by "[Mr.] Lynn Kurtz" <kurtz@[EMAIL PROTECTED]
>
Apr 9, 2008 at 05:56 PM
On Wed, 9 Apr 2008 07:19:59 -0700 (PDT), "yongtze.tan@[EMAIL PROTECTED]
"
<yongtze.tan@[EMAIL PROTECTED]
> wrote:
>I'm having trouble with this proof:
>
>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 contradiction from this?
--Lynn
