Question

Prove that √2 + √5 is irrational number​

Answer 1

Answer:

A rational number can be written in the form of p/q. p,q are integers then (p²-7q²)/2q is a rational number. ... Therefore, √2+√5 is an irrational number. Hence proved.

Answer 2

$\huge\mathbb\green{♡〖Qบęຮτïøň〗 ♡}$

Prove that √2 + √5 is irrational number

$\\\huge{ \mathbb⫷❥\red{ᴀ᭄} \green{ñ}\mathbb\blue{§} \purple{₩} \mathbb \orange{Σ}\pink{ℝ}⫸}\:$

Given: √2+√5

We need to prove√2+√5 is an irrational number.

Proof

Let us assume that √2+√5 is a rational number.

A rational number can be written in the form of p/q where p,q are integers and q≠0

√2+√5 = p/q

On squaring both sides we get,

(√2+√5)² = (p/q)²

√2²+√5²+2(√5)(√2) = p²/q²

2+5+2√10 = p²/q²

7+2√10 = p²/q²

2√10 = p²/q² – 7

√10 = (p²-7q²)/2q

p,q are integers then (p²-7q²)/2q is a rational number.

Then √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

Our assumption is incorrect

√2+√5 is an irrational number.

Hence proved.

$\huge{\red{ᎻᎧᎮᏋ\: ᶦƬ \:Helpร}}$

$\huge\boxed{\fcolorbox{black}{pink}{25Tђαήks=iภbøאָ}}$

$\\\huge\underline{ \mathbb\blue{קleสseﾂ}\:\green{thคήks}\:\orange{⦇էօ⦈}}\:$

$\huge\underline{ \mathbb\red{爪ℽ}\:\purple{❺\:Anຮwerຮ}}$

$\huge\underline{ \mathbb\red{⪨OŇLY⪩}}\:$

$\\\\\huge\underline{ \red{ƑØℓℓõw}\:\blue{ᙢě}}\:$