I am having some problems proving that the following sum is irrational or rational:


This is all I've got for now:

$\log_2(3)=\frac mn \iff 2^{\frac mn}=3 \iff 2^m=3^n$ so $\log_2(3)$ = irrational.

$\log_3(2)=\frac qr \iff 3^{\frac qr}=2 \iff 3^q=2^r$ so $\log_3(2)$ = irrational.

Now I'm having trouble with proving that $\log_2(3)+\log_3(2)$ is irrational. I know that the sum of two irrational numbers isn't directly irrational. Also, both base numbers of the logarithms are primes.

  • $\begingroup$ Can you express $\log_a b$ in terms of the natural logarithm? $\endgroup$ Oct 24 '14 at 20:18
  • $\begingroup$ I have already tried that: $\frac{\ln(3)}{\ln(2)}+\frac{\ln(2)}{\ln(3)}=\frac{\ln^2(2)+\ln^2(3)}{\ln(2)\ln(3)}$. But I can't get any further than this $\endgroup$
    – Adnan
    Oct 24 '14 at 20:28
  • 3
    $\begingroup$ The point is that $\zeta := \log_3 2 = \frac{1}{\log_2 3}$. So if $\zeta + \frac{1}{\zeta} = q \in \mathbb{Q}$, then $\zeta$ would be of the form $\alpha + \beta\sqrt{m}$ with $\alpha,\beta\in\mathbb{Q}$ for some integer $m$. In particular, $\zeta$ would be algebraic. $\endgroup$ Oct 24 '14 at 20:36
  • 2
    $\begingroup$ And $\log_3 2$ is transcendental by Gelfond-Schneider. $\endgroup$ Oct 24 '14 at 20:52
  • $\begingroup$ A related question. See the comment section on @Mookid's now-deleted answer. $\endgroup$
    – Lucian
    Oct 24 '14 at 22:22

Let $x=\log_23$. Note that $\log_32=\frac1x$, so if $x+\frac1x$ were rational we would have $x+\frac1x=q\in\mathbb Q$, or: $x^2-qx+1=0$, meaning that $x$ is an algebraic number (of degree at most $2$). Now we can rewrite $\log_23=x$ a bit and use a strong theorem, the Gelfond-Scheider theorem, to obtain a contradiction: $$2^x=3.$$ Gelfond-Schneider says $$(\text{algebraic and not 0 or 1})^{\text{algebraic and irrational}}=\text{transcendental}.$$ Because $2$ is algebraic and not $0$ or $1$, $x$ is algebraic and irrational but $3$ is not transcendental, we obtain a contradiction.

Note: Using the same method we can prove that $\log_23+\log_32$ is in fact transcendental; this is because the field (yes, it's a field, and that's not trivial) of algebraic numbers (contained in $\mathbb C$) over $\mathbb Q$ is algebraically closed.


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