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

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.

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