       Re: what's wrong with this?!!

• To: mathgroup at smc.vnet.net
• Subject: [mg114923] Re: what's wrong with this?!!
• Date: Thu, 23 Dec 2010 03:52:07 -0500 (EST)
• References: <732015.44033.qm@web26103.mail.ukl.yahoo.com> <FB9C30F3-35B3-4FE6-8D52-F38FE365817D@mimuw.edu.pl>

```The "Last modified" comment on documentation pages tells when
the documentation page, more precisely, the basic design part

The basic design of Reduce has not changed since Version 5.
New solving methods and bug fixes have been added to Reduce
in every version, and new documentation examples have been
added since version 5 as well.

The V5 output

In:= Reduce[
Element[{iP, N}, Integers] && Not[(iP < N)] &&
Exists[{iPP}, (iPP < N) && iP == iPP + 1],
{iP}, Backsubstitution -> True]

Out= (N | iP) \[Element] Integers && N <= iP < 1 + N

describes the correct solution set. It is not the simplest
possible description, since for integers N and iP,
N <= iP < 1 + N is equivalent to iP == N.
The result has been improved in V6.

Out= (N | iP) \[Element] Integers && iP == N

Best regards,

Wolfram Research

Andrzej Kozlowski wrote:
> First, the information that Reduce has not been modified since version 5 (which indeed is what the documentation for Reduce suggests) is clearly false. In fact Reduce can do things in version 7 that it could not do in earlier ones, e.g
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> x == 0
>
> \$Version
>
> "8.0 for Mac OS X x86 (64-bit) (November 6, 2010)"
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> During evaluation of In:= Reduce::nsmet:This system cannot be \
> solved with the methods available to Reduce. >>
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> \$Version
>
> "6.0 for Mac OS X x86 (64-bit) (May 21, 2008)"
>
> So Reduce has certainly been "modified", at least between versions 6 and 7. In addition there have been bug fixes (which I presume does not count as "modification") and some functions called up by Reduce have been modified (which again perhaps does not count as a modification of Reduce).
>
> However, in this case Mathematica 6.03 gives the same answer as Mathematica 8. I can't check what happens with Mathematica 5 as I no longer have it installed.
>
> I am curious about the examples you claim work with Mathematica 5 but not with 8. Perhaps they only appear to work in 5 (or appear not to work in 8)?
>
> Andrzej Kozlowski
>
>
>
> On 22 Dec 2010, at 11:18, olfa mraihi wrote:
>
>> So what's wrong with Reduce in Mathematica 5 to get the output Element[n | iP, Integers] && n<= iP < 1 + n?
>>
>> knowing that Reduce has not been modified in Mathematica 8 so why does it give the correct output which is Element[n | iP, Integers] && iP == n?
>> I 'm using Mathematica 5 and I'm constrained to use it because some of my examples work on Mathematica 5 but not on Mathematica 8.
>>
>> Thank you.
>>
>> --- En date de : Mer 22.12.10, Andrzej Kozlowski <akoz at mimuw.edu.pl> a écrit :
>>
>> De: Andrzej Kozlowski <akoz at mimuw.edu.pl>
>> Objet: Re: [mg114909] what's wrong with this?!!
>> À: "olfa" <olfa.mraihi at yahoo.fr>
>> Cc: mathgroup at smc.vnet.net
>> Date: Mercredi 22 décembre 2010, 9h48
>>
>> The output I get (with Mathematica 8) is:
>>
>> In:= Reduce[Element[{iP, n}, Integers] &&  !iP < n &&
>>    Exists[{iPP}, iPP < n && iP == iPP + 1], {iP},
>>   Backsubstitution -> True]
>>
>> Element[n | iP, Integers] && iP == n
>>
>> Andrzej Kozlowski
>>
>> On 22 Dec 2010, at 08:36, olfa wrote:
>>
>>> Hi mathematica community,
>>>
>>> this is the system to solve:
>>> Reduce[
>>> Element[{iP, N}, Integers] &&
>>> Not[(iP < N)] &&
>>> Exists[{iPP}, (iPP < N) && iP == iPP + 1],
>>> {iP}, Backsubstitution -> True]
>>>
>>> the output is:
>>> (N | iP) =E2=88=88 Integers && N =E2=89=A4 iP < 1 + N
>>>
>>> but it should be:
>>> iP==N
>>>
>>> so what's the problem what is missing to get the correct output?
>>>
>>> thank you very much.
>>>
>>
>
>

```

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