e is transcendental


Napier’s constant e is transcendental.

This theoremMathworldPlanetmath was first proved by Hermite in 1873.  The below proof is near the one given by Hurwitz.  We at first derive a couple of auxiliary results.

Let f(x) be any polynomialPlanetmathPlanetmath of μ and F(x) the sum of its derivativesPlanetmathPlanetmath,

F(x):=f(x)+f(x)+f′′(x)++f(μ)(x). (1)

consider the productPlanetmathPlanetmathΦ(x):=e-xF(x).  The derivative of this is simply


Applying the mean value theorem (http://planetmath.org/MeanValueTheorem) to the function Φ on the interval with end points 0 and x gives


which implies that  F(0)=e-xF(x)+e-ξf(ξ)x.  Thus we obtain the

Lemma 1.F(0)ex=F(x)+xex-ξf(ξ) (ξ is between 0 and x)

When the polynomial f(x) is expanded by the powers of x-a, one gets


comparing this with (1) one gets the

Lemma 2.  The value F(a) is obtained so that the polynomial f(x) is expanded by the powers of x-a and in this the powers x-a, (x-a)2, …, (x-a)μ are replaced respectively by the numbers 1!, 2!, …, μ!.

Now we begin the proof of the theorem.  We have to show that there cannot be any equation

c0+c1e+c2e2++cnen= 0 (2)

with integer coefficients ci and at least one of them distinct from zero.  The proof is indirect.  Let’s assume the contrary.  We can presume that  c00.

For any positive integer ν, lemma 1 gives

F(0)eν=F(ν)+νeν-ξνf(ξν)(0<ξν<ν). (3)

By virtue of this, one may write (2), multiplied by F(0), as

c0F(0)+c1F(1)+c2F(2)++cnF(n)=-[c1e1-ξ1f(ξ1)+2c2e2-ξ2f(ξ2)++ncnen-ξnf(ξn)]. (4)

We shall show that the polynomial f(x) can be chosen such that the left has absolute valueMathworldPlanetmathPlanetmathPlanetmath less than 1.

We choose

f(x):=xp-1(p-1)![(x-1)(x-2)(x-n)]p, (5)

where p is a positive prime numberMathworldPlanetmath on which we later shall set certain conditions.  We must determine the corresponding values F(0), F(1), …, F(n).

For determining F(0) we need, according to lemma 2, to expand f(x) by the powers of x, getting


where A1,A2, are integers, and to replace the powers xp-1, xp, xp+1, … with the numbers (p-1)!, p!, (p+1)!, …  We then get the expression


in which K0 is an integer.

We now set for the prime p the condition  p>n.  Then, n! is not divisible by p, neither is the former addend (-1)npn!p.  On the other hand, the latter addend pK0 is divisible by p.  Therefore:
(α) F(0) is a non-zero integer not divisible by p.

For determining F(1), F(2), …, F(n) we expand the polynomial f(x) by the powers of x-ν, putting  x:=ν+(x-ν).  Because f(x) the factor (x-ν)p, we obtain an of the form


where the Bi’s are integers.  Using the lemma 2 then gives the result


with Kν a certain integer.  Thus:
(β) F(1), F(2), …, F(n) are integers all divisible by p.

So, the left hand of (4) is an integer having the form c0F(0)+pK with K an integer.  The factor F(0) of the first addend is by (α) indivisible by p.  If we set for the prime p a new requirement  p>|c0|,  then also the factor c0 is indivisible by p, and thus likewise the whole addend c0F(0).  We conclude that the sum is not divisible by p and therefore:
(γ) If p in (5) is a prime number greater than n and |c0|, then the left of (4) is a nonzero integer.

We then examine the right hand of (4).  Because the numbers ξ1, …, ξn all are positive (cf. (3)), so the e1-ξ1, …, en-ξn all are <en.  If  0<x<n, then in the polynomial (5) the factors x, x-1, …, x-n all have the absolute value less than n and thus


Because ξ1, …, ξn all are between 0 and n (cf. (3)), we especially have

|f(ξν)|<nn(nn+1)p-1(p-1)!ν=1, 2,,n.

If we denote by c the greatest of the numbers |c0|, |c1|, …, |cn|, then the right hand of (4) has the absolute value less than


But the limit of (nn+1)p-1(p-1)! is 0 as  p, and therefore the above expression is less than 1 as soon as p exeeds some number p0.

If we determine the polynomial f(x) from the equation (5) such that the prime p is greater than the greatest of the numbers n, |c0| and p0 (which is possible since there are infinitely many prime numbers (http://planetmath.org/ProofThatThereAreInfinitelyManyPrimes)), then the having the absolute value <1.  The contradictionMathworldPlanetmathPlanetmath proves that the theorem is right.


  • 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I.  WSOY, Helsinki (1950).
Title e is transcendental
Canonical name EIsTranscendental
Date of creation 2013-03-22 15:10:32
Last modified on 2013-03-22 15:10:32
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 40
Author pahio (2872)
Entry type Theorem
Classification msc 26C05
Classification msc 11J82
Classification msc 11J81
Synonym e is transcendental
Synonym transcendence of e
Related topic NaturalLogBase
Related topic FundamentalTheoremOfTranscendence
Related topic LindemannWeierstrassTheorem
Related topic EIsIrrational
Related topic ErIsIrrationalForRinmathbbQsetminus0
Related topic ExampleOfTaylorPolynomialsForTheExponentialFunction
Related topic ProofThatEIsNotANaturalNumber