Prove that the logarithm of 3 base 10 is irrational
The Fundamental Theorem of Arithmetic is that every integer is a product of primes.
So far I have,
Suppose $\log_{10}(5)$ is rational.
Then suppose $\log_{10}(5) = \frac {p}{q}$ for some positive integers $p$ and $q$
with $\frac {p}{q}$ in lowest terms and $p< q$.
Exponentiating both sides using 10 as the base we get, $5=10^{p/q}$. Take both sides to the qth power. We get $5^q=10^p=2^p*5^p$. Then we get $5^{q-p}=2^p$.
But I'm not sure if this has anything to do with the Fundamental Theorem of Arithmetic.
If you have another way of doing this that would be great too.
Best Answer
$5^{q-p}$ is odd and therefore $p=0$. Hence $5^{q-p}=1$, thus $q-p=0$ therefore $q=0$ contradicting the fact $q$ is nonzero