Lindemann–Weierstrass theorem

The theorem is also known variously as the Hermite–Lindemann theorem and the Hermite–Lindemann–Weierstrass theorem. Charles Hermite first proved the simpler theorem where the {{math|α_i}} exponents are required to be rational integers and linear independence is only assured over the rational integers,{{Harvnb|Hermite|1873|pp=18–24}}.{{Harvnb|Hermite|1874}} a result sometimes referred to as Hermite's theorem.{{Harvnb|Gelfond|2015}}. Although that appears to be a special case of the above theorem, the general result can be reduced to this simpler case. Lindemann was the first to allow algebraic numbers into Hermite's work in 1882.{{Harvnb|Lindemann|1882a}}, {{Harvnb|Lindemann|1882b}}. Shortly afterwards Weierstrass obtained the full result,{{Harvnb|Weierstrass|1885|pp=1067–1086}}, and further simplifications have been made by several mathematicians, most notably by David Hilbert{{Harvnb|Hilbert|1893|pp=216–219}}. and Paul Gordan.{{Harvnb|Gordan|1893|pp=222–224}}.

{{also|e (mathematical constant)|Pi}}

The transcendence of {{math| e }} and {{pi}} are direct corollaries of this theorem.

Suppose {{math| α }} is a non-zero algebraic number; then {{math| {α} }} is a linearly independent set over the rationals, and therefore by the first formulation of the theorem {{math| {e^α} }} is an algebraically independent set; or in other words {{math| e^α }} is transcendental. In particular, {{math| e¹ {{=}} e }} is transcendental. (A more elementary proof that {{math| e }} is transcendental is outlined in the article on transcendental numbers.)

Alternatively, by the second formulation of the theorem, if {{math| α }} is a non-zero algebraic number, then {{math| {0, α} }} is a set of distinct algebraic numbers, and so the set {{math| {e⁰, e^α} {{=}} {1, e^α} }} is linearly independent over the algebraic numbers and in particular {{math| e^α }} cannot be algebraic and so it is transcendental.

To prove that {{pi}} is transcendental, we prove that it is not algebraic. If {{pi}} were algebraic, {{pi}}i would be algebraic as well, and then by the Lindemann–Weierstrass theorem {{math| e^{{pi}}i {{=}} −1 }} (see Euler's identity) would be transcendental, a contradiction. Therefore {{pi}} is not algebraic, which means that it is transcendental.

A slight variant on the same proof will show that if {{math| α }} is a non-zero algebraic number then {{math| sin(α), cos(α), tan(α) }} and their hyperbolic counterparts are also transcendental.

{{math_theorem|name={{math|p}}-adic Lindemann–Weierstrass Conjecture.|math_statement=Suppose {{math| p }} is some prime number and {{math| α₁, ..., α_n }} are p-adic numbers which are algebraic and linearly independent over $\backslash mathbb\{Q\}$, such that {{math| {{!}} α_i {{!}}_p < 1/p }} for all {{math| i; }} then the p-adic exponential functions {{math| exp_p(α₁), . . . , exp_p(α_n) }} are {{math|p}}-adic numbers that are algebraically independent over $\backslash mathbb\{Q\}$. }}

An analogue of the theorem involving the modular function {{math| j }} was conjectured by Daniel Bertrand in 1997, and remains an open problem.{{Harvnb|Bertrand|1997|pp=339–350}}. Writing {{math| q {{=}} e^2{{pi}}iτ }} for the square of the nome and {{math| j(τ) {{=}} J(q), }} the conjecture is as follows. {{math_theorem|name=Modular conjecture|Let {{math| q₁, ..., q_n }} be non-zero algebraic numbers in the complex unit disc such that the {{math| 3n }} numbers

:$\backslash left\; \backslash \{\; J(q\_1),\; J\text{'}(q\_1),\; J$(q_1), \ldots, J(q_n), J'(q_n), J(q_n) \right \}

are algebraically dependent over $\backslash mathbb\{Q\}$. Then there exist two indices {{math| 1 ≤ i < j ≤ n }} such that {{math| q_i }} and {{math| q_j }} are multiplicatively dependent.}}

