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Topic: Power of 2 divisible by 3
Replies: 19   Last Post: Feb 1, 2006 10:37 PM

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 Robert Israel Posts: 11,902 Registered: 12/6/04
Re: Power of 2 divisible by 3
Posted: Feb 1, 2006 12:59 PM

Chip Eastham <hardmath@gmail.com> wrote:

>> Gerry Myerson
>> >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?

>To take a stab at Gerry's question, suppose f:Z-->R is a ring
>homomorphism, not necessarily onto, such that f(2^k) is divisible
>by f(3) in R. Since f(3) = f(2) + f(1), it is pretty inescapable that
>f(3) must also divide f(1) by the following argument:
>
>f(3) divides f(2^k)^2 = f(4)^k, and f(3) divides f(4^k - 1) =
>f(4^k)-f(1)
>
>So the only escape hole I can see that might prevent concluding
>that f(3) is a unit in R, and that f(3) thus already divides f(2) as
>we discovered above, would be a "cheat" that ring homomorphism
>doesn't require mapping the unit 1 of Z to the unit of R.

>I will leave it someone with a more freshly unreasonable
>mind to discover if such a "cheat" is actually feasible.

R need not have a unit element at all, but its subring f(Z)
does (namely f(1)).

For any x and y in Z such that gcd(x,y) = 1 we have a x + b y = 1
for some a and b in Z, and therefore a f(x) + b f(y) = f(1).
So if f(y) divides f(x), it also divides f(1), and therefore
divides f(z) for all z in Z.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada