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e is not a quadratic irrational
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(Proof)
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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
Since $k!\mid n!$ for $k\le n$ , the first two terms are integers. So the third term should be an integer. However,
is less than $1$ by our assumption that $n>\abs{a}+\abs{c}$ . 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 \begin{equation*} \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)!} \end{equation*}is not identically zero. The contradiction completes the proof.
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Cross-references: proof, completes, absolute value, terms, fix, contradiction, converges, Taylor series, coefficients, integer, polynomial, root, quadratic extension, irrational
There is 1 reference to this entry.
This is version 8 of e is not a quadratic irrational, born on 2003-11-21, modified 2005-03-18.
Object id is 5426, canonical name is EIsIrrational.
Accessed 5198 times total.
Classification:
| AMS MSC: | 26E99 (Real functions :: Miscellaneous topics :: Miscellaneous) | | | 11J72 (Number theory :: Diophantine approximation, transcendental number theory :: Irrationality; linear independence over a field) |
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Pending Errata and Addenda
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