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# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.7

• Last Updated : 15 Dec, 2020

### Question 1: For any two sets A and B, prove that: Aâ€˜ â€“ Bâ€˜ = B â€“ A

Solution:

To prove:

Aâ€™ â€“ Bâ€™ = B â€“ A

Firstly show that

Aâ€™ â€“ Bâ€™ âŠ† B â€“ A

Let, x âˆˆ Aâ€™ â€“ Bâ€™

â‡’ x âˆˆ Aâ€™ and x âˆ‰ Bâ€™

â‡’ x âˆ‰ A and x âˆˆ B (since, A âˆ© Aâ€™ = Ï•)

â‡’ x âˆˆ B â€“ A

It is true for all x âˆˆ Aâ€™ â€“ Bâ€™

Therefore,

Aâ€™ â€“ Bâ€™ = B â€“ A

Hence, Proved.

### (iv) A â€“ B = A Î” (A âˆ© B)

Solution:

(i) A âˆ© (Aâ€™ âˆª B) = A âˆ© B

Here, LHS A âˆ© (Aâ€™ âˆª B)

Upon expanding

(A âˆ© Aâ€™) âˆª (A âˆ© B)

We know, (A âˆ© Aâ€™) =Ï•

â‡’ Ï• âˆª (Aâˆ© B)

â‡’ (A âˆ© B)

Therefore,

LHS = RHS

Hence, proved.

(ii) A â€“ (A â€“ B) = A âˆ© B

For any sets A and B we have De-Morganâ€™s law

(A âˆª B)â€™ = Aâ€™ âˆ© Bâ€™, (A âˆ© B) â€˜ = Aâ€™ âˆª Bâ€™

Take LHS

= A â€“ (Aâ€“B)

= A âˆ© (Aâ€“B)â€™

= A âˆ© (Aâ€™ âˆª Bâ€™)â€™) (since, (Bâ€™)â€™ = B)

= A âˆ© (Aâ€™ âˆª B)

= (A âˆ© Aâ€™) âˆª (A âˆ© B)

= Ï• âˆª (A âˆ© B) (since, A âˆ© Aâ€™ = Ï•)

= (A âˆ© B) (since, Ï• âˆª x = x, for any set)

= RHS

Therefore,

LHS = RHS

Hence, proved.

(iii) A âˆ© (A âˆª Bâ€™) = Ï•

Here, LHS A âˆ© (A âˆª Bâ€™)

= A âˆ© (A âˆª Bâ€™)

= A âˆ© (Aâ€™âˆ© Bâ€™) {By Deâ€“Morganâ€™s law}

= (A âˆ© Aâ€™) âˆ© Bâ€™ (since, A âˆ© Aâ€™ = Ï•)

= Ï• âˆ© Bâ€™

= Ï• (since, Ï• âˆ© Bâ€™ = Ï•)

= RHS

Therefore,

LHS = RHS

Hence, proved.

(iv) A â€“ B = A Î” (A âˆ© B)

Here, RHS A Î” (A âˆ© B)

A Î” (A âˆ© B) (since, E Î” F = (Eâ€“F) âˆª (Fâ€“E))

= (A â€“ (A âˆ© B)) âˆª (A âˆ© B â€“A) (since, E â€“ F = E âˆ© Fâ€™)

= (A âˆ© (Aâ€™ âˆª Bâ€™)) âˆª (A âˆ© Aâ€™ âˆ© B) {by using De-Morganâ€™s law and associative law}

= (A âˆ© Aâ€™) âˆª (A âˆ© Bâ€™) âˆª (Ï• âˆ© B) (by using distributive law)

= Ï• âˆª (A âˆ© Bâ€™) âˆª Ï•

= A âˆ© Bâ€™ (since, A âˆ© Bâ€™ = A â€“ B)

= A â€“ B

= LHS

Therefore,

LHS = RHS

Hence, Proved

### Question 3: If A, B, C are three sets such that A âŠ‚ B, then prove that C â€“ B âŠ‚ C â€“ A.

Solution:

Given: A âŠ‚ B

To prove: C â€“ B âŠ‚ C â€“ A

Let’s assume, x âˆˆ C â€“ B

â‡’ x âˆˆ C and x âˆ‰ B

â‡’ x âˆˆ C and x âˆ‰ A

â‡’ x âˆˆ C â€“ A

Thus, x âˆˆ Câ€“B â‡’ x âˆˆ C â€“ A

This is true for all x âˆˆ C â€“ B

Therefore,

C â€“ B âŠ‚ C â€“ A

Hence, proved.

