Does someone know a proof that $\{1,e,e^2,e^3\}$ is linearly independent over $\mathbb{Q}$?

The proof should not use that $e$ is transcendental.

$e:$ Euler's number.

$\{1,e,e^2\}$ is linearly independent over $\mathbb{Q}$

Any hints would be appreciated.

  • 4,928
  • 2
  • 18
  • 42
  • 1
    What known facts about e are you allowed to use? For example in your case it would suffice to use that for all $p,q,r,s\in \Bbb Q, e\neq p+\sqrt{q}+\sqrt[3]{r}+\sqrt[6]{s}.$ – walcher Aug 26 '13 at 23:19
  • How are these vectors? – rurouniwallace Aug 26 '13 at 23:22
  • 5
    @Ataraxia A vector is a member of a vector space and $\Bbb R$ can be a vector space over $\Bbb Q$. – Git Gud Aug 26 '13 at 23:23
  • @GitGud So in this case the vector space is $\mathbb{R}$? – rurouniwallace Aug 26 '13 at 23:28
  • 1
    @Ataraxia If you're not too picky with rigour, yes, over the field $\Bbb Q$. Translating the question to a more elementary version: prove that $$(\forall \lambda _0, \lambda _1, \lambda _2, \lambda _3\in \Bbb Q)(\lambda _0+\lambda _1e+\lambda _2e^2+\lambda _3e^3=0\implies \lambda _0=\lambda _1=\lambda _2=\lambda _3=0)$$ – Git Gud Aug 26 '13 at 23:28
  • @Ataraxia $1,e,e^2,e^3$ are all elements of a vector space, here $\mathbb{R}$ over the field $\mathbb{Q}$... the linear independence question is whether there exists nontrivial $k_i\in\mathbb{Q}$ s.t. $k_0+k_1e+k_2e^2+k_3e^3=0$ – obataku Aug 26 '13 at 23:29
  • 1
    @GitGud Ahh I get it now. Thanks! – rurouniwallace Aug 26 '13 at 23:35
  • 2
    The scalars are members of $\mathbb Q$ and the vector space could be $\mathbb R$, but it could be a much smaller vector space, spanned by the four listed elements. – Michael Hardy Aug 26 '13 at 23:54
  • 1
    How can you not use the fact that $e$ is transcendental? What if it satisfied a polynomial of degree $3$ over $\Bbb Q$? What *do* we get to know? – Ted Shifrin Aug 27 '13 at 00:01
  • 1
    @TedShifrin The goal is to find separate proofs. Prove that $e$ is not root of any polynomial $\mathbb{Q}[x]$ is much stronger than $e$ is not root of a cubic polynomial over $\mathbb{Q}$. – felipeuni Aug 27 '13 at 00:24
  • @Ataraxia: that doesn't suffice. You can't get the roots of a cubic polynomial with Galois group $S_3$ that way. – Qiaochu Yuan Aug 27 '13 at 00:33
  • My cynic suggestion is: imitate the proof of the transcendence of $e$ in your particular case (the proof that uses integrals and products involving arbitrarily large primes, see for example the calculus book from Michael Spivak). – Matemáticos Chibchas Aug 27 '13 at 00:35
  • @MatemáticosChibchas Ok, is a way to prove. But the idea is to find a short proof, as in the proof of $\{1,e,e^2\}$ only series. – felipeuni Aug 27 '13 at 00:54
  • @AndréNicolas Can you please share with us your method? I became intrigued... – Matemáticos Chibchas Aug 27 '13 at 01:20
  • There was an error. – André Nicolas Aug 27 '13 at 01:30
  • From wikipedia (https://en.wikipedia.org/wiki/Proof_that_e_is_irrational) "In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that $e$ is not a root of a third degree polynomial with rational coefficients. In particular, $e^3$ is irrational." The reference quoted is Hurwitz, Adolf (1933) [1891]. *"Über die Kettenbruchentwicklung der Zahl $e$"* – Paramanand Singh Aug 10 '15 at 10:39
  • The paper mentioned in previous comment is available as part of this publication https://archive.org/download/schriftenderphys3132phys/schriftenderphys3132phys.djvu but it is in German I guess. Roughly it says that if $x = (e - 1)/2, x' = (e^{2} - 1)/2$ then we can't have integers $A, B, C, D$ such that $x = (Ax' + B)/(Cx' + D)$ and this is equivalent to linear independence of $1, e, e^{2}, e^{3}$ over $\mathbb{Q}$. The proof is based on analysis of continued fractions of $x, x'$. – Paramanand Singh Aug 10 '15 at 11:05

4 Answers4


Since I've spent enough time thinking about this, yet not getting a proof, I might as well show what I've got. Others can comment on whether or not more can be done.

