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# Show that the following four conditions are equivalent : (i) A ⊂ B (ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A

**Solution:**

To prove (i) is equivalent to (ii):

From (i) A ⊂ B

It means that all elements of A are in B.

i.e., there is no element of A that is NOT in B.

i.e., A – B = Φ (which is (ii))

Thus, (i) ⇔ (ii)

To prove (i) is equivalent to (iii):

From (i) A ⊂ B.

It means that all elements of A are in B.

Then A ∪ B = Set of all elements of A and B = B (which is (iii))

Thus, (i) ⇔ (iii)

To prove (i) is equivalent to (iv):

From (i) A ⊂ B.

It means that all elements of A are in B.

Then A ∩ B = Set of all elements common to A and B = B (which is (iv))

Thus, (i) ⇔ (iv)

We have proved that (i) ⇔ (ii), (i) ⇔ (iii), and (i) ⇔ (iv).

Thus,

(i) ⇔ (ii) ⇔ (iii) ⇔ (iv)

NCERT Solutions Class 11 Maths Chapter 1 Exercise ME Question 4

## Show that the following four conditions are equivalent : (i) A ⊂ B(ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A

**Summary:**

We are asked to prove the given four conditions are equivalent. We have proved that A ⊂ B ⇔ A – B = Φ ⇔ A υ B = B ⇔ A ∩ B = A.

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