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

Given: $5\ +\ 3\sqrt{2}$

To do: Here we have to prove that $5\ +\ 3\sqrt{2}$ is an irrational number.

Solution:

Let us assume, to the contrary, that $5\ +\ 3\sqrt{2}$ is rational.

So, we can find integers a and b ($≠$ 0) such that  $5\ +\ 3\sqrt{2}\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

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

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

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

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

Here, $\frac{a\ -\ 5b}{3b}$ is a rational number but $\sqrt{2}$ is irrational number. 

But, Irrational number  $≠$  Rational number.

This contradiction has arisen because of our incorrect assumption that $5\ +\ 3\sqrt{2}$ is rational.

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Updated on: 10-Oct-2022

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