Question

Prove that √3+ √7 is an irrational number.​

Answer 1

Solution :-

Proof :

Let $\sqrt{3} + \sqrt{7}$ is a rational no.

So,

$\sqrt{3} + \sqrt{7}$ ${= \dfrac{p}{q}}$

(Where p , q € integer , q ≠ 0 , p,q are co-prime) ---- important line ☺️

Now, Squaring both side

$(\sqrt{3} + \sqrt{7})^2$ ${= (\dfrac{p}{q}})^2$

${3 + 7 + 2\sqrt{3}\sqrt{7} = \dfrac{p^2}{q^2}}$

${10 + 2\sqrt{21} = \dfrac{p^2}{q^2}}$

${2\sqrt{21} = \dfrac{p^2}{q^2} - 10}$

Since, Irrational ≠ Rational

This contradict that our assumption is wrong.

Hence, $\sqrt{3} + \sqrt{7}$ is an irrational number.