Question

Let A = {x : x $\in \mathbb{N}$}, B = {x : x = 2n, n$\in \mathbb{N}$}, C = {x : x = 2n – 1, n$\in \mathbb{N}$} and D = {x : x is a prime natural number}. Find A $\cap$ D.

Answer 1

Hint: As we are given a well defined collection of sets so we can easily find the elements of each of the sets by substituting the values of n belonging to natural numbers. We have to convert the given set builder form to roster form using this method.

Complete step-by-step answer:

We will consider the well defined sets individually here. First we will consider the set A = {x : x $\in \mathbb{N}$}. Here we can clearly see that this set contains all elements of natural numbers. Therefore we have A = {1, 2, 3, 4, 5, ...}.
Now we will consider B = {x : x = 2n, n$\in \mathbb{N}$}. Here if we substitute n = 1 we get x = 2. Similarly, we get the elements as a multiple of 2. Therefore, we have B = {2, 4, 6, 8, ...}.
Now we will consider the well defined set C = {x : x = 2n – 1, n$\in \mathbb{N}$}. After substituting n = 1 we have x = 2(1) – 1 or, x = 1. And solving in this manner we will have the C = {1, 3, 5, 7, ...}.
Now we will consider the last set which is D = {x : x is a prime natural number}. It clearly says that the set D is the collection of all prime numbers. Therefore, we have that D = { 2, 3, 5, 7, ...}.
Now for the expression A $\cap$ D we see that we will consider all the common sets between the elements of sets A and D. Thus, we have A $\cap$ D = {1, 2, 3, 4, 5, ...} $\cap$ { 2, 3, 5, 7, ...}. With the help of intersection between these two sets we get the common elements as, A $\cap$ D = {2, 3, 5, 7, ...}.
Hence, A $\cap$ D = {2, 3, 5, 7, ...} or D.

Note: While defining the elements of set D one should be aware not to write 1 in it. As 1 is neither a prime number nor a composite number. Here A $\cap$ D = D as all the elements between these two sets are from D only.