# Class 11 RD Sharma Solutions – Chapter 2 Relations – Exercise 2.2

**Question 1: Given A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, find (A Ã— B) âˆ© (B Ã— C).**

**Solution:**

Given:

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

Let us find: (A Ã— B) âˆ© (B Ã— C)

(A Ã— B) = {1, 2, 3} Ã— {3, 4}

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

(B Ã— C) = {3, 4} Ã— {4, 5, 6}

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

Therefore,

(A Ã— B) âˆ© (B Ã— C) = {(3, 4)}

**Question 2: If A = {2, 3}, B = {4, 5}, C = {5, 6} find A Ã— (B âˆª C), (A Ã— B) âˆª (A Ã— C).**

**Solution:**

Given:

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

Let us find: A x (B âˆª C) and (A x B) âˆª (A x C)

(B âˆª C) = {4, 5, 6}

A Ã— (B âˆª C) = {2, 3} Ã— {4, 5, 6}

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

(A Ã— B) = {2, 3} Ã— {4, 5}

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

(A Ã— C) = {2, 3} Ã— {5, 6}

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

Therefore,

(A Ã— B) âˆª (A Ã— C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

A Ã— (B âˆª C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

**Question 3: If A = {1, 2, 3}, B = {4}, C = {5}, then verify that:**

**(i) A Ã— (B âˆª C) = (A Ã— B) âˆª (A Ã— C)**

**(ii) A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)**

**(iii) A Ã— (B â€“ C) = (A Ã— B) â€“ (A Ã— C)**

**Solution:**

Given:

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

(i) A Ã— (B âˆª C) = (A Ã— B) âˆª (A Ã— C)Let’s assume LHS: (B âˆª C)

(B âˆª C) = {4, 5}

A Ã— (B âˆª C) = {1, 2, 3} Ã— {4, 5}

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

Now, RHS

(A Ã— B) = {1, 2, 3} Ã— {4}

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

(A Ã— C) = {1, 2, 3} Ã— {5}

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

(A Ã— B) âˆª (A Ã— C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

Therefore,

LHS = RHS

(ii) A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)Let’s assume LHS: (B âˆ© C)

(B âˆ© C) = âˆ… (No common element)

A Ã— (B âˆ© C) = {1, 2, 3} Ã— âˆ…

= âˆ…

Now, RHS

(A Ã— B) = {1, 2, 3} Ã— {4}

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

(A Ã— C) = {1, 2, 3} Ã— {5}

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

(A Ã— B) âˆ© (A Ã— C) = âˆ…

Therefore,

LHS = RHS

(iii) A Ã— (B âˆ’ C) = (A Ã— B) âˆ’ (A Ã— C)Let’s assume LHS: (B âˆ’ C)

(B âˆ’ C) = âˆ…

A Ã— (B âˆ’ C) = {1, 2, 3} Ã— âˆ…

= âˆ…

Now, RHS

(A Ã— B) = {1, 2, 3} Ã— {4}

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

(A Ã— C) = {1, 2, 3} Ã— {5}

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

(A Ã— B) âˆ’ (A Ã— C) = âˆ…

Therefore,

LHS = RHS

**Question 4: Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that:**

**(i) A Ã— C âŠ‚ B Ã— D**

**(ii) A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)**

**Solution:**

Given:

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

(i) A x C âŠ‚ B x DLet us consider LHS A x C

A Ã— C = {1, 2} Ã— {5, 6}

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

Now, RHS

B Ã— D = {1, 2, 3, 4} Ã— {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)}

Since, all elements of A Ã— C is in B Ã— D.

Therefore,

We can say A Ã— C âŠ‚ B Ã— D

(ii) A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)Let’s assume LHS A Ã— (B âˆ© C)

(B âˆ© C) = âˆ…

A Ã— (B âˆ© C) = {1, 2} Ã— âˆ…

= âˆ…

Now, RHS

(A Ã— B) = {1, 2} Ã— {1, 2, 3, 4}

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

(A Ã— C) = {1, 2} Ã— {5, 6}

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

Hence, there is no common element between A Ã— B and A Ã— C

(A Ã— B) âˆ© (A Ã— C) = âˆ…

Therefore,

A Ã— (B âˆ© C) = (A Ã— B) âˆ© (A Ã— C)

**Question 5: If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find**

**(i) A Ã— (B âˆ© C)**

**(ii) (A Ã— B) âˆ© (A Ã— C)**

**(iii) A Ã— (B âˆª C)**

**(iv) (A Ã— B) âˆª (A Ã— C)**

**Solution:**

Given:

