I don’t know if this is an original idea, but regardless I’m quite proud of myself for coming up with it. I think it’s very clever. This will be a proof by contradiction.
Suppose, to the contrary, that log10 of 2 is rational. Then log10 2 = a/b where a and b are unknown integers. Recall from high school math that:
log10 2 = a/b
2 = 10^(a/b)
2 = (10^a)^(1/b)
2^b = 10^a
This latter equation, 2^b = 10^a, is logically equivalent to log10 2 = a/b, and also much nicer to look at. Very ho-hum so far.
The trick is to remember that all composite numbers have one and only one prime factorization. Because 2 is prime, 2^b is a prime factorization. But we also know that every number 10^a has 5 as a divisor. Because 5 is prime, it must represented in the prime factorization. But if it is true that some number is equal to 2^b, then 2^b must be that number’s prime factorization. This is a contradiction.
Quod erat demonstrandum. Criticisms welcome.