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avatar+838 

proof \(\sqrt{2}\) is irrational

 Jan 7, 2019
 #1
avatar+100516 
+3

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

 

 

cool cool cool

 Jan 7, 2019
 #2
avatar+838 
+1

thank you very much

YEEEEEET  Jan 7, 2019

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