8
$\begingroup$

It is well known that there exist two irrational numbers $a$ and $b$ such that $a^b$ is rational.

By the way, I've been interested in the following two propositions.

Proposition 1 : For each irrational number $a\gt 0$, there exists an irrational number $b$ such that $a^b$ is rational.

Proposition 2 : For each irrational number $b$, there exists an irrational number $a$ such that $a^b$ is rational.

I got the following :

Proposition 1 is true.

Suppose that both $\frac{\ln 2}{\ln a}$ and $\frac{\ln 3}{\ln a}$ are rational. There exists a set of four non-zero integers $(m_1,m_2,n_1,n_2)$ such that $\frac{\ln 2}{\ln a}=\frac{n_1}{m_1}$ and $\frac{\ln 3}{\ln a}=\frac{n_2}{m_2}$. Since one has $a=2^{m_1/n_1}=3^{m_2/n_2}$, one has $2^{m_1n_2}=3^{m_2n_1}$. This is a contradiction. It follows that either $\frac{\ln 2}{\ln a}$ or $\frac{\ln 3}{\ln a}$ is irrational. Hence, either setting $b=\frac{\ln 2}{\ln a}$ or setting $b=\frac{\ln 3}{\ln a}$ works.

Then, I began to consider if proposition 2 is true.

To prove that proposition 2 is true, it is sufficient to show that for each irrational number $b$, there exists a rational number $c$ such that $c^{1/b}$ is irrational.

This seems true, but I have not been able to prove that. So, my question is the following :

Question : Is proposition 2 true? If yes, how can we show that? If no, what is a counterexample?

Proposition 2 : For each irrational number $b$, there exists an irrational number $a$ such that $a^b$ is rational.

$\endgroup$
  • $\begingroup$ I think it isn't too hard to prove the following statement. If $b>0$ irrational then there exists an $a>0$ irrational such that $a^b$ or $a^{1/b}$ is rational. $\endgroup$ – Leonhardt von M Apr 7 '15 at 12:50
  • $\begingroup$ @LeonhardtvonM: Thank you for your interesting comment. Unfortunately, it is difficult at least for me. Can you elaborate? $\endgroup$ – mathlove Apr 8 '15 at 10:27
  • $\begingroup$ You're right, that is more difficult than I expected. $\endgroup$ – Leonhardt von M Apr 8 '15 at 20:45
  • $\begingroup$ "We may suppose $a > 0$": No, you have to make that a condition in your statement of Proposition 1. If $a < 0$ and $b$ is irrational, then $a^b$ is undefined. (But Proposition 2 is OK.) $\endgroup$ – TonyK Feb 19 '16 at 14:55
  • $\begingroup$ @TonyK: I see. Thank you for pointing that out. $\endgroup$ – mathlove Feb 20 '16 at 5:21
4
$\begingroup$

What you want to prove is: given an irrational number $b$, then one of the following numbers : $$r^{\frac{1}{b}}\ \ \ \ \ \ r\in \Bbb Q $$ is irrational, this seems to be an attainable result for what we know nowadays about the transcendence of numbers, we have for example the following theorem:

Six Exponentials Theorem:Let $(x_1,x_2)$ and $(y_1,y_2,y_3)$ be two sets of complex numbers linearly independent over the rationals. Then at least one of $$e^{x_1y_1},e^{x_1y_2},e^{x_1y_3},e^{x_2y_1},e^{x_2y_2},e^{x_2y_3}$$ is transcendental.

Given an irrational number $x$, let $x_1=1,x_2=x$ and $y_1=\ln(p_1),y_2=\ln(p_2),y_3=\ln(p_3)$ for some primes $p_1,p_2,p_3$ hence using this theorem we have : at least one of: $$p_1,p_2,p_3,p_1^x,p_2^x,p_3^x $$ is irrational which gives us the following well known consequence:

Six Exponentials Theorem (special case). If $x$ is a real number such that $p_1^x$ , $p_2^x$ and $p_3^x$ are rational numbers for three distinct primes $p_1, p_2$ and $p_3$, then $x\in \Bbb Z$

If we use this theorem one $k^{\frac{1}{b}}$ of the numbers $2^{\frac{1}{b}},3^{\frac{1}{b}},5^{\frac{1}{b}}$ is irrational. and hence you can take $a=k^{\frac{1}{b}}$ and we have $a^b$ is an integer among $\{2,3,5\}$ which implies of course that it's rational.

Comment Maybe there is a very simpler answer which does not use this strong theorem.

$\endgroup$
  • $\begingroup$ Thank you very much for this answer. I have a few questions. (1) In the theorem, you wrote $p_1^x \color{red}{1},p_2^x$. The $\color{red}{1}$ is a typo, isn't it? (2) You wrote "rational integers" in the theorem. Aren't they "rational numbers"? (3) I got two links of the theorem. wiki and mathworld. But I cannot get your special case. Can you elaborate on how to get the special case? $\endgroup$ – mathlove Apr 8 '15 at 10:25
  • $\begingroup$ Thanks for the corrections, I added some additional information in my answer and explained how to get the special case from the mathword link. I did not think about a simpler answer, so maybe there is one!! it's very likely. I can not use the methods used for the Six exponentiel theorem because it involves a lot of things I don't overcome and I don't understand well. $\endgroup$ – Elaqqad Apr 8 '15 at 12:18
  • $\begingroup$ Ah, I understand how to get the special case. I'm very glad to know the theorem though understanding the theorem itself seems difficult for me:) Anyway, thank you so much! $\endgroup$ – mathlove Apr 8 '15 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.