, the unique positive real number whose square is 2, is not a rational number.

Proof

Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{a}{b}$ for some integers $a$ and $b$ ; without_loss_of_generality let $\frac{a}{b}$ be in lowest_terms, i.e. $gcd (a, b) = 1$ . We have

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

From the definition of $\sqrt{2}$ ,

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

So $a^{2}$ is a multiple of $2$ . Since $2$ is prime, $a$ must be a multiple of 2; let $a = 2 k$ . Then

$2 b^{2} = (2 k)^{2} = 4 k^{2}$ $b^{2} = 2 k^{2}$

So $b^{2}$ is a multiple of $2$ , and so is $b$ . But then $2 | gcd (a, b)$ , which contradicts the assumption that $\frac{a}{b}$ is in lowest terms! So there isn't any way to express $\sqrt{2}$ as a fraction in lowest terms, and thus there isn't a way to express $\sqrt{2}$ as a ratio of integers at all. That is, $\sqrt{2}$ is irrational.