proof sqrt 2​ is irrational

Let us suppose that there exists  integers a, b such that

√2  =   a / b      where a/b   is reduced to lowest terms  [a and b have no factors in  common ]

So

b√2 = a           square both  sides

2b^2  = a^2

But a^2  must be even......so  ......"a" must be even......

Then   a =  2m    ....so a^2  = 4m...so......

2b^2 = (2m)^2

2b^2 =  4m^2        divide both sides by 2

b^2 = 2m^2      then b must also be even

But....we assumed that a, b had no factors in common.....so.....they can't both be even

So....our aqssumption that

√2 = a / b  must be false, since it leads to a false conclusion