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[parent] example of transcendental number (Example)

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:
$\displaystyle \vert\alpha-\frac{p}{q}\vert> \frac{c}{q^m}$
for all rationals $ p/q$ (with $ q>0$).

Next we use the theorem to construct a transcendental number.

Corollary 1   The real number
$\displaystyle \psi= \sum_{n=1}^\infty \frac{1}{10^{n!}}=0.1100010\ldots$
is transcendental.
Proof. Clearly, the number $ \psi$ is well defined, i.e. the series converges. Indeed,
$\displaystyle \frac{1}{10^{n!}}<\frac{1}{10^n}$
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

$\displaystyle \vert\psi - \frac{p}{q}\vert< \frac{c}{q^m}$
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
$\displaystyle \vert\psi -\frac{p}{q}\vert<\frac{1}{q^{m+k}}<\frac{1}{2^k}\cdot \frac{1}{q^m}<\frac{c}{q^m}$
For all $ j>k+m$ we define a rational number $ p_j/q_j$ by:
$\displaystyle p_j=10^{j!}\sum_{n=1}^j 10^{-n!},\quad q_j=10^{j!}$
then $ p_j$ and $ q_j$ are relatively prime integers and we have:
$\displaystyle \vert\psi - \frac{p_j}{q_j}\vert$ $\displaystyle =$ $\displaystyle \sum_{n=j+1}^\infty \frac{1}{10^{n!}}$  
  $\displaystyle <$ $\displaystyle \frac{1}{10^{(j+1)!}}(1+\frac{1}{10}+\frac{1}{10^2}+\ldots)$  
  $\displaystyle =$ $\displaystyle 10/9\cdot \frac{1}{q_j^{(j+1)}}$  
  $\displaystyle <$ $\displaystyle \frac{1}{q_j^j}$  
  $\displaystyle <$ $\displaystyle \frac{1}{q_j^{(k+m)}}$  

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. $ \qedsymbol$

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



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See Also: Liouville approximation theorem, Roth's theorem

Keywords:  transcendental, algebraic

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Cross-references: similar, inequality, integers, relatively prime, rational number, algebraic, contradiction, comparison test, converges, series, well defined, number, transcendental, real number, transcendental number, rationals, degree, algebraic number, Liouville's approximation theorem, application
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This is version 4 of example of transcendental number, born on 2005-02-16, modified 2005-02-16.
Object id is 6760, canonical name is ExampleOfTranscendentalNumber.
Accessed 2664 times total.

Classification:
AMS MSC: 11J81 (Number theory :: Diophantine approximation, transcendental number theory :: Transcendence )
 11J82 (Number theory :: Diophantine approximation, transcendental number theory :: Measures of irrationality and of transcendence)
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