Question

Show that \(\log_23\) is irrational.
Proof:
Let \(\log_23\) be a rational number. Then \(\log_23=a/b\), where \((a,b)=1\) with \(b\neq0\). It follows that \(2^{a/b}=3\)...

I get stuck there. How could I proceed?

Answer 1

Perhaps:

... \(2^a=3^b\)?

Answer 2

I see how I can reach a contradiction now.

Answer 3

Yeah, and \(2^a = 3^b, b \neq 0\) isn't possible

Answer 4

So, by contradiction \(\log_23\) is rational

Answer 5

irrational*