Are the numbers ##eπ## and ##e+ π## transcendental?

[chwala]

TL;DR Summary

A number is called an algebraic number if it is a solution of a polynomial equation $a_0 z^n+a_1z^{n-1} + ... a_{n-1}z + a_n =0$ where $a_0,a_1 ...a_n$ are integers...otherwise transcendental.

My question is [following the example on the attachment which is apparently clear to me].
1. Are the numbers $eπ$ and $e+ π$ Transcendental?
2. Algebraic numbers can also be rational and not necessarily integers? is that correct?

 

[pasmith]

All rational numbers are algebraic: $n/m$ is the solution of $mz - n = 0$.

 

[fresh_42]

TL;DR Summary: A number is called an algebraic number if it is a solution of a polynomial equation $a_0 z^n+a_1z^{n-1} + ... a_{n-1}z + a_n =0$ where $a_0,a_1 ...a_n$ are integers...otherwise transcendental.

My question is [following the example on the attachment which is apparently clear to me].
1. Are the numbers $eπ$ and $e+ π$ Transcendental?

Presumably. The standard theorems about transcendency don't apply to them, but I haven't checked in detail. More interesting is the question: Who cares? These numbers do not occur naturally and I haven't seen any theorem that needed to know whether $\mathbb{Q}[e,\pi ]$ is of transcendental degree one or two. I guess literally nobody will expect it to be one.

2. Algebraic numbers can also be rational and not necessarily integers? is that correct?

$\mathbb{Z}\subsetneq \mathbb{Q} \subsetneq \mathbb{A}\subsetneq \mathbb{C}$ if $\mathbb{A}$ is the field of all algebraic numbers over the rationals.

 

[chwala]

What i would add on this is that $e$ and $π$ are irrational thus we cannot solve for them in a given equation say,

$z-e- π=0$

as is the case with algebraic terms in any given polynomial.

 

[fresh_42]

What i would add on this is that $e$ and $π$ are irrational thus we cannot solve for them in a given equation say,

$z-e- π=0$ as is the case with polynomials.

You must be careful. $\pi, -\pi$ and $\pi^{-1}$ are transcendent, but neither is $(\pi)+(-\pi)$ nor $(\pi)\cdot (\pi^{-1}).$

 

[pbuk]

What i would add on this is that $e$ and $π$ are irrational

So is $\sqrt 2$ (but it is not of course transcendental, by definition).

You must be careful. $\pi, -\pi$ and $\pi^{-1}$ are transcendent, but neither is $(\pi)+(-\pi)$ nor $(\pi)\cdot (\pi^{-1}).$

And nor is $e^{i \pi}$.

 

[nuuskur]

I believe it is an open problem: are $\frac{\pi}{e}, e\pi, e+\pi$ irrational? Transcendental?
See no.22

It is known that either $\pi+e$ or $e\pi$ is transcendental.

Spoiler

If both are algebraic, then $(\pi+e)^2 - 4e\pi$ is algebraic. So, $\pi-e$ is algebraic, which implies
$$\frac{1}{2}((\pi+e) - (\pi-e)) = e$$
is algebraic, a contradiction.