Given that $\sqrt{2}$ is irrational, prove that$( 5+3\sqrt{2})$ is an irrational number.

Given: $\sqrt{2}$ is irrational number.

To do: To prove that $( 5+3\sqrt{2})$ is an irrational number.

Solution:

Let us assume that is rational.

 Then there exist co-prime positive integers a and b such that 

 $5+3\sqrt{2}=\frac{a}{b}$

$\Rightarrow 3\sqrt{2}=\frac{a}{b}-5$

$\Rightarrow \sqrt{2}=\frac{a-5b}{3b}$

$\sqrt{2}$ is an irrational. [ a, b are integers, $\frac{a-5b}{3b}$ is rational].

This contradicts the fact that $\sqrt{2}$ is irrational.

So our assumption is incorrect.

Hence, $5+3\sqrt{2}$ is an irrational number.
 

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Updated on: 10-Oct-2022

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