e is not a quadratic irrational

We wish to show that e is not a quadratic irrational, i.e. (e) is not a quadratic extension of . 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 ex:


This converges for every x, so e=k=01k! and e-1=k=0(-1)k1k!. Arguing by contradictionMathworldPlanetmathPlanetmath, assume ae2+be+c=0 for integers a, b and c. That is the same as ae+b+ce-1=0.

Fix n>|a|+|c|, then a,cn! and kn, k!n!. Consider

0=n!(ae+b+ce-1) =an!k=01k!+b+cn!k=0(-1)k1k!

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

|k=n+1(a+c(-1)k)n!k!| (|a|+|c|)k=n+1n!k!

is less than 1 by our assumptionPlanetmathPlanetmath that n>|a|+|c|. Since there is only one integer which is less than 1 in absolute valueMathworldPlanetmathPlanetmathPlanetmathPlanetmath, this means that k=n+1(a+c(-1)k)1k!=0 for every sufficiently large n which is not the case because


is not identically zero. The contradiction completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath 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