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 B $\cap$ C.

Answer 1

Hint: The first step is to convert the form of the set given in the question. We will find the values of the well defined sets A, B, C and D by substituting all the elements of natural numbers one by one. By this the elements will be represented in a numerical form of sets.

Complete step-by-step answer:
Now we will start by framing 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 we will consider the union of B and C and write it as B $\cap$ C. Here, we see that we will consider all the common sets between the elements of sets B and C. Thus, we have B $\cap$ C = {2, 4, 6, 8, ...} $\cap$ {1, 3, 5, 7, ...}. With the help of intersection between these two sets we get the common elements as, B $\cap$ C = $\phi$.
Hence, B $\cap$ C is an empty set.

Note: If we were asked to consider n belongs to the whole numbers then also we would have got an empty set. Or actually the set C would have been started from - 1. The empty set can be written as a symbol $\phi$ or {}. One should be aware of writing an empty set like \left\\{ \phi \right\\}. This symbol is not defined and the answer will be wrong.