$~$\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.