{{math_theorem|name=Lindemann–Weierstrass Theorem (Baker's reformulation).|math_statement=If {{math| a₁, ..., a_n }} are algebraic numbers, not all zero, and {{math| α₁, ..., α_n }} are distinct algebraic numbers, then{{citation

| last = Baker | first = Alan

| doi = 10.1017/CBO9781139093835

| isbn = 978-1-107-60379-0

| mr = 2954465

| page = 53

| publisher = Cambridge University Press, Cambridge

| title = A Comprehensive Course in Number Theory

| year = 2012}}

:$a\_1\; e^\{\backslash alpha\_1\}\; +\; a\_2\; e^\{\backslash alpha\_2\}\; +\; \backslash cdots\; +\; a\_n\; e^\{\backslash alpha\_n\}\; \backslash ne\; 0.$}}

The proof relies on two preliminary lemmas. Notice that Lemma B itself is already sufficient to deduce the original statement of Lindemann–Weierstrass theorem.

{{math_theorem|name=Lemma A.|math_statement=Let {{math|c(1), ..., c(r)}} be integers and, for every {{mvar|k}} between {{math|1}} and {{mvar|r}}, let {{math|{γ(k)₁, ..., γ(k)_m(k)} }} be the roots of a non-zero polynomial with integer coefficients $T\_k(x)$. If {{math|γ(k)_i ≠ γ(u)_v }} whenever {{math|(k, i) ≠ (u, v)}}, then

:$c(1)\backslash left\; (e^\{\backslash gamma(1)\_1\}+\backslash cdots+\; e^\{\backslash gamma(1)\_\{m(1)\}\}\; \backslash right\; )\; +\; \backslash cdots\; +\; c(r)\; \backslash left\; (e^\{\backslash gamma(r)\_1\}+\backslash cdots+\; e^\{\backslash gamma(r)\_\{m(r)\}\}\; \backslash right)\; =\; 0$

has only the trivial solution $c(i)=0$ for all $i\; =\; 1,\; \backslash dots,\; r.$}}

Proof of Lemma A. To simplify the notation set:

:$$

\begin{align}

& n_0 =0, & & \\

& n_i =\sum\nolimits_{k=1}^i m(k), & & i=1,\ldots,r \\

& n=n_r, & & \\

& \alpha_{n_{i-1}+j} =\gamma(i)_j, & & 1\leq i\leq r,\ 1\leq j\leq m(i) \\

& \beta_{n_{i-1}+j} =c(i).

\end{align}

Then the statement becomes

:$\backslash sum\_\{k=1\}^n\; \backslash beta\_k\; e^\{\backslash alpha\_k\}\backslash neq\; 0.$

Let {{mvar|p}} be a prime number and define the following polynomials:

: $f\_i(x)\; =\; \backslash frac\; \{\backslash ell^\{np\}\; (x-\backslash alpha\_1)^p\; \backslash cdots\; (x-\backslash alpha\_n)^p\}\{(x-\backslash alpha\_i)\},$

where {{mvar|ℓ}} is a non-zero integer such that $\backslash ell\backslash alpha\_1,\backslash ldots,\backslash ell\backslash alpha\_n$ are all algebraic integers. DefineUp to a factor, this is the same integral appearing in Transcendental number#A proof that e is transcendental, where {{math|β₁ {{=}} 1, ..., β_m {{=}} m.}} The rest of the proof of the Lemma is analog to that proof.

: $I\_i(s)\; =\; \backslash int^s\_0\; e^\{s-x\}\; f\_i(x)\; \backslash ,\; dx.$

Using integration by parts we arrive at

: $I\_i(s)\; =\; e^s\; \backslash sum\_\{j=0\}^\{np-1\}\; f\_i^\{(j)\}(0)\; -\; \backslash sum\_\{j=0\}^\{np-1\}\; f\_i^\{(j)\}(s),$

where $np-1$ is the degree of $f\_i$, and $f\_i^\{(j)\}$ is the j-th derivative of $f\_i$. This also holds for s complex (in this case the integral has to be intended as a contour integral, for example along the straight segment from 0 to s) because

:$-e^\{s-x\}\; \backslash sum\_\{j=0\}^\{np-1\}\; f\_i^\{(j)\}(x)$

is a primitive of $e^\{s-x\}\; f\_i(x)$.

