Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:52:10Z http://mathoverflow.net/feeds/question/105153 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105153/proof-of-the-transcendence-of-the-champernowne-constant-with-thue-siegel-roth Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth Katrin 2012-08-21T12:23:00Z 2012-08-25T11:50:08Z <p>It is well know that the Champernowne Constant </p> <p>0,1234567891011121314151617....</p> <p>is transcendental. This was shown by Kurt Mahler in 1937. But the proof of the transcendence should also work with the famous Thue-Siegel-Roth theorem (http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem), that was proved in 1955. </p> <p>I'm looking for a reference of the transcendence-proof where this theorem is used. </p> http://mathoverflow.net/questions/105153/proof-of-the-transcendence-of-the-champernowne-constant-with-thue-siegel-roth/105158#105158 Answer by quid for Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth quid 2012-08-21T13:19:15Z 2012-08-21T13:19:15Z <p>There are applications of Thue--Siegel--Roth and related results to this question (but it seems that they use in some form or another the rational approximations already present in Mahler's work). </p> <p>For example, Baker in "On Mahler's classification of transcendental numbers." Acta Math. 111 1964 97–120, shows that this constant is not a U-number using a result he describes as "This extends a theorem of LeVeque [..] which itself is a generalisation of Roth's Theorem" </p> <p>However, then for the proof of this application it reads "It is clear from the proof of these results that the hypotheses of Theorem 1 are satisfied..." Where 'the proof of these results' refer to Mahler's.</p> <p>And much more recently, Adamczewski and Bugeaud in "<a href="http://math.univ-lyon1.fr/~adamczew/MesTransI.pdf" rel="nofollow">Mesures de transcendance et aspects quantitatifs de la méthode de Thue-Siegel-Roth-Schmidt.</a>" Proc. Lond. Math. Soc. 101 (2010), no. 1, 1–26 generalise this result of Baker together with a classical result of Ridout to then show that the generalizations of the Champernonwne Constant (any base, any suitable polynomial) are all S or T numbers. See section 3 of the paper, in particular Théorème 3.2. But again the proof uses the rational approximations of the original transcendence proofs (by Mahler); cf. the final paragraph of Section 3.</p> http://mathoverflow.net/questions/105153/proof-of-the-transcendence-of-the-champernowne-constant-with-thue-siegel-roth/105459#105459 Answer by Wadim Zudilin for Proof of the transcendence of the Champernowne Constant with Thue-Siegel-Roth Wadim Zudilin 2012-08-25T11:50:08Z 2012-08-25T11:50:08Z <p>Here is an extract from van der Poorten's "Obituary. Kurt Mahler (1903--1988)" (see p.353 in <em>J. Austral. Math. Soc. Ser. A</em> <strong>51</strong> (1991)):</p> <blockquote> <p>In a more unexpected way, Mahler's arguments led to the following amusing result: Suppose $f$ is a non-constant polynomial taking integer values at the nonnegative integers. Then the concatenated decimal $$\phi=0.f(1)f(2)f(3)\dots$$ is transcendental. In particular Champernowne's normal number $$0.123\dots910111213\dots$$ is transcendental. Mahler's argument relies on the observation that one readily obtains rational approximations to $\phi$ with denominators high powers of the base 10, thus composed of the primes 2 and 5 alone. Perhaps disappointingly, Roth's definitive form of the Thue--Siegel inequalities permits a more immediate argument obviating the need for an appeal to the $p$-adic results.</p> </blockquote> <p>This is to say that Roth's argument is more superior than Mahler's but it appeared some 20 years later...</p>