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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Apr 7, 2015 at 12:50
  • $\begingroup$ @LeonhardtvonM: Thank you for your interesting comment. Unfortunately, it is difficult at least for me. Can you elaborate? $\endgroup$
    – mathlove
    Commented Apr 8, 2015 at 10:27
  • $\begingroup$ You're right, that is more difficult than I expected. $\endgroup$ Commented Apr 8, 2015 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
    Commented Feb 19, 2016 at 14:55
  • $\begingroup$ @TonyK: I see. Thank you for pointing that out. $\endgroup$
    – mathlove
    Commented Feb 20, 2016 at 5:21

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Apr 8, 2015 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
    Commented Apr 8, 2015 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
    Commented Apr 8, 2015 at 12:27

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