Consider the following sum:

:$\backslash begin\{align\}$

J_i &=\sum_{k=1}^n\beta_k I_i(\alpha_k)\\[5pt]

&= \sum_{k=1}^n\beta_k \left ( e^{\alpha_k} \sum_{j=0}^{np-1} f_i^{(j)}(0) - \sum_{j=0}^{np-1} f_i^{(j)}(\alpha_k)\right ) \\[5pt]

&=\left(\sum_{j=0}^{np-1}f_i^{(j)}(0)\right)\left(\sum_{k=1}^n \beta_k e^{\alpha_k}\right)-\sum_{k=1}^n\sum_{j=0}^{np-1} \beta_kf_i^{(j)}(\alpha_k)\\[5pt]

&= -\sum_{k=1}^n \sum_{j=0}^{np-1} \beta_kf_i^{(j)}(\alpha_k)

\end{align}

In the last line we assumed that the conclusion of the Lemma is false. In order to complete the proof we need to reach a contradiction. We will do so by estimating $|J\_1\backslash cdots\; J\_n|$ in two different ways.

First $f\_i^\{(j)\}(\backslash alpha\_k)$ is an algebraic integer which is divisible by p! for $j\backslash geq\; p$ and vanishes for

:$\backslash ell^\{np\}(p-1)!\backslash prod\_\{k\backslash neq\; i\}(\backslash alpha\_i-\backslash alpha\_k)^p.$

This is not divisible by p when p is large enough because otherwise, putting

:$\backslash delta\_i=\backslash prod\_\{k\backslash neq\; i\}(\backslash ell\backslash alpha\_i-\backslash ell\backslash alpha\_k)$

(which is a non-zero algebraic integer) and calling $d\_i\backslash in\backslash mathbb\; Z$ the product of its conjugates (which is still non-zero), we would get that p divides $\backslash ell^p(p-1)!d\_i^p$, which is false.

So $J\_i$ is a non-zero algebraic integer divisible by (p − 1)!. Now

:$J\_i=-\backslash sum\_\{j=0\}^\{np-1\}\backslash sum\_\{t=1\}^r\; c(t)\backslash left(f\_i^\{(j)\}(\backslash alpha\_\{n\_\{t-1\}+1\})\; +\; \backslash cdots\; +\; f\_i^\{(j)\}(\backslash alpha\_\{n\_t\})\backslash right).$

Since each $f\_i(x)$ is obtained by dividing a fixed polynomial with integer coefficients by $(x-\backslash alpha\_i)$, it is of the form

:$f\_i(x)=\backslash sum\_\{m=0\}^\{np-1\}g\_m(\backslash alpha\_i)x^m,$

where $g\_m$ is a polynomial (with integer coefficients) independent of i. The same holds for the derivatives $f\_i^\{(j)\}(x)$.

Hence, by the fundamental theorem of symmetric polynomials,

:$f\_i^\{(j)\}(\backslash alpha\_\{n\_\{t-1\}+1\})+\backslash cdots+f\_i^\{(j)\}(\backslash alpha\_\{n\_t\})$

is a fixed polynomial with rational coefficients evaluated in $\backslash alpha\_i$ (this is seen by grouping the same powers of $\backslash alpha\_\{n\_\{t-1\}+1\},\backslash dots,\backslash alpha\_\{n\_t\}$ appearing in the expansion and using the fact that these algebraic numbers are a complete set of conjugates). So the same is true of $J\_i$, i.e. it equals $G(\backslash alpha\_i)$, where G is a polynomial with rational coefficients independent of i.

Finally $J\_1\backslash cdots\; J\_n=G(\backslash alpha\_1)\backslash cdots\; G(\backslash alpha\_n)$ is rational (again by the fundamental theorem of symmetric polynomials) and is a non-zero algebraic integer divisible by $(p-1)!^n$ (since the $J\_i$'s are algebraic integers divisible by $(p-1)!$). Therefore

:$|J\_1\backslash cdots\; J\_n|\backslash geq\; (p-1)!^n.$

However one clearly has:

:$|I\_i(\backslash alpha\_k)|\; \backslash leq$

where {{mvar|F_i}} is the polynomial whose coefficients are the absolute values of those of f_i (this follows directly from the definition of $I\_i(s)$). Thus

:$|J\_i|\backslash leq\; \backslash sum\_\{k=1\}^n\; \backslash left\; |\backslash beta\_k\backslash alpha\_k\; \backslash right\; |e^$

and so by the construction of the $f\_i$'s we have $|J\_1\backslash cdots\; J\_n|\backslash le\; C^p$ for a sufficiently large C independent of p, which contradicts the previous inequality. This proves Lemma A. ∎

{{math_theorem|name=Lemma B.|math_statement=If b(1), ..., b(n) are integers and γ(1), ..., γ(n), are distinct algebraic numbers, then

:$b(1)e^\{\backslash gamma(1)\}+\backslash cdots+\; b(n)e^\{\backslash gamma(n)\}\; =\; 0$

has only the trivial solution $b(i)=0$ for all $i\; =\; 1,\; \backslash dots,\; n.$}}

Proof of Lemma B: Assuming

:$b(1)e^\{\backslash gamma(1)\}+\backslash cdots+\; b(n)e^\{\backslash gamma(n)\}=\; 0,$

we will derive a contradiction, thus proving Lemma B.

