Question

Prove that √2+2√5 is irrational.​

Answer 1

Heya friend...

here is your answer!

$let \: \: \sqrt{2} + 2 \sqrt{5} = p \div q$

by squaring both sides,

$(\sqrt{2} + 2 \sqrt{5}) ^{2} = (p \div q) ^{2}$

$( \sqrt{2}) {}^{2} + 2( \sqrt{2})(2 \sqrt{5} + (2 \sqrt{5}) {}^{2} \\ = p {}^{2} \div q {}^{2}$

$2 + 4 \sqrt{10} + 20 = p {}^{2} \div q {}^{2}$

$22 + 4 \sqrt{10} = p {}^{2} \div q {}^{2}$

$4 \sqrt{10} = (p {}^{2} \div q {}^{2}) - 22$

$\sqrt{10} = p {}^{2} - 22q {}^{2} \div 4q {}^{2}$

p and q are integers then,

$p {}^{2} - 22q {}^{2} \div q {}^{2}$

is a rational number.

Then,

$\sqrt{10}$

is also a rational number.

But this contradicts the fact that

$\sqrt{10}$

is an irrational number.

Therefore, our supposition is wrong.

so,

$\sqrt{2} + 2 \sqrt{5}$

is an irrational number.