Question

if A=(x:x is a natural number) B=(x:x is an even natural number) C= is an odd natural number) D=(x:x is a prime number )
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Answer 1

Answer:

(i) A∩D $=$ D

(ii) A∩B $=$  B

(iii) B∩C $=$ ∅

(iv) C∩D $=D-\{2\}$

Step-by-step explanation:

Given A $=\{x:\ x\ is\ a\ natural\ number\}$

          B $=\{x:\ x\ is\ an\ even\ number\}$

          C $=\{x:\ x\ is \ an\ odd\ number\}$

          D $=\{x:\ x\ is\ a\ prime\ number\}$

We are asked to find the intersections

(i) A∩D         (ii)A∩B           (iii) B∩C            (iv)C∩D

We know the intersection of two sets contains only the elements which are present in both the sets at the same time.

That is,  suppose A and B are two sets, then

A∩B $=\{x:\ xEA\ and\ xEB\}$

Therefore here,

(i) A∩D $=$ D .

      Because every prime number is a natural number, but every natural number cannot be prime. Therefore the set D is contained in A. Hence their intersection is nothing but D itself.

(ii) A∩B $=$ B

       As said in the earlier case, every even number is a natural number, therefore the set B is contained in the set A. Hence their intersection is nothing but B itself.

(iii) B∩C $=$ ∅ (the null set)

       There does not exist any number which is both even and odd. Therefore there is no element in the intersection of B and C. Hence it results the  null set.

(iv) C∩D $=D-\{2\}$

        Because every prime number except $2$ is an odd number. an even number other than $2$ cannot be a prime number as it has the factor two .

Therefore every element in D other than $2$ is contained in C. Hence the whole set D excluding the number $2$ (because two is not in C) is in their intersection.

Hence the answer

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Answer 2

Answer:

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Explanation:

A={x:xis a natural number}={1,2,3,4,5.....}

B={x:x is an even natural number}={2,4,6,8.....}

C={x:x is an odd natural number}={1,3,5,7,9.......}

D={x:xis a prime number}={2,3,5,7....}

(i) A∩B={x:x is a even natural number}=B

(ii) A∩C={x:x is an odd natural number}=C

(iii) A∩D={x:x is a prime number}=D

(iv) B∩C=ϕ

(v) B∩D={2}

(vi) C∩D={x:xis odd prime number}