Is a number whose infinite decimal part is the sequence of even numbers, transcendental? How about a number whose infinite decimal part is the odd numbers? Would the odds be more difficult to prove since they contain almost the entire sequence of primes?
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8$\begingroup$ Both are transcendental, and neither is really harder to establish than the other. $\endgroup$– Andrés E. CaicedoCommented Apr 30, 2017 at 0:43
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11$\begingroup$ An excellent reference for results on transcendental number theory is MR2077395 (2005f:11145) Reviewed. Burger, Edward B.(1-WLMS); Tubbs, Robert(1-CO), Making transcendence transparent. An intuitive approach to classical transcendental number theory. Springer-Verlag, New York, 2004. x+263 pp. ISBN: 0-387-21444-5. $\endgroup$– Andrés E. CaicedoCommented Apr 30, 2017 at 0:57
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11$\begingroup$ The methods required to deal with the numbers you ask about are explained in sections 1.6 and 1.7 of that book. The result follows from the highly nontrivial Roth's theorem, for which there are several very good (but somewhat sophisticated) references (the book does not include a complete proof of this result). $\endgroup$– Andrés E. CaicedoCommented Apr 30, 2017 at 1:03
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4$\begingroup$ For a proof of Roth's theorem, I suggest for instance Chapter 6 of MR2216774 (2007a:11092) Reviewed. Bombieri, Enrico(1-IASP); Gubler, Walter(D-DORT), Heights in Diophantine geometry. New Mathematical Monographs, 4. Cambridge University Press, Cambridge, 2006. xvi+652 pp. ISBN: 978-0-521-84615-8; 0-521-84615-3. $\endgroup$– Andrés E. CaicedoCommented Apr 30, 2017 at 1:03
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5$\begingroup$ @dhy, if you shift it left 2 places and subtract (that is to say, if you multiply by 99), the 101214161820...949698 part becomes 020202...0202. Shift left 3 places and see what happens to 100102104...994996998. $\endgroup$– Gerry MyersonCommented Apr 30, 2017 at 6:56
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1 Answer
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In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a non-constant polynomial such that $p(n) \in \mathbb{N}$ for every $n\in \mathbb{N}$, then the number
$$0.p(1)p(2)p(3)p(4)\ldots,$$
which is formed concatenating after the decimal point the values of $p(1), p(2), p(3), \ldots$ (in that order), is a transcendental and non-Liouville number.
answered Apr 30, 2017 at 6:23
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2$\begingroup$ Non-Liouville! That's wonderful!! I'm glad to know these are transcendental. $\endgroup$– user10290Commented Apr 30, 2017 at 7:40
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4$\begingroup$ There are only countably many polynomials which take natural values at all naturals (since any polynomial is entirely determined by its values on the naturals), so yes, countable. $\endgroup$ Commented Apr 30, 2017 at 8:35
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3$\begingroup$ @ErinCarmody: The proof the normality of these numbers was the subject matter of a 1952 paper of H. Davenport & P. Erdös: renyi.hu/~p_erdos/1952-04.pdf $\endgroup$ Commented Apr 30, 2017 at 21:17
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4$\begingroup$ In this case, the underlying polynomial may be taken as p(x) = 42... $\endgroup$ Commented Apr 30, 2017 at 22:50
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5$\begingroup$ @MateenUlhaq: What polynomial? There is no polynomial whose values at natural numbers are 4,2,4,2,... $\endgroup$– BurakCommented May 1, 2017 at 19:46