pi is transcendental

Assume that $\pi\in\overline{\mathbb{Q}}$. Since $\overline{\mathbb{Q}}$ is a field, then $i\pi\in\overline{\mathbb{Q}}$ too. Let $P(x)\in \mathbb{Z}[x]$ be a minimal degree polynomial such that $i\pi$ is a root of $P(x)=0$. If $\deg P=d$, then $P(x)=0$ has $d$ roots $\alpha_1\,(=i\pi),\alpha_2,\cdots,\alpha_d$, that is,

$$P(x)=a(x-\alpha_1)\cdots(x-\alpha_d).$$

Next let's make a equation $\sum e^{\beta_k}=0$, where $\beta_k$ is a linear combination of $\{\alpha_k\}$. Define

$$\begin{eqnarray*} &D&\stackrel{\mathrm{def}}{=} \{\delta_1\alpha_1+\cdots+\del... ...ta\neq0\},\\ & &=\{\beta_1,\cdots,\beta_n\}.\quad (\sharp B=n) \end{eqnarray*}$$

Note that $\sharp D=2^d,\;1\leqq n \leqq 2^d-1$. According to Euler's formula,

$$e^{\alpha_1}=e^{i\pi}=\cos\pi+i\sin\pi=-1.$$

So

$$\begin{eqnarray*} % latex2html id marker 316&&(e^{\alpha_1}+1)\cdots(e^{\alph... ...\\ &\therefore&\displaystyle\tsum_{k=1}^n e^{\beta_k}+2^d-n=0. \end{eqnarray*}$$

The following part of the proof is similar to that of the case $e$. Let

$$\begin{eqnarray*} &J&\stackrel{\mathrm{def}}{=}\tsum_{k=1}^n I(\beta_k) =\tsum_{k=1}^n \int_0^{\beta_k} e^{\beta_k-u}f(u)du, \end{eqnarray*}$$

where

$$\begin{eqnarray*} &f(x)&\stackrel{\mathrm{def}}{=} a^{np}x^{p-1}(x-\beta_1)^p\... ...a_1\vert)^p\cdots(x+\vert\beta_n\vert)^p,\\ &m&=\deg f=p-1+np, \end{eqnarray*}$$

where $a$ is the coefficient of the highest order term of $P(x)$. Then according to the equality (2) of the lemma 2

$$\begin{eqnarray*} &J&=\tsum_{k=1}^n \left(e^{\beta_k}\tsum_{j=0}^m f^{(j)}(0) ... ...=0}^m f^{(j)}(0) -\tsum_{j=0}^m \tsum_{k=1}^n f^{(j)}(\beta_k). \end{eqnarray*}$$

Let $\{\sigma_1,\cdots,\sigma_r\}$ be $r$ elementary symmetric polynomials on $\{\alpha_1,\cdots,\alpha_r\}$, then

$$a\sigma_1,\cdots,a\sigma_r \in\mathbb{Z}. \quad (\because P(x)\in\mathbb{Z}[x])$$

And

$$\begin{eqnarray*} &&g(x)\stackrel{\mathrm{def}}{=} (ax-a\beta_1)\cdots(ax-a\be... ...hbb{Z}[ax],\\ &&\text{where } f(x)=x^{p-1}\left(g(x)\right)^p, \end{eqnarray*}$$

because $n$ elementary symmetric polynomials on $\{a\beta_1,\cdots,a\beta_n\}$ are symmetric polynomials on $\{a\alpha_1,\cdots,a\alpha_r\}$, that is, on $\{a\sigma_1,\cdots,a\sigma_r\}$.

Then according to the lemma 3 (the 2nd case, see (4)),

$$\begin{eqnarray*} &&f^{(j)}(0)\begin{cases} =(p-1)!(-a)^{np}(\beta_1\cdots\bet... ...{cases}\\ &&\tsum_{k=1}^n f^{(j)}(\beta_k) \equiv 0 \pmod{p!}. \end{eqnarray*}$$

So if $p$ is a prime and $p>\max\{\vert a^n\beta_1\cdots\beta_n\vert,2^d-n\}$, then $(p-1)!\mid J$ and $p\nmid J$. Therefore

$$\begin{eqnarray*} &&\vert J\vert\geqq (p-1)!. \end{eqnarray*}$$

On the other hand, according to the inequality (3) of the lemma 2

$$\begin{eqnarray*} &\vert J\vert&\leqq \tsum_{k=1}^n \vert I(\beta_k)\vert \\ ... ...ert\max_{1\leqq k\leqq n}\{\vert\beta_k\vert\}\right)^{(n+1)p}. \end{eqnarray*}$$

The lower and upper bound of $\vert J\vert$ contradict to the lemma 1 (see (1)). Therefore $\pi$ is transcendental.

Appendix : The set of algebraic numbers e and pi are transcendental e is transcendental