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

(N/A) Let us assume,to the contrary,that $\sqrt{2}$ is rational.
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 shows that $a^2$ is divisible by $2$,and consequently,$a$ is also divisible by $2$ (by the Fundamental Theorem of Arithmetic).
So,we can write $a = 2k$ for some integer $k$.
Substituting this into the equation $a^2 = 2b^2$,we get $(2k)^2 = 2b^2$,which simplifies to $4k^2 = 2b^2$,or $b^2 = 2k^2$.
This shows that $b^2$ is divisible by $2$,and consequently,$b$ is also divisible by $2$.
Thus,both $a$ and $b$ have at least $2$ as a common factor,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption that $\sqrt{2}$ is rational is false,and $\sqrt{2}$ must be irrational.