 9

Are there any common elements? No.

A $\cap$ B = $\phi$ (the empty set).

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(v) A = {1, 2, 3}, B = $\phi$

$A \cap B$ contains elements that are in both A and B.

Elements in A: 1, 2, 3

Elements in B: The empty set has no elements.

Are there any common elements? No, since B has no elements.

A $\cap$ B = $\phi$.

Note that $A \cap \phi = \phi$, which is a general property of the intersection operation.

The given sets are:

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

B = {7, 9, 11, 13}

C = {11, 13, 15}

D = {15, 17}

---

The intersection of sets contains common elements. The union of sets contains all unique elements.

---

(i) A ∩ B

Elements common to A and B: {7, 9, 11}.

A ∩ B = {7, 9, 11}.

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(ii) B ∩ C

Elements common to B and C: {11, 13}.

B ∩ C = {11, 13}.

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(iii) A ∩ C ∩ D

This is the intersection of A, C, and D. We can find $(A \cap C) \cap D$ or $A \cap (C \cap D)$.

First, find A ∩ C:

Elements common to A and C: {11}. So, A ∩ C = {11}.

Now, find $(A \cap C) \cap D$: {11} ∩ {15, 17}.

There are no common elements.

A ∩ C ∩ D = $\phi$.

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(iv) A ∩ C

Elements common to A and C: {11}.

A ∩ C = {11}.

---

(v) B ∩ D

Elements common to B and D: There are no common elements.

B ∩ D = $\phi$.

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(vi) A ∩ (B ∪ C)

First, find B ∪ C:

B ∪ C contains elements in B or C: {7, 9, 11, 13} ∪ {11, 13, 15} = {7, 9, 11, 13, 15}.

Now, find A ∩ (B ∪ C):

{3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}.

Elements common to these two sets: {7, 9, 11}.

A ∩ (B ∪ C) = {7, 9, 11}.

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(vii) A ∩ D

Elements common to A and D: There are no common elements.

A ∩ D = $\phi$.

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(viii) A ∩ (B ∪ D)

First, find B ∪ D:

B ∪ D contains elements in B or D: {7, 9, 11, 13} ∪ {15, 17} = {7, 9, 11, 13, 15, 17}.

Now, find A ∩ (B ∪ D):

{3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}.

Elements common to these two sets: {7, 9, 11}.

A ∩ (B ∪ D) = {7, 9, 11}.

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(ix) ( A ∩ B ) ∩ ( B ∪ C )

We already found A ∩ B = {7, 9, 11}.

We already found B ∪ C = {7, 9, 11, 13, 15}.

Now, find the intersection of these two results:

{7, 9, 11} ∩ {7, 9, 11, 13, 15}.

Elements common to these two sets: {7, 9, 11}.

( A ∩ B ) ∩ ( B ∪ C ) = {7, 9, 11}.

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(x) ( A ∪ D) ∩ ( B ∪ C)

First, find A ∪ D:

A ∪ D contains elements in A or D: {3, 5, 7, 9, 11} ∪ {15, 17} = {3, 5, 7, 9, 11, 15, 17}.

We already found B ∪ C = {7, 9, 11, 13, 15}.

Now, find the intersection of these two results:

{3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15}.

Elements common to these two sets: {7, 9, 11, 15}.

( A ∪ D) ∩ ( B ∪ C) = {7, 9, 11, 15}.

The given sets are:

A = {x : x is a natural number} = {1, 2, 3, 4, 5, ...}

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, ...}

---

We need to find the intersection of the given pairs of sets. The intersection contains elements common to both sets.

---

(i) A ∩ B

A $\cap$ B contains elements that are natural numbers and are also even natural numbers.

Every even natural number is also a natural number. So, the elements common to both sets are simply all the even natural numbers.

A $\cap$ B = {x : x is an even natural number} = B.

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(ii) A ∩ C

A $\cap$ C contains elements that are natural numbers and are also odd natural numbers.

Every odd natural number is also a natural number. So, the elements common to both sets are simply all the odd natural numbers.

A $\cap$ C = {x : x is an odd natural number} = C.

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(iii) A ∩ D

A $\cap$ D contains elements that are natural numbers and are also prime numbers.

Every prime number is also a natural number. So, the elements common to both sets are simply all the prime numbers.

A $\cap$ D = {x : x is a prime number} = D.

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(iv) B ∩ C

B $\cap$ C contains elements that are even natural numbers and are also odd natural numbers.

An integer cannot be both even and odd simultaneously.

There are no elements common to sets B and C.

B $\cap$ C = $\phi$.

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(v) B ∩ D

B $\cap$ D contains elements that are even natural numbers and are also prime numbers.

The elements of B are {2, 4, 6, 8, ...}.

The elements of D are {2, 3, 5, 7, 11, ...}.

The only number that is both even and prime is 2.

B $\cap$ D = {2}.

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(vi) C ∩ D

C $\cap$ D contains elements that are odd natural numbers and are also prime numbers.

The elements of C are {1, 3, 5, 7, 9, 11, ...}.

The elements of D are {2, 3, 5, 7, 11, ...}.

The common elements are prime numbers that are odd.

Excluding 2 (which is even) from the set of prime numbers, we get all other prime numbers, which are odd.

C $\cap$ D = {3, 5, 7, 11, 13, ...}. This is the set of all odd prime numbers.

Two sets are said to be disjoint if their intersection is the empty set, i.e., they have no common elements.

Sets A and B are disjoint if $A \cap B = \phi$.

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

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

Let Set 2 = {x : x is a natural number and 4 ≤ x ≤ 6}. The natural numbers between 4 and 6, inclusive, are 4, 5, and 6. So, Set 2 = {4, 5, 6}.

Find the intersection of Set 1 and Set 2:

{1, 2, 3, 4} $\cap$ {4, 5, 6}.

The common element is 4.

The intersection is {4}.

Since the intersection is not the empty set ($\{4\} \neq \phi$), the sets are not disjoint.

This pair of sets is not disjoint.

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

Let Set 1 = {a, e, i, o, u}.

Let Set 2 = {c, d, e, f}.

Find the intersection of Set 1 and Set 2:

{a, e, i, o, u} $\cap$ {c, d, e, f}.

The common element is 'e'.

The intersection is {e}.

Since the intersection is not the empty set ($\{e\} \neq \phi$), the sets are not disjoint.

This pair of sets is not disjoint.

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

Let Set 1 = {x : x is an even integer} = {..., -4, -2, 0, 2, 4, ...}.

Let Set 2 = {x : x is an odd integer} = {..., -3, -1, 1, 3, 5, ...}.

Find the intersection of Set 1 and Set 2:

{..., -4, -2, 0, 2, 4, ...} $\cap$ {..., -3, -1, 1, 3, 5, ...}.

An integer is either even or odd; it cannot be both simultaneously.

There are no elements common to the set of even integers and the set of odd integers.

The intersection is the empty set $\phi$.

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

This pair of sets is disjoint.

The given sets are:

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}

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The difference between two sets X and Y, denoted by $\mathbf{X - Y}$, is the set of all elements that are in X but are not in Y.

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(i) A – B

We need the elements that are in set A but not in set B.

The elements in A are {3, 6, 9, 12, 15, 18, 21}.

The elements common to A and B are {12}.

Removing the common elements from A, we get:

A – B = {3, 6, 9, 15, 18, 21}.

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(ii) A – C

We need the elements that are in set A but not in set C.

The elements in A are {3, 6, 9, 12, 15, 18, 21}.

The elements common to A and C are {6, 12}.

Removing the common elements from A, we get:

A – C = {3, 9, 15, 18, 21}.

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(iii) A – D

We need the elements that are in set A but not in set D.

The elements in A are {3, 6, 9, 12, 15, 18, 21}.

The elements common to A and D are {15}.

Removing the common elements from A, we get:

A – D = {3, 6, 9, 12, 18, 21}.

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(iv) B – A

We need the elements that are in set B but not in set A.

The elements in B are {4, 8, 12, 16, 20}.

The elements common to B and A are {12}.

Removing the common element from B, we get:

B – A = {4, 8, 16, 20}.

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(v) C – A

We need the elements that are in set C but not in set A.

The elements in C are {2, 4, 6, 8, 10, 12, 14, 16}.

The elements common to C and A are {6, 12}.

Removing the common elements from C, we get:

C – A = {2, 4, 8, 10, 14, 16}.

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(vi) D – A

We need the elements that are in set D but not in set A.

The elements in D are {5, 10, 15, 20}.

The elements common to D and A are {15}.

Removing the common element from D, we get:

D – A = {5, 10, 20}.

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(vii) B – C

We need the elements that are in set B but not in set C.

The elements in B are {4, 8, 12, 16, 20}.

The elements common to B and C are {4, 8, 12, 16}.

Removing the common elements from B, we get:

B – C = {20}.

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(viii) B – D

We need the elements that are in set B but not in set D.

The elements in B are {4, 8, 12, 16, 20}.

