$\sqrt 2$, 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 $\sqrt 2$,

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

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

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

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.