PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: High
[parent] e is not a quadratic irrational (Proof)

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$:

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

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\le 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\le n$, the first two terms are integers. So the third term should be an integer. However,
$\displaystyle \left\lvert \sum_{k=n+1}^\infty (a+c(-1)^k)\frac{n!}{k!}\right\rvert$ $\displaystyle \le (\left\lvert a\right\rvert +\left\lvert c\right\rvert )\sum_{k=n+1}^\infty \frac{n!}{k!}$    
  $\displaystyle =(\left\lvert a\right\rvert +\left\lvert c\right\rvert )\sum_{k=n+1}^\infty \frac{1}{(n+1)(n+2)\dotsb k}$    
  $\displaystyle \le (\left\lvert a\right\rvert +\left\lvert c\right\rvert )\sum_{k=n+1}^\infty (n+1)^{n-k}$    
  $\displaystyle =(\left\lvert a\right\rvert +\left\lvert c\right\rvert )\sum_{t=1}^\infty (n+1)^{-t}$    
  $\displaystyle =(\left\lvert a\right\rvert +\left\lvert c\right\rvert )\frac{1}{n}$    

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

is not identically zero. The contradiction completes the proof.



"e is not a quadratic irrational" is owned by mathcam. [ full author list (3) | owner history (1) ]
(view preamble)

View style:

See Also: e is irrational, $e^r$ is irrational for $r\in\mathbb{Q}\setminus\{0\}$, e is transcendental


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: 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 4085 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)
Pending Errata and Addenda
None.
[ View all 4 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)