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 $e^{-1}={\sum }_{k=0}^{\mathrm{\infty }}{(-1)}^k\frac{1}{k!}$. Arguing by 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.

Title	e is not a quadratic irrational
Canonical name	EIsNotAQuadraticIrrational
Date of creation	2013-03-22 14:04:06
Last modified on	2013-03-22 14:04:06
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	11
Author	mathcam (2727)
Entry type	Proof
Classification	msc 11J72
Classification	msc 26E99
Related topic	EIsIrrationalProof
Related topic	ErIsIrrationalForRinmathbbQsetminus0
Related topic	EIsTranscendental