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(i) Given: A = {2, 3, 5} and B = {5, 7} 

To find: A × B 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {2, 3, 5} and B = {5, 7}. So,

 A × B = (2, 3, 5) × (5, 7) 

= {(2, 5), (3, 5), (5, 5), (2, 7), (3, 7), (5, 7)} 

(ii) Given: A = {2, 3, 5} and B = {5, 7} 

To find: B × A 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {2, 3, 5} and B = {5, 7}. So, 

B × A = (5, 7) × (2, 3, 5) 

= {(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)} 

(iii) Given: A = {2, 3, 5} and B = {2, 3, 5} 

To find: A × A 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e.

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {2, 3, 5} and A = {2, 3, 5}. So, 

A × A = (2, 3, 5) × (2, 3, 5) 

= {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)}

(iv) Given: B = {5, 7} 

To find: B × B 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, B = {5, 7} and B = {5, 7}. So, 

B × B = (5, 7) × (5, 7) 

= {(5, 5), (5, 7), (7, 5), (7, 7)}

Given: P = {a, b} and Q = {x, y, z} 

To show: P × Q ≠ Q × P 

Now, firstly we find the P × Q and Q × P 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, P = (a, b) and Q = (x, y, z). So, 

P × Q = (a, b) × (x, y, z) 

= {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)} 

Q × P = (x, y, z) × (a, b) 

= {(x, a), (y, a), (z, a), (x, b), (y, b), (z, b)} 

Since by the definition of equality of ordered pairs .i.e. the corresponding first elements are equal and the second elements are also equal, but here the pair (a, x) is not equal to the pair (x, a) 

∴ P× Q ≠ Q × P 

Hence proved

Given: A = {9, 1} and B = {1, 2, 3} 

To show: A × B ≠ B × A 

Now, firstly we find the A × B and B × A 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q}

Here, A = (9, 1) and B = (1, 2, 3). So,

A × B = (9, 1) × (1, 2, 3) 

= {(9, 1), (9, 2), (9, 3), (1, 1), (1, 2), (1, 3)} 

B × A = (1, 2, 3) × (9, 1) 

= {(1, 9), (2, 9), (3, 9), (1, 1), (2, 1), (3, 1)} 

Since by the definition of equality of ordered pairs .i.e. the corresponding first elements are equal and the second elements are also equal, but here, the pair (9, 1) is not equal to the pair (1, 9) 

∴ A × B ≠ B × A 

Hence proved

Since, (a, 0), (b, 1), (c, 0) are the elements of A × B. 

∴ a, b, c Є A and 0, 1 Є B 

It is given that n(A) = 3 and n(B) = 2 

∴ a, b, c Є A and n(A) = 3 

⇒ A = {a, b, c}

and 0, 1 Є B and n(B) = 2 

⇒ B = {0, 1}

Here, 

A = { 1,3,5} and B = {2,4} 

Also, 

R = { (x,y) : x,y ϵ A×B and x>y}

= {(3,2) ,(5,2),(5,4)}

(d) \(2^{n^2}\)

Set A has n elements ⇒ n(A) = n 

⇒ A × A has n × n = n² elements 

∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

(c) Only transitive

Let N be the set of natural numbers. Then 

R = {(a, b) : a < b, a, b ∈N}

A natural number is not less than itself 

⇒ (a, a)∉R where a ∈N 

⇒ R is not reflexive 

• V a, b ∈N, (a, b) ∈R ⇒ a < b \(\not\Rightarrow\) b < a ⇒ (b, a) ∉R 

⇒ R is not symmetric. 

• V a, b, c ∈N, (a, b) ∈R and (b, c) ∈R 

⇒ a < b and b < c 

⇒ a < c (a, c) ∈R 

⇒ R is transitive.

(c) Only symmetric

• V a ∈ R, a² + a² ≠ 1 ⇒ (a, a) ∉R ⇒ R is not reflexive. 

For example, 0² + 0² = 0, 1² + 1² = \(\big(\frac{1}{2}\big)^2\) + \(\big(\frac{1}{2}\big)^2\) = \(\frac{1}{2}\)

and so on. 

