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!