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.
Best Answer
As has been pointed out in comments and in another answer, $10^{a/b}\neq 10^{a}10^{\frac{1}b}$. This is a rather subtle error, however there's a notable warning flag that could alert you to it: Your proof does not use the hypothesis that $a$ and $b$ are integers. This is a serious issue, because it means you've proved the (false) statement that $\log(2)$ cannot be written as a fraction $\frac{a}b$ - even if we let $a$ and $b$ be real, but: $$\frac{\log(2)}1=\log(2)$$