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