• V a, b ∈R, (a, b) ∈R ⇒ a² + b² = 1 ⇒ b² + a² = 1 

⇒ (b, a) ∈R 

⇒ R is symmetric 

• V a, b, c ∈R, (a, b) ∈R and (b, c) ∈R 

⇒ a² + b² = 1 and b² + c² = 1 which does not necessarily mean 

a² + c² = 1 ⇒ (a, c) ∉R 

For example, let a = 0, b = 1, c = 0 

0² + 1² = 1 and 1² + 0² = 1 

But 0² + 0² ≠ 1.

(a) 1

For the relation R to become transitive: 

(1, 2) ∈R and (2, 3) ∈R should imply (1, 3) ∈R 

∴ Minimum one ordered pair (1, 3) should be added to R.

Given a and b are integers, i.e. a,b ∈ Z.

∴ Domain of R = Set of all first elements in the relation.

= Values of ‘a’ which are in the relation.

= Z (Integers)

Range of R=Set of all second elements in the relation.

= Values of ‘b’ which are in the relation.

= Z(Integers)

Since, a²+ b² = 25 and a, b are integers;

⇒ R= {(5,0), (0,5), (-5,0), (0, -5), (3,4), (4,3), (-3, -4), (-4, -3),

(-3,4), (4, -3), (-4,3), (3, -4)}

⇒ Domain of R= {-5, -4, -3,0,3,4,5}

We will prove this using an example.

Let A = {a, b, c} be a set and

R = {(a, a) (b, b) (c, c) (a, b) (b, a)} and

S = {(a, a) (b, b) (c, c) (b, c) (c, d)} are two relations on A

Clearly R and S are transitive relation on A

Now,

R ⋃ S = {(a, a) (b, b) (c, c) (a, b) (b, a) (b, c) (c, b)}

Here, (a, b) ∈ R ⋃ S and (b, c) ∈ R ⋃ S

but (a, c) ∉ R ⋃ S

∴ R ⋃ S is not transitive

(b) Reflexive and transitive but not symmetric

Let A = {1, 2, 3, 4} 

• ∵ (1, 1), (2, 2), (3, 3) and (4, 4) ∈R ⇒ R is reflexive 

• ∵ (1, 2) ∈ R but (2, 1) ∉ R ; (1, 3) ∈ R and (3, 1) ∉ R ; (3, 2) ∈R and (2, 3) ∉ R ⇒ R is not symmetric 

• (1, 3) ∈ R and (3, 2) ∈ R and (1, 2) ∈ R 

⇒ R is transitive.

A. symmetric

Think of line as a vector quantity:

As ℓ₁⊥ ℓ₂;

And ℓ₂⊥ ℓ₁

Hence R is symmetric.

Also Given a relation R over straight lines such that ℓ₁⊥ ℓ₂

As ℓ₁⊥ ℓ_2:

⇒ ℓ₁. ℓ₂=0(DOT PRODUCT)

∵ cos θ =0 as θ =90°;

⇒ This thing is possible if ℓ₁ and ℓ₂ are symmetric.

E.g. A= 2i-4j and B=-4i-2j

⇒ A. B=-8+8=0

(d) An equivalence relation

Let R = {(x, y) : line x is parallel to line y, x y ∈ set of coplanar straight lines}. 

• Every line is parallel to itself. So, if x ∈S, then (x, x) ∈R 

⇒ R is reflexive 

• If (x, y) ∈R ⇒ x | | y ⇒ y | | x ⇒ (y, x) ∈R 

⇒ R is symmetric 

• (x, y) ∈R and (y, z) ∈R 

⇒ x | | y and y | | z 

⇒ x | | z ⇒ (x, z) ∈R 

⇒ R is transitive 

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

(b) {5, 6, 7, 8, 9, 10}

2. The range of the given relation is defined by the second element, i.e., (x + 5) in the ordered pair (x + 1, x + 5) defining the relation.

∵ x ∈{0, 1, 2, 3, 4, 5}, therefore Range = {0 + 5, 1 + 5, 2 + 5, 3 + 5, 4 + 5, 5 + 5} = {5, 6, 7, 8, 9, 10}.

(d) R = {(a, b) : b/a is odd}

• Since (2, 6) ∈R, the relation “a and b are odd” does not exist. 

