question_answer

Given that \[\sqrt{2}\] is irrational, prove that \[\left( 5+3\sqrt{2} \right)\]is an irrational number.

Answer:

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.