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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.