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

(i) A Ã— (B âˆ© C)(B âˆ© C) = {4}

A Ã— (B âˆ© C) = {1, 2, 3} Ã— {4}

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

(ii) (A Ã— B) âˆ© (A Ã— C)(A Ã— B) = {1, 2, 3} Ã— {3, 4}

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

(A Ã— C) = {1, 2, 3} Ã— {4, 5, 6}

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

(A Ã— B) âˆ© (A Ã— C) = {(1, 4), (2, 4), (3, 4)}

(iii) A Ã— (B âˆª C)(B âˆª C) = {3, 4, 5, 6}

A Ã— (B âˆª C) = {1, 2, 3} Ã— {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 Ã— B) âˆª (A Ã— C)(A Ã— B) = {1, 2, 3} Ã— {3, 4}

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

(A Ã— C) = {1, 2, 3} Ã— {4, 5, 6}

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

(A Ã— B) âˆª (A Ã— 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)}

**Question 6: Prove that:**

**(i) (A âˆª B) Ã— C = (A Ã— C) = (A Ã— C) âˆª (B Ã— C)**

**(ii) (A âˆ© B) Ã— C = (A Ã— C) âˆ© (B Ã— C)**

**Solution:**

(i) (A âˆª B) Ã— C = (A Ã— C) = (A Ã— C) âˆª (B Ã— C)Let (x, y) be an arbitrary element of (A âˆª B) Ã— C

(x, y) âˆˆ (A âˆª B) C

Since, (x, y) are elements of Cartesian product of (A âˆª B) Ã— C

x âˆˆ (A âˆª B) and y âˆˆ C

(x âˆˆ A or x âˆˆ B) and y âˆˆ C

(x âˆˆ A and y âˆˆ C) or (x âˆˆ Band y âˆˆ C)

(x, y) âˆˆ A Ã— C or (x, y) âˆˆ B Ã— C

(x, y) âˆˆ (A Ã— C) âˆª (B Ã— C) â€¦ (1)

Let (x, y) be an arbitrary element of (A Ã— C) âˆª (B Ã— C).

(x, y) âˆˆ (A Ã— C) âˆª (B Ã— C)

(x, y) âˆˆ (A Ã— C) or (x, y) âˆˆ (B Ã— C)

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

(x âˆˆ A or x âˆˆ B) and y âˆˆ C

x âˆˆ (A âˆª B) and y âˆˆ C

(x, y) âˆˆ (A âˆª B) Ã— C â€¦ (2)

From 1 and 2, we get: (A âˆª B) Ã— C = (A Ã— C) âˆª (B Ã— C)

(ii) (A âˆ© B) Ã— C = (A Ã— C) âˆ© (B Ã— C)Let (x, y) be an arbitrary element of (A âˆ© B) Ã— C.

(x, y) âˆˆ (A âˆ© B) Ã— C

Since, (x, y) are elements of Cartesian product of (A âˆ© B) Ã— C

x âˆˆ (A âˆ© B) and y âˆˆ C

(x âˆˆ A and x âˆˆ B) and y âˆˆ C

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

(x, y) âˆˆ A Ã— C and (x, y) âˆˆ B Ã— C

(x, y) âˆˆ (A Ã— C) âˆ© (B Ã— C) â€¦ (1)

Let (x, y) be an arbitrary element of (A Ã— C) âˆ© (B Ã— C).

(x, y) âˆˆ (A Ã— C) âˆ© (B Ã— C)

(x, y) âˆˆ (A Ã— C) and (x, y) âˆˆ (B Ã— C)

(x âˆˆA and y âˆˆ C) and (x âˆˆ Band y âˆˆ C)

(x âˆˆA and x âˆˆ B) and y âˆˆ C

x âˆˆ (A âˆ© B) and y âˆˆ C

(x, y) âˆˆ (A âˆ© B) Ã— C â€¦ (2)

From 1 and 2, we get: (A âˆ© B) Ã— C = (A Ã— C) âˆ© (B Ã— C)

**Question 7: If A Ã— B âŠ† C Ã— D and A âˆ© B âˆˆ âˆ…, Prove that A âŠ† C and B âŠ† D.**

**Solution:**

Given:

A Ã— B âŠ† C x D and A âˆ© B âˆˆ âˆ…

A Ã— B âŠ† C x D denotes A Ã— B is subset of C Ã— D that is every element A Ã— B is in C Ã— D.

And A âˆ© B âˆˆ âˆ… denotes A and B does not have any common element between them.

A Ã— B = {(a, b): a âˆˆ A and b âˆˆ B}

Therefore,

We can say (a, b) âŠ† C Ã— D [Since, A Ã— B âŠ† C x D is given]

a âˆˆ C and b âˆˆ D

a âˆˆ A = a âˆˆ C

A âŠ† C

And

b âˆˆ B = b âˆˆ D

B âŠ† D

Hence proved.

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