TITLE: Alternate proof for "$\log_{10}{2}$ is irrational"
QUESTION [6 upvotes]: I need to prove that $\log_{10}{2}$ is irrational. I understand the way this proof was done using contradiction to show that the even LHS does not equal the odd RHS, but I did it a different way and wanted to check its validity! 
Prove by contradiction: Suppose that $\log{2}$ is rational - that is, it can be written as
$$\log{2} = \frac{a}{b}$$ where $a$ and $b$ are integers. Then
$$2 = 10^{\frac{a}{b}}$$
$$2 = 10^a10^{\frac{1}{b}}$$
$$\frac{2}{10^{a}} = 10^{\frac{1}{b}}$$
Log both sides: 
$$\log(\frac{2}{10^{a}}) = \frac{1}{b}$$
$$\log{2} - \log(10^a) = \frac{1}{b}$$
$$\log{2} = \frac{1}{b} + a$$
$$\log{2} = \frac{ab+1}{b}$$
However we assumed that $\log(2)=\frac{a}{b}$ and thus we have a contradiction.

REPLY [5 votes]: Others have already pointed out that your proof was wrong. A different way to see that your proof is wrong is as follows: if I would replace 2 in your proof everywhere by 10, I would get the result that log(10) is also irrational.
Any proof you have for the irrationality of log(2) should not work for log(10). (If you take 10 as the base of your logarithm.)