Show $\log 15$ is irrational and that $\log 3$+ $\log 5$ is also irrational.

Question

I am trying to show that $\log 15$ and $\log 3$ + $\log 5$ is irrational.

For $\log 15$ I feel like I have no issues showing this is irrational.

By contradiction assume $\log 15$ is rational then

$\log 15 = \frac{m}{n}$ where $m,n\in\mathbb{N}$ and $n\not= 0$

$15 = 10^{\frac{m}{n}}\Rightarrow 15^n = 10^m \Rightarrow 3^n5^n = 5^m2^m$

The left side of the equation has an odd multiple of $5$ but the right side has an even multiple of $5$ and we have arrived at a contradiction.

But for $\log 3$ + $\log 5$ I do not even know where to begin, I am assuming I cannot use the fact that $\log 15 = \log 3$ + $\log 5$ as that would be trivial. Should I show they are individually irrational? But that leads me to my next thought as the sum of two irrational numbers is not always irrational (even though in this case I know the sum is irrational). And then would that lead me to breaking it into a case by case where I assume one is rational and the other is not, etc. I am still an undergrad so I do not know much advanced number theory so I would appreciate a more intuitive approach if possible.

Answer 1

Where does the restriction end? Without using the log identity directly, I can still assume $\log 3 + \log 5 = \frac mn$ and then

$$10^{\log 3+\log 5} = 10^{m/n} \implies 10^{\log 3}\cdot 10^{\log 5} = 10^{m/n} \implies 3\cdot 5 = 10^{m/n}$$

Answer 2

Since $\log_{10}3+\log_{10}5=\log_{10}(3\times5)=\log15$ and since $\log_{10}15$ is irractional…

On the other hand, the problem with the equality $3^n5^n=5^m2^m$ lies in the fact that $3^n5^n$ is odd, whereas $5^m2^m$ is even.

Answer 3

The sum of two irrational numbers can be rational. Take $a$ irrational, and $1-a$, which is also irrational (in fact $-a$ is even simpler, but I guess that this would not satisfy the readers). So there is no way to prove that the sum of two general irrationals is irrational, as this is false.

If you are not allowed to use that $\log 3+\log 5=\log 15$ and only that $\log 3$ and $\log 5$ are irrational, I don't see a way to obtain a proof. You could claim that these irrationals are "special" in that they are logarithms of integers, but this brings you back to the sum/product property. If they have no special property, then no proof.