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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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    – Verbeia
    Aug 16, 2012 at 23:14
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    $\begingroup$ @NasserM.Abbasi I think this is because the documentation is usually written by the same person that coded the functionality. While an economical approach, it's not necessarily the best for expository value: the author will already have internalized the purpose and semantics of the code, so describing it to others in an unbiased way is that much harder for them. I don't blame the developers for that, of course--I'm a terrible example of this myself. $\endgroup$ Aug 17, 2012 at 1:03
  • $\begingroup$ @OleksandrR. I think you just described my papers :D $\endgroup$
    – rm -rf
    Aug 17, 2012 at 4:14

1 Answer 1

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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]
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    $\begingroup$ Congrats on 10k! $\endgroup$
    – rcollyer
    Aug 17, 2012 at 4:23
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    $\begingroup$ Worth noting that this issue has been fixed in Mathematica 9, where Solve now gives the correct result. $\endgroup$ Dec 9, 2012 at 18:00
  • $\begingroup$ @OleksandrR. Thanks, updated. $\endgroup$
    – Artes
    Dec 10, 2012 at 2:18

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