# 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

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

###### 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}. 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.

Title example of transcendental number ExampleOfTranscendentalNumber 2013-03-22 15:02:45 2013-03-22 15:02:45 alozano (2414) alozano (2414) 7 alozano (2414) Example msc 11J82 msc 11J81 LiouvillesTheorem RothsTheorem