# Prove that $\frac{2\ +\ \sqrt{3}}{5}$ is an irrational number, given that  $\sqrt{3}$ is an irrational number.

Given: $\frac{2\ +\ \sqrt{3}}{5}$

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

Solution:

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

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

Where a and b are co-prime.

Now,

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

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

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

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

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

But, Irrational number  $≠$  Rational number.

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

So, this proves that  $\frac{2\ +\ \sqrt{3}}{5}$  is an irrational number.

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Updated on: 10-Oct-2022

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