# Alternative irrationality proofs

There’s a standard proof that $\sqrt{2}$ is irrational:

- Suppose $\sqrt{2} = \frac{p}{q}$, where $p$ and $q$ are coprime ((every fraction can be written this way));
- Then $q\sqrt{2} = p$, so $2q^2 = p^2$. Hence, $p^2$ is even – which can only be the case if $p$ is even.
- So, letting $p = 2r$, $2q^2 = 4r^2$, or $q^2 = 2r^2$. That means $q$ is also even – contradicting the assumption that $p$ and $q$ are coprime.

All well and good, whoever discovered that deserves a nice swim in the Mediterranean.

But that’s not the end of the story. As mathematicians, we are – or should be – always looking for different ideas, different tools, different ways to approach problems.

There are other ways to prove that $\sqrt{2}$ is irrational. My dear friend @colinthemathmo tooted one such:

A “new to me” proof that $\sqrt{2}$ is irrational, found by Sergey Markelov while still in high school.

In the decimal system, a square of an integer may only end in 0, 1, 4, 5, 6, or 9, whereas twice a square may only end with 0,2,8. So if $a^2=2b^2$, both $a$ and $b$ must end with 0.

This triggers an infinite descent which proves that this is impossible, and so $a^2=2b^2$ has no solutions in [positive] integers, hence 2 is never the square of a rational.

That’s quite lovely! It’s even quicker if you work base 3 rather than base 10:

- In the ternary system, a square of an integer may only end in 0 or 1, and twice a square may only end in 0 or 2. So, if $a^2 = 2b^2$, both $a$ and $b$ must end with 0.
- This triggers an infinite descent which proves that this is impossible, and so $a^2=2b^2$ has no solutions in positive integers, hence 2 is never the square of a rational.

(As I write this up, I notice that it also works in base 2! Continual refinement. Lovely stuff.)

However! Those proofs all feel morally the same. I have a couple of other proofs I like.

(1) Factors

The first is based on the fundamental theorem of arithmetic: every positive integer can be written in exactly one way as $\br{2^{k_2}} \br{3^{k_3}} \br{5^{k_5}} \br{7^{k_7}}\dots$, where the bases are the prime numbers and the exponents are non-negative integers. ((Removing the last restriction gives you the positive rationals, but we’re not doing that today.))

For a square number, $p^2$, all of the $k_p$s must be even.

But for $2p^2$, $k_2$ is odd, so $2p^2$ is not a square.

(2) Continued fractions

(This is not obviously based on continued fractions, but it’s where I got the idea. It is not a new proof – but rediscovering things is one of the joys of maths.)

- Suppose $\sqrt{2} = \frac{p}{q}$ with the fraction in lowest form. Since $1 < \sqrt{2} < 2$, $q < p < 2q$.
- Consider $\frac{2q - p}{p-q} = \frac{2q - q\sqrt{2}}{q\sqrt{2}-q}$
- $\dots = \frac{2-\sqrt{2}}{\sqrt{2}-1} = \sqrt{2}$
- But $2q - p < p$ (because $p>q$) and $q-p < q$, contradicting the assumption that the fraction is in lowest form.

When I tweeted this, I was sent proofs without words by @hartkp and @daveinstpaul, the tidier of which I include here:

— David Radcliffe (@daveinstpaul) May 21, 2021

Do you have any other favourite irrationality-of-$\sqrt{2}$ proofs? I’d love to hear about them!

@hartkp has obliged with the following (I paraphrase):

Suppose $m$ is the smallest positive integer such that $m\sqrt{2}$ is an integer. Consider $n = m\sqrt{2}-m$. By supposition, this is a positive integer, and smaller than $m$ because $1 <\sqrt{2}<2$. However, $n\sqrt{2} = 2m - m\sqrt{2}$, which (again by supposition) is a positive integer. This contradicts our supposition, which must therefore be false.

- Edited 2022-07-25 to add @hartkp’s proof and correct a twitter handle.