# Proof that the square root of 2 is irrational

Let the number whose square is $2$ $=$ $\sqrt{2}$

Suppose that $\sqrt{2}$ is the ratio of two whole numbers with no common fators $a$ and $b$:

$\sqrt{2} = \frac{a}{b}$

Squaring both sides of the equation gives:

$2 = \frac{a^2}{b^2}$

Multiplying both sides by $b^2$ gives:

$2b^2 = a^2$

$2b^2$ must be even because $2$ is a factor. Therefore $a^2$ must be even.

The square root of any even number is even. Therefore $a$ must be even.

If $a$ is even it must contain $2$ as a factor. Therefore $a = 2d$ where $d$ is some whole number.

Substituting $2d$ for $a$ we get:

$2b^2 = (2d)^2$ $2b^2 = 4d^2$ $b^2 = 2d^2$

Applying the same logic we used to show that $a$ is even we can see that $b$ is also even.

If $a$ and $b$ are both even $2$ must be a common factor. Yet we eliminated common factors at the start.

This contradiction prooves that $\sqrt{2}$ cannot be the ratio of two whole numbers and must be irrational.

## Sources

- Mathematics for the Nonmathematician (pp 66-68) - Morris Kline