For every integer $x$ the equation Mod[x, 1] == 0 holds. While
Simplify[Mod[x, 1] == 0, Element[x,Integers]] gives True,
Reduce[Mod[x, 1] == 0, x, Integers] gives False. Why?
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3 
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1 Answer
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3 Reduce works fine for a slightly more sophisticated expression, e.g. :
Reduce[ ForAll[ x, x ∈ Integers, Mod[ x, 1] == 0], x] True
however there is a bug in Solve :
Solve[ Mod[x, 1] == 0, x, Integers] {}
therefore it is not surprising we have an analogical issue in Reduce :
Reduce[ Mod[x, 1] == 0, x, Integers] False
Seemingly there has not been much clamor therefore it has not been a high priority to improve it.
One can work around these problems :
Reduce[ Mod[ a x, a] == 0 && a == 1, x, Integers] C[1] ∈ Integers && a == 1 && x == C[1]
or simply
Reduce[ Mod[ x, 1] == a, x, Integers] C[1] ∈ Integers && a == 0 && x == C[1]
Solve[ Mod[ x, 1] == a, x, Integers] {{x -> ConditionalExpression[C[1], C[1] ∈ Integers && a == 0]}}
Edit
The above problems with Reduce and Solve were found in Mathematica 8. Before there had been :
ver. 7
Solve[ Mod[x, 1] == 0, x, Integers] Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >> {{x -> InverseFunction[Mod, 1, 2][0, 1]}}
Reduce[ Mod[x, 1] == 0, x, Integers] False
Now these bugs have been fixed :
ver. 9
Solve[ Mod[x, 1] == 0, x, Integers] {{x -> ConditionalExpression[C[1], C[1] ∈ Integers]}}
Reduce[ Mod[x, 1] == 0, x, Integers] C[1] ∈ Integers && x == C[1]
- 2$\begingroup$ Congrats on 10k! $\endgroup$rcollyer– rcollyer2012-08-17 04:23:34 +00:00Commented Aug 17, 2012 at 4:23
- 1$\begingroup$ Worth noting that this issue has been fixed in Mathematica 9, where
Solvenow gives the correct result. $\endgroup$Oleksandr R.– Oleksandr R.2012-12-09 18:00:14 +00:00Commented Dec 9, 2012 at 18:00 - $\begingroup$ @OleksandrR. Thanks, updated. $\endgroup$Artes– Artes2012-12-10 02:18:28 +00:00Commented Dec 10, 2012 at 2:18