Prove that $\sqrt2 + \sqrt3$ is irrational.

Given: $\sqrt2\ +\ \sqrt3$

To do: Here we have to prove that $\sqrt2\ +\ \sqrt3$ is an irrational number.

Solution:

Let us assume, to the contrary, that $\sqrt2\ +\ \sqrt3$ is rational.

So, we can find integers a and b ($≠$ 0) such that  $\sqrt2\ +\ \sqrt3\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$\sqrt2\ +\ \sqrt3\ =\ \frac{a}{b}$

$\sqrt3\ =\ \frac{a}{b}\ -\ \sqrt2$

Squaring both sides:

$(\sqrt{3} )^{2} \ =\ \left(\frac{a}{b} \ -\ \sqrt{2}\right)^{2}$

$3\ =\ \left(\frac{a}{b}\right)^{2} \ +\ 2\ -\ 2\sqrt{2}\left(\frac{a}{b}\right)$

$3\ =\ \frac{a^{2}}{b^{2}} \ +\ 2\ -\ 2\sqrt{2}\left(\frac{a}{b}\right)$

$2\sqrt{2}\left(\frac{a}{b}\right) \ =\ \frac{a^{2}}{b^{2}} \ +\ 2\ -\ 3$

$2\sqrt{2}\left(\frac{a}{b}\right) \ =\ \frac{a^{2}}{b^{2}} \ -\ 1$

$2\sqrt{2}\left(\frac{a}{b}\right) \ =\ \frac{a^{2} \ -\ b^{2}}{b^{2}}$

$\sqrt{2} \ =\ \frac{a^{2} \ -\ b^{2}}{b^{2}} \ \times \ \frac{b}{2a}$

$\sqrt{2} \ =\ \frac{a^{2} \ -\ b^{2}}{2ab}$

Here, $\frac{a^{2} \ -\ b^{2}}{2ab}$ 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 $\sqrt2\ +\ \sqrt3$ is rational.

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Updated on: 10-Oct-2022

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