Question

prove that root3- root2 is an irrational number.​

Answer 1

Step-by-step explanation:

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.