# An Irrational Proof

One of my first adventures into formal proof-based mathematics was a proof that the square root of 2 is irrational, that is, can not be expressed in the form $\frac{a}{b}$ where $a$ and $b$ are integers.

It didn’t feel good at the time. Many readers are likely familiar with it but let’s delve into the details (before we present another [better?] proof).

Proof. Assume that we can write $\sqrt{2} = a/b$ where $a$ and $b \neq 0$ are integers with no factors in common other than one.

Then we can square both sides and rearrange to get

$2b^2 = a^2.$

From the above, we see that $a^2$ is even. It follows that $a$ must also be even and so we can write $a = 2k$ where $k$ is an integer.

Substituting this into the above equation yields

$2b^2 = 4k^2$

which further simplifies to

$b^2 = 2k^2$

Therefore, $b^2$ is even and so $b$ must also be even.

Thus, we have shown that $a$ and $b$ are both even, but this is a contradiction as we assumed that they have no factors in common.

I’m not sure why it didn’t sit with me particularly well. It seemed to me like we made an assumption that had very little to do with the square root of 2 itself. I can’t really put my finger on it.

Theodor Estermann provided a nicer proof with a contradiction that feels slightly better to me. Harley Flanders generalised the argument to $\sqrt{m}$ where $m$ is not a square number. We present his proof as follows:

Proof. We let $n$ denote the integer part of $\sqrt{m}$, that is, $n$ is the integer satisfying

$n < \sqrt{m} < n+1.$

We will prove that $\sqrt{m} - n$ is irrational as it follows then that $\sqrt{m}$ is irrational.

Let’s assume that $\sqrt{m} - n$ is rational. Then we may write it as

$\sqrt{m} - n = \frac{p}{q}$

where $0 < p < q$ and the denominator $q$ is as small as possible. This is fine to assume, as given all of the rational representations of $\sqrt{m} - n$ there must be one with the smallest possible denominator (see the well-ordering principle).

Then, we have that

$\frac{q}{p} = \frac{1}{\sqrt{m} - n}.$

We do the high-schooler thing and multiply the top and bottom of this fraction by $\sqrt{m} + n$ to get

$\frac{q}{p} = \frac{\sqrt{m} + n}{m - n^2}.$

We can also recover the number of interest, $\sqrt{m} - n$ in the numerator of this new fraction:

$\frac{q}{p} = \frac{(\sqrt{m}-n) + 2n}{m - n^2}.$

Now, we can rearrange for $\sqrt{m}-n$. Doing this gives us that (try it yourself!):

$\sqrt{m} - n = \frac{(m-n^2)q-2np}{p}.$

This is a contradiction as we assumed that $q$ was the smallest possible denominator, but here we have found a fraction with $p$ as the denominator, and we know that $p < q$.