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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg86675] Re: Does Mathematica think Sqrt[2] is rational?
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
  • Date: Tue, 18 Mar 2008 04:48:46 -0500 (EST)
  • References: <frkv66$109$1@smc.vnet.net>

magma <maderri2 at gmail.com> wrote:
> This might be already well known, but...
>
> Consider the following expression
>
> expr = ForAll[{x, y}, x^2 != 2 y^2]
>
> If we test it over the Reals, it should be false, since x=Sqrt[2] and
> y=1 falsify it.
> Mathematica agrees with us
>
> Reduce[expr, {x, y}, Reals]
>
> However if we test it over the Integers or the Rationals it should be
> true, as a demonstration that Sqrt[2] is not rational.
> Mathematica , unfortunately,  does not agree with us
>
> Reduce[expr, {x, y}, Integers]
>
> Reduce[expr, {x, y}, Rationals]
>
> So it appears that Mathematica thinks that there are 2 integers x, y
> such that x^2 = 2 y^2.
> Let's consider the negation of the previous expression
>
> expr2 = Exists[{x, y}, x^2 == 2 y^2]
>
> Mathematica correctly finds out that expr is true over the reals
>
> Reduce[expr2, {x, y}, Reals]
>
> But gives a wrong answer over the integers or the rationals
>
> Reduce[expr2, {x, y}, Integers]
>
> Reduce[expr2, {x, y}, Rationals]

> Conclusion : Mathematica does not seem to realize that Sqrt[2] is
> irrational.
>
> Am I missing something or WRI has been infiltrated by
> Pythagoreans :-)  ?

You seem to have missed the fact that {x,y} = {0,0} gives a solution of
x^2 == 2 y^2 which is both _integer_ and _rational_, as well as real.

David


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