Question

How to prove that root 3 plus root 7 is irrational?

Answer 1

ur answer
plz mark brainliest

Answer 2

Answer:

√3+√7 is irrational.

Step-by-step explanation:

Let us assume that √3+√7 is rational.

That is , we can find coprimes a and b (b≠0) such that $\sqrt{3}+\sqrt{7}=\sqrt{a}{b}$

Therefore,

$\sqrt{7}=\frac{a}{b}-\sqrt{3}$

Squaring on both sides ,we get

$7=\frac{a^{2}}{b^{2}}+3-2\times \frac{a}{b}\times \sqrt{3}$

Rearranging the terms ,

$2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}}{b^{2}}+3-7\\=\frac{a^{2}}{b^{2}}-4$

$\implies 2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}$

$\implies \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}\times \frac{b}{2a}\\=\frac{a^{2}-4b^{2}}{2ab}$

Since, a and b are integers , $\frac{(a^{2}-4b^{2})}{2ab}$ is rational ,and so √3 also rational.

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that √3+√7 is rational.

Hence, √3+√7 is irrational.

•••♪