5

From Vladimir Zorich Analysis I:

Let us call an irrational number $a \in \mathbb{R}$ well approximated by rational numbers if for any natural number $n, N \in \mathbb{N}$ there exists a rational number $\frac{p}{q}$ such that $\lvert a - \frac{p}{q} \rvert < \frac{1}{Nq^n}$.

(a) Construct an example of a well-approximated irrational number.

(b) Prove that a well-approximated irrational number cannot be algebraic, that is, it is transcendental (Liouville's theorem).

The answer to the question (a) is $a=\sum_{n \geq 1}{\frac{1}{10^{n!}}}$, where $a$ is called Liouville's constant.

As for (b):

I've got stuck almost immediately due to lack of my knowledge of number theory. I have to show for any polynomial $P(x)=\sum_{k=0}^n{a_{k}x^k}$:

$$\forall l \in \mathbb{R}, \lvert l - \frac{p}{q} \rvert < \frac{1}{Nq^n}, P(l) \neq 0$$

But suppose:

$$P(l)=0$$

Then we can assume few points:

  • $P(x)$ might have a set of rational roots $S \subset \mathbb{Q}$, such that $l \notin S$
  • There exists $\frac{p}{q} \notin S$ such that $P(\frac{p}{q}) \neq 0$
  • $P(\frac{p}{q})=\sum_{k=0}^na_k\frac{p^k}{q^k}=a_1 \frac{p}{q} + a_{2} \frac{p^2}{p^2} + ... + a_{k} \frac{p^k}{q^k}$. Therefore it is seen that $P(\frac{p}{q})=\frac{1}{q^n}c$ for some $c \in \mathbb{R}$ and thus $P(\frac{p}{q}) \geq \frac{1}{q^n}$
  • Since $P(x)=0$, then $\lvert P(x) - P(\frac{p}{q}) \rvert \geq \frac{1}{q^n}$
  • By triangle inequality, $\lvert l - \frac{p}{q} \rvert < \frac{1}{Nq^n} \implies \lvert l \rvert + \lvert \frac{p}{q} \rvert < \frac{1}{Nq^n} \implies \frac{p}{q} < \frac{1}{Nq^n} + \lvert l \rvert$

Question:

Can I utilize the points above to construct a simple elegant proof for Liouville's theorem? Perhaps constraints from rational root theorem can be used in some way?

Thank you!

By the way: I've been able to find a single proof from small research, but I find it slightly implicit and complex and struggle to understand the essence of some parts.

Also, I do wonder what kind of solution would be most suitable from real-analysis perspective, since Zorich only briefly touched definition of transcendal numbers and $q$-nary approximations.

ShellRox
  • 977
  • 2
    The "by the way" part you mentioned is the Liouville's Lemma (one example how to use it here). – rtybase Sep 09 '19 at 19:03
  • 3
    @rtybase I forgot to mention the proof using mean value theorem in my answer. It's the most elegant proof I've found, I guess I will utilize that if I can't find any simpler proof. Thank you! – ShellRox Sep 09 '19 at 19:09
  • 2
    Also have a look at this book in preview mode, the beginning of the Chapter 1. – rtybase Sep 09 '19 at 19:10
  • 1
    The proof that @rtybase posted is the standard proof (in my opinion). Maybe you are also interested in the optimal improvement of Liouville's theorem, which you can find under the name "Roth's theorem". That theorem is not that easy to prove though. – Con Sep 09 '19 at 20:48
  • 1
    @ThorWittich Yes I believe the best way is to just write out a standard proof. I appreciate your reference for Roth's theorem, its proof is most probably out of my league but it is a very motivating example. – ShellRox Sep 09 '19 at 21:07

0 Answers0