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 \(a \in \mathbb{Z}\), \(b \in \mathbb{Z}_{> 0}\) and \(\gcd(a, b) = 1\).

\[\begin{align*} \sqrt{d} &= \frac{a}{b} \\ d &= \frac{a^{2}}{b^{2}} \\ db^{2} &= a^{2} \\ \end{align*}\]

Therefore \(d \mid a^2\), since \(d\) is squarefree, we can thus conclude by this result that \(d \mid a\). As such, we write \(a = dc\).

\[\begin{align*} db^{2} &= (dc)^{2} \\ db^{2} &= d^2c^{2} \\ b^{2} &= dc^{2} \\ \end{align*}\]

By a similar argument to the above, we now also may conclude that \(d \mid b\), 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 \(n = k \ell^2\) 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 \(\sqrt{n} = \sqrt{k} \ell\). Then, given that \(\ell\) 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 \(n = k \ell^2\) 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, \(n = \ell^2\) and \(n\) is a square. In all other cases, \(\sqrt{n}\) must be irrational.