We wish to show that $e$ is not a quadratic irrational, i.e. $\mathbb{Q}(e)$ is not a quadratic extension of $\mathbb{Q}$ . To do this, we show that it can not be the root of any quadratic polynomial with integer coefficients.

We begin by looking at the Taylor series for $e^x$ : \begin{equation*} e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}. \end{equation*} This converges for every $x\in\mathbb{R}$ , so $e=\sum_{k=0}^{\infty}\frac{1}{k!}$ and $e^{-1}=\sum_{k=0}^{\infty}(-1)^k\frac{1}{k!}$ . Arguing by contradiction, assume $ae^2+be+c=0$ for integers $a$ , $b$ and $c$ . That is the same as $ae+b+ce^{-1}=0$ .

Fix $n>\abs{a}+\abs{c}$ , then $a,c\mid n!$ and $\forall k\le n$ , $k!\mid n!\;$ . Consider