Question

Q. prove that √n is irrational.

Answer 1

Answer on Question #62407 – Math – Combinatorics | Number Theory

Question

Prove that $\sqrt{n}$ is irrational

Solution

We will prove by contradiction that if $n$ is natural and it is not square of natural number, then $\sqrt{n}$ is irrational.

Suppose there exists some $n$ which is not complete square, but $\sqrt{n}$ is rational.

Then we can write

$$\sqrt{n} = \frac{p}{q},$$

where $p$ and $q$ are coprime natural numbers.

Squaring both sides we have

$$n = \frac{p^2}{q^2} \Rightarrow nq^2 = p^2.$$

It follows from the last equality that $p^2$ is divisible by $q$ and it is coprime to $q$. So $q$ is obviously must be 1.

Then

$$\sqrt{n} = p \Rightarrow n = p^2,$$

which contradicts the condition 'n is not complete square'.

Therefore, assumption $\sqrt{n}$ is rational' was false.

So we proved that $\sqrt{n}$ is irrational for those $n$ which are not complete squares.

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