example of transcendental number

The following is a classical application of Liouville’s approximation theorem. For completeness, we state Liouville’s result here:

Theorem 1.

For any algebraic number $\\alpha$ with degree $m>1$ , there exists a constant $c=c(\\alpha)>0$ such that:

|\\alpha-\\frac{p}{q}|>\\frac{c}{q^{m}}

for all rationals $p/q$ (with $q>0$ ).

Next we use the theorem to construct a transcendental number.

Corollary 1.

The real number

\\psi=\\sum_{n=1}^{\\infty}\\frac{1}{10^{n!}}=0.1100010\\ldots

is transcendental.

Proof.

Clearly, the number $\\psi$ is well defined, i.e. the series converges. Indeed,

\\frac{1}{10^{n!}}<\\frac{1}{10^{n}}

and $\\sum_{n=1}^{\\infty}10^{-n}=1/9$ . Thus, by the comparison test, the series converges and $0<\\psi<1/9$ .

Suppose, for a contradiction, that $\\psi$ is algebraic of degree $m$ . We will construct infinitely many rationals $p/q$ such that

|\\psi-\\frac{p}{q}|<\\frac{c}{q^{m}}

where $c=c(\\psi)$ is the constant given by the theorem above. Let $k\\in\\mathbb{N}$ be such that $1/2^{k}<c$ . Then, in fact, we will show that there are infinitely many rationals $p/q$ with $q\\geq 2$ such that

|\\psi-\\frac{p}{q}|<\\frac{1}{q^{m+k}}<\\frac{1}{2^{k}}\\cdot\\frac{1}{q^{m}}<\\frac%{c}{q^{m}}

For all $j>k+m$ we define a rational number $p_{j}/q_{j}$ by:

p_{j}=10^{j!}\\sum_{n=1}^{j}10^{-n!},\\quad q_{j}=10^{j!}

then $p_{j}$ and $q_{j}$ are relatively prime integers and we have:

$\\displaystyle\|\\psi-\\frac{p_{j}}{q_{j}}\|$	$\\displaystyle=$	$\\displaystyle\\sum_{n=j+1}^{\\infty}\\frac{1}{10^{n!}}$
	$\\displaystyle<$	$\\displaystyle\\frac{1}{10^{(j+1)!}}(1+\\frac{1}{10}+\\frac{1}{10^{2}}+\\ldots)$
	$\\displaystyle=$	$\\displaystyle 10/9\\cdot\\frac{1}{q_{j}^{(j+1)}}$
	$\\displaystyle<$	$\\displaystyle\\frac{1}{q_{j}^{j}}$
	$\\displaystyle<$	$\\displaystyle\\frac{1}{q_{j}^{(k+m)}}$

where in the last inequality we have used the fact that $j>k+m$ . Therefore, all rationals $\\{p_{j}/q_{j}\\}_{j=k+m+1}^{\\infty}$ satisfy the desired inequality, which leads to the contradiction with the theorem above. Thus $\\psi$ cannot be algebraic and it must be transcendental.∎

Many other similar transcendental numbers can be constructed in this fashion.