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Re: Does Mathematica think Sqrt[2] is rational?


Ops..
I have overlooked the trivial solution x=0 and y=0 in the logical
equation.

If we consider

Reduce[ForAll[{x, y}, x != 0 && Element[x | y, Integers], x^2 != 2
y^2]]

as suggested by Andrzej, Mathematica answers correctly.

Oh well.. the pythagoreans did not come back after all.

So Mathematica correctly realizes that Sqrt[2] is irrational.


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