       • To: mathgroup at smc.vnet.net
• Subject: [mg117690] Re: About C[i]
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 30 Mar 2011 04:07:50 -0500 (EST)

```olfa wrote:
> Hi Mathematica community,
>
> I have this solution set:
> ((C | C) \[Element] Integers && i == C && k == 7 C &&
>    kP == 0 && iP == C + C) ||((C | C) \[Element] Integers
> &&
>  i == C && k == 2 + 7 C &&   kP == 2 && iP == C + C)
>
> it means that it is an infinite solution set.
> I need to expess it without C and C since C==k/7||(k-2)/7
> and C==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.

In:= Eliminate[
Reduce[(i == C && k == 7 C && kP == 0 &&
iP == C + C) || (i == C && k == 2 + 7 C && kP == 2 &&
iP == C + C), {i, k, kP, iP, C, C}], {C, C}]

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

To answer the last question, no, it does not look like a finite set.
Both k and i can take on infinitely many values, and they parametrize
the solution set.

Daniel Lichtblau
Wolfram Research

```

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