Question

prove that
$\sqrt{2}$
is a irrational number

Answer 1

$\sqrt{2}$
in decimal place sis non terminating and non recurring (repeating) so it is a irrational number

Answer 2

Hello Mate!

Let the root 2 be a rational number in form of p/q where p/q are in simplest form.

_/2 = p/q
2 = p^2 / q^2
2q^2 = p^2

Here p^2 is an even number
p is an even number

lets introduce a natural number m

2m = p
2^2m^2 = p^2
4m^2 = 2q^2
2m^2 = q^2

Here q^2 is an even number
q is an even number

So we get that that p and q both are even integer but it can not be possible as p/q was in its simplest form so we arrise at a contradiction that root 2 is irrational number.

Hope it helps☺!✌