The elements common to B and D are {20}.

Removing the common element from B, we get:

B – D = {4, 8, 12, 16}.

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(ix) C – B

We need the elements that are in set C but not in set B.

The elements in C are {2, 4, 6, 8, 10, 12, 14, 16}.

The elements common to C and B are {4, 8, 12, 16}.

Removing the common elements from C, we get:

C – B = {2, 6, 10, 14}.

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(x) D – B

We need the elements that are in set D but not in set B.

The elements in D are {5, 10, 15, 20}.

The elements common to D and B are {20}.

Removing the common element from D, we get:

D – B = {5, 10, 15}.

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(xi) C – D

We need the elements that are in set C but not in set D.

The elements in C are {2, 4, 6, 8, 10, 12, 14, 16}.

The elements common to C and D are {10}.

Removing the common element from C, we get:

C – D = {2, 4, 6, 8, 12, 14, 16}.

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(xii) D – C

We need the elements that are in set D but not in set C.

The elements in D are {5, 10, 15, 20}.

The elements common to D and C are {10}.

Removing the common element from D, we get:

D – C = {5, 15, 20}.

The given sets are:

X = {a, b, c, d}

Y = {f, b, d, g}

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(i) X – Y

$X - Y$ is the set of elements which are in X but not in Y.

Elements in X: a, b, c, d.

Elements in Y: f, b, d, g.

Elements in X that are also in Y (intersection): b, d.

Remove {b, d} from {a, b, c, d}.

X – Y = {a, c}.

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(ii) Y – X

$Y - X$ is the set of elements which are in Y but not in X.

Elements in Y: f, b, d, g.

Elements in X: a, b, c, d.

Elements in Y that are also in X (intersection): b, d.

Remove {b, d} from {f, b, d, g}.

Y – X = {f, g}.

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(iii) X ∩ Y

$X \cap Y$ is the set of elements which are common to both X and Y.

Elements in X: a, b, c, d.

Elements in Y: f, b, d, g.

The elements common to both sets are b and d.

X ∩ Y = {b, d}.

Given sets are:

R = the set of all real numbers ($\mathbb{R}$)

Q = the set of all rational numbers ($\mathbb{Q}$)

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A real number is a number that can be found on the number line. It includes rational numbers and irrational numbers.

A rational number is a number that can be expressed as a fraction $\frac{p}{q}$ of two integers, where $p$ is an integer and $q$ is a non-zero integer ($q \neq 0$). Examples include $\frac{1}{2}$, $-3$, $0.75$ (which is $\frac{3}{4}$), $0.\overline{3}$ (which is $\frac{1}{3}$).

An irrational number is a real number that cannot be expressed as a simple fraction $\frac{p}{q}$. Their decimal expansions are non-terminating and non-repeating. Examples include $\sqrt{2}$, $\pi$, $e$.

---

We need to find R – Q.

$R - Q = \{x : x \in R \text{ and } x \notin Q\}$.

This set contains all real numbers that are not rational numbers.

By definition, the set of real numbers is composed of rational numbers and irrational numbers.

Real Numbers = Rational Numbers $\cup$ Irrational Numbers.

Also, the set of rational numbers and the set of irrational numbers are disjoint (a number cannot be both rational and irrational).

The set of real numbers that are not rational numbers is precisely the set of irrational numbers.

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The set of irrational numbers is often denoted by I or $\mathbb{R} \setminus \mathbb{Q}$.

R – Q = {x : x is a real number and x is not a rational number}

R – Q = {x : x is an irrational number}

R – Q = The set of irrational numbers.

Two sets are disjoint if their intersection is the empty set ($\phi$). This means they have no elements in common.

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(i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.

Let Set 1 = {2, 3, 4, 5} and Set 2 = {3, 6}.

Find the intersection: {2, 3, 4, 5} $\cap$ {3, 6}.

The common element is 3. The intersection is {3}.

Since the intersection {3} is not the empty set $\phi$, the sets are not disjoint.

The statement is False. Justification: The intersection of the two sets is {3}, which is not the empty set.

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(ii) { a, e, i, o, u } and { a, b, c, d }are disjoint sets.

Let Set 1 = {a, e, i, o, u} and Set 2 = {a, b, c, d}.

Find the intersection: {a, e, i, o, u} $\cap$ {a, b, c, d}.

The common element is 'a'. The intersection is {a}.

Since the intersection {a} is not the empty set $\phi$, the sets are not disjoint.

The statement is False. Justification: The intersection of the two sets is {a}, which is not the empty set.

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(iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.

Let Set 1 = {2, 6, 10, 14} and Set 2 = {3, 7, 11, 15}.

Find the intersection: {2, 6, 10, 14} $\cap$ {3, 7, 11, 15}.

There are no common elements between the two sets.

The intersection is the empty set $\phi$.

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

The statement is True. Justification: The intersection of the two sets is $\phi$, the empty set.

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(iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.

Let Set 1 = {2, 6, 10} and Set 2 = {3, 7, 11}.

Find the intersection: {2, 6, 10} $\cap$ {3, 7, 11}.

There are no common elements between the two sets.

The intersection is the empty set $\phi$.

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

The statement is True. Justification: The intersection of the two sets is $\phi$, the empty set.

Given the universal set U and set A:

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

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

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The complement of a set A (with respect to a universal set U), denoted by A' or $A^c$ or $\bar{A}$, is the set of all elements in U that are not in A.

In set-builder form, $A' = \{x : x \in U \text{ and } x \notin A\}$.

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To find A', we look at the elements in U and exclude those that are present 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 1, 3, 5, 7, and 9 from U.

The remaining elements in U are 2, 4, 6, 8, 10.

---

Thus, A' consists of the elements 2, 4, 6, 8, and 10.

$A' = {2, 4, 6, 8, 10}$.

Given the universal set U and set A:

U = The set of all students of Class XI of a coeducational school.

A = The set of all girls in Class XI.

---

We need to find A', the complement of set A with respect to the universal set U.

$A' = \{x : x \in U \text{ and } x \notin A\}$.

This means A' is the set of all elements in U that are not in A.

U consists of all students (both boys and girls) in Class XI.

A consists only of the girls in Class XI.

We are looking for students in Class XI who are not girls.

In a coeducational school, students who are not girls are boys.

---

Therefore, A' is the set of all boys in Class XI.

$A'$ = The set of all boys in Class XI.

Given sets are:

U = {1, 2, 3, 4, 5, 6}

A = {2, 3}

B = {3, 4, 5}

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First, find A′ and B′.

A′ is the set of elements in U but not in A.

A′ = U - A = {1, 2, 3, 4, 5, 6} - {2, 3}.

Removing 2 and 3 from U, we get {1, 4, 5, 6}.

A′ = {1, 4, 5, 6}.

B′ is the set of elements in U but not in B.

B′ = U - B = {1, 2, 3, 4, 5, 6} - {3, 4, 5}.

Removing 3, 4, and 5 from U, we get {1, 2, 6}.

B′ = {1, 2, 6}.

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Next, find A′ ∩ B′.

A′ ∩ B′ is the set of elements common to A′ and B′.

A′ = {1, 4, 5, 6}

B′ = {1, 2, 6}

The common elements are 1 and 6.

A′ ∩ B′ = {1, 6}.

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Now, find A ∪ B.

A ∪ B is the set of elements in A or in B or both.

A = {2, 3}

B = {3, 4, 5}

Combining unique elements: 2, 3, 4, 5.

A ∪ B = {2, 3, 4, 5}.

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Finally, we need to show that (A ∪ B)′ = A′ ∩ B′.

First, find (A ∪ B)′.

(A ∪ B)′ is the complement of the set (A ∪ B) with respect to U.

(A ∪ B)′ = U - (A ∪ B) = {1, 2, 3, 4, 5, 6} - {2, 3, 4, 5}.

Removing 2, 3, 4, and 5 from U, we get {1, 6}.

(A ∪ B)′ = {1, 6}.

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Comparing the results:

A′ ∩ B′ = {1, 6}

(A ∪ B)′ = {1, 6}

Since both sets contain exactly the same elements, we have shown that (A ∪ B)′ = A′ ∩ B′.

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This example demonstrates De Morgan's Law for union.

The given sets are:

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}

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The complement of a set S (S') with respect to U is U - S.

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(i) A′

A′ = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4}.

Removing 1, 2, 3, 4 from U, we get {5, 6, 7, 8, 9}.

A′ = {5, 6, 7, 8, 9}.

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(ii) B′

B′ = U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}.

Removing 2, 4, 6, 8 from U, we get {1, 3, 5, 7, 9}.

B′ = {1, 3, 5, 7, 9}.

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(iii) (A ∪ C)′

First, find A ∪ C.

A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}.

Now, find the complement of A ∪ C:

(A ∪ C)′ = U - (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6}.

Removing 1, 2, 3, 4, 5, 6 from U, we get {7, 8, 9}.

(A ∪ C)′ = {7, 8, 9}.

