Prove that $6+\sqrt{2}$ is an irrational number.

(N/A) Assume that $6+\sqrt{2}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $6+\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{b} - 6$.
Since $a$ and $b$ are integers,$\frac{a}{b} - 6 = \frac{a-6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the established fact that $\sqrt{2}$ is an irrational number.
This contradiction has arisen due to our incorrect assumption that $6+\sqrt{2}$ is rational.
Therefore,we conclude that $6+\sqrt{2}$ is an irrational number.