Question

If $n$is a natural number then root $n$is ?

Answer 1

Solve for the required value

Let $n$be a natural number.

Case I: $n$is a perfect square such as $(1,4,9,16,25,...etc.)$

Natural numbers are numbers that start at $1$and end at $\infty .$In other words, they are set of whole numbers excluding $0.$

Then $\sqrt{n}$is a natural number.$\sqrt{1},\sqrt{4},\sqrt{9},\sqrt{16},\sqrt{25},etc.$ are all natural numbers.

Case II: $\sqrt{n}$is not a prefect square such as $(2,3,5,6,7,8,...etc.)$

An irrational number is a type of real number which cannot be represented as a simple fraction.

$\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},\sqrt{7}etc.$ are all irrational.

Hence, $\sqrt{n}$ is either a natural number of an irrational number..