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(iv) (A ∪ B)′

First, find A ∪ B.

A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}.

Now, find the complement of A ∪ B:

(A ∪ B)′ = U - (A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8}.

Removing 1, 2, 3, 4, 6, 8 from U, we get {5, 7, 9}.

(A ∪ B)′ = {5, 7, 9}.

---

(v) (A′)′

The complement of the complement of a set is the set itself. $(A')' = A$.

We found A′ = {5, 6, 7, 8, 9}.

(A′)′ = U - A′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {5, 6, 7, 8, 9}.

Removing 5, 6, 7, 8, 9 from U, we get {1, 2, 3, 4}.

(A′)′ = {1, 2, 3, 4}. This is indeed set A.

---

(vi) (B – C)′

First, find B – C.

B – C is the set of elements in B but not in C.

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

Elements common to B and C: {4, 6}.

Removing {4, 6} from B {2, 4, 6, 8}, we get {2, 8}.

B – C = {2, 8}.

Now, find the complement of B – C:

(B – C)′ = U - (B – C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 8}.

Removing 2 and 8 from U, we get {1, 3, 4, 5, 6, 7, 9}.

(B – C)′ = {1, 3, 4, 5, 6, 7, 9}.

The universal set is U = {a, b, c, d, e, f, g, h}.

The complement of a set S (S') with respect to U is the set of all elements in U that are not in S, i.e., $S' = U - S$.

---

(i) A = {a, b, c}

A′ = U - A = {a, b, c, d, e, f, g, h} - {a, b, c}.

Removing a, b, c from U, we get d, e, f, g, h.

A′ = {d, e, f, g, h}.

---

(ii) B = {d, e, f, g}

B′ = U - B = {a, b, c, d, e, f, g, h} - {d, e, f, g}.

Removing d, e, f, g from U, we get a, b, c, h.

B′ = {a, b, c, h}.

---

(iii) C = {a, c, e, g}

C′ = U - C = {a, b, c, d, e, f, g, h} - {a, c, e, g}.

Removing a, c, e, g from U, we get b, d, f, h.

C′ = {b, d, f, h}.

---

(iv) D = { f, g, h, a}

D′ = U - D = {a, b, c, d, e, f, g, h} - {f, g, h, a}.

Removing f, g, h, a from U, we get b, c, d, e.

D′ = {b, c, d, e}.

The universal set U is the set of all natural numbers, U = N = {1, 2, 3, 4, 5, ...}.

The complement of a set A, denoted by $\mathbf{A'}$ or $\mathbf{A^c}$, is the set of all elements in the universal set U that are not in set A. Mathematically, $\mathbf{A' = U - A}$.

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

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

The complement A' consists of all natural numbers that are not even.

A′ = U - A = {x : x is a natural number} - {x : x is an even natural number}.

The natural numbers that are not even are the odd natural numbers.

A′ = {x : x is an odd natural number}.

In roster form, A′ = {1, 3, 5, 7, ...}.

---

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

A = {1, 3, 5, 7, ...}

The complement A' consists of all natural numbers that are not odd.

A′ = U - A = {x : x is a natural number} - {x : x is an odd natural number}.

The natural numbers that are not odd are the even natural numbers.

A′ = {x : x is an even natural number}.

In roster form, A′ = {2, 4, 6, 8, ...}.

---

(iii) Let A = {x : x is a positive multiple of 3}

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

The complement A' consists of all natural numbers that are not multiples of 3.

A′ = U - A = {x : x is a natural number} - {x : x is a multiple of 3}.

A′ = {x : x ∈ N and x is not a multiple of 3}.

---

(iv) Let A = { x : x is a prime number }

A = {2, 3, 5, 7, 11, 13, ...}

The complement A' consists of all natural numbers that are not prime. Remember that the set of natural numbers starts from 1.

A′ = U - A = {x : x is a natural number} - {x : x is a prime number}.

The natural numbers that are not prime include 1 (which is neither prime nor composite) and all composite numbers (natural numbers greater than 1 that are not prime).

A′ = {x : x is a natural number and x is not a prime number}.

In roster form, A′ = {1, 4, 6, 8, 9, 10, 12, 14, 15, 16, ...}.

---

(v) Let A = {x : x is a natural number divisible by 3 and 5}

A number divisible by both 3 and 5 is divisible by their least common multiple, which is $LCM(3, 5) = 15$.

So, A = {x : x is a natural number and x is a multiple of 15}.

A = {15, 30, 45, 60, ...}

The complement A' consists of all natural numbers that are not multiples of 15.

A′ = U - A = {x : x is a natural number} - {x : x is a multiple of 15}.

A′ = {x : x ∈ N and x is not divisible by 15}.

---

(vi) Let A = { x : x is a perfect square }

A = {1², 2², 3², 4², 5², ...} = {1, 4, 9, 16, 25, ...}

The complement A' consists of all natural numbers that are not perfect squares.

A′ = U - A = {x : x is a natural number} - {x : x is a perfect square}.

A′ = {x : x ∈ N and x is not a perfect square}.

---

(vii) Let A = { x : x is a perfect cube}

A = {1³, 2³, 3³, 4³, ...} = {1, 8, 27, 64, ...}

The complement A' consists of all natural numbers that are not perfect cubes.

A′ = U - A = {x : x is a natural number} - {x : x is a perfect cube}.

A′ = {x : x ∈ N and x is not a perfect cube}.

---

(viii) Let A = { x : x + 5 = 8 }

First, we find the element(s) in set A by solving the equation $x + 5 = 8$ for a natural number $x$.

$x + 5 = 8$

$x = 8 - 5$

$x = 3$

Since 3 is a natural number, the set A contains the element 3.

A = {3}.

The complement A' consists of all natural numbers except 3.

A′ = U - A = {1, 2, 3, 4, 5, ...} - {3}.

A′ = {x : x ∈ N and x $\neq$ 3}.

In roster form, A′ = {1, 2, 4, 5, 6, 7, ...}.

---

(ix) Let A = { x : 2x + 5 = 9}

First, we find the element(s) in set A by solving the equation $2x + 5 = 9$ for a natural number $x$.

$2x + 5 = 9$

$2x = 9 - 5$

$2x = 4$

$x = \frac{4}{2}$

$x = 2$

Since 2 is a natural number, the set A contains the element 2.

A = {2}.

The complement A' consists of all natural numbers except 2.

A′ = U - A = {1, 2, 3, 4, 5, ...} - {2}.

A′ = {x : x ∈ N and x $\neq$ 2}.

In roster form, A′ = {1, 3, 4, 5, 6, 7, ...}.

---

(x) Let A = { x : x ≥ 7 }

A = {natural numbers greater than or equal to 7} = {7, 8, 9, 10, ...}.

The complement A' consists of all natural numbers that are not greater than or equal to 7. These are the natural numbers less than 7.

A′ = U - A = {1, 2, 3, 4, 5, 6, 7, 8, ...} - {7, 8, 9, 10, ...}.

A′ = {1, 2, 3, 4, 5, 6}.

A′ = {x : x ∈ N and x < 7}.

Alternatively, A′ = {x : x is a natural number less than 7}.

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(xi) Let A = { x : x ∈ N and 2x + 1 > 10 }

First, we find the natural numbers $x$ that satisfy the inequality $2x + 1 > 10$.

$2x + 1 > 10$

$2x > 10 - 1$

$2x > 9$

$x > \frac{9}{2}$

$x > 4.5$

The natural numbers $x$ that are greater than 4.5 are 5, 6, 7, 8, ...

So, A = {5, 6, 7, 8, ...}.

The complement A' consists of all natural numbers that are not in A. These are the natural numbers that are less than or equal to 4.5.

A′ = U - A = {1, 2, 3, 4, 5, 6, 7, ...} - {5, 6, 7, 8, ...}.

A′ = {1, 2, 3, 4}.

A′ = {x : x ∈ N and x ≤ 4}.

Alternatively, A′ = {x : x is a natural number less than 5}.

The given sets are:

Universal Set $\mathbf{U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}}$

Set $\mathbf{A = \{2, 4, 6, 8\}}$

Set $\mathbf{B = \{2, 3, 5, 7\}}$

---

We need to verify two properties related to complements, union, and intersection of sets using the given sets U, A, and B. These properties are known as De Morgan's Laws.

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(i) Verify $\mathbf{(A \cup B)' = A' \cap B'}$

Let's calculate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately.

LHS: $\mathbf{(A \cup B)'}$

First, find the union of sets A and B: $\mathbf{A \cup B}$. The union contains all elements that are in A or in B or in both.

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

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

Next, find the complement of $\mathbf{(A \cup B)}$ with respect to the universal set U. This means finding the elements in U that are not in $A \cup B$.

$(A \cup B)' = U - (A \cup B)$

$(A \cup B)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 4, 5, 6, 7, 8\}$

Removing the elements {2, 3, 4, 5, 6, 7, 8} from U, we are left with {1, 9}.

