Prove that $\sqrt{2}$ is an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{2} = \frac{a}{b}$.
Squaring both sides,we get $2 = \frac{a^2}{b^2}$,which implies $a^2 = 2b^2$.
This means $a^2$ is divisible by $2$,so $a$ must also be divisible by $2$ (by the Fundamental Theorem of Arithmetic).
Let $a = 2k$ for some integer $k$.
Substituting this into $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This means $b^2$ is divisible by $2$,so $b$ must also be divisible by $2$.
Since both $a$ and $b$ are divisible by $2$,they have a common factor of $2$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{2}$ is an irrational number.