proof of Lindemann-Weierstrass theorem and that e and π are transcendental

This article provides a proof of the Lindemann-Weierstrass theoremMathworldPlanetmath, using a method similarPlanetmathPlanetmath to those used by Ferdinand von Lindemann and Karl Weierstrass. This material is taken from [1] and expanded for clarity.

Before attacking the general case, we first use the same methods to prove two earlier theorems, namely that both e and π are transcendentalPlanetmathPlanetmath. These proofs introduce the methods to be used in the more general theorem. At the end, we present some trivial but important corollaries.

Both e and π were both known to be irrational in the 1700’s (Euler showed the former; Lambert the latter). But e was not shown to be transcendental until 1873 (by Hermite, see [3] and [4]), and Lindemann showed π to be transcendental as well in the late 1870’s. He also sketched a proof of the general theorem, which was fleshed out by Weierstrass and Hilbert among others in the late 1800’s.

The following construct is used in all three proofs. Suppose f(x) is a real polynomial, and let


Integrating by parts (, we get


Continuing, and integrating by parts a total of m=degf times, we get

I(t)=etj=0mf(j)(0)-j=0mf(j)(t) (1)

where f(j)(x) is the jth derivativePlanetmathPlanetmath of f.

If f(x)=aixi, let F(x)=|ai|xi; i.e. (, the polynomialMathworldPlanetmathPlanetmath whose coefficients are the absolute valuesMathworldPlanetmathPlanetmathPlanetmathPlanetmath of those for f. Then using trivial bounds ( on the integrand, we get

|I(t)|0t|et-xf(x)|𝑑x|t|e|t|F(|t|). (2)

We now proceed to prove the theorems. The proofs of all three are similar, although the proof for e is the easiest. The steps of the proofs are as follows:

  • Assume the theorem is false, and write down an equation in exponentialPlanetmathPlanetmath form that shows that the number in question is algebraic (for π, we will use eiπ=-1 to write the equation in that form).

  • Define a polynomial or set of polynomials f, and an associated number J (or a sequenceMathworldPlanetmath of numbers) that is a linear combinationMathworldPlanetmath of the values of I at the exponents in question. The motivation for the choice of f used in each theorem is not given in Hermite’s proof. An excellent exposition of how these definitions are relevant to the problem is given in [2]. In essence, the ratio of the terms of I(t) in the proof that e is transcendental relates to a Padé approximation to et; this approximation is better the larger degf gets.

  • Analyze J to show that it is integralPlanetmathPlanetmath ( and nonzero, and derive a lower boundMathworldPlanetmath on J.

  • Use equation (2) to derive a trivial upper bound on J.

  • Note that the upper bound is lower than the lower bound, disproving the original assumptionPlanetmathPlanetmath.

The “magic” in the proof consists of finding the appropriate choice of f, as well as in defining the transformation I to begin with. Once that is done, the work in the proof is in showing J integral (which is harder for the more general theorems) and in deriving the lower bound. But the outline of the proof remains the same across all three theorems.

Theorem 1.

e is transcendental.


Suppose not, so that e is algebraic. Then e satisfies some integer polynomial aixi=0 with a00. This means that


Let p be a (sufficiently large) prime, define a polynomial of degree m=(n+1)p-1 by


and let


We derive two sets of inconsistentPlanetmathPlanetmath bounds on J, thus showing that the original hypothesis is false and e is transcendental.

First, apply equation (1) to J:

J =k=0nakI(k)

where the last equality follows because of the assumed linear dependence of the powers of e.

Consider the values of f(j)(k). If j<p-1, then none of the factors in f vanish by differentiationMathworldPlanetmath, so that f(j)(k)=0 for all k. If j=p-1, then only the initial factor xp-1 can vanish in any term in f(j)(x), and so f(j)(k)=0 if k>0. In the case where j=p-1,k=0, the only term in the derivative that contributes a nonzero value is the term where xp-1 is differentiated each time; that term gives


Finally, if jp, then the terms in f(j)(k) that are nonzero are those terms in which (x-k)p has been differentiated away; in these cases, the terms have a leading coefficient that is thus a multipleMathworldPlanetmath of p!.

