Question

$prove \: that\sqrt{2} \: is \: irrational$
​

Answer 1

Step-by-step explanation:

Let us assume on the contrary that √2 is a rational number. Then, there exist positive integers a and b such that

$\sqrt{2} = \frac{a}{b}$

where, a and b, are co-prime i.e. their HCF is 1

$({ \sqrt{2} })^{2} = ({ \frac{a}{b} })^{2}$

$2 = \frac{ {a}^{2} }{{b}^{2} }$

$2 {b}^{2} = {a}^{2}$

$here \: 2 \: is \: factor \: of \: {a}^{2} \: because \: {a}^{2} = 2 {b}^{2}$

$so \: 2 \: \: is \: a \: factor \: f \: (a) \: also \\ a = 2c \: for \: some \: integer \: c$

$2 {b}^{2} = 4 {c}^{2} \: \: (coz \: 2 {b}^{2} = {a}^{2} )$

${b}^{2} = 2 {c}^{2} \\ here \: we \: obtain \: that \: 2 \: is \: factor \: of \: \\ b \: also$

So, we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no comon factor other than 1. This means that our supposition is wrong.

Hence,

√2 is an irrational number.

*PLEASE MARK AS BRAINLIEST*

Answer 2

Check out a channel called kpop staniac

Support this channel

Like , subscribe, share

pls help, my friend has been working on tht edit for days. pls share....