Prove that the following number is irrational.
$\frac{3}{2\sqrt{5}}$

Given: $\frac{3}{2\sqrt{5}}$

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

Solution:

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

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

Where a and b are co-prime.

Now,

$\frac{3}{2\sqrt{5}}\ =\ \frac{a}{b}$

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

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

But, Rational number  $≠$  Irrational number.

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

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Updated on: 10-Oct-2022

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