Irrationality of $e$

I saw this proof and it looks exciting, but I don't know why this part is correct:

$$(2k - 1)!e^{-1} \in \mathbb{Z}$$

  • 5,978
  • 3
  • 13
  • 29

2 Answers2


First, note that $e \not = 0$; indeed, it is strictly greater than $1$ by examining the series of strictly positive terms.

Now, if $e$ is rational, then there is some $K$ such that $e^{-1} (2K-1)!$ is integer (since $e^{-1}$ has a denominator); and incidentally, it's true that for all $k \geq K$, we have $e^{-1} (2k-1)!$ rational.

Patrick Stevens
  • 34,379
  • 5
  • 38
  • 88

This is a proof by contradiction that $e^{-1}$ is irrational. The author presumes that $e^{-1}$ is rational, in which case $e^{-1}=x/y$ for some integers $x$ and $y$. If this is true, then there is some $k$ such that $(2k-1)!e^{-1}$ is an integer- we just need to choose $k=y$, because then $(2k-1)!$ has a factor of $y$, so the denominator is cancelled out. This reaches a contradiction, because then that integer would be strictly between $0$ and $1/2$, proving $e^{-1}$ to be irrational. This does not prove $e^{-1}(2k-1)!$ is an integer, because it made a false assumption.

Just to be clear, "$e^{-1}(2k-1)!$ is an integer" is false. Let $p$ be irrational, $q$ be a nonzero rational, and $r$ be rational. Assume $pq=r$. Then $p=r/q$, meaning that $p$ is rational, a contradiction (you can multiply those ratios of integers to get another ratio of integers). Then setting $p=e^{-1}$, q=(2k-1)!, r=pq$, we can see that their product must be irrational, i.e. not an integer.

Kevin Long
  • 5,041
  • 3
  • 19
  • 33