Show that $2 − \sqrt{3}$ is an irrational number.

Given: $2\ −\ \sqrt{3}$

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

Solution:

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

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

Where a and b are co-prime.

Now,

$2\ −\ \sqrt{3}\ =\ \frac{a}{b}$

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

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

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

But, Rational number  $≠$  Irrational number.

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

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Updated on: 10-Oct-2022

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