Question

prove that n(A union B)=n(A) +n(B)-n (A intersection B). 'A' is the set of prime numbers less than 10. 'B' is the set of odd numbers less than 10​

Answer 1

SOLUTION

GIVEN

'A' is the set of prime numbers less than 10.

'B' is the set of odd numbers less than 10

TO PROVE

n(A ∪ B) = n(A) + n(B) - n (A ∩ B)

EVALUATION

Here it is given that

A' is the set of prime numbers less than 10.

∴ A = { 2 , 3 , 5 , 7 }

So n(A) = 4

'B' is the set of odd numbers less than 10

∴ B = { 1 , 3 , 5 , 7 , 9 }

So n(B) = 5

Again

A ∪ B = { 1 , 2 , 3 , 5 , 7 , 9 }

So n(A ∪ B) = 6

Also A ∩ B = { 3 , 5 , 7 }

So n(A∩B) = 3

We have to verify

n(A ∪ B) = n(A) + n(B) - n (A ∩ B)

LHS = n(A ∪ B) = 6

RHS = n(A) + n(B) - n (A ∩ B) = 4 + 5 - 3 = 6

Hence LHS = RHS

Hence verified

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