Let us choose a polynomial with integer coefficients which vanishes on all the $\backslash gamma(k)$'s and let $\backslash gamma(1),\backslash ldots,\backslash gamma(n),\backslash gamma(n+1),\backslash ldots,\backslash gamma(N)$ be all its distinct roots. Let b(n + 1) = ... = b(N) = 0.

The polynomial

:$P(x\_1,\backslash dots,x\_N)=\backslash prod\_\{\backslash sigma\backslash in\; S\_N\}(b(1)\; x\_\{\backslash sigma(1)\}+\backslash cdots+b(N)\; x\_\{\backslash sigma(N)\})$

vanishes at $(e^\{\backslash gamma(1)\},\backslash dots,e^\{\backslash gamma(N)\})$ by assumption. Since the product is symmetric, for any $\backslash tau\backslash in\; S\_N$ the monomials $x\_\{\backslash tau(1)\}^\{h\_1\}\backslash cdots\; x\_\{\backslash tau(N)\}^\{h\_N\}$ and $x\_1^\{h\_1\}\backslash cdots\; x\_N^\{h\_N\}$ have the same coefficient in the expansion of P.

Thus, expanding $P(e^\{\backslash gamma(1)\},\backslash dots,e^\{\backslash gamma(N)\})$ accordingly and grouping the terms with the same exponent, we see that the resulting exponents $h\_1\backslash gamma(1)+\backslash dots+h\_N\backslash gamma(N)$ form a complete set of conjugates and, if two terms have conjugate exponents, they are multiplied by the same coefficient.

So we are in the situation of Lemma A. To reach a contradiction it suffices to see that at least one of the coefficients is non-zero. This is seen by equipping {{math|C}} with the lexicographic order and by choosing for each factor in the product the term with non-zero coefficient which has maximum exponent according to this ordering: the product of these terms has non-zero coefficient in the expansion and does not get simplified by any other term. This proves Lemma B. ∎

We turn now to prove the theorem: Let a(1), ..., a(n) be non-zero algebraic numbers, and α(1), ..., α(n) distinct algebraic numbers. Then let us assume that:

: $a(1)e^\{\backslash alpha(1)\}+\backslash cdots\; +\; a(n)e^\{\backslash alpha(n)\}\; =\; 0.$

We will show that this leads to contradiction and thus prove the theorem. The proof is very similar to that of Lemma B, except that this time the choices are made over the a(i)'s:

For every i ∈ {1, ..., n}, a(i) is algebraic, so it is a root of an irreducible polynomial with integer coefficients of degree d(i). Let us denote the distinct roots of this polynomial a(i)₁, ..., a(i)_d(i), with a(i)₁ = a(i).

Let S be the functions σ which choose one element from each of the sequences (1, ..., d(1)), (1, ..., d(2)), ..., (1, ..., d(n)), so that for every 1 ≤ i ≤ n, σ(i) is an integer between 1 and d(i). We form the polynomial in the variables $x\_\{11\},\backslash dots,x\_\{1d(1)\},\backslash dots,x\_\{n1\},\backslash dots,x\_\{nd(n)\},y\_1,\backslash dots,y\_n$

: $Q(x\_\{11\},\backslash dots,x\_\{nd(n)\},y\_1,\backslash dots,y\_n)=\backslash prod\backslash nolimits\_\{\backslash sigma\backslash in\; S\}\backslash left(x\_\{1\backslash sigma(1)\}y\_1+\backslash dots+x\_\{n\backslash sigma(n)\}y\_n\backslash right).$

Since the product is over all the possible choice functions σ, Q is symmetric in $x\_\{i1\},\backslash dots,x\_\{id(i)\}$ for every i. Therefore Q is a polynomial with integer coefficients in elementary symmetric polynomials of the above variables, for every i, and in the variables y_i. Each of the latter symmetric polynomials is a rational number when evaluated in $a(i)\_1,\backslash dots,a(i)\_\{d(i)\}$.

