I try to prove that the number $$\log_2 5 +\log_3 5$$ is irrational. But I have no idea how to do it. Any hints are welcome.

I'm not sure if this is helpful, but we do have$$ \log_2 5 + \log_3 5 = \frac{\log 5}{\log 2} + \frac{\log5}{\log 3} = \frac{\log5(\log 3 + \log 2)}{\log2 \log 3} = \frac{\log 5 \log 6}{\log2 \log 3} $$ where $\log$ can be given the base of your choosing. – Ben Grossmann Oct 22 '14 at 18:41

We can also rewrite it as $$ \frac{1}{\log_5 2} + \frac{1}{\log_5 3} $$ – Ben Grossmann Oct 22 '14 at 18:43

1Anybody finds in the web the lightest reference to this matter? It seems like nobody had thought never about it. – ajotatxe Oct 22 '14 at 21:10

@ajotatxe I couldn't agree more. I found a few proofs and used a step or two to make my own proof, but ultimately I was only to get that each component is irrational, rather than their sum. – daOnlyBG Oct 22 '14 at 22:09

OK, @FisiaiLusia, I'm curious: where did you find this problem? – daOnlyBG Oct 23 '14 at 00:45

@daOnlyBG, Somebody asks me about it. – Oct 23 '14 at 07:38

@FisiaiLusia And does that somebody know the answer? – ajotatxe Oct 23 '14 at 08:05

I think, he don't know. – Oct 23 '14 at 09:00

You know, I don't think there's an algebraic way of proving this. One might have to use a theorem or lemma found in analysis or measure theory. There's only so many operations/identities you could take advantage of, and after working on this consistently for the past day, I can't seem to find anything that works. – daOnlyBG Oct 23 '14 at 15:59

1Crossposted on MO: http://mathoverflow.net/questions/185540/provingtheirrationalityofthisnumber – Qiaochu Yuan Nov 10 '14 at 07:25
2 Answers
Questions of this form fall under the general heading of transcendental number theory. Conditional on a major conjecture in this field called Schanuel's conjecture, this number is not only irrational but transcendental. In fact, conditional on Schanuel's conjecture much more is true:
(Conditional) Theorem: The logarithms $\log 2, \log 3, \log 5, \dots$ of the primes are algebraically independent.
Essentially this means that any rational function of the logarithms of the primes is transcendental unless you can simplify it, as a rational function, to a rational number. The conclusion follows because your expression can be written $\frac{\log 5}{\log 2} + \frac{\log 5}{\log 3}$.
Proof. By unique prime factorization, the logarithms of the primes are linearly independent over $\mathbb{Q}$. If $p_1, p_2, \dots$ is an enumeration of the primes, then by Schanuel's conjecture it follows that $\mathbb{Q}(\log p_1, \log p_2, \dots \log p_k)$ has transcendence degree at least $k$, hence exactly $k$, for all $k$. $\Box$
 359,788
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If both of them are rational then...
$$ \log_2(5)=\frac{m_1}{n_1} \implies 2^{m_1/n_1}= 5 \implies 2^{m_1}=5^{n_1} $$ $$ \log_3(5)=\frac{m_2}{n_2}\implies 3^{m_2/n_2}=5 \implies 3^{m_2}=5^{n_2} $$
The rightmost equation would only be true if both $n,m$ are $0$, but that contradicts the leftmost equation ($n_{1/2}\in\mathbb{N_{/0}},m_{1/2}\in\mathbb{Z}$)
For $n,m > 1$ the rightmost line can't be true because it contradicts prime factorization. (due to this fact we can also ignore the following cases: $0<n_{1/2}<1<m_{1/2};0<n_{1/2},m_{1/2}<1; 0<m_{1/2}<1<n_{1/2}$
So the sum of two irrational numbers ($i_1, i_2$) is rational if $i_1+i_2=\frac{m}{n} $ for $\frac{m}{n}$ being the rational sum.
$$ \frac{m}{n}=\log_3(5)+\log_2(5) \implies 2^{m/n}=2^{\log_3(5)+\log_2(5)}\implies $$ $$ 2^{m/n}=5\times2^{\log_3(5)}\implies 2^m=5^{n}\times2^{n\log_3(5)} $$
And here is where we have the next contradiction: The left side is always rational but $2^{n\log_3(5)}$ is irrational and so is the product on the right side of the last equation (because the product of a rational and irrational number is always irrational).
Therefore the statement is false and the sum has to be irrational. That was a tricky one!
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Do you mean to suggest $m,n$ can be taken to be the same for both $\log_2 5$ and $\log_3 5$? You seem pretty proficient with MathJax/LaTeX, so I'm sure you can best Edit the Answer to suit your intentions. – hardmath Nov 10 '14 at 03:40

@hardmath thank you for that! No this is the first time I used MathJax/LaTeX :/ But you are right, that was very ambiguous  I fixed it.@integrator: how do I do the fractions in powers? – GregLSteiner Nov 10 '14 at 03:50


4This argument is incomplete. You still need to prove that $2^{n \log_3(5)}$ is irrational. – Qiaochu Yuan Nov 10 '14 at 07:07

That's a pretty big assertion there at the end about $2^{n\log_3{5}}$ – Thomas Andrews Dec 23 '15 at 19:21