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 $e^{x}$:

 $e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}.$

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 $n>\left\lvert a\right\rvert+\left\lvert c\right\rvert$, then $a,c\mid n!$ and $\forall k\leq n$, $k!\mid n!\;$. Consider

 $\displaystyle 0=n!(ae+b+ce^{-1})$ $\displaystyle=an!\sum_{k=0}^{\infty}\frac{1}{k!}+b+cn!\sum_{k=0}^{\infty}(-1)^% {k}\frac{1}{k!}$ $\displaystyle=b+\sum_{k=0}^{n}(a+c(-1)^{k})\frac{n!}{k!}+\sum_{k=n+1}^{\infty}% (a+c(-1)^{k})\frac{n!}{k!}$

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

 $\displaystyle\left\lvert\sum_{k=n+1}^{\infty}(a+c(-1)^{k})\frac{n!}{k!}\right\rvert$ $\displaystyle\leq(\left\lvert a\right\rvert+\left\lvert c\right\rvert)\sum_{k=% n+1}^{\infty}\frac{n!}{k!}$ $\displaystyle=(\left\lvert a\right\rvert+\left\lvert c\right\rvert)\sum_{k=n+1% }^{\infty}\frac{1}{(n+1)(n+2)\cdots k}$ $\displaystyle\leq(\left\lvert a\right\rvert+\left\lvert c\right\rvert)\sum_{k=% n+1}^{\infty}(n+1)^{n-k}$ $\displaystyle=(\left\lvert a\right\rvert+\left\lvert c\right\rvert)\sum_{t=1}^% {\infty}(n+1)^{-t}$ $\displaystyle=(\left\lvert a\right\rvert+\left\lvert c\right\rvert)\frac{1}{n}$

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

 $\sum_{k=n+1}^{\infty}(a+c(-1)^{k})\frac{1}{k!}-\sum_{k=n+2}^{\infty}(a+c(-1)^{% k})\frac{1}{k!}=(a+c(-1)^{n+1})\frac{1}{(n+1)!}$

is not identically zero. The contradiction completes the proof.

Title e is not a quadratic irrational EIsNotAQuadraticIrrational 2013-03-22 14:04:06 2013-03-22 14:04:06 mathcam (2727) mathcam (2727) 11 mathcam (2727) Proof msc 11J72 msc 26E99 EIsIrrationalProof ErIsIrrationalForRinmathbbQsetminus0 EIsTranscendental