Is $\frac{\pi}{e}$ irrational? Can we make nonconstant polynomial functions that $\frac{f(\pi)}{g(e)}$ = rational number?

Question

If:

$f(\pi) = a_n *\pi^n + a_{n-1} *\pi^{n-1}... a_1 *\pi$

and

$g(e) = b_n *e^n + b_{n-1} *e^{n-1}... b_1 *e$

$f(\pi) <> 0$, and $g(e) <> 0$

where $a_n$'s and $b_n$'s are whole or rational numbers. the powers also can be whole or rational numbers. for example: $f(\pi) = \frac{1}{3} *\pi^\frac{4}{7}+\pi^\frac{1}{10}$

Is it possible that $\frac{f(\pi)}{g(e)} = \frac{n}{m}$ (rational number)

IMO If we have whole powers it is impossible to get a rational result, but if we can use rational powers I'm not sure.

Answer 1

Your first question is very difficult to proof maybe. It is still not known whether this is true, if you still want to have a look, you may see https://www.reddit.com/r/math/comments/42zosg/is_pie_irrational/

But here is an interesting thing i found on this site. If you consider the polynomial $x^2+(\pi+\frac{1}{e})x+\frac{\pi}{e}=(x-\pi)(x-\frac{1}{e})$. This polynomial must have at least one irrational coefficient. Now, $\frac{e}{\pi}$ is in this, and also, a number is transcedental if its reciprocal is.

Answer 2

Let's clean up the notation, generalize a little, and turn this into an "easier" problem.

Let $f(x) = \sum_{i=1}^M a_i x^{p_i}$ and $g(x) = \sum_{i=1}^N b_i x^{q_i}$ be given, with all the $a_i \neq 0$, $b_i \neq 0$, $p_i$, and $q_i$ rational (in reduced form). We wish to determine if any choices of the $a_i$, $b_i$, $p_i$, and $q_i$ allow that $\dfrac{f(\pi)}{g(\mathrm{e})}$ is rational. Let's call $a_i$, $b_i$, $p_i$, and $q_i$ the "determining variables" to ease the reading of the following.

Let $g$ be the least common multiple of the denominators of the $p_i$. Set $r_i = g p_i$ and observe that all the $r_i$ are integers. Let $r$ be the minimum of the $r_i$. (Note that some of the $r_i$ may be negative, to $r$ may be negative.) Then $$ f(x) = (x^{r/g})\sum_{i=1}^M a_i (x^{1/g})^{r_i - r} $$ is a product of a rational power of $x$ and a polynomial (in the usual sense). Likewise, letting $h$ be the least common multiple of the denominators of the $q_i$, setting $s_i = h q_i$, and setting $s$ to be the minimum of the $s_i$, we have $$ g(x) = (x^{s/h})\sum_{i=1}^N b_i (x^{1/h})^{s_i - s} \text{,} $$ also a product of a rational power of $x$ and a polynomial.

Suppose that $ \dfrac{f(\pi)}{g(\mathrm{e})} = \dfrac{m}{n}$ for integers $m$ and $n$ with $n \neq 0$. Then $$ \frac{n f(\pi)}{m g(\mathrm{e})} = 1 \text{.} $$ But \begin{align*} n f(x) &= n (x^{r/g})\sum_{i=1}^M a_i (x^{1/g})^{r_i - r} \\ &= (x^{r/g})\sum_{i=1}^M (n a_i) (x^{1/g})^{r_i - r} \text{,} \end{align*} and similarly for $m g(x)$. This means if there are rational choices of the determining variables making $f(\pi)/g(\mathrm{e})$ rational, then we can replace the $a_i$ and $b_i$ with integer multiples of themselves and make $f(\pi)/g(\mathrm{e}) = 1$. Therefore, we only need to determine if there are choices of the determining variables that make $f(\pi)/g(\mathrm{e}) = 1$, which is equivalent to $f(\pi) = g(\mathrm{e})$.

Suppose $f(\pi) = g(\mathrm{e})$. Then $$ (\pi^{r/g})\sum_{i=1}^M a_i (\pi^{1/g})^{r_i - r} = (\mathrm{e}^{s/h})\sum_{i=1}^N b_i (\mathrm{e}^{1/h})^{s_i - s} \text{.} $$ This is equivalent to $$ \mathbb{Q}(\sqrt[g]{\pi}) \cap \mathbb{Q}(\sqrt[h]{\mathrm{e}}) \varsupsetneq \mathbb{Q} $$

Generically answering this sort of question (algebraic dependence of two transcendental numbers that do not both come from "the same thing") is very, very hard. We don't even currently know if $\pi$ and $\mathrm{e}$ are algebraically independent, so we can't answer the "easier" version of this: $$ \mathbb{Q}(\pi) \cap \mathbb{Q}(\mathrm{e}) \varsupsetneq \mathbb{Q} \text{?} $$