LHS = $\{1, 9\}$.

RHS: $\mathbf{A' \cap B'}$

First, find the complement of set A with respect to U: $\mathbf{A'}$. This means finding the elements in U that are not in A.

$A' = U - A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\}$

$A' = \{1, 3, 5, 7, 9\}$

Next, find the complement of set B with respect to U: $\mathbf{B'}$. This means finding the elements in U that are not in B.

$B' = U - B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 3, 5, 7\}$

$B' = \{1, 4, 6, 8, 9\}$

Now, find the intersection of A' and B': $\mathbf{A' \cap B'}$. The intersection contains the elements that are common to both A' and B'.

$A' \cap B' = \{1, 3, 5, 7, 9\} \cap \{1, 4, 6, 8, 9\}$

The common elements are 1 and 9.

RHS = $\{1, 9\}$.

Comparing the LHS and RHS:

LHS = $\{1, 9\}$

RHS = $\{1, 9\}$

Since LHS = RHS, the property $(A \cup B)' = A' \cap B'$ is verified for the given sets.

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(ii) Verify $\mathbf{(A \cap B)' = A' \cup B'}$

Let's calculate the Left Hand Side (LHS) and the Right Hand Side (RHS) separately.

LHS: $\mathbf{(A \cap B)'}$

First, find the intersection of sets A and B: $\mathbf{A \cap B}$. The intersection contains the elements that are common to both A and B.

$A \cap B = \{2, 4, 6, 8\} \cap \{2, 3, 5, 7\}$

The common element is 2.

$A \cap B = \{2\}$

Next, find the complement of $\mathbf{(A \cap B)}$ with respect to the universal set U. This means finding the elements in U that are not in $A \cap B$.

$(A \cap B)' = U - (A \cap B)$

$(A \cap B)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2\}$

Removing the element {2} from U, we are left with {1, 3, 4, 5, 6, 7, 8, 9}.

LHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$.

RHS: $\mathbf{A' \cup B'}$

We have already calculated the complements A' and B' in part (i):

$A' = \{1, 3, 5, 7, 9\}$

$B' = \{1, 4, 6, 8, 9\}$

Now, find the union of A' and B': $\mathbf{A' \cup B'}$. The union contains all elements that are in A' or in B' or in both.

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

Combining unique elements: 1, 3, 5, 7, 9 (from A') and 1, 4, 6, 8, 9 (from B'). The unique elements are 1, 3, 4, 5, 6, 7, 8, 9.

RHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$.

Comparing the LHS and RHS:

LHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$

RHS = $\{1, 3, 4, 5, 6, 7, 8, 9\}$

Since LHS = RHS, the property $(A \cap B)' = A' \cup B'$ is verified for the given sets.

We will draw Venn diagrams to represent the specified set operations. A Venn diagram uses a rectangle to represent the universal set (U) and circles (usually overlapping) within the rectangle to represent subsets. The shaded region in each diagram represents the set being described.

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(i) (A ∪ B)′

This represents the complement of the union of sets A and B. The union A ∪ B is the region that includes all elements in A or B or both. The complement (A ∪ B)′ includes all elements in the universal set U that are not in A ∪ B. This is the region outside both circles A and B, but inside the rectangle representing U.

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(ii) A′ ∩ B′

A′ represents the complement of set A (elements in U but not in A). B′ represents the complement of set B (elements in U but not in B). The intersection A′ ∩ B′ represents the elements that are in both A′ and B′. This means elements that are not in A AND not in B. This is the region outside both circles A and B, but inside the rectangle representing U.

Note that the Venn diagrams for (i) and (ii) are identical, visually representing De Morgan's Law: $(A \cup B)' = A' \cap B'$.

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(iii) (A ∩ B)′

This represents the complement of the intersection of sets A and B. The intersection A ∩ B is the region where the circles A and B overlap (elements common to both A and B). The complement (A ∩ B)′ includes all elements in the universal set U that are not in A ∩ B. This is the entire region within U EXCEPT the overlapping area of A and B.

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(iv) A′ ∪ B′

A′ represents the complement of set A. B′ represents the complement of set B. The union A′ ∪ B′ represents the elements that are in A′ or in B′ or both. This means elements that are not in A OR not in B. This is the entire region within U EXCEPT the part that is inside both A and B (which is A ∩ B).

Note that the Venn diagrams for (iii) and (iv) are identical, visually representing the other De Morgan's Law: $(A \cap B)' = A' \cup B'$.

Given the universal set U and set A:

U = The set of all triangles in a plane.

A = The set of all triangles with at least one angle different from $60^\circ$.

---

We need to find A', the complement of set A with respect to the universal set U.

$A' = \{x : x \in U \text{ and } x \notin A\}$.

This means A' is the set of all elements in U that are not in A.

The elements of U are triangles in a plane.

Set A consists of triangles where at least one angle is *not* $60^\circ$.

So, A' must consist of triangles where the condition "at least one angle different from $60^\circ$" is false.

The negation of "at least one angle different from $60^\circ$" is "all angles are equal to $60^\circ$".

---

A triangle in U belongs to A' if and only if all its angles are equal to $60^\circ$.

A triangle with all angles equal to $60^\circ$ is called an equilateral triangle.

Thus, A' is the set of all equilateral triangles in the plane.

---

A′ = The set of all equilateral triangles.

Let U be the universal set.

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(i) A ∪ A′ = . . .

A ∪ A′ is the union of set A and its complement A'. A' contains all elements in U that are not in A.

The union of A and A' will contain all elements that are either in A or not in A (but in U).

This covers all elements in the universal set U.

A ∪ A′ = U.

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(ii) $\phi$′ ∩ A = . . .

First, find $\phi$′. The complement of the empty set ($\phi$) with respect to the universal set U is the set of all elements in U that are not in $\phi$. Since $\phi$ has no elements, $\phi$' contains all elements of U.

$\phi$' = U.

Now, find $\phi$′ ∩ A = U ∩ A.

The intersection of the universal set U and any set A contains the elements that are common to both U and A. Since A is a subset of U, the elements common to both are simply the elements of A.

U ∩ A = A.

$\phi$′ ∩ A = A.

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(iii) A ∩ A′ = . . .

A ∩ A′ is the intersection of set A and its complement A'. A' contains all elements in U that are not in A.

The intersection of A and A' will contain elements that are both in A and not in A.

There are no such elements.

A ∩ A′ = $\phi$ (the empty set).

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(iv) U′ ∩ A = . . .

First, find U′. The complement of the universal set U with respect to itself is the set of elements in U that are not in U. There are no such elements.

U′ = $\phi$ (the empty set).

Now, find U′ ∩ A = $\phi$ ∩ A.

The intersection of the empty set $\phi$ and any set A contains the elements that are common to both. Since $\phi$ has no elements, there are no common elements.

$\phi$ ∩ A = $\phi$.

U′ ∩ A = $\phi$.

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Summary of the answers:

(i) A ∪ A′ = U

(ii) $\phi$′ ∩ A = A

(iii) A ∩ A′ = $\phi$

(iv) U′ ∩ A = $\phi$

Let A be the set of letters needed to spell the word “CATARACT”.

The letters in the word "CATARACT" are C, A, T, A, R, A, C, T.

In a set, we only list the unique elements. The unique letters are C, A, T, R.

So, A = {C, A, T, R}.

---

Let B be the set of letters needed to spell the word “TRACT”.

The letters in the word "TRACT" are T, R, A, C, T.

In a set, we only list the unique elements. The unique letters are T, R, A, C.

So, B = {T, R, A, C}.

---

We need to show that set A and set B are equal.

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

Set A = {C, A, T, R}

Set B = {T, R, A, C}

Comparing the elements:

The elements in A are C, A, T, R.

The elements in B are T, R, A, C.

Both sets contain the letters C, A, T, and R.

---

Since set A and set B contain exactly the same elements, they are equal.

Therefore, the set of letters needed to spell “CATARACT” and the set of letters needed to spell “TRACT” are equal.

A = {C, A, T, R}

B = {C, A, T, R}

A = B.

Let the given set be S = {-1, 0, 1}.

The number of elements in set S is $n=3$.

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

In this case, the number of subsets is $2^3 = 8$.

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We need to list all 8 subsets. A subset is formed by taking any combination of elements from the original set, including no elements and all elements.

Subsets can be categorized by the number of elements they contain:

1. Subset with 0 elements (the empty set):

$\phi$ = { }

2. Subsets with 1 element:

We take each element individually and form a set.

{-1}

{0}

{1}

3. Subsets with 2 elements:

We take pairs of elements and form sets.

{-1, 0}

{-1, 1}

{0, 1}

4. Subset with 3 elements (the set itself):

{-1, 0, 1}

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Listing all the subsets together:

The subsets of {–1, 0, 1} are:

$\phi$, {-1}, {0}, {1}, {-1, 0}, {-1, 1}, {0, 1}, {-1, 0, 1}.

