# Can the irrationality of the square root of 2 be proved by using Dirichlet's theorem on primes in an arithmetic progression?

The title says it all. I intend to answer the question myself, in the affirmative. (I would have left the body blank, but the system requires me to post at least 30 characters.)

Here is an alternative proof (this is ridiculous, but fun!). Suppose that $(a/b)^2 = 2$. This equation can be seen as an equation in $\mathbf F_p$, whenever $p \nmid b$. On the other hand, Gauss shows that for an odd prime $p$,

$$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/2}.$$

This is $-1$ if $p \equiv \pm 3 \mod 8$ and $+1$ if $p \equiv \pm 1 \mod 8$.

By Dirichlet's theorem on primes in arithmetic progressions, there exists an odd prime $p$ such that $\left(\frac{2}{p}\right) = -1$ and $p \nmid b$. But this contradicts the fact that $(a/b)^2 = 2$ in $\mathbf F_p$.

• >this is ridiculous, but fun! -- yes, and here is a link to where other such is discussed: mathoverflow.net/questions/42512/… – user27325 Nov 2 '13 at 16:45
• "[T]he ridiculous is one of the best methods to shatter the iron confines of pedestrian thought[.]" --Sherlock Holmes --quoted by Dr. John Watson, on p. 16 of The SHERLOCK HOLMES Puzzle Collection (2011), ISBN 978-1-86200-884-7 – user27325 Nov 10 '13 at 19:17
• @EsperantoSpeaker1 Cool quote! :) – Bruno Joyal Nov 10 '13 at 23:02

Yes.

Dirichlet's theorem states that every proper arithmetic progression contains infinitely many primes.

Theorem. The square root of $2$ is irrational.

Proof: Suppose that $a$ and $b$ are relatively-prime positive integers such that $(a/b)^2 = 2$. Then, by inspection, $b > 1$, and so $b^2$ is composite. Since $a$ and $b$ are relatively prime, so are $a^2$ and $b^2$. By Dirichlet's theorem, there exists a positive integer $n$ and a prime $p$ such that $a^2 + nb^2 = p$. Then, dividing through by $b^2$, we have $a^2/b^2 + n = p/b^2$. Then, since $a^2/b^2 = (a/b)^2$, we have $(a/b)^2 + n = p/b^2$. Then, since $(a/b)^2 = 2$, we have $2 + n = p/b^2$. However, the left side is an integer, but the right side cannot be an integer, since it is the ratio of a prime to a composite number. This contradiction shows that the square root of $2$ must be irrational. Q.E.D.

• Cute! With somewhat (!) less machinery, one can say that there are integers $x$ and $y$ such that $a^2 x+b^2 y=1$, and use essentially the same argument. – André Nicolas Nov 2 '13 at 2:50