• Since (1, 5) and (1, 7) ∈R, the relation “a and b are even” does not exist. 

• None of the ordered pairs in R are prime numbers.

• \(\frac{5}{1}\) = 5, \(\frac{7}{1}\) = 7, \(\frac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

(d) an equivalence relation

Given, a R b ⇒ sin²a + cos²b = 1 

Reflexive: a R a ⇒ sin² a + cos² a = 1 ∀ a ∈ R (True) 

Symmetric: a R b ⇒ sin² a + cos² b = 1 

⇒ 1 – cos² a + 1 – sin² b = 1 

⇒ sin² b + cos² a = 1 

⇒ b R a ∀ a, b ∈ R (True) 

Transitive: a R a and b R c 

⇒ sin² a + cos² b = 1 and sin² b + cos² c = 1 

∴ Adding these two equations we get 

sin² a + cos² b + sin² b + cos² c = 2 

⇒ sin² a + cos² c = 1 ⇒ a R c (True)

∴ R is an equivalence relation.

(a) Reflexive

Given, {(\(x\), y) : 2\(x\) – y = 10} 

Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 

∴ Point (10, 10) ∈ N ⇒ R is reflexive.

(c) S is an equivalence relation but R is not an equivalence relation

R = {(x, y) | x, y ∈ R, x = wy, w is a rational number} 

Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number) 

Hence R is reflexive. 

Symmetric: x R y \(\not\Rightarrow\) y R x as 

0 R 1 ⇒ 0 = (0) . 1 where 0 is a rational number but 

1 R 0 ⇒ 1 = (w) 0 which is not true for any rational number. 

∴ R is not an equivalence relation

S = \(\bigg\{\bigg(\frac{m}{n},\frac{p}{q}\bigg)\bigg|\) m, n, p, q, ∈ I, n, q, ≠ 0 and qm = pn\(\bigg\}\)

Reflexive \(\frac{m}{n}S\frac{m}{n}\) ⇒ mn = nm (True)

Symmetric \(\frac{m}{n}S\frac{p}{q}\) ⇒ mq = pn ⇒ pn = mq ⇒ \(\frac{p}{q}S\frac{m}{n}\) (True)

Transitive \(\frac{m}{n}S\frac{p}{q}\) and \(\frac{p}{q}S\frac{r}{s}\)

⇒ mq = pn and ps = qr 

⇒ mq.ps = pn.qr ⇒ ms = nr

⇒ \(\frac{m}{n}\) = \(\frac{r}{s}\) ⇒ \(\frac{m}{n}\) S \(\frac{r}{s}\) (True)

∴ S is an equivalence relation.

(i) Given: A = {-2, 2} and B = {0, 3, 5} 

To find: A × B 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {-2, 2} and B = {0, 3, 5}. So, 

A × B = {(-2, 0), (-2, 3), (-2, 5), (2, 0), (2, 3), (2, 5)} (ii) 

Given: A = {-2, 2} and B = {0, 3, 5} 

To find: B × A 

By the definition of the Cartesian product,

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {-2, 2} and B = {0, 3, 5}. So, 

B × A = {(0, -2), (0, 2), (3, -2), (3, 2), (5, -2), (5, 2)}

(iii) Given: 

A = {-2, 2} 

To find: A × A 

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, A = {-2, 2} and A = {-2, 2}.So, 

A × A = {(-2, -2), (-2, 2), (2, -2), (2, 2)} 

(iv) Given: B = {0, 3, 5} 

To find: B × B

By the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q} 

Here, B = {0, 3, 5} and B = {0, 3, 5}. So, 

B × B = {(0, 0), (0, 3), (0, 5), (3, 0), (3, 3), (3, 5), (5, 0), (5, 3), (5, 5)}

Given: A = {All books in a library of a school} 

R = {(x, y): x and y have the same number of pages} 

Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A 

Symmetric: Since books x and y have the same number of pages, so (x, y) ∈ R. 

Since books y and x have the same number of pages, so (y, x) ∈ R. 

⇒ R is symmetric on A. 

Transitivity: Books x, y, z have the same number of pages ⇒ (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. 

Hence, R is an equivalence relation.