We are given the condition $A \cup B = A \cap B$. We need to show that this condition implies that A = B.

To show that A = B, we need to show that A $\subset$ B and B $\subset$ A.

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Part 1: Show A $\subset$ B

Let $x$ be an arbitrary element of set A. So, $x \in A$.

By the definition of union, if $x \in A$, then $x \in A \cup B$.

We are given that $A \cup B = A \cap B$.

Since $x \in A \cup B$ and $A \cup B = A \cap B$, it follows that $x \in A \cap B$.

By the definition of intersection, if $x \in A \cap B$, then $x \in A$ and $x \in B$.

From $x \in A$ and $x \in B$, we conclude that $x \in B$.

So, we started with an arbitrary element $x \in A$ and showed that $x \in B$.

This means that every element of A is also an element of B.

Therefore, A $\subset$ B.

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Part 2: Show B $\subset$ A

Let $y$ be an arbitrary element of set B. So, $y \in B$.

By the definition of union, if $y \in B$, then $y \in A \∪ B$.

We are given that $A \cup B = A \cap B$.

Since $y \in A \cup B$ and $A \cup B = A \cap B$, it follows that $y \in A \cap B$.

By the definition of intersection, if $y \in A \cap B$, then $y \in A$ and $y \in B$.

From $y \in A$ and $y \in B$, we conclude that $y \in A$.

So, we started with an arbitrary element $y \in B$ and showed that $y \in A$.

This means that every element of B is also an element of A.

Therefore, B $\subset$ A.

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Since we have shown that A $\subset$ B and B $\subset$ A, by the definition of set equality, it implies that A = B.

Thus, if A ∪ B = A ∩ B, then A = B.

To determine the subset relationships, we first need to clearly list the elements of each set.

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Set A = { x : x ∈ R and x satisfy $x^2 – 8x + 12 = 0$ }

This set contains the real numbers that are solutions to the equation $x^2 – 8x + 12 = 0$.

We solve the quadratic equation by factoring:

$x^2 – 8x + 12 = 0$

We look for two numbers that multiply to +12 and add up to -8. These numbers are -2 and -6.

So, we can factor the equation as:

$(x - 2)(x - 6) = 0$

This equation is true if either $(x - 2) = 0$ or $(x - 6) = 0$.

If $x - 2 = 0$, then $x = 2$.

If $x - 6 = 0$, then $x = 6$.

The roots are 2 and 6. Both 2 and 6 are real numbers.

So, Set A = {2, 6}.

---

The other sets are already given in roster form:

Set B = {2, 4, 6}

Set C = {2, 4, 6, 8, 10, ...} (This represents the set of even natural numbers greater than or equal to 2).

Set D = {6}

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Now, we check which set is a subset of which other set. A set P is a subset of set Q ($P \subset Q$) if every element of P is also an element of Q.

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Let's compare set A = {2, 6} with the other sets:

Is A $\subset$ B? Elements of A are 2 and 6. Both 2 and 6 are present in B = {2, 4, 6}. Yes, A $\subset$ B.

Is A $\subset$ C? Elements of A are 2 and 6. Both 2 and 6 are present in C = {2, 4, 6, 8, ...}. Yes, A $\subset$ C.

Is A $\subset$ D? Elements of A are 2 and 6. Element 2 is in A but not in D = {6}. No, A $\not\subset$ D.

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Let's compare set B = {2, 4, 6} with the other sets (we already checked A $\subset$ B, so B $\not\subset$ A unless A=B):

Is B $\subset$ A? Elements of B are 2, 4, and 6. Element 4 is in B but not in A = {2, 6}. No, B $\not\subset$ A.

Is B $\subset$ C? Elements of B are 2, 4, and 6. All are present in C = {2, 4, 6, 8, ...}. Yes, B $\subset$ C.

Is B $\subset$ D? Elements of B are 2, 4, and 6. Elements 2 and 4 are in B but not in D = {6}. No, B $\not\subset$ D.

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Let's compare set C = {2, 4, 6, 8, ...} with the other sets (we already checked A $\subset$ C and B $\subset$ C, so C cannot be a subset of A or B unless C=A or C=B, which is false):

Is C $\subset$ A? C contains elements like 4, 8, etc., which are not in A = {2, 6}. No, C $\not\subset$ A.

Is C $\subset$ B? C contains elements like 8, 10, etc., which are not in B = {2, 4, 6}. No, C $\not\subset$ B.

Is C $\subset$ D? C contains elements like 2, 4, etc., which are not in D = {6}. No, C $\not\subset$ D.

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Let's compare set D = {6} with the other sets (we already checked A $\not\subset$ D, B $\not\subset$ D, C $\not\subset$ D):

Is D $\subset$ A? The element of D is 6. 6 is present in A = {2, 6}. Yes, D $\subset$ A.

Is D $\subset$ B? The element of D is 6. 6 is present in B = {2, 4, 6}. Yes, D $\subset$ B.

Is D $\subset$ C? The element of D is 6. 6 is present in C = {2, 4, 6, 8, ...}. Yes, D $\subset$ C.

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Summarizing the subset relationships we found:

• A $\subset$ B
• A $\subset$ C
• B $\subset$ C
• D $\subset$ A
• D $\subset$ B
• D $\subset$ C

These are the subset relationships among the given sets.

We will examine each statement to determine if it is true or false and provide justification.

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(i) If $x \in A$ and $A \in B$, then $x \in B$

This statement says that if $x$ is an element of set A, and set A is an element of set B, then $x$ must be an element of set B.

Let's consider a specific example to see if this holds true in all cases.

Let set A = {1}. Here, $x = 1$ is an element of A, so $x \in A$.

Let set B = {A, 2}. Here, set A itself is an element of B, so $A \in B$. Set B is actually the set {{1}, 2}.

Now, according to the statement, if $x \in A$ and $A \in B$, then $x \in B$. This means if $1 \in A$ and $A \in B$, then $1 \in B$.

Let's check if 1 is an element of B = {{1}, 2}. The elements listed in B are the set {1} and the number 2. The number 1 itself is not an element of B.

So, the statement is False.

Example (Counterexample): Let A = {1} and B = {{1}, 2}. Here, $1 \in A$ and $A \in B$ (since {1} is an element of B), but $1 \notin B$.

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(ii) If $A \subset B$ and $B \in C$, then $A \in C$

This statement says that if set A is a subset of set B, and set B is an element of set C, then set A must be an element of set C.

Let's use an example.

Let set A = {1}.

Let set B = {1, 2}. Here, every element of A (which is just 1) is also in B, so $A \subset B$.

Let set C = {B, 3}. Here, set B itself is an element of C, so $B \in C$. Set C is actually the set {{1, 2}, 3}.

Now, according to the statement, if $A \subset B$ and $B \in C$, then $A \in C$. This means if {1} $\subset$ {1, 2} and {1, 2} $\in$ {{1, 2}, 3}, then {1} $\in$ {{1, 2}, 3}.

Let's check if A = {1} is an element of C = {{1, 2}, 3}. The elements listed in C are the set {1, 2} and the number 3. The set {1} is not an element of C.

So, the statement is False.

Example (Counterexample): Let A = {1}, B = {1, 2}, and C = {{1, 2}, 3}. Here, $A \subset B$ and $B \in C$, but $A \notin C$.

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(iii) If $A \subset B$ and $B \subset C$, then $A \subset C$

This statement says that if A is a subset of B, and B is a subset of C, then A must be a subset of C.

This is a fundamental property of subsets called transitivity.

Let's prove that this is True.

Assume that A $\subset$ B and B $\subset$ C are true.

We want to show that A $\subset$ C. By the definition of a subset, this means we need to show that every element that is in A is also in C.

Let $x$ be any element such that $x \in A$.

Since we assumed $A \subset B$ and we know $x \in A$, the definition of a subset tells us that $x$ must also be an element of B. So, $x \in B$.

Now we know $x \in B$. Since we also assumed $B \subset C$ and we know $x \in B$, the definition of a subset tells us that $x$ must also be an element of C. So, $x \in C$.

We started with an arbitrary element $x$ in A ($x \in A$) and showed that this element $x$ must also be in C ($x \in C$).

Therefore, by the definition of a subset, A is a subset of C.

The statement is True.

Proof: Let $x \in A$. Since $A \subset B$, $x \in B$. Since $B \subset C$, $x \in C$. Thus, every element of A is an element of C, so $A \subset C$.

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(iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$

This statement says that if A is not a subset of B, and B is not a subset of C, then A must not be a subset of C.

Let's try to find an example where A is not a subset of B, B is not a subset of C, but A *is* a subset of C.

Let set A = {1}.

Let set B = {2}. Is A $\not\subset$ B? Yes, because 1 is in A but 1 is not in B.

Let set C = {1, 3}. Is B $\not\subset$ C? Yes, because 2 is in B but 2 is not in C.

Now let's check if A $\not\subset$ C. A = {1}, C = {1, 3}. Is A a subset of C? Yes, because the only element in A (which is 1) is also in C.