### (v) (A â€“ B) âˆª (A âˆ© B) = A

Solution:

(i) (A âˆª B) â€“ B = A â€“ B

Let’s assume LHS (A âˆª B) â€“ B

= (Aâ€“B) âˆª (Bâ€“B)

= (Aâ€“B) âˆª Ï• (since, Bâ€“B = Ï•)

= Aâ€“B (since, x âˆª Ï• = x for any set)

= RHS

Hence, proved.

(ii) A â€“ (A âˆ© B) = A â€“ B

Let’s assume LHS A â€“ (A âˆ© B)

= (Aâ€“A) âˆ© (Aâ€“B)

= Ï• âˆ© (A â€“ B) (since, A-A = Ï•)

= A â€“ B

= RHS

Hence, proved.

(iii) A â€“ (A â€“ B) = A âˆ© B

Let’s assume LHS A â€“ (A â€“ B)

Let, x âˆˆ A â€“ (Aâ€“B) = x âˆˆ A and x âˆ‰ (Aâ€“B)

x âˆˆ A and x âˆ‰ (A âˆ© B)

= x âˆˆ A âˆ© (A âˆ© B)

= x âˆˆ (A âˆ© B)

= (A âˆ© B)

= RHS

Hence, proved.

(iv) A âˆª (B â€“ A) = A âˆª B

Let’s assume LHS A âˆª (B â€“ A)

Let, x âˆˆ A âˆª (B â€“A) â‡’ x âˆˆ A or x âˆˆ (B â€“ A)

â‡’ x âˆˆ A or x âˆˆ B and x âˆ‰ A

â‡’ x âˆˆ B

â‡’ x âˆˆ (A âˆª B) (since, B âŠ‚ (A âˆª B))

This is true for all x âˆˆ A âˆª (Bâ€“A)

âˆ´ A âˆª (Bâ€“A) âŠ‚ (A âˆª B) —(equation 1)

Contrarily,

Let x âˆˆ (A âˆª B) â‡’ x âˆˆ A or x âˆˆ B

â‡’ x âˆˆ A or x âˆˆ (Bâ€“A) (since, B âŠ‚ (A âˆª B))

â‡’ x âˆˆ A âˆª (Bâ€“A)

âˆ´ (A âˆª B) âŠ‚ A âˆª (Bâ€“A) —(equation 2)

From equation 1 and 2 we get,

A âˆª (B â€“ A) = A âˆª B

Hence, proved.

(v) (A â€“ B) âˆª (A âˆ© B) = A

Let’s assume LHS (A â€“ B) âˆª (A âˆ© B)

Let, x âˆˆ A

Then either x âˆˆ (Aâ€“B) or x âˆˆ (A âˆ© B)

â‡’ x âˆˆ (Aâ€“B) âˆª (A âˆ© B)

âˆ´ A âŠ‚ (A â€“ B) âˆª (A âˆ© B) —(equation 1)

Contrarily,

Let x âˆˆ (Aâ€“B) âˆª (A âˆ© B)

â‡’ x âˆˆ (Aâ€“B) or x âˆˆ (A âˆ© B)

â‡’ x âˆˆ A and x âˆ‰ B or x âˆˆ B

â‡’ x âˆˆ A

(Aâ€“B) âˆª (A âˆ© B) âŠ‚ A —(equation 2)

âˆ´ From equation (1) and (2), We get

(Aâ€“B) âˆª (A âˆ© B) = A

Hence, proved

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