# Prove that $\sqrt{p} + \sqrt{q}$ is irrational, where $p$ and $q$ are primes.

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Given:

$p$ and $q$ are primes.

To do:

We have to provethat  $\sqrt{p}\ +\ \sqrt{q}$  is an irrational number.

Solution:

Let us assume, to the contrary, that  $\sqrt{p}\ +\ \sqrt{q}$  is rational.

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

Where $a$ and $b$ are co-prime.

Now,

$\sqrt{p}\ +\ \sqrt{q}\ =\ \frac{a}{b}$

$\sqrt{p}\ =\ \frac{a}{b}\ -\ \sqrt{q}$

Squaring both sides:

$(\sqrt{p})^2\ =\ (\frac{a}{b}\ -\ \sqrt{q})^2$

$p\ =\ (\frac{a}{b})^2\ +\ (\sqrt{q})^2\ -\ 2(\frac{a}{b})(\sqrt{q})$

$p\ =\ \frac{a^2}{b^2}\ +\ q\ -\ 2(\frac{a}{b})(\sqrt{q})$

$2(\frac{a}{b})(\sqrt{q})\ =\ \frac{a^2}{b^2}\ +\ q\ -\ p$

$\sqrt{q}\ =\ (\frac{b}{2a})(\frac{a^2}{b^2}\ +\ q\ -\ p)$

Here,  $(\frac{b}{2a})(\frac{a^2}{b^2}\ +\ q\ -\ p)$  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  $\sqrt{p}\ +\ \sqrt{q}$  is rational.

So, this proves that  $\sqrt{p}\ +\ \sqrt{q}$  is an irrational number.

Updated on 10-Oct-2022 13:27:08