Combinations of Transcendental Numbers are still transcendental numbers?

Question

We know there are numbers like $\pi$, $e$, $\phi$ or also $\zeta(3)$ which are transcendental numbers.

I was wondering if combinations of transcendental numbers are still transcendental numbers, like $e + \pi$ or $\frac{1}{\phi}$ or whatever.

Answer 1

$e+\pi$ and $e\pi$ both can not be rational, because if both be rational then we have $x^2-(e+\pi)x+e\pi=0$ has roots $\pi$ and $e$ and leads to contradiction. Also if $z$ is transcendental, then $\frac{1}{z}$ is transcendental, if not then we have $\sum_{i=0}^na_i(\frac{1}{z})^i=0$ for some $a_i\in \mathbb{Z},a_n\not = 0$ and this means $\sum_{i=0}^na_{n-i}z^i=0$, if there exists a $j<n$ such that $a_j\not =0$, this leads to contradiction because it shows that $z$ is algebraic and if for all $j<n$ we have $a_j=0$, then we have $\frac{a_n}{z}=0$ and it contradicts by $a_n\not =0$.

Answer 2

Hint:

$\pi $ and $1-\pi$ are transcendental. What about $\pi+(1-\pi)$?

Answer 3

Polynomials over the rationals in one transcendental number are transcendental, almost (but not quite) by definition. $e^5 + 3e^3 - 2e + 2$ is transcendental.

In two transcendental numbers... well of course $\pi + (1-\pi)$ is rational.

And as for $e + \pi$, who knows?