Does Mathematica think Sqrt[2] is rational?

*To*: mathgroup at smc.vnet.net*Subject*: [mg86668] Does Mathematica think Sqrt[2] is rational?*From*: magma <maderri2 at gmail.com>*Date*: Mon, 17 Mar 2008 00:22:31 -0500 (EST)

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] Finally, if we try the inequality without the quantifiers, the results are not very illuminating with the Reals, but still correct Reduce[x^2 != 2 y^2, {x, y}, Reals] While with Integers Mathematica misses the point entirely Reduce[x^2 != 2 y^2, {x, y}, Integers] Conclusion : Mathematica does not seem to realize that Sqrt[2] is irrational. Am I missing something or WRI has been infiltrated by Pythagoreans :-) ?

**Follow-Ups**:**Re: Does Mathematica think Sqrt[2] is rational?***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>