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 $\cap$ B.

Answer 1

Hint: We will apply substitution here. In the place of n we will substitute the values which satisfies n as in different sets. This is how we can convert the given set builder form to roster form and then solve the question.

Complete step-by-step answer:
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, moving on to the expression A $\cap$ B we see that we will take the common sets between the elements of sets A and B. Thus, we have A $\cap$ B = {1, 2, 3, 4, 5, ...} $\cap$ {2, 4, 6, 8, ...}. This results into A $\cap$ B = {2, 4, 6, 8, ....}.
Hence, A $\cap$ B = {2, 4, 6, 8, ....} or B only.

Note: As we are given sets A, B, C and D, we can say that these are well defined collections of numbers. The symbol of dots after elements in the sets means that these sets are never ending. So, we also should not write these sets till five terms or any number of terms. These should be written as infinite terms. The values of x are supposed to be elements of natural numbers, so we must be careful while writing sets as 0 is not a natural number.