Prove that $\sqrt3 + \sqrt5$ is irrational.

Given: 

$\sqrt3\ +\ \sqrt5$

To do: 

We have to prove that $\sqrt3\ +\ \sqrt5$ is an irrational number.

Solution:

Let us assume, to the contrary, that $\sqrt3\ +\ \sqrt5$ is rational.

So, we can find integers a and b ($≠$ 0) such that  $\sqrt5\ +\ \sqrt3\ =\ \frac{a}{b}$.

Where a and b are co-prime.

Now,

$\sqrt5\ +\ \sqrt3\ =\ \frac{a}{b}$

$\sqrt3\ =\ \frac{a}{b}\ -\ \sqrt5$

Squaring both sides:

$(\sqrt{3} )^{2} \ =\ \left(\frac{a}{b} \ -\ \sqrt{5}\right)^{2}$

$3\ =\ \left(\frac{a}{b}\right)^{2} \ +\ 5\ -\ 2\sqrt{5}\left(\frac{a}{b}\right)$

$3\ =\ \frac{a^{2}}{b^{2}} \ +\ 5\ -\ 2\sqrt{5}\left(\frac{a}{b}\right)$

$2\sqrt{5}\left(\frac{a}{b}\right) \ =\ \frac{a^{2}}{b^{2}} \ +\ 5\ -\ 3$

$2\sqrt{5}\left(\frac{a}{b}\right) \ =\ \frac{a^{2}}{b^{2}} \ +\ 2$

$2\sqrt{5}\left(\frac{a}{b}\right) \ =\ \frac{a^{2} \ +\ 2b^{2}}{b^{2}}$

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

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

Here, $\frac{a^{2} \ +\ 2b^{2}}{2ab}$ 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 $\sqrt3\ +\ \sqrt5$ is rational.

So, this proves that $\sqrt3\ +\ \sqrt5$ is an irrational number.

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Updated on: 10-Oct-2022

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