Question

Prove that root2 +root3 is irrational

Answer 1

hope it will help you

Answer 2

$\huge\star\mathfrak\purple{{HEY MATE....!!}}$

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

PROOF:

let us assume that $\sqrt{2} +\sqrt{3}$ is not an irrational.

$\sqrt{2} +\sqrt{3}$ is rational

$\sqrt{2} +\sqrt{3}$ =$\frac{p}{q}$ where p,q is = z ,q is not equal to" 0".

squaring on both sides.

$(\sqrt{2} +\sqrt{3})^2$ = $(\frac{p}{q})^2$

$(\sqrt{2})^2 +(\sqrt{3})^2$ + 2 × $\sqrt{2}$ +$\sqrt{3}$ = $\frac{{p}^2}q}}^2$

5+2$\sqrt{6}$= $\frac{{p}^2}q}}^2$

$\sqrt{6}$ = $\frac{p^2-5{q}^2}{2{q}^2}$

LHS :

$\sqrt{6}$ is an irrational

because $\sqrt{6}$ is not a perfect square

RHS:

$\frac{p^2-5{q}^2}{2{q}^2}$

becomes rational where p,q is = Z and q is not equal to "0"

✴.LHS is not equal to RHS

✴.It is contradiction to our assumption.

✴.therefore our assumption is wrong

$\sqrt{2} +\sqrt{3}$ is not a rational

✴.It is an irrational.

$<marquee>#Be Brainly...✔ </marquee>$