We found an example where the premises ($A \not\subset B$ and $B \not\subset C$) are true, but the conclusion ($A \not\subset C$) is false (since A $\subset$ C is true). This contradicts the statement.

So, the statement is False.

Example (Counterexample): Let A = {1}, B = {2}, and C = {1, 3}. Here, $A \not\subset B$ (since $1 \notin B$) and $B \not\subset C$ (since $2 \notin C$), but $A \subset C$.

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(v) If $x \in A$ and $A \not\subset B$, then $x \in B$

This statement says that if $x$ is an element of A, and A is not a subset of B, then $x$ must be an element of B.

If $A \not\subset B$, it means that there is at least one element in A that is not in B. It doesn't mean that *all* elements in A are not in B, and it certainly doesn't mean that a specific element $x \in A$ *must* be in B.

Consider an example:

Let set A = {1, 2}.

Let set B = {1, 3}. Is A $\not\subset$ B? Yes, because element 2 is in A but not in B.

Let's pick an element $x$ from A. Let $x = 2$. So, $x \in A$.

According to the statement, if $x \in A$ and $A \not\subset B$, then $x \in B$. This means if $2 \in A$ and $A \not\subset B$, then $2 \in B$.

Let's check if 2 is an element of B = {1, 3}. No, 2 is not in B.

So, the statement is False.

Example (Counterexample): Let A = {1, 2} and B = {1, 3}. Let $x = 2$. Here, $x \in A$ and $A \not\subset B$ (since $2 \notin B$), but $x \notin B$.

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(vi) If $A \subset B$ and $x \notin B$, then $x \notin A$

This statement says that if set A is a subset of set B, and $x$ is an element that is not in B, then $x$ must also not be in A.

Intuitively, if A is "inside" B, and $x$ is "outside" of B, then $x$ must also be "outside" of A. This makes sense.

Let's prove that this is True.

Assume that A $\subset$ B and $x \notin B$ are true.

We want to show that $x \notin A$.

We can use a proof by contradiction. Assume, for the sake of argument, that $x \in A$.

We are given that A $\subset$ B. By the definition of a subset, if an element is in A, it must also be in B.

Since we assumed $x \in A$ and we know $A \subset B$, it must follow that $x \in B$.

However, we were initially given that $x \notin B$.

This leads to a contradiction: $x \in B$ and $x \notin B$ cannot both be true.

Our assumption that $x \in A$ must therefore be false.

So, $x \notin A$.

The statement is True.

Proof: Assume $A \subset B$ and $x \notin B$. We want to show $x \notin A$. Suppose $x \in A$. Since $A \subset B$, $x$ must also be in B, i.e., $x \in B$. But this contradicts the given information that $x \notin B$. Therefore, the assumption $x \in A$ is false, so $x \notin A$.

Solution:

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Given three sets A, B, and C such that:

1. $\mathbf{A \cup B = A \cup C}$

2. $\mathbf{A \cap B = A \cap C}$

We need to show that B = C.

To show that two sets are equal, we must prove that each set is a subset of the other. That is, we must show that every element of B is an element of C ($B \subset C$) and every element of C is an element of B ($C \subset B$).

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Part 1: Show $\mathbf{B \subset C}$

Let $x$ be an arbitrary element of set B. So, $x \in B$.

By the definition of union, if $x \in B$, then $x \in A \cup B$.

We are given the condition $A \cup B = A \cup C$.

Since $x \in A \cup B$ and $A \cup B$ is the same set as $A \cup C$, it follows that $x \in A \cup C$.

By the definition of union, if $x \in A \cup C$, then $x \in A$ or $x \in C$.

We consider these two possibilities:

Case 1: $x \in A$.

In this case, $x$ is in A, and we started with the assumption that $x$ is in B ($x \in B$). So, $x$ is an element common to both A and B. By the definition of intersection, $x \in A \cap B$.

We are given the condition $A \cap B = A \cap C$.

Since $x \in A \cap B$ and $A \cap B$ is the same set as $A \cap C$, it follows that $x \in A \cap C$.

By the definition of intersection, if $x \in A \cap C$, then $x \in A$ and $x \in C$.

From this, we directly conclude that $x \in C$.

Case 2: $x \in C$.

In this case, $x$ is already in C, so the requirement that $x$ must be in C is satisfied.

In both cases (whether $x \in A$ or $x \notin A$), we have concluded that if $x \in B$, then $x \in C$.

So, we started with an arbitrary element $x \in B$ and showed that $x \in C$.

This means that every element of B is also an element of C. Therefore, B $\subset$ C.

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Part 2: Show $\mathbf{C \subset B}$

Let $y$ be an arbitrary element of set C. So, $y \in C$.

By the definition of union, if $y \in C$, then $y \in A \cup C$.

We are given the condition $A \cup C = A \cup B$.

Since $y \in A \cup C$ and $A \cup C$ is the same set as $A \cup B$, it follows that $y \in A \cup B$.

By the definition of union, if $y \in A \cup B$, then $y \in A$ or $y \in B$.

We consider these two possibilities:

Case 1: $y \in A$.

In this case, $y$ is in A, and we started with the assumption that $y$ is in C ($y \in C$). So, $y$ is an element common to both A and C. By the definition of intersection, $y \in A \cap C$.

We are given the condition $A \cap C = A \cap B$.

Since $y \in A \cap C$ and $A \cap C$ is the same set as $A \cap B$, it follows that $y \in A \cap B$.

By the definition of intersection, if $y \in A \cap B$, then $y \in A$ and $y \in B$.

From this, we directly conclude that $y \in B$.

Case 2: $y \in B$.

In this case, $y$ is already in B, so the requirement that $y$ must be in B is satisfied.

In both cases (whether $y \in A$ or $y \notin A$), we have concluded that if $y \in C$, then $y \in B$.

So, we started with an arbitrary element $y \in C$ and showed that $y \in B$.

This means that every element of C is also an element of B. Therefore, C $\subset$ B.

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Since we have shown that B $\subset$ C (from Part 1) and C $\subset$ B (from Part 2), by the definition of set equality, it implies that B = C.

Thus, if $A \cup B = A \cup C$ and $A \cap B = A \cap C$, then $B = C$.

Solution:

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To show that the four conditions are equivalent, we need to prove that each condition implies the others. We can do this by showing a chain of implications, for example, (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (i).

Let's recall the definitions:

A $\subset$ B means that every element of A is an element of B.

A – B = $\phi$ means that the set of elements in A that are not in B is empty.

A ∪ B = B means that the union of A and B is equal to B.

A ∩ B = A means that the intersection of A and B is equal to A.

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Step 1: Show (i) $\implies$ (ii)

Assume (i) A $\subset$ B is true.

We want to show that (ii) A – B = $\phi$ is true.

The set A – B contains elements that are in A but not in B. If A – B were not empty, it would contain at least one element, say $x$. If $x \in A – B$, then by definition, $x \in A$ and $x \notin B$.

However, we are assuming $A \subset B$. By definition of a subset, if $x \in A$, then $x$ must also be in B ($x \in B$).

This contradicts the statement $x \notin B$. Therefore, our assumption that A – B is not empty must be false.

Thus, A – B = $\phi$.

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Step 2: Show (ii) $\implies$ (iii)

Assume (ii) A – B = $\phi$ is true.

We want to show that (iii) A ∪ B = B is true. To show set equality, we must show $A \cup B \subset B$ and $B \subset A \cup B$.

The inclusion $B \subset A \cup B$ is always true by the definition of union (any element in B is automatically in the union of A and B).

Now, let's show $A \cup B \subset B$. Let $x$ be an arbitrary element such that $x \in A \cup B$.

By the definition of union, $x \in A$ or $x \in B$.

If $x \in B$, then $x$ is already in B, and the requirement is satisfied.

If $x \in A$, then consider if $x$ is also in B or not. If $x \in A$ and $x \notin B$, then $x \in A – B$. But we assumed $A – B = \phi$, which means there are no elements in A – B. So, it's impossible for an element $x \in A$ to also be outside of B ($x \notin B$). This means if $x \in A$, it must be that $x \in B$.

So, in both possibilities ($x \in A$ or $x \in B$), we conclude that $x \in B$.

Thus, every element of A ∪ B is an element of B, which means $A \cup B \subset B$.

Since $A \cup B \subset B$ and $B \subset A \cup B$, we have A ∪ B = B.

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Step 3: Show (iii) $\implies$ (iv)

Assume (iii) A ∪ B = B is true.

We want to show that (iv) A ∩ B = A is true. To show set equality, we must show $A \cap B \subset A$ and $A \subset A \cap B$.

The inclusion $A \cap B \subset A$ is always true by the definition of intersection (any element in the intersection of A and B must be in A).

Now, let's show $A \subset A \cap B$. Let $x$ be an arbitrary element such that $x \in A$.

By the definition of union, if $x \in A$, then $x \in A \cup B$.