Now assume p>n; then f(p-1)(0) is a multiple of (p-1)! but not of p!. Putting together the above computations, we get


where M,N are integers and pM. Assume also p>a0; then Np+a0M must be nonzero and thus |J|(p-1)!.

On the other hand, obviously F(k)(2n)m and thus


where a=max(|a1|,,|an|) and c is some constant that does not depend on p.

But for p large enough, (p-1)!>cp no matter what c is, so these two bounds on J are contradictory. ∎

Theorem 2.

π is transcendental.


Again suppose not. Then iπ is also algebraic. Suppose the minimal polynomialPlanetmathPlanetmath, f, of iπ has degree d, say


with a00, and let θ1=iπ,θ2,,θd be the conjugatesPlanetmathPlanetmath of iπ (the other roots of f). Then since eiπ=-1, we have


Note that since θi is algebraic for each i, then adθi is an algebraic integerMathworldPlanetmath.

Each term in this productMathworldPlanetmathPlanetmath can be written as a power of e, where the exponent is of the form


and each ϵi is either 0 or 1. Denote by α1,,αn those exponents that are nonzero. Note that at least one exponent is zero, and thus n<2d. We then have

(2d-n)+eα1+eαn=0 (3)

We will show that if we define f by


with p a (sufficiently large) prime, then


satisfies the same incompatible bounds as in the previous theorem.

Let m=degf=(n+1)p-1. As before, we see that

J =k=1nI(αk)

where the last equality follows from equation (3).

The remainder of the proof is quite similar to the above proof for e, except that we must first show that the sum over k above is an integer; this was clear in the previous theorem.

Consider the inner sum over k. It is clear that this is a symmetric polynomialMathworldPlanetmath with integer coefficients in adα1,,adαn. The adαi are algebraic integers since the adθi are. By the fundamental theorem of symmetric polynomials, it follows that the inner sum is in fact a polynomial in the elementary symmetric functions on the adαi. Since the αi are the nonzero elements of the β, we see that the sum is also a polynomial in the elementary symmetric functions of the adβ, and thus is a symmetric polynomial with integer coefficients in the adθi. Again applying the fundamental theorem of symmetric polynomials, we see that the sum over k must be a polynomial in the elementary symmetric functions of the adθi. But these elementary symmetric functions are simply the coefficients of the minimal polynomial of adiπ, which are integers. Thus the inner sum is an integer.

By argumentsMathworldPlanetmath identical to those in the previous theorem, we have that f(j)(αk)=0 when j<p and thus that f(j)(αk) is an integral multiple of p!; that f(j)(0) is an integral multiple of p! when jp-1; and that


is an integral multiple of (p-1)! but is not divisible by p if p is chosen to exceed adα1αn. Thus if also p>n-2d, we have that J is nonzero and divisible by (p-1)! and thus |J|(p-1)!. But again, similar to the proof in the previous theorem, we have that


and again these estimates are contradictory. ∎

The Lindemann-Weierstrass theorem generalizes both these two statements and their proofs.

Theorem 3.

If α1,,αn are algebraic and distinct, and if β1,,βn are algebraic and non-zero, then


Note that the facts that e and π are transcendental follow trivially from this theorem. For example, if e were algebraic, then e is the root of a polynomial βixi where βi, in contradictionMathworldPlanetmathPlanetmath to the theorem.


The proof follows the same general lines as above, but there are additional complexities introduced by the arbitrary αi. In the proof of the transcendality of π, we were able to use facts about the relationship of the exponents in the proof; no such relationship is available to us in this more general setting.

Again start by supposing

β1eα1+βneαn=0 (4)

where the αi,βi are as given.

Claim we can assume, without loss of generality, that βi. For if not, take all the expressions formed by substituting for one or more of the βi one of its conjugates, and multiply those by the equation above. The result is a new expression of the same form (with different αi), but where the coefficients are rational numbers. Clear denominators, proving the claim.

