# Irrationality of Square Root of 2 by Infinite Cover

[Kac & Ulam, pp. 6-7] gave a spectacular restatement of the irrationality of $\sqrt{2}$:

Consider all rational numbers in the interval from 0 to 1, excluding 0. Each number can be written in a unique way as a fraction $a/b$ where $a$ and $b$ have no divisors in common. Imagine now $a/b$ as a centre of an interval of length $1/2b^{2};\;$ in other words, cover $a/b\;$ by the interval with endpoints $a/b - 1/4b^{2}\;$ and $a/b + 1/4b^{2}.\;$ Since the rational numbers form a dense set (i.e., in every interval no matter how small there are always rational numbers) and, since the sum of lengths of all covering intervals is found to be infinite, it would seem that, having so generously covered all rational numbers, we have automatically covered all numbers. However, we shall show that $\displaystyle\frac{\sqrt{2}}{2}\;$ remains uncovered! In fact the number $|b^{2} - 2a^{2}|\;$ being an integer must be at least $1;\;$ it cannot be 0 since $\sqrt{2}\;$ is irrational.

Hence, successively,

$\displaystyle \frac{|b^{2}-2a^{2}|}{2b^2}\ge\frac{1}{2b^2},$

$\displaystyle \left|\frac{\sqrt{2}}{2}-\frac{a}{b}\right|\cdot\left|\frac{\sqrt{2}}{2}+\frac{a}{b}\right|\ge\frac{1}{2b^2},$

$\displaystyle \left|\frac{\sqrt{2}}{2}-\frac{a}{b}\right|\ge \frac{1}{2b^2}\frac{1}{\left|\frac{\sqrt{2}}{2}+\frac{a}{b}\right|}\gt \frac{1}{2b^2}\frac{1}{2},$

$\displaystyle \left|\frac{\sqrt{2}}{2}-\frac{a}{b}\right|\gt\frac{1}{4b^2}.$

and the assertion that 2/2 is not covered follows.

### References

1. M. Kac and S. M. Ulam, Mathematics and Logic, Dover Publications, NY, 1968, 6-7