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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.