For each irrational number $b$, does there exist an irrational number $a$ such that $a^b$ is rational? 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.

 A: 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.
