Question

prove that √2+√3 is an irrational​

Answer 1

Let's assume that $\sf {{\sqrt{2}} + {\sqrt{3}}}$ is rational .

Then there exist co prime integers p , q

q ≠ 0 such that

$\sf {{\sqrt{2}} + {\sqrt{3}} = {\dfrac{p}{q}}}$

⇒$\sf{{\dfrac{p}{q} - {\sqrt{3}} = {\sqrt{2}}}}$

Squaring on both sides , we get

⇒ $\sf{({\dfrac{p}{q}} - {\sqrt{3}}})^2 = ({\sqrt{2}})^2$

⇒$\sf{{\dfrac{p^2}{q^2}} -2{\dfrac{p}{q}}{\sqrt{3}} + ({\sqrt{3}})^2 = 2 }$

⇒ $\sf{{\dfrac{p^2}{q^2}} - 2{\dfrac{p}{q}}{\sqrt{3}} + 3 = 2 }$

⇒ $\sf{{\dfrac{p^2}{q^2}} + 1 = 2 {\dfrac{p}{q}}{\sqrt{3}}}$

⇒ $\sf{{\dfrac{p^2+q^2}{q^2}} \times{\dfrac{q}{2p}} = {\sqrt{3}}}$

⇒ $\sf{{\dfrac{p^2+q^2}{2pq}} = {\sqrt{3}}}$

Since p , q are integers , $\sf{{\dfrac{p^2+q^2}{2pq}} = {\sqrt{3}}}$ is a rational number ⇒$\sf{{\sqrt{3}}}$ is a rational number .

This contradicts the fact that $\sf{{\sqrt{3}}}$ is irrational .

Thus , our assumption is incorrect .

So , $\sf {{\sqrt{2}} + {\sqrt{3}}}$ is irrational .