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Q8 of 145 Page 47

If A ⊂ B, show that (B’ – A’) = ϕ.

As A ⊂ B the set A is inside set B



Hence A ∪ B = B


Taking compliment


⇒ (A ∪ B)’ = B’


Using De-Morgan’s law (A ∪ B)’ = A’ ∩ B’


⇒ A’ ∩ B’ = B’ …(i)


Now we know that


B’ = (B’ – A’) + (A’ ∩ B’)



Using (i)


⇒ B’ = (B’ – A’) + B’


⇒ (B’ – A’) = B’ – B’


⇒ (B’ – A’) = 0


⇒ (B’ – A’) = {ϕ}


Hence proved


More from this chapter

All 145 →
6

If A and B are two sets such than n(A) = 54, n(B) = 39 and n(B – A) = 13 then find n(A ∪ B).

Hint n(B) = n(B – A) + n(A ∩ B) ⇒ n(A ∩ B) = (39 – 13) = 26.


7

If A ⊂ B, prove that B’ ⊂ A’.

9

Let A = {x : x = 6n N) and B = {x : x = 9n, n ϵ N}, find A ∩ B.

10

If A = {5, 6, 7}, find P(A).

Questions · 145
1. Sets
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