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Can we find smallest positive $x$ such that $\pi^x$ is rational? Is this possible if $x$ is rational?

My attempt: I don't know how to start!

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closed as off-topic by Saad, Namaste, B. Mehta, Delta-u, Ethan Bolker May 19 '18 at 21:12

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  • $\begingroup$ What are your thoughts on this question? Do you think such an $x$ exists? What do you know about $\pi$ already? $\endgroup$ – B. Mehta Apr 16 '18 at 3:07
  • $\begingroup$ I know high school mathematics, and learning undergraduate mathematics. I am not sure if such x exists or not. About $\pi$ i know that it can be represented as infinite series $\endgroup$ – kayush Apr 16 '18 at 3:10
  • $\begingroup$ For future reference, here's how to write a good question! $\endgroup$ – B. Mehta Apr 16 '18 at 3:11
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This approach uses a little calculus:

Consider some properties of the function

$\phi(x) = \pi^x, \; x \in \Bbb R; \tag 1$

we have:

$\phi(0) = \pi^0 = 1; \tag 2$

$\phi(1) = \pi^1 = \pi; \tag 3$

also, $\phi(x)$ is differentiable; indeed,

$\phi(x) = \pi^x = e^{(\ln \pi)x}, \tag 4$

whence

$\phi'(x) = (\ln \pi)e^{(\ln \pi)x} > 0 \; \text{for} \; x \in \Bbb R; \tag 5$

we thus see that $\phi(x) = \pi^x$ is monotonically increasing for all $x$.

It now follows from the intermediate value theorem that $\phi(x)$ takes on every rational value $r \ge 1$. Since $\phi(x) = \pi^x$ is strictly increasing and there is no smallest rational larger than $1$, there is no least $x > 0$ for which $\phi(x) = \pi^x$ is rational.

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Note that

$$\pi ^{\log_\pi x } = x $$ so for every positive rational $x$ you have a solution.

Since positive rationals do not have a minimum, your equation does not have a minimum.

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There is no smallest $x$. For any number of the form $$x = \frac{\ln(r)}{\ln(\pi)} = \log_\pi(r)$$ where $r$ is rational, it follows that $\pi^x = r$.

$x$ increases with $r$. The restriction $x > 0$ implies that $r > 1$, and there is no smallest rational above $1$. (Proof: Cut in half $r$'s distance from $1$.)

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Well, $\pi^x$ will yield $1$ when $x=0$ and get larger as $x$ gets larger. So we want to minimize $n$ in $\pi^x=1+n$, which can be rewritten as $x=\log_\pi(1+n)$. Really, there is no "smallest $x$" because although we want to minimize $n$ but maintain it above $0$ we will always be able to halve it.

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Let $x=x_0$ be the smallest positive number such that $\pi^{x}$ is rational.

We have, $\pi^{x_0}>1$. Let $\displaystyle x_1=\log_\pi\left(\frac{\pi^{x_0}+1}{2}\right)$.

Then

$$0<\log_\pi\left(\frac{1+1}{2}\right)<x_1<\log_\pi\left(\frac{2\pi^{x_0}}{2}\right)=x_0$$

$x_1$ is a smaller positive number satisfying the condition.

This leads to a contradiction and hence such a smallest number does not exist.

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In the case $x$ is rational:

No. We know that $\pi$ is a transcendental number, which means there isn't any polynomial with integer coefficients $f$ such that $f(\pi) = 0$. But, if $\pi^{\frac{q}{r}} = \frac{a}{b}$ for some integers $q,r,a,b$, then $b^r \pi^q = a^r$, so $f(x) = b^r x^q - a^r$ is an integer polynomial solved by $\pi$.

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