• To: mathgroup at smc.vnet.net
• Subject: [mg117707] Re: About C[i]
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Wed, 30 Mar 2011 04:10:59 -0500 (EST)

```s1 = ((C[1] | C[2]) \[Element] Integers && i == C[2] && k == 7 C[1] &&
kP == 0 &&
iP == C[1] + C[2]) || ((C[1] | C[2]) \[Element] Integers &&
i == C[2] && k == 2 + 7 C[1] && kP == 2 && iP == C[1] + C[2]);
s2 = s1 /. C[2] -> i /. Thread[C[1] -> {k/7, (k - 2)/7}];
s3 = DeleteCases[s2, Element[_, _], Infinity]

(kP == 0 && iP == i + k/7) || (k == 2 + k && kP == 2 && iP == i + k/7)

The second alternative is impossible, however, since k == 2+k is
impossible.

Hence, C[1] == k/7 doesn't fit the solution.

Bobby

On Tue, 29 Mar 2011 06:57:39 -0500, olfa <olfa.mraihi at yahoo.fr> wrote:

>
> Hi Mathematica community,
>
> I have this solution set:
> ((C[1] | C[2]) \[Element] Integers && i == C[2] && k == 7 C[1] &&
>    kP == 0 && iP == C[1] + C[2]) ||((C[1] | C[2]) \[Element] Integers
> &&
>  i == C[2] && k == 2 + 7 C[1] &&   kP == 2 && iP == C[1] + C[2])
>
> it means that it is an infinite solution set.
> I need to expess it without C[1] and C[2] since C[1]==k/7||(k-2)/7
> and C[2]==i . consequently the solution set would be expressed as
> (kP==0 && iP==k/7+i)||(kP==2&&iP==(k-2)/7+i)
>  is it possible? how?can we say that is now transformed into finite
> set?
>
> thank you very much for your help.
>

--
DrMajorBob at yahoo.com

```

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