The evaluated polynomial $Q(a(1)\_1,\backslash dots,a(n)\_\{d(n)\},e^\{\backslash alpha(1)\},\backslash dots,e^\{\backslash alpha(n)\})$ vanishes because one of the choices is just σ(i) = 1 for all i, for which the corresponding factor vanishes according to our assumption above. Thus, the evaluated polynomial is a sum of the form

: $b(1)e^\{\backslash beta(1)\}+\; b(2)e^\{\backslash beta(2)\}+\; \backslash cdots\; +\; b(N)e^\{\backslash beta(N)\}=\; 0,$

where we already grouped the terms with the same exponent. So in the left-hand side we have distinct values β(1), ..., β(N), each of which is still algebraic (being a sum of algebraic numbers) and coefficients $b(1),\backslash dots,b(N)\backslash in\backslash mathbb\; Q$.

The sum is nontrivial: if $\backslash alpha(i)$ is maximal in the lexicographic order, the coefficient of $e^\{|S|\backslash alpha(i)\}$ is just a product of a(i)_j's (with possible repetitions), which is non-zero.

By multiplying the equation with an appropriate integer factor, we get an identical equation except that now b(1), ..., b(N) are all integers. Therefore, according to Lemma B, the equality cannot hold, and we are led to a contradiction which completes the proof. ∎

Note that Lemma A is sufficient to prove that e is irrational, since otherwise we may write e = p / q, where both p and q are non-zero integers, but by Lemma A we would have qe − p ≠ 0, which is a contradiction. Lemma A also suffices to prove that {{pi}} is irrational, since otherwise we may write {{pi}} = k / n, where both k and n are integers) and then ±i{{pi}} are the roots of n²x² + k² = 0; thus 2 − 1 − 1 = 2e⁰ + e^i{{pi}} + e^−i{{pi}} ≠ 0; but this is false.

Similarly, Lemma B is sufficient to prove that e is transcendental, since Lemma B says that if a₀, ..., a_n are integers not all of which are zero, then

: $a\_ne^n+\backslash cdots+a\_0e^0\backslash ne\; 0.$

Lemma B also suffices to prove that {{pi}} is transcendental, since otherwise we would have 1 + e^i{{pi}} ≠ 0.

Baker's formulation of the theorem clearly implies the first formulation. Indeed, if $\backslash alpha(1),\backslash ldots,\backslash alpha(n)$ are algebraic numbers that are linearly independent over $\backslash Q$, and

:$P(x\_1,\; \backslash ldots,\; x\_n)=\; \backslash sum\; b\_\{i\_1,\backslash ldots,\; i\_n\}\; x\_1^\{i\_1\}\backslash cdots\; x\_n^\{i\_n\}$

is a polynomial with rational coefficients, then we have

:$P\backslash left(e^\{\backslash alpha(1)\},\backslash dots,e^\{\backslash alpha(n)\}\backslash right)=\; \backslash sum\; b\_\{i\_1,\backslash dots,i\_n\}\; e^\{i\_1\; \backslash alpha(1)\; +\; \backslash cdots\; +\; i\_n\; \backslash alpha(n)\},$

and since $\backslash alpha(1),\backslash ldots,\backslash alpha(n)$ are algebraic numbers which are linearly independent over the rationals, the numbers $i\_1\; \backslash alpha(1)\; +\; \backslash cdots\; +\; i\_n\; \backslash alpha(n)$ are algebraic and they are distinct for distinct n-tuples $(i\_1,\backslash dots,i\_n)$. So from Baker's formulation of the theorem we get $b\_\{i\_1,\backslash ldots,i\_n\}=0$ for all n-tuples $(i\_1,\backslash dots,i\_n)$.

