On Apr 10, 3:56 am, "[Mr.] Lynn Kurtz" <ku...@[EMAIL PROTECTED]
> wrote:
> On Wed, 9 Apr 2008 07:19:59 -0700 (PDT), "yongtze....@[EMAIL PROTECTED]
"
>
> <yongtze....@[EMAIL PROTECTED]
> wrote:
> >I'm having trouble with thisproof:
>
> >if n is an integer, and if log_2_n is rational,proof that log_2_n
> >must be integer.
>
> >Can anyone help? the base of the log is 2.
>
> Suppose log_2(n) = p/q where p, q integers and the fraction is reduced
> to lowest terms.
>
> Then 2^(p/q) = n so 2^p = n^q
>
> What can n have for factors? Can you get a contradictionfrom this?
>
> --Lynn
from 2^p = n^q, we can see that for the since all p, q, and n are
integer: n must be an even integer and it must be in the from of (2^k)
where k is an integer.
therefore 2^p = (2^k)^q => 2^p = 2^(kq)
so p = kq => p/q = k.
but from our assumption that p/q has been reduced to lowest terms so k
can't be an integer. <- here is the contradiction.
is this right?
Thank Lynn.
|