Given: A = {1, 2, 3} and B = {2, 4} 

To find: A × B, B × A, A × A, (A × B) ∩ (B × A) 

Now, 

A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)} 

B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)} 

A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} 

B × B = {(2, 2), (2, 4), (4, 2), (4, 4)} 

Intersection of two sets represents common elements of both the sets.

So,

(A × B) ∩ (B × A) = {(2, 2)}

Reflexive: If A is a set, then A ⊂ A.               False 

Symmetric: If A and B are sets and A ⊂ B, then B ⊂ A.        False 

Transitive: If A, B and C are sets and if A ⊂ B and B ⊂ C then A ⊂ C         True 

Hence ‘⊂’ is not an equivalence relation since it possesses only the transitive property.

(i) Transitive

Consider A, B and C ∈ S, where (A, B) and (B, C) ∈ R

We get

(A, B) ∈ R => A ⊂ B ……. (1)

(B, C) ∈ R => B ⊂ C …….. (2)

Using both the equations we get

A ⊂ C => (A, C) ∈ R

Hence, R is a transitive relation S.

(ii) Non reflexive

We know that

A ⊄ A where (A, A) ∈ R

Hence, R is non reflexive.

(iii) Non symmetric

Consider A ⊂ B where (A, B) ∈ R

We know that B ⊄ A

So (B, A) ∉ R

We get (A, B) ∈ R and (B, A) ∉ R

Hence, R is non symmetric.

Given : A = {2, 3}, B = {4, 5} and C = {5, 6} 

To find : (A × B) ∪ (A × C) 

Since, (B ∪ C) = {4, 5, 6} 

∴A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} 

(A × B) = {(2, 4), (2, 5), (3, 4), (3, 5)} 

(A × C) = {(2, 5), (2, 6), (3, 5), (3, 6)} 

∴ (A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(i) Given as

R = {a, b): a ∈ N, a < 5, b = 4}

Since, the natural numbers less than 5 are 1, 2, 3 and 4

a = {1, 2, 3, 4} and b = {4}

R = {(1, 4), (2, 4), (3, 4), (4, 4)}

Therefore,

The domain of relation R = {1, 2, 3, 4}

The range of relation R = {4}

(ii) Given as

S = {a, b): b = |a - 1|, a ∈ Z and |a| ≤ 3}

Here, Z denotes integer which can be positive as well as negative

Then, |a| ≤ 3 and b = |a - 1|

∴ a = {-3, -2, -1, 0, 1, 2, 3}

For, a = -3, -2, -1, 0, 1, 2, 3 we get,

S = {(-3, |-3 – 1|), (-2, |-2 – 1|), (-1, |-1 – 1|), (0, |0 – 1|), (1, |1 – 1|), (2, |2 – 1|), (3, |3 – 1|)}

S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)}

S = {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}

b = 4, 3, 2, 1, 0, 1, 2

Therefore,

The domain of relation S = {0, -1, -2, -3, 1, 2, 3}

The range of relation S = {0, 1, 2, 3, 4}

Given as

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) A x C ⊂ B x D

Let us consider the LHS A x C

A × C = {1, 2} × {5, 6}

= {(1, 5), (1, 6), (2, 5), (2, 6)}

Then, RHS

B × D = {1, 2, 3, 4} × {5, 6, 7, 8}

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

Here, all elements of A × C is in B × D.

∴We can say that A × C ⊂ B × D

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let us consider the LHS A × (B ∩ C)
(B ∩ C) = ∅

A × (B ∩ C) = {1, 2} × ∅

= ∅

Then, RHS

(A × B) = {1, 2} × {1, 2, 3, 4}

= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

(A × C) = {1, 2} × {5, 6}

= {(1, 5), (1, 6), (2, 5), (2, 6)}

Here, there is no common element between A × B and A × C

(A × B) ∩ (A × C) = ∅

Thus, A × (B ∩ C) = (A × B) ∩ (A × C)

Given: (A) = 5 and n(B) = 4 

To find: [(A × B) ∩ (B×A)] 

n (A × B) = n(A) × n(B) = 5 x 4 = 20 

n (A ∩ B) = 3 

(given: A and B has 3 elements in common) 

