# Thread: Proving (URGENT)

1. ## Proving (URGENT)

Prove that if n is an element of Z(the integers) and log2n is rational, then log2n is an integer.

2. Consider this:
$log_2 (n)^k = k log_2 (n)$

When n = 2, $log_2 (2) = 1$

If n is not a power of 2, then
$log_2 n = \frac{ln (n)}{ln (2)}$

is ln 2 rational?

3. Originally Posted by vincentngtf
Suppose that $\log_2 n = p/q$ where $p/q$ is a positive rational number. This means $2^{p/q} = n\implies \left( 2^{p/q} \right)^q = n^q\implies 2^p = n^q$. Now the fundamental theorem of arithmetic allows us to factorize $n$, the thing is that all factors much be $2$'s because the LHS is made out of two, thus, $n=2^m$. This means, $2^p = 2^{mq}$. Thus, $p=mq$ which means $q|p$ which tells us that $p/q$ is an integer.