• # question_answer 7) Given that $\sqrt{2}$ is irrational, prove that $\left( 5+3\sqrt{2} \right)$is an irrational number.

 Given, $\sqrt{2}$ is irrational number. Let $\sqrt{2}=m$ Suppose, $5+3\text{ }\sqrt{2}$ is a rational number. So,       $5+3\text{ }\sqrt{2}=\frac{a}{b}$                    $(a\ne b,b\ne 0)$ $3\sqrt{2}=\frac{a}{b}-5$ $3\sqrt{2}=\frac{a-5b}{b}$ or         $\sqrt{2}=\frac{a-5b}{3b}$ So,       $\frac{a-5b}{3b}=m$ But $\frac{a-5b}{3b}$ is rational number, so m is rational number which contradicts the fact that $m=\sqrt{2}$ is irrational number. So, our supposition is wrong. Hence, $5+3\sqrt{2}$ is also irrational.