- Thread starter
- #1

Attempt:

Proof by contradiction: Assume $\log_p(q)$ is rational.

Suppose $\log_p(q) = \dfrac{m}{n}$ where $m,n \in \mathbb{Z}$ and $\gcd(m,n) = 1$.

Then, $p^{\frac{m}{n}} = q$ which implies $p^m = q^n$.

- Thread starter KOO
- Start date

- Thread starter
- #1

Attempt:

Proof by contradiction: Assume $\log_p(q)$ is rational.

Suppose $\log_p(q) = \dfrac{m}{n}$ where $m,n \in \mathbb{Z}$ and $\gcd(m,n) = 1$.

Then, $p^{\frac{m}{n}} = q$ which implies $p^m = q^n$.

- Admin
- #2

- Jan 26, 2012

- 4,202

Almost there! Can $p^m=q^n$ happen for any two distinct primes?

Attempt:

Proof by contradiction: Assume $\log_p(q)$ is rational.

Suppose $\log_p(q) = \dfrac{m}{n}$ where $m,n \in \mathbb{Z}$ and $\gcd(m,n) = 1$.

Then, $p^{\frac{m}{n}} = q$ which implies $p^m = q^n$.