Irrationality of Roots of Non Squares

This note deals with the standard techniques for proving the following fact.

Theorem

The square root of any positive integer which is not a square is irrational.

We first prove a special case.

Lemma

The square root of any squarefree integer greater than $1$ is irrational.

Proof

Let $d$ be squarefree. We prove that $\sqrt{d}$ is irrational by contradiction, that is, we assume it can be written in the form $\frac{a}{b}$ where $a \in Z$ , $b \in Z_{> 0}$ and $gcd (a, b) = 1$ .

\begin{aligned} \sqrt{d} & = \frac{a}{b} \\ d & = \frac{a^{2}}{b^{2}} \\ d b^{2} & = a^{2} \end{aligned}

Therefore $d ∣ a^{2}$ , since $d$ is squarefree, we can thus conclude by this result that $d ∣ a$ . As such, we write $a = d c$ .

\begin{aligned} d b^{2} & = (d c)^{2} \\ d b^{2} & = d^{2} c^{2} \\ b^{2} & = d c^{2} \end{aligned}

By a similar argument to the above, we now also may conclude that $d ∣ b$ , and thus $gcd (a, b) \geq d > 1$ , a contradiction.

It should be noted that the above proof can be completed without the condition that $gcd (a, b) = 1$ simply by noting that after dividing out a common factor of $d$ from $a$ and $b$ , we are left with the same initial construction and the argument of the proof can repeat. Therefore, we have proven that $a$ and $b$ are both infinitely divisible by $d$ , which is a contradiction since the only integer satisfying this property is $0$ and $b \neq 0$ .

The above method does not work for examples such as $12$ , however $\sqrt{12}$ is irrational. This means a different approach is needed to prove this. We now handle the general case.

Lemma

For a positive integer $n = k ℓ^{2}$ written with the squarefree part taken out, $\sqrt{n}$ is irrational if and only if $\sqrt{k}$ is irrational.

Proof

Taking the square root of either side of the squarefree factorisation of $n$ we get $\sqrt{n} = \sqrt{k} ℓ$ . Then, given that $ℓ$ is rational, and the product of a non-zero rational number with an irrational number is irrational, we can conclude that $\sqrt{n}$ is rational if and only if $\sqrt{n}$ is.

The main result then follows as a direct consequence.

Proof

For any integer $n = k ℓ^{2}$ written as above, the only case in which the squareroot of a squarefree integer $k$ can be rational is if $k = 1$ . In this case, $n = ℓ^{2}$ and $n$ is a square. In all other cases, $\sqrt{n}$ must be irrational.