Question

prove that under root 2+under root 5 is a irrational.​

Answer 1

Step-by-step explanation:

if u want √2 and √5 to prove use contradiction method.

Answer 2

Step-by-step explanation:

${\large{\mathfrak{\underline{\pink{Question :-}}}}}$

Prove that (√2 + √5 ) is irrational number

${\large{\mathfrak{\underline{\pink{Answer :-}}}}}$

Given :

• √2+√5

To prove :

• √2+√5 is an irrational number.

Proof :

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

As we know,

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

Therefore

${\bold{\sf{⟼ \: \: \: \: \: √2 \: + \: √5 \: = \: \frac{p}{q}}}}$

On squaring both sides we get,

${\bold{\sf{⟼ \: \: \: \: \: (√2 \: + \: √5)² \: = \: (\frac{p}{q})²}}}$

${\bold{\sf{⟼ \: \: \: \: \: √2² \: + \: √5² \: + \: 2(√5)(√2) \: = \: \frac{p²}{q²}}}}$

${\bold{\sf{⟼ \: \: \: \: \: 2 \: + \: 5 \: + \: 2√10 \: = \: \frac{p²}{q²}}}}$

${\bold{\sf{⟼ \: \: \: \: \: 7 \: + \: 2√10 \: = \: \frac{p²}{q²}}}}$

${\bold{\sf{⟼ \: \: \: \: \: 2√10 \: = \: \frac{p²}{q²} \: – \: 7}}}$

${\bold{\sf{⟼ \: \: \: \: \: \: √10 \: = \: \frac{(p² \: - \: 7q²)}{2q}}}}$

$\\ {\bold{\sf{\underline{⟼ \: \: \: \: \: p, \: q \: are \: integers \: then \: \frac{(p²-7q²)}{2q} \: is \: a \: rational \: number.}}}}$

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.

${\large{\mathfrak{\fbox{\orange{Hence \: proved}}}}}$