# Irrationality Proof

1. Aug 18, 2009

### jgens

1. The problem statement, all variables and given/known data

Prove that $\log_{10}(2)$ is irrational.

2. Relevant equations

N/A

3. The attempt at a solution

Suppose not, then $\log_{10}(2) = p/q$ where p and q are integers. This implies that $2 = 10^{p/q}$ or similarly, $2^q = 10^p$. However, this is a contradiction since each number's prime factorization is unique - $2^q$ contains only 2's as prime factors while $10^p$ contains both 2's and 5's. Therefore, our assumption that $\log_{10}(2)$ was rational must have been incorrect. This completes the proof.

I'm really bad at these irrationality proofs so I was wondering if someone could comment on the validity of my method. Thanks!

2. Aug 18, 2009

### HallsofIvy

Staff Emeritus
That looks like a perfectly valid proof to me!

3. Aug 18, 2009

### jgens

Swell! Thank you very much!

4. Aug 19, 2009

### Unit

this is really clever!
i would have had no idea what to have done.

5. Aug 19, 2009

### Hurkyl

Staff Emeritus
This comment is probably somewhat pedantic, but I think it's worth saying anyways.

$2^q = 10^p$ is not quite a contradiction -- it can be satisfied when p=q=0. Of course, it's easy to derive a contradiction from that possibility.

6. Aug 19, 2009

### jgens

Perhaps it's a bit pedantic but I definately should have considered that case. Thanks for your input Hurkyl!