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

The proof should not use that $\pi$ is transcendental.

$\{1,e,e^2,e^3\}$ is linearly independent over $\mathbb{Q}$

Any hints would be appreciated.

  • $\begingroup$ Can it be a proof of the transcendence of $\pi$? :P $\endgroup$ – Patrick Da Silva Aug 27 '13 at 6:02
  • $\begingroup$ Well, the proof is equivalent to $\pi$ being neither rational nor a quadratic irrational. $\endgroup$ – Hagen von Eitzen Aug 27 '13 at 6:04
  • 3
    $\begingroup$ @Hagen von Eitzen ; it's possible that OP is looking for a weaker proof than a proof of transcendence, i.e. an argument which shows precisely that $[\mathbb Q(\pi) : \mathbb Q] > 2$ (in the language of Galois theory...), that's why I ask. $\endgroup$ – Patrick Da Silva Aug 27 '13 at 6:05
  • $\begingroup$ @PatrickDaSilva Agreed - not rational and not a quadratic irrational is still far away from transcendental. $\endgroup$ – Hagen von Eitzen Aug 27 '13 at 6:20
  • $\begingroup$ Hermite's proof (en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational) shows that $\pi^2$ is irrational. Yes, I know this isn't what you want. $\endgroup$ – Robert Israel Aug 27 '13 at 6:26

Edit_1: Indeed, something is wrong. I found the mistake. The main trick used below is not correct: $$e^{2\pi^2i}=(e^{2\pi i})^\pi=1^\pi=1=e^{2k\pi i}$$ But then $2\pi^2 = 2k\pi$ for some $k\in\Bbb Z$, so that $\pi=k\in \Bbb Z$, which is not possible. I am not exactly sure what is the error. I believe it should be due to $1^\pi =1$, i.e. this is only true in $\Bbb R$ but we are in $\Bbb C$.

I thought of a possible solution, but I am not sure if I "cheated" somewhere. I guess it is best to post my thoughts.

Suppose $$\alpha\pi^2+\beta\pi+\gamma=0$$ for some $\alpha,\beta,\gamma\in \Bbb Q$. Our goal is to prove that $\alpha=\beta=\gamma=0$. Let $D$ be the common denominator of $\alpha,\beta,\gamma$, then multiplying by $D$ we can assume that $$a\pi^2+b\pi+c=0$$ for some $a,b,c\in \Bbb Z$. Then: \begin{align*} 2(a\pi+b)\pi i &= -2c i\\ e^{2(a\pi+b)\pi i} &= e^{-2c i}\\ (e^{2\pi i})^{a\pi +b} &= e^{-2c i}\\ 1^{a\pi+b} &= e^{-2c i}\\ 1 &= e^{-2c i}\\ 2k\pi i &= -2c i, \quad k\in \Bbb Z\\ -k\pi &= c \end{align*} One possibility is for $k\neq 0$. Then $$\pi=-\frac{c}{k}\in\Bbb Q$$ However, suppose first that $\pi$ is irrational. Then we must have $k=0$ and hence $c=0$. Therefore the equation becomes \begin{align*} a\pi^2+b\pi &=0\\ a\pi+b &= 0 \end{align*} since $\pi$ is irrational, once again we must have $a=0$ and $b=0$, which shows that $\{1,\pi,\pi^2\}$ is linearly independent over $\Bbb Q$. Therefore the problem reduces to proving that $\pi$ is irrational. We can refer to the standard proofs for this fact.

  • $\begingroup$ Yes, there is a mistake as you pointed out. The problem is that for $x,y \in \mathbb{C}$, the expression $x^y$ is in general multivalued. Since the exponential function is a bona fide, single-valued function on $\mathbb{C}$, you could replace the multivalued $e^x$ by $\exp (x)$. But for example, $1^\pi$ can be assigned the value $\exp(2n\pi^2i)$ for any $n \in \mathbb{Z}$. $\endgroup$ – A Kubiesa Aug 12 '14 at 23:01

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