Question

Prove that 2√3+√5 is irrational number

Answer 1

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Answer 2

Answer:

2√3+√5 is irrational.

Step-by-step explanation:

Let us suppose that 2√3+√5 is rational.

$Let\: 2\sqrt{3}+\sqrt{5}=\frac{a}{b},\\where\:a,b\:are\: integers\\ \:and\:b≠0$

$Therefore, \:2\sqrt{3}=\frac{a}{b}-\sqrt{5}.$

Squaring on both sides, we get

$12=\frac{a^{2}}{b^{2}}+5-2\times \frac{a}{b}\times \sqrt{5}$

Rearranging

$\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}+5-12$

$\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}-7$

$\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}$

$\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}\times \frac{b}{2a}$

After cancellation, we get

$\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{2ab}$

Since , a ,b are integers , $</p><p>\frac{a^{2}-7b^{2}}{2ab}$

is rational,and so √5 is rational.

This contradicts the fact that √5 is irrational.

Hence , 2√3+√5 is irrational.

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