Prove Irrationality of \log_{10}(2)

[jgens]

Homework Statement

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

Homework Equations

N/A

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!

 

[HallsofIvy]

That looks like a perfectly valid proof to me!

 

[jgens]

Swell! Thank you very much!

 

[Unit]

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

 

[Hurkyl]

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.

 

[jgens]

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