# Prove that $3 + 2\sqrt5$ is irrational.

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Given:

$3\ +\ 2\sqrt{5}$

To do:

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

Solution:

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

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

Where a and b are co-prime.

Now,

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

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

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

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

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

But, irrational number $≠$ Rational number.

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

So, this proves that $3\ +\ 2\sqrt{5}$ is an irrational number.
Updated on 10-Oct-2022 13:19:30