Prove at least one of $\pi+e$, $\pi e$ is transcendental. I understand to prove at least one of them irrational you would compose a function by which $\pi$ and $e$ are roots $((x-\pi)(x-e))$, and show that at least one coefficient cannot be rational because $\pi$ and $e$ are transcendental.
From here I developed $3$ cases;
$1)$ $(\pi+e)$ is rational and $(\pi e)$ is irrational or
$2)$ $(\pi+e)$ is irrational and $(\pi e)$ is rational or
$3)$ $(\pi+e)$ is irrational and $(\pi e)$ is irrational
I compose a new function, $(x-(\pi+e))(x-(\pi e))$, which produces $x^2+(\pi+e+\pi e)x+(\pi+e)(\pi e)$
For cases $1$ and $2$, they are transcendental, but I am unsure of how to show it for case $3.$
 A: The trick is that if $\alpha$ satisfies a polynomial $p(x)$ with algebraic coefficients (not just rational coefficients), then $\alpha$ must be algebraic.  This can be proven with some linear algebra: let $K$ be the field generated by the coefficients of $p$.  Since the coefficients are algebraic, $K$ has finite dimension over $\mathbb{Q}$.  Since the vector space formed by the powers of $\alpha$ is finite-dimensional over $K$, it is also finite-dimensional over $\mathbb{Q}$.  That is, some linear combination of the powers of $\alpha$ with rational coefficients vanishes, so $\alpha$ is a root of a polynomial with rational coefficients.
So now if $\pi e$ and $\pi+e$ were both algebraic, then the polynomial $p(x)=(x-\pi)(x-e)$ would have algebraic coefficients.  This implies that its roots ($\pi$ and $e$) would be algebraic, which is a contradiction.
A: If $\pi+e$ and $\pi e$ were both algebraic, then $(\pi+e)^2-4\pi e$ would be algebraic, and thefore $\pi-e$ would be algebraic. But then $\frac{1}{2}((\pi+e)-(\pi-e))$ would be algebraic, that is, $e$ would be algebraic, which is not the case.
