Question

prove that root2+root7 is irrational.........its urgents plzz answer it fast i will mark as brainlist​

Answer 1

Hey fam!

To prove:- (√2+√7) is irrational

Proof:-

First of all, If possible,

Let, (√2+√7) is rational.

Then, it can be written in the form of a/b where a and b are integers and b ≠0.

So,

(√2+√7)=a/b

Squaring both sides we get,

(√2+√7)^2 =a^2/b^2

=> 2+7+2√14 = $\frac{ {a}^{2} }{ {b}^{2} }$

=>9 + 2√14 = a^2/b^2

=>2√14 = $\frac{ {a}^{2} }{ {b}^{2} } - 9$

=>√14 = $(\frac{ {a}^{2} }{ {b}^{2} } - 9) \frac{1}{2}$

Here, a and b are rational, 1/2 is rational and 9 is rational.

So,

$(\frac{ {a}^{2} }{ {b}^{2} } - 9) \frac{1}{2}$is also, rational and √14 is also rational.

But,

This is contradiction because √14 is irrational. Therefore Our assumption was wrong.

So,

(√2+√7) is irrational.

-Hence proved-

Cheers!

Answer 2

Sia 91 has written the correct answer you can write it