proof \(\sqrt{2}\) is irrational

 Jan 7, 2019

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 ]




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



cool cool cool

 Jan 7, 2019

thank you very much

YEEEEEET  Jan 7, 2019

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