Question

If ${N_a}$ is equal to $\\{ an:n \in N\\}$, then find ${N_5} \cap {N_7}$
1). ${N_7}$
2). $N$
3). ${N_{35}}$
4). ${N_5}$

Answer 1

First, we have to find the values of ${N_5}$ and ${N_7}$ by using the conditions given in the question. To find ${N_5}$ we have to multiply 5 with n (n is the natural numbers), we will get so many values of it. Then similarly with ${N_7}$ we will get so many values. Then, we have to find ${N_5} \cap {N_7}$ and for that we will find common values from both ${N_5}$ and ${N_7}$.

Complete step-by-step solution:
Given: {N_a} = \left\\{ {an:n \in N} \right\\}--------(1)
This tells us that n is the natural number.
Now, we will find values of ${N_5}$ by replacing a from 5 in equation (1)
{N_5} = \left\\{ {5n:n \in N} \right\\}
{N_5} = \left\\{ {5n:n = 1,2,3,4,5,6,7................} \right\\}
Now, we will multiply 5 with all the natural numbers and find all the possible values.
{N_5} = \left\\{ {5,10,15,20,25,30,35,....................} \right\\}
These are the possible values of ${N_5}$.
Similarly, we will find the values of ${N_7}$
{N_7} = \left\\{ {7n:n \in N} \right\\}
{N_7} = \left\\{ {7n:n = 1,2,3,4,5,6,7,..............} \right\\}
{N_7} = \left\\{ {7,14,21,28,35,42,49,...............} \right\\}
Now, we have to find ${N_5} \cap {N_7}$ and for this we have to find common values in${N_5}$ and ${N_7}$.
5 and 7 are prime numbers which means they don’t have any common factors. So, we will find the least common factor of them because its multiples will be the common factors in ${N_5}$ and ${N_7}$.
Least common factor of 5 and 7 is 35. So, all the multiples of 35 will be common in both ${N_5}$and ${N_7}$.
{N_5} \cap {N_7} = \left\\{ {35,70,105,140,..........} \right\\}
{N_5} \cap {N_7} = \left\\{ {35n:n \in N} \right\\}
${N_5} \cap {N_7} = {N_{35}}$
So, option 3. Is the correct answer.

Note: Natural numbers are a part of the number system, including all the positive integers from 1 to infinity. The least common factor of two integers is the smallest positive integer that is divisible by both the integers.