Question

Let $A=\left\{ x:x\in N \right\},B=\left\{ x:x=2n,n\in N \right\}$. Find $A\cap B$.

Answer 1

Hint:To solve this question we will first convert all the sets from the set builder form to the roster form. And then to find $A\cap B$, we will find the common elements of both the sets, A and B from the roster form and then describe it in the set builder form.

Complete step-by-step answer:
In this question, we have to find the set $A\cap B$ when $A=\left\{ x:x\in N \right\},B=\left\{ x:x=2n,n\in N \right\}$.
First, we will convert all the required sets from set builder form to roster form. Let us consider set A, that is, $A=\left\{ x:x\in N \right\}$. We know that N represents natural numbers and natural numbers are {1, 2, 3, 4, 5 …..}. So we can write, $A=\left\{ 1,2,3,4,5,6...... \right\}$.
Similarly we have been given that $B=\left\{ x:x=2n,n\in N \right\}$, where N represents natural numbers. So, we can write the possible values of 2n as follows. For n = 1, we have $x=2\times 1=2$, for n = 2, we have $x=2\times 2=4$ and so on till infinity. So, we can write it as {2, 4, 6, 8…..}. Therefore, we can write the set B as $B=\left\{ 2,4,6,8,10...... \right\}$.
Now, we are asked to find the value of $A\cap B$. So, we can write it as follows.
$A\cap B=\left\{ 1,2,3,4,5...... \right\}\cap \left\{ 2,4,6,8,10...... \right\}$
Here, we can see that A and B have {2, 4, 6, 8……} as the common terms and we know that $'\cap '$ represents the set which contains the elements both A and B contain. So, we can write,
$A\cap B=\left\{ 2,4,6,8...... \right\}$
And we know that $\left\{ 2,4,6,8...... \right\}=B$. So, we get, $A\cap B=B$.
Hence, we can say that $A\cap B=B$ where, $A=\left\{ x:x\in N \right\}$ and $B=\left\{ x:x=2n,n\in N \right\}$.

Note: In this question, the possible mistake that one can make is by considering $\left( A\cap B \right)$ as $\left( A\cup B \right)$ in a hurry which will result in the wrong answer. Also, we need to remember that N represents natural numbers and that the natural numbers start from 1 and not 0.