Solution to the decimal problem

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In this post I will discuss the solution of the problem I posted a week ago. Firstly I would thank Prof. Purusottam Rath for pointing out that this problem has already been solved. In 1961, Kurt Mahler published the solution in Lectures on Diophantine Approximations. In fact, M=0.12345678.... is called Mahler’s number since Mahler  showed it to be transcendental. For complete proof refer, Section 1.6 of “Making transcendence transparent: an intuitive approach to classical transcendental number theory“, by Edward Burger and Robert Tubbs, Springer-Verlag (2004). But I can give an outline of proof here.

The most basic type of transcendental numbers are the Liouville’s Numbers, these numbers satisfy following theorem (proved here on pp. 19):

Liouville’s Theorem: Let \alpha be a real number . Suppose there exists an infinite sequence of rational numbers p_n/q_n satisfying the inequality \displaystyle{\left|\alpha - \frac{p_n}{q_n}\right|<\frac{1}{q_n^n}}. Then \alpha is transcendental.

To be able to apply this theorem we use truncation procedure i.e. obtain the approximations of the number \alpha  by truncating the decimal expansion of \alpha immediately before each long run of zeros, and using this to get the desired inequality.

For Mahler number, Liouville’s theorem alone is not sufficient. Since, if we attempt truncation procedure, we will see that the number of decimal digits before each run of zeros far exceeds the length of the run. For example, a run of 2 zeros occurs after 189 digits. But, using Liouville’s theorem we can prove a partial result:

(1) There exists an infinite sequence of rational numbers p_n/q_n satisfying the inequality \displaystyle{\left|M - \frac{p_n}{q_n}\right|<\frac{1}{q_n^{4.5}}}. Hence, Mahler number M is either a transcendental number or an algebraic number of degree at least 5.

Since we need a stronger inequality, we will use following theorem (proved here on pp. 54), which states that:

Thue-Seigel-Roth Theorem: Let \alpha be an irrational algebraic number. Then for any \varepsilon > 0 there exists a constant c(\alpha, \varepsilon) depending on \alpha  such that \displaystyle{\frac{c(\alpha,\varepsilon)}{q^{2+\varepsilon}}<\left|\alpha - \frac{p}{q}\right|}

From this theorem we conclude that

(2) If M is algebraic of degree d\geq 2 (I showed in previous post that it is irrational) then for \varepsilon =0.5 there exists a constant c such that for all p_n/q_n, \displaystyle{\frac{c}{q_n^{2.5}}<\left|M-\frac{p_n}{q_n}\right|}

Using (1) and (2) we conclude that for all n,

\displaystyle{0<c<\frac{1}{q_n^2}}

But as q_n\rightarrow \infty, this inequality cannot hold. Hence M is transcendental.

This number, 0.123456789101112... is also known as Champernowne’s constant. It is an example of what we call Normal numbers. I shall discuss more about it in future posts.

 

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3 responses »

  1. Pingback: Not all numbers are computable | Gaurish4Math

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