In order to calculate n [(A × B) ∩ (B × A)], 

We will assume,

A = (x, x, x, y, z) and B = (x, x, x, p) 

So, we have 

(A × B) = {(x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (x, x), (x, x), (x, x), (x, p), (y, x), (y, x), (y, x), (y, p), (z, x), (z, x), (z, x), (z, p)} 

(B × A) = {(x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (x, x), (x, x), (x, x), (x, y), (x, z), (p, x), (p, x), (p, x), (p, y), (p, z)} 

[(A × B) ∩ (B × A)] = {(x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x), (x, x)} 

∴ We can say that n [(A × B) ∩ (B × A)] = 9

We know, 

(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) 

Here A and B have an element in common i.e., n(A ∩ B) = 1 = (B ∩ A) 

So, n((A × B) ∩ (B × A)) = n((A ∩ B) × (B ∩ A)) = n(A ∩ B) × n(B ∩ A) = 1 × 1 = 1 

That means, A × B and B × A have an element in common if and only if A and B have an element in common. [Proved] 

Given,

A = {1, 2, 3}, B = {4} and C = {5} 

(i) To prove : A × (B ∪ C) = (A × B) ∪ (A × C) 

LHS : (B ∪ C) = {4, 5} 

Therefore, 

A × (B ∪ C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} 

RHS : (A × B) = {(1, 4), (2, 4), (3, 4)} 

(A × C) = {(1, 5), (2, 5), (3, 5)} 

(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)} 

∴ LHS = RHS 

(ii) To prove: A × (B ∩ C) = (A × B) ∩ (A × C) 

LHS : (B ∩ C) = ∅ 

(No common element) 

A × (B ∩ C) = ∅ 

RHS : (A × B) = {(1, 4), (2, 4), (3, 4)} 

(A × C) = {(1, 5), (2, 5), (3, 5)} 

(A × B) ∩ (A × C) = ∅ 

∴ LHS = RHS 

(iii) To prove : A × (B − C) = (A × B) − (A × C) 

LHS : (B − C) = ∅ 

A × (B − C) = ∅ 

RHS : (A × B) = {(1, 4), (2, 4), (3, 4)} 

(A × C) = {(1, 5), (2, 5), (3, 5)} 

(A × B) − (A × C) =∅ 

∴ LHS = RHS

Given:

A, B and C three sets are given. 

Need to prove: A × (B ∪ C) = (A × B) ∪ (A × C) 

Let us consider, (x, y)^∈A × (B ∪ C) 

⇒ x∈A and y∈(B ∪ C)

⇒ x∈A and (y∈B or y∈C) 

⇒ (x∈A and y∈B) or (x∈A and y∈C) 

⇒ (x, y)∈(A × B) or (x, y)∈(A × C) 

⇒ (x, y)∈(A × B) ∪ (A × C)

From this we can conclude that, 

⇒ A × (B ∪ C) ⊆ (A × B) ∪ (A × C) ---- (1) 

Let us consider again, (a, b)∈(A × B) ∪ (A × C) 

⇒ (a, b)∈(A × B) or (a, b)∈(A × C) 

⇒ (a∈A and b∈B) or (a∈A and b∈C) 

⇒ a∈A and (b∈B or b∈C) 

⇒ a∈A and b∈(B ∪ C) 

⇒ (a, b) ∈A × (B ∪ C)

From this, we can conclude that, 

⇒ (A × B) ∪ (A × C) ⊆ A × (B ∪ C) ---- (2)

Now by the definition of the set we can say that, from (1) and (2), 

A × (B ∪ C) = (A × B) ∪ (A × C) [Proved]

Given as

Here, A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}

(i) A × (B ∩ C)

(B ∩ C) = {4}

A × (B ∩ C) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(ii) (A × B) ∩ (A × C)

(A × B) = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(A × C) = {1, 2, 3} × {4, 5, 6}

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

(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}

(iii) A × (B ∪ C)

Since, (B ∪ C) = {3, 4, 5, 6}

A × (B ∪ C) = {1, 2, 3} × {3, 4, 5, 6}

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

(iv) (A × B) ∪ (A × C)

(A × B) = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(A × C) = {1, 2, 3} × {4, 5, 6}

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

Thus, (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

Given: A, B and C three sets are given. 

