Question

prove √2+√5 is irrational(class 10 )​

Answer 1

Your answer is following below-:::

Let √2+√5 be a rational number.

A rational number can be written in the form of p/q.

√2+√5=p/q

Squaring on both sides,

(√2+√5)²=(p/q)²

[√2²+√5²+2(√2)(√5)]=p²/q²

(2+5+2√10)=p²/q²

2√10+7=p²/q²

2√10=p²/q²-7

2√10=(p²-7q²)/q²

√10=(p²-7q²)/2q²

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.

So,our supposition is false.

Therefore, √2+√5 is an irrational number.

Hence proved.

Answer 2

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

Then their exact co-prime positive integers a and b such that

$\sqrt{2} + \sqrt{5} = \frac{a}{b} \\ \frac{a}{b} - \sqrt{2} = \sqrt{5} \\squaring \: on \: both \: sides \\ (\frac{a}{b} - \sqrt{2} ) ^{2} = (\sqrt{5} ) ^{2} \\ \frac{a^{2} }{ {b}^{2} } - \frac{2a}{b} \sqrt{2} + 2 = 5 \\ \frac{ {a}^{2} }{ {b}^{2} } - 3 = \frac{2a}{b} \sqrt{2} \\ \frac{ {a}^{2} - 3 {b}^{2} }{2ab} = \sqrt{2} \\ a \: and \: b \: are \: integers \\ then \: \frac{ {a}^{2} - 3 {b}^{2} }{2ab} \: is \: a \: rational \\ therefore \: \sqrt{2} \: is \: a \: rational \: number.$

This contradict the fact that √2 is irrational. So our assumption is wrong.

Hence,√2+√5 is irrational