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

Since $x \in A \cup B$ and $A \cup B$ is the same set as B, it follows that $x \in B$.

So, we have shown that if $x \in A$, then $x \in B$.

If $x \in A$ and we also know $x \in B$, then by the definition of intersection, $x \in A \cap B$.

Thus, every element of A is an element of A ∩ B, which means $A \subset A \cap B$.

Since $A \cap B \subset A$ and $A \subset A \cap B$, we have A ∩ B = A.

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Step 4: Show (iv) $\implies$ (i)

Assume (iv) A ∩ B = A is true.

We want to show that (i) A $\subset$ B is true. To do this, we must show that every element of A is an element of B.

Let $x$ be an arbitrary element such that $x \in A$.

We assumed that $A = A \cap B$.

Since $x \in A$ and A is the same set as $A \cap B$, it follows that $x \in A \cap B$.

By the definition of intersection, if $x \in A \cap B$, then $x$ must be in A and $x$ must be in B. This means $x \in A$ and $x \in B$.

From this, we directly conclude that $x \in B$.

So, we started with an arbitrary element $x \in A$ and showed that $x \in B$.

This proves that every element of A is an element of B. Therefore, A $\subset$ B.

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We have shown the chain of implications: (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (iv) $\implies$ (i).

This proves that all four conditions are equivalent to each other.

Solution:

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We are given the condition that A $\subset$ B. This means that every element of set A is also an element of set B.

We want to show that C – B $\subset$ C – A. By the definition of a subset, this means we must show that every element that is in the set C – B is also in the set C – A.

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Let $x$ be an arbitrary element such that $x \in C – B$.

By the definition of set difference, if $x \in C – B$, then $x$ must be in set C and $x$ must not be in set B.

So, we know two things about the element $x$:

1. $x \in C$

2. $x \notin B$

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Now, let's consider what it means for an element to be in C – A. By the definition of set difference, $x \in C – A$ means $x \in C$ and $x \notin A$.

From our starting point ($x \in C – B$), we already know that $x \in C$ (point 1 above).

We need to show that $x \notin A$ must also be true, given our initial conditions ($x \in C – B$, which means $x \notin B$, and the given $A \subset B$).

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We are given that $A \subset B$. By the definition of a subset, if an element is in A, then it must be in B. The contrapositive of this statement is also true: if an element is not in B, then it cannot be in A. (This was shown to be true in Question 2, part (vi)).

Since we know that $x \notin B$ (from point 2 above) and we are given $A \subset B$, we can apply this property. Because $x$ is not in the larger set B, $x$ cannot be in the smaller set A.

Therefore, we conclude that $x \notin A$.

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Combining the two facts we have established about $x$:

We know $x \in C$.

We know $x \notin A$.

By the definition of set difference, having $x \in C$ and $x \notin A$ means that $x \in C – A$.

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We started with an arbitrary element $x \in C – B$ and successfully showed that this element $x$ must also be in C – A ($x \in C – A$).

This proves that every element of C – B is an element of C – A.

Therefore, C – B $\subset$ C – A.

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Formal Proof:

Assume $A \subset B$.

Let $x \in C – B$.

By definition of set difference, $x \in C$ and $x \notin B$.

We are given $A \subset B$. If $x \notin B$, then it is impossible for $x$ to be in $A$ (because if $x$ were in A, it would have to be in B as well, by the definition of $A \subset B$). Thus, $x \notin A$.

So, we have $x \in C$ and $x \notin A$.

By definition of set difference, $x \in C – A$.

Since every element $x \in C – B$ is also an element $x \in C – A$, we have shown that C – B $\subset$ C – A.

Solution:

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We need to prove two set identities. To show that two sets are equal, we need to prove that every element of the first set is in the second set, and every element of the second set is in the first set (i.e., show subset inclusion in both directions).

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Part 1: Show $\mathbf{A = ( A \cap B ) \cup ( A – B )}$

Show $\mathbf{A \subset ( A \cap B ) \cup ( A – B )}$

Let $x$ be an arbitrary element such that $x \in A$.

For any element $x$ and any set B, $x$ is either in B or $x$ is not in B. So, we consider two cases:

Case 1: $x \in B$.

If $x \in A$ (our starting assumption) 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, $x$ is also in the union of $A \cap B$ with any other set, including $A – B$. So, $x \in (A \cap B) \cup (A – B)$.

Case 2: $x \notin B$.

If $x \in A$ (our starting assumption) 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, $x$ is also in the union of $A – B$ with any other set, including $A \cap B$. So, $x \in (A \cap B) \cup (A – B)$.

In both cases, if $x \in A$, we have shown that $x \in (A \cap B) \cup (A – B)$.

Therefore, A $\subset$ (A ∩ B) ∪ (A – B).

Show $\mathbf{( A \cap B ) \cup ( A – B ) \subset A}$

Let $y$ be an arbitrary element such that $y \in (A \cap B) \cup (A – B)$.

By the definition of union, if $y$ is in the union, then $y$ must be in the first set or the second set (or both).

So, $y \in (A \cap B)$ or $y \in (A – B)$.

Case 1: $y \in A \cap B$.

By the definition of intersection, if $y \in A \cap B$, then $y$ must be in A and $y$ must be in B. So, $y \in A$ and $y \in B$.

This directly implies that $y \in A$.

Case 2: $y \in A – B$.

By the definition of set difference, if $y \in A – B$, then $y$ must be in A and $y$ must not be in B. So, $y \in A$ and $y \notin B$.

This directly implies that $y \in A$.

In both cases, if $y \in (A \cap B) \cup (A – B)$, we have shown that $y \in A$.

Therefore, (A ∩ B) ∪ (A – B) $\subset$ A.

Since we have shown both inclusions, we conclude that A = ( A ∩ B ) ∪ ( A – B ).

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Part 2: Show $\mathbf{A \cup ( B – A ) = ( A \cup B )}$

Show $\mathbf{A \cup ( B – A ) \subset ( A \cup B )}$

Let $x$ be an arbitrary element such that $x \in A \cup (B – A)$.

By the definition of union, $x \in A$ or $x \in (B – A)$.

Case 1: $x \in A$.

If $x \in A$, then by the definition of union, $x$ is also in the union of A with any other set, including B. So, $x \in A \cup B$.

Case 2: $x \in B – A$.

By the definition of set difference, if $x \in B – A$, then $x \in B$ and $x \notin A$.

If $x \in B$, then by the definition of union, $x$ is also in the union of A and B. So, $x \in A \cup B$.

In both cases, if $x \in A \cup (B – A)$, we have shown that $x \in A \cup B$.

Therefore, A ∪ (B – A) $\subset$ (A ∪ B).

Show $\mathbf{( A \cup B ) \subset A \cup ( B – A )}$

Let $y$ be an arbitrary element such that $y \in A \cup B$.

By the definition of union, $y \in A$ or $y \in B$.

Case 1: $y \in A$.

If $y \in A$, then by the definition of union, $y$ is also in the union of A with any other set, including $B – A$. So, $y \in A \cup (B – A)$.

Case 2: $y \in B$.

If $y \in B$, we consider two subcases: either $y$ is also in A, or $y$ is not in A.

Subcase 2a: $y \in A$. If $y \in B$ and $y \in A$, then $y \in A \cup (B – A)$ (because $y \in A$, same as Case 1).

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, $y$ is also in the union of A and $B – A$. So, $y \in A \cup (B – A)$.

In all possibilities ($y \in A$ or $y \in B$), we conclude that $y \in A \cup (B – A)$.

Therefore, (A ∪ B) $\subset$ A ∪ (B – A).

Since we have shown both inclusions, we conclude that A ∪ ( B – A ) = ( A ∪ B ).

Solution:

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We will use basic properties of set operations to show the given identities. These identities are known as Absorption Laws.

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(i) Show $\mathbf{A \cup ( A \cap B ) = A}$

We need to show that A ∪ (A ∩ B) is equal to A. We can do this using element-wise proof.

Show $\mathbf{A \cup ( A \cap B ) \subset A}$

Let $x$ be an arbitrary element such that $x \in A \cup (A \cap B)$.

By the definition of union, $x \in A$ or $x \in (A \cap B)$.

Case 1: $x \in A$. This case directly shows $x \in A$.

Case 2: $x \in A \cap B$. By the definition of intersection, if $x \in A \cap B$, then $x \in A$ and $x \in B$. This also implies $x \in A$.

In both cases, if $x \in A \cup (A \cap B)$, then $x \in A$.

Thus, $\mathbf{A \cup ( A \cap B ) \subset A}$.

Show $\mathbf{A \subset A \cup ( A \cap B )}$

Let $x$ be an arbitrary element such that $x \in A$.

By the definition of union, if $x \in A$, then $x$ is automatically in the union of A with any other set, including A ∩ B.

Thus, if $x \in A$, then $x \in A \cup (A \cap B)$.

Therefore, $\mathbf{A \subset A \cup ( A \cap B )}$.

