Question

prove that root 2+root 7 is irrational ​

Answer 1

√2+√7=0

√2=-√7

we know that√2 is irrational

So √2+√7 is irrational

Answer 2

Let us assume that √2+√7 is not an irrational

√2+√7 becomes rational

√2+√7 = p/q where p,q belongs to integers and q ≠ 0

Squaring on both sides

(√2+√7)2 = (p/q)2

(√2)2 + (√7)2 + 2 × √2 × √7 = p2 / q2

2+7+2√14 = p2/q2

9+2√14 = p2 / q2

2√14 = p2 / q2 - 9

2√14 = p2 - 9q2 / q2

√14 = p2 - 9q2 / 2q2

LHS :- √14 is a irrational because '14' is not a perfect square .

RHS :- p2 - 9q2 / 2q2 becomes rational where p,q are integers and q ≠ 0

But LHS ≠ RHS

It is contradiction to our assumption

Therefore Our assumption is wrong

√2 + √7 is an irrational .