Your problem is solved if you can show that for any integers $a, b, c$, we have $$\sum^\infty_{n=0} \frac{1}{n!} (a + b 2^n + c 3^n)$$ irrational (using taylor series).

WLOG, assume $c>0$. Pick $N$ large so that $(a+b2^n+c3^n) > 0$ for all $N \geq 0$. Then our problem is equivalent to showing that the series with strictly positive terms $$\sum^\infty_{n=N} \frac{1}{n!} (a + b 2^n + c 3^n)$$ is irrational. Suppose it was not and equal to $p/q$. Now we try to mimic the proof of irrationality of $e$.


$$x = q!\left(\frac{p}{q} - \sum_{n=N}^{q} \frac{1}{n!} (a + b 2^n + c 3^n) \right).$$ One easily sees by distributing that it is an integer, and because our original series contains only positive terms, $x>0$.

Note that we can also write $$x = \sum_{n=q+1}^\infty \frac{q!}{n!} (a + b 2^n + c 3^n).$$

Now if $b=c=0$, then using $q!/n! < 1/(q+1)^{n-q}$ gives a geometric series bound that gives $x < 1/q$. Then we can get $x<1$ which is a contradiction that $x$ is an integer.

The terms $2^n$ and $3^n$ grow too fast for this same trick to work. You'd get bounds of $2^q/q$ and $3^q/q$ respectively. Since $q!/n! < 1/(q+1)^(n-q)$ is not tight, it is still possible that we can get our sum under 1. Or maybe we can monkey with our original definition for $x$.

I think what really needs to be copied are proofs of the irrationality of $e^2$ and $e^3$, but I am not aware of such proofs. Googling, I found a very algebraic proof of the irrationality of $e^2$, but I didn't read it carefully. This suggests proofs of the irrationality of $e^2$ may not easily generalize, and hence you aren't really proving that $e$ is transcendental at the same time.

  • 14,161
  • 2
  • 33
  • 75
  • Has this approach been hinted at in the comments? – abnry Aug 27 '13 at 00:56
  • 2
    For what it's worth, multiplying by $e^{-1}$, you could replace $a+b2^n+c3^n$ with $a(-1)^n+b+c2^n$ – Jonas Meyer Aug 27 '13 at 01:40
  • Well, that would solve the problem for $\{1, e, e^2\}$. – abnry Aug 27 '13 at 01:41
  • I mean $a_0+a_1e+a_2e^2+a_3e^3=0$ if and only if $a_0e^{-1}+a_1+a_2e+a_3e^2=0$, so it is enough to show that $ae^{-1}+be+ce^2$ cannot be rational unless $a=b=c=0$. Maybe replacing $3$ with $-1$ is easier; I don't know. – Jonas Meyer Aug 27 '13 at 01:47
  • Yes, and in the case $a_0 + a_1 e + a_2 e^2 = 0$, we only need to show that $a_1 e^{-1} + a_2 e$ is irrational, which yields $c=0$ and the series $\sum q!/n! [a_1 (-1)^n + a_2]$, which is manageable. – abnry Aug 27 '13 at 01:49
  • Oh, thanks, I see what you meant. – Jonas Meyer Aug 27 '13 at 01:50
  • Come to think of it, for any finite collection $e^{a_n}$ with $\sup a_n - \inf a_n \leq 2$, the collection must be linearly independent. – abnry Aug 27 '13 at 02:16
  • 1
    The alternative proofs are the only those which were given by Joseph Liouville and presented in my blog (which OP has linked in the question). The technique used by Liouville can't be adapted for $e^{3}$ in the sense that the terms $2^{n}/n!$ can be bounded by dividing with suitable power of $2$, but the same can't be done for $3^{n}/n!$. – Paramanand Singh Aug 29 '13 at 04:34
  • 1
    The technique of dividing by a power of $2$ is explained in the posts http://paramanands.blogspot.com/2013/03/another-proof-that-e-squared-is-irrational.html and http://paramanands.blogspot.com/2013/03/proof-that-e-squared-is-not-a-quadratic-irrationality.html where it is proved that $1, e^{2}, e^{4}$ are linearly independent over $\mathbb{Q}$. – Paramanand Singh Aug 29 '13 at 04:43

Below is my attempt which is too long for a comment and may be saveable, (doubt it).

Consider the differential equation $y^{(4)}-6y^{(3)}+11y''-6y'=\textbf 0$, where $\bf 0$ is the null function over some non-trivial interval $I$ containing $1$.