Now assume that the first formulation of the theorem holds. For $n=1$ Baker's formulation is trivial, so let us assume that $n>1$, and let $a(1),\backslash ldots,a(n)$ be non-zero algebraic numbers, and $\backslash alpha(1),\backslash ldots,\backslash alpha(n)$ distinct algebraic numbers such that:

:$a(1)e^\{\backslash alpha(1)\}\; +\; \backslash cdots\; +\; a(n)e^\{\backslash alpha(n)\}\; =\; 0.$

As seen in the previous section, and with the same notation used there, the value of the polynomial

:$Q(x\_\{11\},\backslash ldots,x\_\{nd(n)\},y\_1,\backslash dots,y\_n)=\backslash prod\backslash nolimits\_\{\backslash sigma\backslash in\; S\}\backslash left(x\_\{1\backslash sigma(1)\}y\_1+\backslash dots+x\_\{n\backslash sigma(n)\}y\_n\backslash right),$

at

:$\backslash left\; (a(1)\_1,\backslash ldots,a(n)\_\{d(n)\},e^\{\backslash alpha(1)\},\backslash ldots,e^\{\backslash alpha(n)\}\; \backslash right)$

has an expression of the form

: $b(1)e^\{\backslash beta(1)\}+\; b(2)e^\{\backslash beta(2)\}+\; \backslash cdots\; +\; b(M)e^\{\backslash beta(M)\}=\; 0,$

where we have grouped the exponentials having the same exponent. Here, as proved above, $b(1),\backslash ldots,\; b(M)$ are rational numbers, not all equal to zero, and each exponent $\backslash beta(m)$ is a linear combination of $\backslash alpha(i)$ with integer coefficients. Then, since $n>1$ and $\backslash alpha(1),\backslash ldots,\backslash alpha(n)$ are pairwise distinct, the $\backslash Q$-vector subspace $V$ of $\backslash C$ generated by $\backslash alpha(1),\backslash ldots,\backslash alpha(n)$ is not trivial and we can pick $\backslash alpha(i\_1),\backslash ldots,\backslash alpha(i\_k)$ to form a basis for $V.$ For each $m=1,\backslash dots,M$, we have

:$\backslash begin\{align\}$

\beta(m) = q_{m,1} \alpha(i_1) + \cdots + q_{m,k} \alpha(i_k), && q_{m,j} = \frac{c_{m,j}}{d_{m,j}}; \qquad c_{m,j}, d_{m,j} \in \Z.

\end{align}

For each $j=1,\backslash ldots,\; k,$ let $d\_j$ be the least common multiple of all the $d\_\{m,j\}$ for $m=1,\backslash ldots,\; M$, and put $v\_j\; =\; \backslash tfrac\{1\}\{d\_j\}\; \backslash alpha(i\_j)$. Then $v\_1,\backslash ldots,v\_k$ are algebraic numbers, they form a basis of $V$, and each $\backslash beta(m)$ is a linear combination of the $v\_j$ with integer coefficients. By multiplying the relation

:$b(1)e^\{\backslash beta(1)\}+\; b(2)e^\{\backslash beta(2)\}+\; \backslash cdots\; +\; b(M)e^\{\backslash beta(M)\}=\; 0,$

by $e^\{N(v\_1+\; \backslash cdots\; +\; v\_k)\}$, where $N$ is a large enough positive integer, we get a non-trivial algebraic relation with rational coefficients connecting $e^\{v\_1\},\backslash cdots,e^\{v\_k\}$, against the first formulation of the theorem.

• Gelfond–Schneider theorem
• Baker's theorem; an extension of Gelfond–Schneider theorem
• Schanuel's conjecture; if proven, it would imply both the Gelfond–Schneider theorem and the Lindemann–Weierstrass theorem

