21
$\begingroup$

Is it true that $\log_2 13$ is irrational?

Let $x=\log_2 13\implies 2^x=13$.

So, it will be an irrational number, if not,$$x=\frac p q$$

and $$2^{\frac p q}=13$$

$$\implies 2^p=13^{q}$$

Since, $13$ is a prime number, $2^p$ divides $13^q$.

So, $2$ divides $13$, which is absurd.

Is this reason worthy? Can you give some other proofs for this?

$\endgroup$
  • 13
    $\begingroup$ That's a nice and elegant proof. You can extend it to $$\log_a b \notin \mathbb Q \qquad \forall a,b\in\mathbb N\text{ s.t. } \gcd(a,b) = 1$$ $\endgroup$ – AlexR Jan 16 '15 at 14:22
  • $\begingroup$ Thank you.. Is there any other proofs??? $\endgroup$ – David Jan 16 '15 at 14:25
  • 2
    $\begingroup$ I don't immediately see how else to tackle the problem, since $2^x = 13$ is an exponential function, so solving it would resort to $\log$ of some basis, but since we want to prove a property of $\log$, that's not going to help us much. Maybe you can reduce it to some other $\log_a b \notin\mathbb Q$ statement this way. $\endgroup$ – AlexR Jan 16 '15 at 14:29
  • $\begingroup$ You could use Gelfond-Schneider theorem, but it's not really necessary to resort to such heavy artillery. $\endgroup$ – Jean-Claude Arbaut Apr 13 '15 at 8:19
3
$\begingroup$

You are done in your solution at the step where you concluded that $2^p = 13^q$. You only need to quote Fundamental Theorem of Arithmetic after that step.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.