Can transcendental to the power transcendental be rational?

Question

Can a transcendental number to the power of a transcendental number be a rational number?

Answer 1

This is a bit easier to see the other way around. Let $q\in\Bbb Q$ be positive ($q\ne 1$), and consider the map $x\mapsto q^x$. This is a bijection from $\Bbb R$ to $(0,\infty)$, and if you restrict it to the transcendental numbers, you still have an uncountable image. That image must contain uncountably many transcendental numbers, because we only have countably many algebraic numbers.

Thus, we have $q^{t_1}=t_2$ for some transcendental numbers $t_1$ and $t_2$. The number $1/t_1$ is also transcendental, and $t_2^{1/t_1}=q$.

In fact, this shows that, if a positive rational number is not $1$, then it can be written as a transcendental power of a transcendental number in uncountably many different ways.

Answer 2

Not only it can, it pretty much has to. And without even giving an example, here's why.

Consider the algebraic numbers, this is a field closed under taking roots of polynomials. But not under exponentiation and logarithms (where the base of the logarithm is in the field, of course).

Using basic model theoretic and set theoretic arguments, we can show that there is a smallest $F\subseteq\Bbb R$ such that:

1. $F$ is a field closed under the partial operator $E(x,y)$.
2. If $x,y\in F$ then $E(x,y)=x^y$ whenever that is defined in $\Bbb R$.
3. We can even require that $F$ is a real-closed field, namely every odd-degreed polynomial with coefficients in $F$ has a root in $F$.

But more importantly,

4. $F$ is countable.

Consider now any $x\in\Bbb R\setminus F$, then for every nonzero $y\in F$ it follows that $x^y\notin F$ either. Otherwise, $x=E(x^y,\frac1y)=x^{y\cdot\frac1y}=x$. Moreover, because $F$ is a real-closed field, it follows that in contains all the algebraic reals.

Therefore if $x\notin F$ and $1\neq q\in\Bbb Q^+$, then the unique $y$ such that $x^y=q$ is such that $y\notin F$ as well, in which case $y$ has to be transcendental. And such $y$ always exists because $y\mapsto x^y$ is a continuous function which is a surjection onto $\Bbb R^+$.

Now. Because $F$ is only countable, it follows that "almost all" (in terms of cardinality, measure, and category) the transcendental numbers $x$ in $\Bbb R$ has infinitely many transcendental numbers $y$ such that $x^y\in\Bbb Q$.

Answer 3

This proves $e$ is transcendental. If we prove $\ln2$ is transcendental, we are done, since $e^{\ln2}$ is then an example of $\text{transc}^{\text{transc}}=\text{rat}$. $\ln2=\frac{1}{\log_2e}$. So if $\log_2e$ is transcendental, so is $\ln2$. That is because algebraic numbers are a field, so if $a$ is algebraic, $\frac{1}{a}$ is too, and if either isn't, neither is. Now, $2^{\log_2e}=e$ by definition. If $\log_2e$ were rational, $\log_2e=\frac{p}{q}$ for $p,q\in\mathbb{Z}$. But then $2^{\log_2e}=2^{\frac{p}{q}}=\sqrt[q]{2^p}$, which satisfies $x^q-2^p=0$, and thus is algebraic. But $2^{\log_2e}=e$ which is transcendental, contradiction. So $\log_2e$ is not rational. Unfortunately, I cannot prove $\log_2e$ is not an algebraic irrational. Someone tried it with the Gelfond-Schneider theorem, but it applies perfectly here if $\log_2e$ is algebraic and irrational, since $2^{\log_2e}=e$ which is transcendental, as the theorem would conclude. Indeed, that guy deleted his answer. The theorem is pretty interesting, but useless here. All that comes to mind now is to try by contradiction. So we have to suppose there exist $a_i\in\mathbb{Q},n\in\mathbb{N}$ such that: $$\sum_{i=0}^na_i(\log_2e)^i=0.$$ That is the translation of "algebraic", since "over $\mathbb{Q}$" is implied there. We know $\log_2e\not\in\mathbb{Q}$. Without loss of generality, we can suppose $a_i\in\mathbb{Z}$, since if $a_i\in\mathbb{Q}\smallsetminus\mathbb{Z}$ then there exists $q\in\mathbb{Z}:qa_i\in\mathbb{Z}$. So we have an integer coefficient linear combination of powers of $\log_2e$ which is zero. Naturally there must be at least one nonzero coefficient with $i$ at least 1, otherwise the assertion is either trivial (if they are all zero) or impossible (if the only nonzero coefficient is $a_0$). In short, we have $a_i\in\mathbb{Z},n\in\mathbb{N}$ such that $a_i$ is nonzero for at least one $i>0$ and: $$\sum_{i=0}^na_i(\log_2e)^i=0.$$ If no power of $\log_2e$ is rational, this seems rather impossible. We would then have a sum of irrationals giving 0. But I can't actually prove it is impossible. Now if this logarithm has a rational power, it satisfies $x^t-\frac{p}{q}=0$. This implies $qx^t=p$. Let us call $\alpha:=\log_2e$. We have supposed $q\alpha^t=p$. OK, I give up. Sorry, I cannot prove $\alpha$ can't be algebraic. I hope someone can point me to a reference.

Answer 4

The first thought that came to my mind was $e^{\pi i}=-1$. Of course, this doesn't work if you require each transcendental number be real.