Question

prove that root2 +root 3 is irrational

Answer 1

Refer the above attachment

Hope it helps you.

Regards,
Shobana

Answer 2

Solution:

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

Let √2+√3=$\frac{a}{b}$,

where a,b are integers and b≠0

Therefore,

$\sqrt{2}=\frac{a}{b}-\sqrt{3}$

On Squaring both sides , we get

$2=\frac{a^{2}}{b^{2}}+3-2\times\frac{a}{b}\times\sqrt{3}$

Rearranging the terms ,

$\frac{2a}{b}\times\sqrt{3}=\frac{a^{2}}{b^{2}}+3-2$

$= \frac{a^{2}}{b^{2}}+1$

$\sqrt{3}=\frac{a^{2}+b^{2}}{2ab}$

Since , a,b are integers ,

$\frac{a^{2}+b^{2}}{2ab}$ is rational, and so √3 is rational.

This contradicts the fact √3 is irrational.

Hence, √2+√3 is irrational.

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