Question

if n is a natural number then root N is

Answer 1

Solution:

The set of natural number is

lN = {1, 2, 3, 4, 5, 6, 7, ...}

Let, n be any natural number.

Then √n = m ( say )

• When n = 1, m = 1, a rational number

• When n = 2, m = √2, an irrational number

• When n = 3, m = √3, an irrational number

• When n = 4, n = 2, a rational number

and so on . . .

We can proceed in this way to check out more natural numbers.

Therefore, the root of any natural number can be either a rational number or an irrational number.

Note: In rational numbers, some natural numbers can be found also but only in the case where 'n' is a perfect square. It is not mentioned separately in the answer.

▪ Rational number - a real number which is expressible as x/y where x, y are integers with non-zero y

Example: 0/1, 2/3, 5/7, ..., 1, 2, ...

▪ Irrational number - a real number which cannot be expressed as a fraction in terms of integers.

Example: √2, √3, √5, 1 + √7, ...

Answer 2

Answer:

Can be natural number or irrational number.

Step-by-step explanation:

Given, n is natural number.

If n is perfect square then $\sqrt{n}$ will be natural number.

For example if n is 4 then  $\sqrt{4}$ = 2 is irrational number.

If n is not perfect square then  $\sqrt{n}$,will be irrational number.

For example if n is 2 then $\sqrt{2}$ will be irrational.