{{Reflist|2}}

• {{Citation | last1=Baker | first1=Alan |author-link=Alan Baker (mathematician) | title=Transcendental number theory | url={{Google books|SmsCqiQMvvgC|Transcendental number theory|plainurl=yes}} | publisher=Cambridge University Press | edition=2nd | series=Cambridge Mathematical Library | isbn=978-0-521-39791-9 | mr=0422171 | year=1990}}
• {{Citation |last=Bertrand |first=D. |author-link=Daniel Bertrand |year=1997 |title=Theta functions and transcendence |journal=The Ramanujan Journal |volume=1 |issue=4 |pages=339–350 |doi=10.1023/A:1009749608672 |s2cid=118628723 }}
• {{Citation | last=Gelfond | first=A.O. | author-link=Alexander Gelfond | translator-last=Boron | translator-first=Leo F. | translator-link=Leo F. Boron | orig-year=1960 | year=2015 | title=Transcendental and Algebraic Numbers | publisher=Dover Publications |location=New York |series=Dover Books on Mathematics | isbn=978-0-486-49526-2 |mr=0057921 | url={{Google books|408wBgAAQBAJ|Transcendental and Algebraic Numbers|plainurl=yes}} }}
• {{Citation | last=Gordan | first=P. | author-link=Paul Gordan | year=1893 | title=Transcendenz von {{math|e}} und {{math|π}}. | journal=Mathematische Annalen | volume=43 | issue=2–3 | pages=222–224 | url=https://gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002254557&physid=PHYS_0223 | doi=10.1007/bf01443647| s2cid=123203471 }}
• {{Citation | last=Hermite | first=C. | author-link=Charles Hermite | year=1873 | title=Sur la fonction exponentielle. | journal=Comptes rendus de l'Académie des Sciences de Paris | volume=77 | pages=18–24 | url=http://gallica.bnf.fr/ark:/12148/bpt6k3034n/f18.image}}
• {{Citation | last=Hermite | first=C. | author-link=Charles Hermite | year=1874 | title=Sur la fonction exponentielle. | publisher=Gauthier-Villars | place=Paris | url=https://archive.org/details/surlafonctionexp00hermuoft}}
• {{Citation | last=Hilbert | first=D. | author-link=David Hilbert | year=1893 | title=Ueber die Transcendenz der Zahlen {{math|e}} und {{math|π}}. | journal=Mathematische Annalen | volume=43 | issue=2–3 | pages=216–219 | url=https://gdz.sub.uni-goettingen.de/index.php?id=11&PID=GDZPPN002254565 | doi=10.1007/bf01443645 | s2cid=179177945 | access-date=2018-12-24 | archive-url=https://web.archive.org/web/20171006113711/https://gdz.sub.uni-goettingen.de/index.php?id=11&PID=GDZPPN002254565 | archive-date=2017-10-06 | url-status=dead }}
• {{Citation | last=Lindemann | first=F. | author-link=Ferdinand Lindemann | year=1882 | title=Über die Ludolph'sche Zahl. | journal=Sitzungsberichte der Königlich Preussischen Akademie der Wissenschaften zu Berlin | volume=2 | pages=679–682 | url=https://archive.org/details/sitzungsberichte1882deutsch/page/679 |ref={{Harvid|Lindemann|1882a}}}}
• {{Citation | last=Lindemann | first=F. | author-link=Ferdinand Lindemann | year=1882 | title=Über die Zahl {{math|π}}. | journal=Mathematische Annalen | volume=20 | issue=2 | pages=213–225 | url=https://gdz.sub.uni-goettingen.de/id/PPN235181684_0020?tify=%7B%22view%22:%22info%22,%22pages%22:%5B227%5D%7D | ref={{Harvid|Lindemann|1882b}} | doi=10.1007/bf01446522 | s2cid=120469397 | access-date=2018-12-24 | archive-url=https://web.archive.org/web/20171006120026/https://gdz.sub.uni-goettingen.de/index.php?id=11&PID=GDZPPN002246910 | archive-date=2017-10-06 | url-status=dead }}
• {{cite book |doi=10.1007/978-1-4939-0832-5_19|chapter=Baker's Theorem |title=Transcendental Numbers |year=2014 |last1=Murty |first1=M. Ram |last2=Rath |first2=Purusottam |pages=95–100 |isbn=978-1-4939-0831-8|url={{Google books|-4jkAwAAQBAJ|page=95|plainurl=yes}}}}
• {{Citation | last=Weierstrass | first=K. | author-link=Karl Weierstrass | year=1885 | title=Zu Lindemann's Abhandlung. "Über die Ludolph'sche Zahl". | journal=Sitzungsberichte der Königlich Preussischen Akademie der Wissen-schaften zu Berlin | volume=5 | pages=1067–1085 | url=https://books.google.com/books?id=jhlEAQAAMAAJ&pg=PA1067}}

• {{Citation | last1=Jacobson | first1=Nathan |author-link=Nathan Jacobson | title=Basic Algebra | volume=I | url={{Google books|JHFpv0tKiBAC&printsec|Basic Algebra I|plainurl=yes}} | publisher=Dover Publications | edition=2nd | isbn=978-0-486-47189-1 | orig-year=1985 | year=2009 }}

• {{MathWorld|title=Hermite-Lindemann Theorem|urlname=Hermite-LindemannTheorem}}
• {{MathWorld|title=Lindemann-Weierstrass Theorem|urlname=Lindemann-WeierstrassTheorem}}

{{DEFAULTSORT:Lindemann-Weierstrass theorem}}

Category:Articles containing proofs

Category:E (mathematical constant)

Category:Exponentials

Category:Pi

Category:Theorems in number theory

Category:Transcendental numbers