Question

show that root2÷root3 is irrational
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Answer 1

Answer:

Answer Expert Verified

Since , a,b are integers , is rational, and so √3 is rational. This contradicts the fact √3 is irrational. Hence, √2+√3 is irrational.

Answer 2

Answer:

Solution:-

$\sf \: Let \: us \: suppose \: that \: √2+√3 \: is \: rational.$

$\sf \: Let √2+√3=\frac{a}{b}$

$\sf \: where \: a,b \: are \: integers \: and \: b≠0$

Therefore,

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

On Squaring both sides , we get

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

Rearranging the terms ,

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

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

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

Since , a,b are integers ,

$\pink{\huge{ = \sf\frac{a^{2}+b^{2}}{2ab}}}$

is rational, and so √3 is rational.

This contradicts the fact √3 is irrational.

Hence, √2+√3 is irrational.