e is not a quadratic irrational

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 $e^x$:

$$e^x=\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}.$$

This converges for every $x\in ℝ$, so $e={\sum }_{k=0}^{\mathrm{\infty }}\frac{1}{k!}$ and $e^{-1}={\sum }_{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{1}{k!}$. Arguingby contradiction, assume $a{e}^2+be+c=0$ for integers$a$, $b$ and $c$. That is the same as $ae+b+c{e}^{-1}=0$.

Fix $n>\left|a\right|+\left|c\right|$, then $a,c\mid n!$ and $\forall k\le n$, $k!\mid n!$.Consider

$0=n!(ae+b+c{e}^{-1})$$=an!{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \frac{1}{k!}}+b+cn!{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{(-1)}^k{\displaystyle \frac{1}{k!}}$

$=b+{\displaystyle \sum _{k=0}^n}(a+c{(-1)}^k){\displaystyle \frac{n!}{k!}}+{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}(a+c{(-1)}^k){\displaystyle \frac{n!}{k!}}$

Since $k!\mid n!$ for $k\le n$, the first two terms are integers. So the third term should be an integer. However,

$\left|{\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}(a+c{(-1)}^k){\displaystyle \frac{n!}{k!}}\right|$$\le (\left|a\right|+\left|c\right|){\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{\displaystyle \frac{n!}{k!}}$

$=(\left|a\right|+\left|c\right|){\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{(n+1)(n+2)\mathrm{\cdots }k}}$

$\le (\left|a\right|+\left|c\right|){\displaystyle \sum _{k=n+1}^{\mathrm{\infty }}}{(n+1)}^{n-k}$

$=(\left|a\right|+\left|c\right|){\displaystyle \sum _{t=1}^{\mathrm{\infty }}}{(n+1)}^{-t}$

$=(\left|a\right|+\left|c\right|){\displaystyle \frac{1}{n}}$

is less than $1$ by our assumption that $n>\left|a\right|+\left|c\right|$. Since there is only one integer which is less than $1$ in absolute value, this means that ${\sum }_{k=n+1}^{\mathrm{\infty }}(a+c{(-1)}^k)\frac{1}{k!}=0$ for every sufficiently large $n$ which is not the case because

$$\sum _{k=n+1}^{\mathrm{\infty }}(a+c{(-1)}^k)\frac{1}{k!}-\sum _{k=n+2}^{\mathrm{\infty }}(a+c{(-1)}^k)\frac{1}{k!}=(a+c{(-1)}^{n+1})\frac{1}{(n+1)!}$$

is not identically zero. The contradiction completes the proof.