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For Pythagoras' theorem, a^2+b^2=c^2, prove that a+b>c

Guest Jun 2, 2017
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Let us first assume that  a + b  = c

 

Then

 

a^2   +  b^2  =  c^2

a^2  + b^2   =  ( a + b)^2

a^2  + b^2  =  a^2 + 2ab + b^2

0  =  2ab      which is impossible  since  a , b   are positive 

 

Next, assume that  a +  b  < c

Then.....there must be some positive n  such that  a + b  + n  =  c

So

a^2 + b^2  = c^2

a^2 + b^2   = (a + b + n)^2

a^2  + b^2  = a^2 + 2ab + 2an + b^2 + 2bn + n^2

0  =  2ab + 2an + 2bn + n^2      which is also impossible since a,b and n are positive

 

So....

a + b  = c   is false

a + b < c  is false

 

And the only thing left is that   a + b > c

 

 

cool cool cool

CPhill  Jun 2, 2017
edited by CPhill  Jun 2, 2017

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