Since we have shown both inclusions, $\mathbf{A \cup ( A \cap B ) = A}$.

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(ii) Show $\mathbf{A \cap ( A \cup B ) = A}$.

We need to show that A ∩ (A ∪ B) is equal to A. We can do this using element-wise proof.

Show $\mathbf{A \cap ( A \cup B ) \subset A}$

Let $x$ be an arbitrary element such that $x \in A \cap (A \cup B)$.

By the definition of intersection, if $x \in A \cap (A \cup B)$, then $x$ must be in the first set and $x$ must be in the second set.

So, $x \in A$ and $x \in (A \cup B)$.

The condition "$x \in A$ and $x \in (A \cup B)$" directly implies $x \in A$ (since $x \in A$ is one of the conditions).

Thus, $\mathbf{A \cap ( A \cup B ) \subset A}$.

Show $\mathbf{A \subset A \cap ( A \cup B )}$

Let $x$ be an arbitrary element such that $x \in A$.

We want to show that $x \in A \cap (A \cup B)$. By the definition of intersection, this means we need to show that $x \in A$ and $x \in A \cup B$.

We already know $x \in A$ (our starting assumption).

By the definition of union, if $x \in A$, then $x$ is automatically in the union of A with any other set, including B. So, $x \in A \cup B$.

Since we have both $x \in A$ and $x \in A \cup B$, by the definition of intersection, $x \in A \cap (A \cup B)$.

Therefore, $\mathbf{A \subset A \cap ( A \cup B )}$.

Since we have shown both inclusions, $\mathbf{A \cap ( A \cup B ) = A}$.

Solution:

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We are asked to show that the statement "if $A \cap B = A \cap C$, then $B = C$" is not always true. To do this, we need to provide a counterexample. A counterexample is a set of specific sets A, B, and C for which the condition $A \cap B = A \cap C$ is true, but the conclusion $B = C$ is false.

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Let's choose simple sets for A, B, and C.

Suppose A contains some elements, and B and C share some elements with A, but B and C are different from each other.

Let A = {1, 2}.

Let B = {1, 3}.

Let C = {1, 4}.

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Now, let's calculate $A \cap B$ and $A \cap C$ for these sets.

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

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

The intersection $A \cap C$ contains the elements that are common to both set A and set C.

$A \cap C = \{1, 2\} \cap \{1, 4\} = \{1\}$.

For our chosen sets, we have $A \cap B = \{1\}$ and $A \cap C = \{1\}$. So, the condition $A \cap B = A \cap C$ is true.

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Now, let's check if the conclusion $B = C$ is true for these sets.

Set B = {1, 3}.

Set C = {1, 4}.

Set B contains the element 3, which is not in Set C. Set C contains the element 4, which is not in Set B.

Since sets B and C do not contain exactly the same elements, B is not equal to C ($B \neq C$).

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We have found a specific example (A={1,2}, B={1,3}, C={1,4}) where the statement "$A \cap B = A \cap C$" is true, but the statement "$B = C$" is false.

Therefore, the statement "A ∩ B = A ∩ C implies B = C" is not always true.

Solution:

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Given three sets A, B, and X such that:

1. $\mathbf{A \cap X = \phi}$

2. $\mathbf{B \cap X = \phi}$

3. $\mathbf{A \cup X = B \cup X}$

We need to show that A = B.

We can use the hints provided and properties of sets to prove this.

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Consider set A. From set properties, we know the Absorption Law which states that for any sets Y and Z, $Y \cap (Y \cup Z) = Y$. Let Y = A and Z = X.

So, we can write: $A = A \cap (A \cup X)$.

Now, we use the given condition (3), $A \cup X = B \cup X$, and substitute $B \cup X$ in place of $A \cup X$ in the equation for A:

$A = A \cap (B \cup X)$

Next, we apply the Distributive Law, which states that for any sets Y, Z, and W, $Y \cap (Z \cup W) = (Y \cap Z) \cup (Y \cap W)$. Let Y = A, Z = B, and W = X.

Applying the distributive law:

$A = (A \cap B) \cup (A \cap X)$

Now, we use the given condition (1), $A \cap X = \phi$, and substitute $\phi$ into the equation:

$A = (A \cap B) \cup \phi$

We know the property that for any set S, the union of S with the empty set is S itself: $S \cup \phi = S$. Let S = $A \cap B$.

Applying this property:

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

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Now, let's consider set B and follow a similar process. Using the Absorption Law with Y = B and Z = X:

$B = B \cap (B \cup X)$.

Now, we use the given condition (3), $B \cup X = A \cup X$, and substitute $A \cup X$ in place of $B \cup X$ in the equation for B:

$B = B \cap (A \cup X)$.

Next, apply the Distributive Law with Y = B, Z = A, and W = X:

$B = (B \cap A) \cup (B \cap X)$.

Now, we use the given condition (2), $B \cap X = \phi$, and substitute $\phi$ into the equation:

$B = (B \cap A) \cup \phi$.

Using the property $S \cup \phi = S$. Let S = $B \cap A$.

Applying this property:

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

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We have found that $A = A \cap B$ and $B = B \cap A$.

We also know that the intersection operation is commutative, meaning $A \cap B = B \cap A$ for any sets A and B.

Since $A = A \cap B$ and $B = B \cap A$, and $A \cap B = B \cap A$, it follows that A = B.

Therefore, if $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$ for some set X, then $A = B$.

Solution:

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We need to find three sets A, B, and C that satisfy the following two conditions:

1. The intersection of each pair of sets is non-empty: $\mathbf{A \cap B \neq \phi}$, $\mathbf{B \cap C \neq \phi}$, and $\mathbf{A \cap C \neq \phi}$.

2. The intersection of all three sets is the empty set: $\mathbf{A \cap B \cap C = \phi}$.

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Let's think about what these conditions mean for the elements in the sets. Condition 1 means there must be at least one element shared between A and B, at least one element shared between B and C, and at least one element shared between A and C. Condition 2 means there should be no element that is present in A, B, AND C simultaneously.

We can achieve this by putting elements only in the pairwise overlapping regions, but not in the central region where all three sets overlap.

Let's assign elements to the pairwise intersections:

• To make $A \cap B \neq \phi$, let's put an element, say 'p', into $A \cap B$. To ensure it's not in $A \cap B \cap C$, this element 'p' should be in A and B, but not in C.
• To make $B \cap C \neq \phi$, let's put an element, say 'q', into $B \cap C$. To ensure it's not in $A \cap B \cap C$, this element 'q' should be in B and C, but not in A.
• To make $A \cap C \neq \phi$, let's put an element, say 'r', into $A \cap C$. To ensure it's not in $A \cap B \cap C$, this element 'r' should be in A and C, but not in B.

Now, let's define the sets based on these requirements for elements 'p', 'q', and 'r':

• A must contain 'p' and 'r', and must not contain 'q'. Let's define A = {p, r}.
• B must contain 'p' and 'q', and must not contain 'r'. Let's define B = {p, q}.
• C must contain 'q' and 'r', and must not contain 'p'. Let's define C = {q, r}.

Let's check the conditions with these example sets:

• $A \cap B$: The elements common to A = {p, r} and B = {p, q} are {p}. $A \cap B = \{p\}$. This is non-empty. Condition 1 (for A $\cap$ B) is satisfied.
• $B \cap C$: The elements common to B = {p, q} and C = {q, r} are {q}. $B \cap C = \{q\}$. This is non-empty. Condition 1 (for B $\cap$ C) is satisfied.
• $A \cap C$: The elements common to A = {p, r} and C = {q, r} are {r}. $A \cap C = \{r\}$. This is non-empty. Condition 1 (for A $\cap$ C) is satisfied.

All parts of Condition 1 are satisfied.

• $A \cap B \cap C$: We can find this by finding the intersection of $A \cap B$ with C. $A \cap B \cap C = (A \cap B) \cap C = \{p\} \cap \{q, r\}$. The intersection of {p} and {q, r} is the empty set, because there are no common elements. $A \cap B \cap C = \phi$. Condition 2 is satisfied.

So, the sets A = {p, r}, B = {p, q}, C = {q, r} satisfy the requirements. We can replace 'p', 'q', 'r' with any distinct elements.

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Example Sets:

Let's use the elements 1, 2, and 3 for 'p', 'q', and 'r' respectively.

Let p = 1, q = 2, r = 3.

A = {r, p} = {3, 1}

B = {p, q} = {1, 2}

C = {q, r} = {2, 3}

Let's verify again:

$A \cap B = \{3, 1\} \cap \{1, 2\} = \{1\} \neq \phi$.

$B \cap C = \{1, 2\} \cap \{2, 3\} = \{2\} \neq \phi$.

$A \cap C = \{3, 1\} \cap \{2, 3\} = \{3\} \neq \phi$.

$A \cap B \cap C = \{1\} \cap \{2, 3\} = \phi$.

These sets satisfy the conditions.