√2 is irrational

Introduction

We can construct a right-angled triangle whose three sides are whole numbers:

Is it possible to construct a square whose side b and diagonal a a are both measured by integers whole numbers? The figure below shows half of the square, which is an isosceles right-angled triangle.

According to the Pythagorean theorem

\[ \begin{equation*} \begin{aligned} b^2+b^2&=a^2\\ 2b^2&=a^2\\ 2&=\left(\frac{a}{b}\right)^2\\ \frac{a}{b}&=\sqrt{2} \end{aligned} \end{equation*} \]

If it’s possible to find integers a, b  that verify this equality, we say that √2 is a rational number. Otherwise we say that √2 is an irrational number.

Demonstration of ” √2 irrational ”

Let’s assume for the sake of argument that √2 is rational: then \(\sqrt{2}=\frac{a}{b}\) where a, b are positive integers. It is possible to simplify the fraction \(\frac{a}{b}\) until a, b are prime to each other (i.e. the fraction \(\frac{a}{b}\) can no longer be simplified).

\[ \begin{equation*} \begin{aligned} \sqrt{2}&=\frac{a}{b}\\ \sqrt{2} \, b&=a\\ 2 \, b^2&=a^2 \end{aligned} \end{equation*} \]

Since a² is even, a is even and   a = 2 p   where p is a positive integer.

\[ \begin{equation*} \begin{aligned} 2\,b^2&=(2\,p)^2\\ 2\,b^2&=4\,p^2\\ b^2&=2\, p^2 \end{aligned} \end{equation*} \]

Since b² is even, b is even. Consequently, it is possible to simplify the fraction \(\frac{a}{b}\) by 2, which contradicts the assumption that a, b are mutually prime.

Since the hypothesis « √2 is rational » leads to a contradiction, the opposite is true, i.e. « √2 is irrational ».