Next, claim we can assume that the αi are a complete set of conjugates, and that if αi,αj are conjugates, then βi=βj. To see this, choose an irreduciblePlanetmathPlanetmathPlanetmath ( integral polynomial having α1,,αn as roots; let αn+1,,αN be the remaining roots, and define βn+1=βN=0. Then clearly we have


(Note the similarity with the proof for π). There are N! factors in this product, so expanding the product, it is a sum of terms of the form


with integral coefficients, and h1++hN=N!. Clearly the set of all such exponents forms a complete set of conjugates. By symmetryPlanetmathPlanetmath considerations, we see that the coefficients of two conjugate terms are equal. Also, the product is not identically zero. To see this, consider the term in the product formed by multiplying together, from each factor, the nonzero terms with the largest exponents in the lexicographic order on . Since the αi are unique (because the polynomial is irreducible), there is only one term with this largest exponent, and it has a nonzero coefficient by construction.

Finally, order the terms so that the conjugates of a particular αi appear together. That is, for the remainder of the proof we may assume that


with the βi,and that there are integers 0=n0<n1<<nr=n chosen so that, foreach 0t<n we have

αnt+1,,αnt+1 form a complete set of conjugates

Now, since αi,βi are algebraic, we can choose l such that lαi,lβi are algebraic integers. Let

fi(x)=lnp((x-α1)(x-αn))p(x-αi), 1in

where again p is a (large) prime. We will develop contradictory estimates for |J1Jn|, where

Ji=β1Ii(α1)++βnIi(αn), 1in

and Ii is the integral ( associated with fi, as above (see equation (1)).

Using equations (1) and (4), we see that

Ji =k=1nβkIi(αk)

Arguing similarly to the foregoing proofs, we see that fi(j)(αk) is an algebraic integer divisible by p! unless j=p-1 and k=i. In this particular case, we have that


and so again, if p is large enough, this is divisible by (p-1)! but not by p!. Thus Ji is a nonzero algebraic integer divisible by (p-1)! but not by p!. As before, we can prove that Ji0.

Ji can be written as follows:


Note that by construction, fi(x) can be written as a polynomial whose coefficients are polynomials in αi, and the integral coefficients of those polynomials are integers independent of i. Thus, noting that the αi form a complete set of conjugates and using the fundamental theorem on symmetric polynomials as in the previous proof, we see that the product of the Ji is in fact a rational number. But it is an algebraic integer, hence an integer. Thus J1Jn, and it is divisible by ((p-1)!)n. Thus |J1Jn|(p-1)!. But the same estimate as in the previous proofs shows that for each i,


which as before is cp for some sufficiently large c. These estimates are again in contradiction, proving the theorem.

Note that Theorems 1 and 2 are trivial corollaries of Theorem 3, as one would expect. (To prove that π is transcendental, note that if it were algebraic, then eiπ=-1 would be transcendental).

Here are some other more or less trivial corollaries.

Corollary 4.

If α0 is algebraic, then eiα is transcendental.


If it were algebraic, say eiα=β, then we have


in contradiction to the above theorem since α0. ∎

Corollary 5.

If α0 is algebraic, then cosα and sinα are both transcendental.


Recall that cosα+isinα=eiα, which is transcendental. If either cosα or sinα were algebraic, then the other would be as well (and thus their sum would be) since cos2α+sin2α=1. Therefore, both cosα and sinα are transcendental. ∎

Corollary 6.

If α>0 is algebraic with α1, then lnα is transcendental.


If β=lnα, then eβ=α. By Corollary 4, since α is algebraic, β cannot be. ∎


  • 1 A. Baker, Transcendental NumberMathworldPlanetmath Theory, Cambridge University Press, 1990.
  • 2 H. Cohn, A Short Proof of the Simple Continued FractionMathworldPlanetmath Expansion of e, American Mathematical Monthly, Jan. 2006, pp. 57-62.
  • 3 C. Hermite, Sur la fonction exponentielle, Compte Rendu Acad. Sci. 77 (1873) 18-24, 74-79, 226-233, and 285-293; also in Œuvres, v. 3, Gauthier-Villiers, Paris, 1912 pp. 150-181.
  • 4 U. G. Mitchell, M. Strain, Osiris, Vol. 1, Jan. 1936, pp. 476-496.
Title proof of Lindemann-Weierstrass theorem and that e and π are transcendental
Canonical name ProofOfLindemannWeierstrassTheoremAndThatEAndpiAreTranscendental
Date of creation 2013-03-22 17:07:55
Last modified on 2013-03-22 17:07:55
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