The theory of ODE tells us that a basis of solutions is $$\{\underbrace{x\mapsto 1}_{\large \varphi_0}, \underbrace{x\mapsto e^x}_{\large \varphi _1}, \underbrace{x\mapsto e^{2x}}_{\large \varphi _2}, \underbrace{x\mapsto e^{3x}}_{\large \varphi _3}\}$$

This implies that $$(\forall \lambda _0,\lambda _1, \lambda _2, \lambda _3\in \Bbb Q)\left[(\forall x\in I)\left(\sum \limits_{k=0}^3\lambda_k\varphi_k(x)=0\right)\implies \lambda _0=\lambda _1=\lambda _2=\lambda _3=0\right] \tag {*}$$

Now if we could somehow prove that $(*)$ would also hold for the intersection of all such intervals $I$ (containing $1$), what we want would follow. But I have no hope of this being doable.

Git Gud
  • 30,699
  • 11
  • 58
  • 117

I thought to add an answer instead of giving long comments.

From Wikipedia we have the following quote

"In 1891, Hurwitz explained how it is possible to prove along the same line of ideas that $e$ is not a root of a third degree polynomial with rational coefficients. In particular, $e^{3}$ is irrational."

The reference quoted is Hurwitz, Adolf (1933) [1891]. "Über die Kettenbruchentwicklung der Zahl $e$".

Luckily after much searching I was able to find this reference in an old journal available on internet archive. Here Hurwitz analyzes the simple continued fractions of numbers related with $e$ and notices that most of them have terms in an arithmetic progression (after a certain point).

He then proves the following theorem:

If simple continued fractions of two positive numbers $x, y$ have terms which are in arithmetic progression (after a certain point) then we can't have a non-trivial bi-linear relation of the form $$y = \frac{Ax + B}{Cx + D}$$ with integer coefficients $A, B, C, D$ unless the terms in their continued fraction belong to the same arithmetic progression.

Then Hurwitz notes that $$x = \frac{e - 1}{2} = [0, 1, 6, 10, 14, 18,\ldots]$$ and $$y = \frac{e^{2} - 1}{2} = [3, 5, 7, 9, \ldots]$$ where notation $$[a_{0}, a_{1}, a_{2}, \ldots]$$ represents the continued fraction $$a_{0} + \dfrac{1}{a_{1} + \dfrac{1}{a_{2} + \dfrac{1}{a_{3} + \cdots}}}$$ And clearly both of them have terms belonging to arithmetic progressions ($6, 10, 14, \ldots$ and $3, 5, 7, 9, \ldots$ respectively) but these are not the terms belonging to same AP and hence there is no non-trivial bi-linear relation of type $$y = \frac{Ax + B}{Cx + D}$$ with integer coefficients $A, B, C, D$.

Now it follows easily that $1, e, e^{2}, e^{3}$ are linearly independent over $\mathbb{Q}$. If it was not the case then we have integers $a, b, c, d$ not all $0$ such that $$ae^{3} + be^{2} + ce + d = 0$$ Using $e^{2} = 2y + 1$ and $e = 2x + 1$ we get $$a(2x + 1)(2y + 1) + b(2y + 1) + c(2x + 1) + d = 0$$ which leads to $$Axy + Bx + Cy + D = 0$$ with $A, B, C, D$ as integers or $$y = -\frac{Bx + D}{Ax + C}$$ and this is not allowed by the theorem of Hurwitz mentioned above.

Unfortunately I could not understand the proof of his theorem on continued fractions (because the whole paper/journal is in German). With reasonable effort and Google Translate I was able to understand the gist of the paper and I have presented the same in this answer. I have asked for the proof of Huzwitz theorem on MSE.

Paramanand Singh
  • 79,254
  • 12
  • 118
  • 274

Using algebra, let $D$ be the differentiate operator for $C^{\infty}$ functions.

So with $$f_n(x)=e^{\lambda_n x}$$

Then $$\forall i \in \{1,...,n\}, \forall x \in \mathbb{R}, D(f_n(x))=\lambda_n \cdot f_n(x) $$

If all $\lambda_i$ are differentant $n$ is the space's dimension then the familly $\left(f_i\right)_{1\le i \le n}$ is free. (because $f_i$ is the eigenvector associates to $\lambda_i$ eigenvalue.)

We conclude with $\lambda_i=i$ ($i$ must be $0$ if you need it) and $x=1$.

  • 1,580
  • 1
  • 13
  • 31
  • And why doesn't that hold for $x=0$ as well? – WimC Aug 10 '15 at 17:47
  • 0 times somthings = 0, and i don't really understand what you mean, i'm french sorry. – ParaH2 Aug 10 '15 at 18:32
  • $f_n(0)=1$ for all $n$ and these values are clearly dependent. Why should $x=1$ be different? – WimC Aug 10 '15 at 19:04
  • How yeah you right, I did it too fast. I will think about it, I think my answer must work, I just need to know how. I'm a chemist I may do some mistakes. If you have any idea tell me. :) – ParaH2 Aug 10 '15 at 19:17