1.12 Transcendental numbers

A complex number is said to be algebraic or transcendental according as it is algebraic or transcendental over $\mathbb{Q}$ . First we provide a little history.

1844: Liouville showed that certain numbers, now called Liouville numbers, are transcendental.

1873: Hermite showed that $e$ is transcendental.

1874: Cantor showed that the set of algebraic numbers is countable, but that $\mathbb{R}$ is not countable. Thus most numbers are transcendental (but it is usually very difficult to prove that any particular number is transcendental).⁹⁹ 9 By contrast, when we suspect that a complex number is algebraic, it is usually possible to prove this, but not always easily.

1882: Lindemann showed that $\pi$ is transcendental.

1934: Gel’fond and Schneider independently showed that $\alpha^{\beta}$ is transcendental if $\alpha$ and $\beta$ are algebraic, $\alpha\neq 0,1$ , and $\beta\notin\mathbb{Q}$ . (This was the seventh of Hilbert’s famous problems.)

2020: Euler’s constant

\gamma\overset{\smash{\lower 1.31395pt\hbox{{def}}}}{=}\lim_{n\rightarrow% \infty}\left(\sum_{k=1}^{n}1/k-\log n\right)

has not yet been proven to be transcendental or even irrational (see Lagarias, J., Euler’s constant. BAMS 50 (2013), no. 4, 527–628, arXiv:1303:1856).

2020: The numbers $e+\pi$ and $e-\pi$ are surely transcendental, but again they have not even been proved to be irrational!

Proposition 1.32

The set of algebraic numbers is countable.

Proof.

Every algebraic number is a root of a polynomial

a_{0}X^{n}+a_{1}X^{n-1}+\cdots+a_{n},\quad a_{0},\ldots,a_{n}\in\mathbb{Z}{}.

For a fixed $N\in\mathbb{N}{}$ , there are only finitely many such polynomials with $n\leq N$ and $|$ $a_{0}|,\ldots,|a_{n}|\leq N$ , and each polynomial has only finitely many roots. Thus, the set of algebraic numbers is a countable union of finite sets $\bigcup_{N\geq 1}A(N)$ , and any such union is countable — for example, choose a bijection from some segment $[0,n(1)]$ of $\mathbb{N}{}$ onto $A(1)$ , extend it to a bijection from a segment $[0,n(2)]$ onto $A(2)$ , and so on. _□

A typical Liouville number is $\sum_{n=0}^{\infty}\frac{1}{10^{n!}}$ — in its decimal expansion there are increasingly long strings of zeros. Since its decimal expansion is not periodic, the number is not rational. We prove that the analogue of this number in base $2$ is transcendental.

Theorem 1.33

The number $\alpha=\sum\frac{1}{2^{n!}}$ is transcendental.

Proof.

¹⁰¹⁰ 10 This proof, which I learnt from David Masser, also works for

{\textstyle\sum}\frac{1}{a^{n!}}

for every integer

a\geq 2

Suppose not, and let

f(X)=X^{d}+a_{1}X^{d-1}+\cdots+a_{d},\quad a_{i}\in\mathbb{Q},

be the minimal polynomial of $\alpha$ over $\mathbb{Q}$ . Thus $[\mathbb{Q}[\alpha]\colon\mathbb{Q}]=d$ . Choose a nonzero integer $D$ such that $D\cdot f(X)\in\mathbb{Z}[X]$ .

Let $\Sigma_{N}=\sum_{n=0}^{N}\frac{1}{2^{n!}}$ , so that $\Sigma_{N}\rightarrow\alpha$ as $N\rightarrow\infty$ , and let $x_{N}=f(\Sigma_{N})$ . As $\alpha$ is not rational, $f(X)$ , being irreducible of degree $>1$ , has no rational root. Since $\Sigma_{N}\neq\alpha$ , it can’t be a root of $f(X)$ , and so $x_{N}\neq 0$ . Obviously, $x_{N}\in\mathbb{Q}$ ; in fact $(2^{N!})^{d}Dx_{N}\in\mathbb{Z}$ , and so

equation (3) (3)

|(2^{N!})^{d}Dx_{N}|\geq 1\text{.}

From the fundamental theorem of algebra (see LABEL:ag5 below), we know that $f$ splits in $\mathbb{C}{}[X]$ , say,

f(X)=\prod_{i=1}^{d}(X-\alpha_{i}),\quad\alpha_{i}\in\mathbb{C},\quad\alpha_{1% }=\alpha,

and so

|x_{N}|=\prod_{i=1}^{d}|\Sigma_{N}-\alpha_{i}|\leq|\Sigma_{N}-\alpha_{1}|(% \Sigma_{N}+M)^{d-1},\quad\text{where }M=\max_{i\neq 1}\{1,|\alpha_{i}|\}\text{.}

But

|\Sigma_{N}-\alpha_{1}|=\sum_{n=N+1}^{\infty}\frac{1}{2^{n!}}\leq\frac{1}{2^{(% N+1)!}}\left(\sum_{n=0}^{\infty}\frac{1}{2^{n}}\right)=\frac{2}{2^{(N+1)!}}.

Hence

|x_{N}|\leq\frac{2}{2^{(N+1)!}}\cdot(\Sigma_{N}+M)^{d-1}

and

|(2^{N!})^{d}Dx_{N}|\leq 2\cdot\frac{2^{d\cdot N!}D}{2^{(N+1)!}}\cdot(\Sigma_{% N}+M)^{d-1}

which tends to $0$ as $N\rightarrow\infty$ because $\frac{2^{d\cdot N!}}{2^{(N+1)!}}=\left(\frac{2^{d}}{2^{N+1}}\right)^{N!}\rightarrow 0$ . This contradicts (3). _□