A very short proof of $e$ is irrational I was talking with my friend and he came up with this very short proof

Given $x\in \mathbb{R}$, if $xy \notin \mathbb{Z}$ for any $y\in \mathbb{Z}$, then $x$ is irrational. Since $e = \sum \frac{1}{n!}$, we see that $ey \notin \mathbb{Z}$ for any $y \in \mathbb{Z}$. So $e$ is irrational.

Is his argument correct? If not, why?
 A: When your friend says : we $see \; that$, he has a very keen eye!
The shortest proof I know is the following. 
Suppose $e = \frac{p}{q}$ with $p$ and $q$ two strictly positive integers. Now: 
\begin{equation*}
q!p - \sum_{k=0}^q \frac{qq!}{k!} = q \sum_{k=q+1}^{+\infty} \frac{1}{(q+1)...k}
\end{equation*}
the term on the right is strictly positive but strictly inferior to 1. Infact, insofar as, when $k \geq 2 +q$, $(q+1)\cdots k > (q+1)^{k-q}$, the term on the right is strictly inferior to $\frac{q}{q+1} \times \frac{1}{1 - \frac{1}{q+1}}= 1$.
Yet, the term on the left is an integer. We hence derive a contradiction.
A: The simplest proof along these lines that I am familiar with:  
$e$ is rational if and only if $e^{-1}$ is rational so assume that $e^{-1}=\frac ab$.  we note that $$\frac ab=e^{-1}=\sum_{n=0}^{\infty} \frac {\left( -1\right)^n}{n!}$$ is an alternating series with decreasing terms.  Thus the error involved in estimating the sum by a truncated series is bounded by, and has the same sign as, the first omitted term.  Assuming, say, that $b$ is odd, we deduce $$1-\frac 1{1!}+\cdots-\frac 1{b!}<\frac ab <1-\frac 1{1!}+\cdots+\frac 1{(b+1)!}$$  multiplying by $b!$ then shows that $$b!-\frac {b!}{1!}+\cdots-\frac {b!}{b!}<b!\times \frac ab <b!-\frac {b!}{1!}+\cdots+\frac {b!}{(b+1)!}$$ Of course, the left hand sum defines an integer and $b!\times \frac ab$ is clearly an integer, and the right hand sum differs from the left by $\frac {b!}{(b+1)!}=\frac 1{b+1}$, a contradiction.  The case $b$ even is similar (or just work with $b+1$ in the above.
A: The definition is not quite right. For any $x \in \mathbb{R}$, including irrational $x$, there is always some $y \in \mathbb{Z}$ such that $xy \in \mathbb{Z}$, namely $y=0$. 
However, if instead of $y \in \mathbb{Z}$ we have $y \in \mathbb{Z} \setminus \{0\}$ the definition is fine.
There's no obvious reason to believe that there is no $y \in \mathbb{Z} \setminus \{0\}$ such that $$y \sum_{n \geq 0} \frac{x^{n}}{n!}$$ is integral.
Perhaps your friend thinks that because it's an infinite sum, the result of the multiplication cannot be integral. This is a mistaken assumption, since, for example,  
$$\sum_{n\geq1} \frac{1}{2^n}$$
is an integer. 
A: Perhaps you can take the fractional part of $n! \times e$  that number is
$$ 0< n! e - \text{whole number } = \frac{1}{n+1}
+ \frac{1}{(n+1)(n+2)}+\dots <1$$
if $e$ is rational  this number  has to be integer eventually. However, this error term is strictly between $0$ and $1$ always. So $e \notin \mathbb{Q}$.
https://en.m.wikipedia.org/wiki/Proof_that_e_is_irrational
