Prove that $\frac{1}{\sqrt{2}}$ is an irrational number.

(N/A) First,rationalize the denominator: $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Assume,for the sake of contradiction,that $\frac{\sqrt{2}}{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\frac{\sqrt{2}}{2} = \frac{a}{b}$.
This implies $\sqrt{2} = \frac{2a}{b}$.
Since $a$ and $b$ are integers,$\frac{2a}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the well-known fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $\frac{1}{\sqrt{2}}$ is rational must be false.
Hence,$\frac{1}{\sqrt{2}}$ is an irrational number.