Prove that the following number is irrational: $6+\sqrt{2}$.

(N/A) Let us assume,to the contrary,that $6+\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $6+\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{b} - 6$.
This simplifies to $\sqrt{2} = \frac{a - 6b}{b}$.
Since $a$ and $b$ are integers,$\frac{a - 6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $6+\sqrt{2}$ is rational is false.
Hence,$6+\sqrt{2}$ is an irrational number.