# Geometric proof that if n is a non-perfect square, then √n is irrational.

I know there is a geometric proof of the irrationality of √2. I thought maybe this one could be generalized for √n when n is a non-perfect square, but I could not find something like that anywhere.

Does anyone know if such a geometric proof exist? I'm researching on different kinds of proof for this theorem, but could only find the algebraic ones.

• There is a nice one for $\sqrt{5}$ and relatives (golden section). Have not seen a general argument. One can imitate $\sqrt{2}$ geometric proof for $\sqrt{n}$ has continued fraction expansion with short cycle length. – André Nicolas Jun 24 '13 at 21:32
• The usual idea for $\sqrt{2}$ is based on the Pythagorean theorem and $2=1^2+1^2$. However on average very few integers can be expressed as the sum of two squares (all primes congruent to 3 mod 4 must appear an even number of times). – vadim123 Jun 24 '13 at 21:35

However, this can be done for $\sqrt{n^2 + 1}$ and $\sqrt{n^2 - 1}$ given any integer $n > 1$.