Re: Power of 2 divisible by 3
In article <1138749336.761546.107350@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Chip Eastham" <hardmath@xxxxxxxxx> wrote:
> Shmuel (Seymour J.) Metz wrote:
> > In <1138647285.161615.119780@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, on
> > 01/30/2006
> > at 10:54 AM, XleonardXcobetX@xxxxxxxxx said:
> >
> > >I cannot seem to find a power of 2 divisible by 3.
> >
> > That's because you're working in Z, the ring of integers. Try it in,
> > e.g., Z/5, where 2^3 is 3 (8=3+5).
>
> In Z/5Z, already 2 is divisible by 3. No need to raise to a power...
OK, so is there any reasonable mathematical system where there is
an entity recognizable as a 2 that is not divisible by an entity
recognizable as a 3, but where some power of that 2 is divisible
by that 3?
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
