Question

Let $A=\left\{ x:x\in N,x\text{ is a multiple of 3} \right\}$ and $B=\left\{ x:x\in N,x\text{ is a multiple of 5} \right\}$. Write $A\cap B$.

Answer 1

Hint:In order to solve this question we will first convert the sets A and B from the set builder form to the roster form. And then we will compare a few terms of both the sets and then we will observe a common relation among all the equal terms and hence we will find $A\cap B$.

Complete step-by-step answer:
In this question, we have been given that there are two sets, A and B such that $A=\left\{ x:x\in N,x\text{ is a multiple of 3} \right\}$ and $B=\left\{ x:x\in N,x\text{ is a multiple of 5} \right\}$. And we are asked to write $A\cap B$.
For this, first we will convert both the sets, A and B from the set builder form to the roster form. Let us consider set A, we are given that $A=\left\{ x:x\in N,x\text{ is a multiple of 3} \right\}$ and we know that multiples of 3 are {3, 6, 9, 12, 15……}. So, we can write set A as follows.
$A=\left\{ 3,6,9,12,15,18,21,24,27,30,33...... \right\}$
Similarly, we have been given that $B=\left\{ x:x\in N,x\text{ is a multiple of 5} \right\}$ and we know that multiples of 5 are {5, 10, 15, 20, 25, 30, 35……}. So, we can write set B as follows.
$B=\left\{ 5,10,15,20,25,30,35,40...... \right\}$
Now, if we compare set A and set B, we can see that the terms common in both the sets are {15, 30, 45……}. So, we can write $A\cap B$ as follows.
$A\cap B=\left\{ 15,30,45...... \right\}$ and we can write 15 = 15(1), 30 = 15(2), 45 = 15(3) and so on.
From this, we can say that $A\cap B$ contains all the elements which are multiples of 15 and they are natural numbers. So, we can write $A\cap B=\left\{ x:x\in N,x\text{ is a multiple of 15} \right\}$.
Hence, we can say that, for $A=\left\{ x:x\in N,x\text{ is a multiple of 3} \right\}$ and $B=\left\{ x:x\in N,x\text{ is a multiple of 5} \right\}$, $A\cap B=\left\{ x:x\in N,x\text{ is a multiple of 15} \right\}$.

Note: We can also solve this question by using the concept of common multiple or LCM, that is, the terms which are common in both A and B will be the only terms which are multiple of 3 and 5 both, that is, the multiples of LCM of 3 and 5, which is 15. So, $A\cap B=\left\{ x:x\in N,x\text{ is a multiple of 15} \right\}$.