Prove that $Ae + B/e$ is irrational. 
Prove that $Ae + B/e$ is irrational. Here A and B are non-zero integers.

We know that the sum of two irrational numbers need not be an irrational number.
For example $\sqrt2+1$ and $1-\sqrt2$ are irrational numbers. But there sum $(\sqrt2+1)+ (1-\sqrt2)=2\enspace (2 \in \mathbb{Q})$ is not an irrational number.
But for this problem, I am finding difficulty even how to begin with, which will slowly lead me to the proof.
 A: Suppose that $Ae + B/e$ is rational. Then $Ae + B/e = p/q$ for some $p,q \in \mathbb{Z}$. Then $Ae^2 - \frac{p}{q} e + B = 0$. Clearing denominators gives us $e$ as a root of $Aqx^2 - px + Bq$ which is impossible since $e$ is transcendental.
A: Have you seen the proof that $e$ is irrational by using the series for $e$?  If so, only a fairly small modification is needed.  If $Ae+B/e$ is rational, then by clearing denominators we can get
$$ae+be^{-1}=c$$
for some integers $a,b,c$.  Writing
$$e=1+\frac1{1!}+\cdots+\frac1{n!}+R_n\ ,\qquad
  e^{-1}=1-\frac1{1!}+\cdots\pm\frac1{n!}+S_n$$
leads to
$$cn!-a(n!+\cdots+1)-b(n!-\cdots\pm1)=(aR_n+bS_n)n!\ .$$
The left hand side is an integer, and if you can show that $0<|{\rm RHS}|<1$ then you have a contradiction.  You may need to consider various cases, e.g., whether $n$ is even or odd.
All this and more in my recent book!  (Moderators feel free to remove if free plugs are not allowed.)
A: An elementary proof, without using the (deep) theorem that $e$ is transcendental. Suppose $A,B,C\in\Bbb Q$ and $Ae+B/e=C.$ We show that $A=B=0.$
Take $D\in\Bbb Z^+$ such that $A'=AD,B'=BD,C'=CD$ are in $\Bbb Z.$
For $n\in\Bbb Z^+$ we have $n!C'=n!eA'+n!(1/e)B'=E_n+\sum_{j=n+1}^{\infty}(A'+(-1)^jB')n!/j!$ where $E_n\in\Bbb Z.$ So we have $\sum_{j=n+1}^{\infty}(A'+(-1)^jB')n!/j!\in\Bbb Z.$
For brevity let $\sum_{j=n+1}^{\infty}(A'+(-1)^jB')n!/j!=R_n.$
We have  $(n+1)R_n-R_{n+1}=A'+(-1)^{n+1}B'.$ We also have $\lim_{n\to\infty}R_n=0$ but  $R_n\in\Bbb Z,$ so $R_n=0$ for all but finitely many $n.$ So take $n$ large enough that $R_n=R_{n+1}=R_{n+2}=0.$ Then $$A'+(-1)^{n+1}B'=(n+1)R_n-R_{n+1}=0$$ $$A'+(-1)^{n+2}B'=(n+2)R_{n+1}-R_{n+2}=0$$ which imply $A'=B'=0.$