Need to prove: A × (B ∩ C) = (A × B) ∩ (A × C) 

Let us consider, (x, y)∈A × (B ∩ C) 

⇒ x∈A and y∈(B ∩ C) 

⇒ x∈A and (y∈B and y∈C) 

⇒ (x∈A and y∈B) and (x∈A and y∈C)

⇒ (x, y)∈(A × B) and (x, y)∈(A × C) 

⇒ (x, y)∈(A × B) ∩ (A × C) 

From this we can conclude that, 

⇒ A × (B ∩ C) ⊆ (A × B) ∩ (A × C) ..... (1) 

Let us consider again, (a, b)∈(A × B) ∩ (A × C) 

⇒ (a, b)∈(A × B) and (a, b)∈(A × C) 

⇒ (a∈A and b∈B) and (a∈A and b∈C) 

⇒ a∈A and (b∈B and b∈C) 

⇒ a∈A and b∈(B ∩ C) 

⇒ (a, b)∈A × (B ∩ C) 

From this, we can conclude that, 

⇒ (A × B) ∩ (A × C) ⊆ A × (B ∩ C) .... (2) 

Now by the definition of the set we can say that, from (1) and (2), 

A × (B ∩ C) = (A × B) ∩ (A × C) [Proved]

Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 

(i) n(A × B) = n(A) × n(B) 

⇒ n(A × B) = 3 × 4 

⇒ n(A × B) = 12 

(ii) n(B × A) = n(B) × n(A) 

⇒ n(B × A) = 4 × 3 

⇒ n(B × A) = 12

(iii) n((A × B) ∩ (B × A)) = n(A × B) + n(B × A) – n((A × B) ∪ (B × A)) 

n((A × B) ∩ (B × A)) = n(A × B) + n(B × A) – n(A × B) + n(B × A)

n((A × B) ∩ (B × A)) = 0

(i) (A ∪ B) x C = (A x C) = (A x C) ∪ (B x C)

Suppose (x, y) be an arbitrary element of (A ∪ B) × C

(x, y) ∈ (A ∪ B) C

Here, (x, y) are elements of Cartesian product of (A ∪ B) × C

x ∈ (A ∪ B) and y ∈ C

(x ∈ A or x ∈ B) and y ∈ C

(x ∈ A and y ∈ C) or (x ∈ Band y ∈ C)

(x, y) ∈ A × C or (x, y) ∈ B × C

(x, y) ∈ (A × C) ∪ (B × C) … (1)

Suppose (x, y) be an arbitrary element of (A × C) ∪ (B × C).

(x, y) ∈ (A × C) ∪ (B × C)

(x, y) ∈ (A × C) or (x, y) ∈ (B × C)

(x ∈ A and y ∈ C) or (x ∈ B and y ∈ C)

(x ∈ A or x ∈ B) and y ∈ C

x ∈ (A ∪ B) and y ∈ C

(x, y) ∈ (A ∪ B) × C … (2)

From the equation 1 and 2, we get: (A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) x C = (A x C) ∩ (B x C)

Suppose (x, y) be an arbitrary element of (A ∩ B) × C.

(x, y) ∈ (A ∩ B) × C

Here, (x, y) are elements of Cartesian product of (A ∩ B) × C

x ∈ (A ∩ B) and y ∈ C

(x ∈ A and x ∈ B) and y ∈ C

(x ∈ A and y ∈ C) and (x ∈ Band y ∈ C)

(x, y) ∈ A × C and (x, y) ∈ B × C

(x, y) ∈ (A × C) ∩ (B × C) … (1)

Suppose (x, y) be an arbitrary element of (A × C) ∩ (B × C).

(x, y) ∈ (A × C) ∩ (B × C)

(x, y) ∈ (A × C) and (x, y) ∈ (B × C)

(x ∈A and y ∈ C) and (x ∈ Band y ∈ C)

(x ∈A and x ∈ B) and y ∈ C

x ∈ (A ∩ B) and y ∈ C

(x, y) ∈ (A ∩ B) × C … (2)

From the equation (1) and (2), we get: (A ∩ B) × C = (A × C) ∩ (B × C)

To prove : (A ∩ B) × C = (A × C) ∩ (B×C) 

Proof : Let (x, y) be an arbitrary element of (A ∩ B) × C. 

⇒ (x, y) ∈ (A ∩ B) × C 

Since, 

(x, y) are elements of Cartesian product of (A ∩ B)× C 

⇒ x ∈ (A ∩ B) and y ∈ C 

⇒ (x ∈ A and x ∈B) and y ∈ C 

⇒ (x ∈ A and y ∈ C) and (x ∈ Band y ∈ C) 

⇒ (x, y) ∈ A × C and (x, y) ∈ B × C 

⇒ (x, y) ∈ (A × C) ∩ (B × C) …1 

Let (x, y) be an arbitrary element of (A × C) ∩ (B × C). 

⇒ (x, y) ∈ (A × C) ∩ (B × C) 

⇒ (x, y) ∈ (A × C) and (x, y) ∈ (B × C) 

⇒ (x ∈A and y ∈ C) and (x ϵ Band y ∈ C) 

⇒ (x ∈A and x ∈ B) and y ∈ C 

⇒ x ∈ (A ∩ B) and y ∈ C 

⇒ (x, y) ∈ (A ∩ B) × C …2 

From 1 and 2, we get : 

(A ∩ B) × C = (A × C) ∩ (B × C)

Given as

Here, A × B ⊆ C x D and A ∩ B ∈ ∅

A × B ⊆ C x D denotes A × B is subset of C × D that is every element A × B is in C × D.

And A ∩ B ∈ ∅ denotes A and B does not have any common element between them.

Now, A × B = {(a, b): a ∈ A and b ∈ B}

∴We can say that (a, b) ⊆ C × D [Since, A × B ⊆ C x D is given]

a ∈ C and b ∈ D

a ∈ A = a ∈ C

A ⊆ C

And

b ∈ B = b ∈ D

B ⊆ D

Thus proved.

Given,

 A × B ⊆ C x D and A ∩ B ∈ ∅ 

To prove : A ⊆ C and B ⊆ D 

A × B ⊆ C x D denotes A × B is subset of C × D that is every element A × B is in C × D 

And A ∩ B ∈ ∅ denotes A and B does not have any common element between them. 

A × B = {(a, b): a ∈ A and b ∈ B} 

Since, 

A × B ⊆ C x D (Given) 

∴We can say (a, b) C × D 

⇒ a ∈ C and b ∈ D 

⇒ A ∈ C and B ∈ D 

(A and B does not have common elements)

Given as

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

The relation from A to B can be defined as:

A × B = {1, 2, 3} × {4, 5, 6}

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

(i) {(1, 6), (3, 4), (5, 2)}

No, it is not a relation from A to B. Thus, the given set is not a subset of A × B as (5, 2) is not a part of the relation from A to B.

(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}

Yes, it is a relation from A to B. Thus, the given set is a subset of A × B.

(iii) {(4, 2), (4, 3), (5, 1)}

No, it is not a relation from A to B. Thus, the given set is not a subset of A × B.

(iv) A × B

Here, A × B is a relation from A to B and can be defined as:

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

We know that 1, 2, 3 ∈ A and (1, 1), (2, 2), (3, 3) ∈ R for each a ∈ A, (a, a) ∈ R

Hence, R is reflexive.

We know that (1, 2) ∈ R but (2, 1) ∉ R

Hence, R is not symmetric.

We know that

(1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R

Hence, R is not transitive

Given,

A = {1, 2, 3}, B = {4, 5, 6} 

A relation from A to B can be defined as : 

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)} 

i. {(1, 6), (3, 4),(5, 2)} 

No, it is not a relation from A to B. 

The given set is not a subset of A × B as (5, 2) is not a part of the relation from A to B. 

ii. {(1, 5), (2, 6), (3, 4),(3, 6)} 

Yes, it is a relation from A to B. 

The given set is a subset of A × B. 

iii. {(4, 2), (4, 3), (5, 1)} 

No, it is not a relation from A to B. 

The given set is not a subset of A × B as (4, 2), (4, 3), (5, 1) are not a part of the relation from A to B. 

iv. A × B 

A × B is a relation from A to B can be defined as: 

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