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?

8$\begingroup$ Both are transcendental, and neither is really harder to establish than the other. $\endgroup$ – Andrés E. Caicedo Apr 30 '17 at 0:43

11$\begingroup$ An excellent reference for results on transcendental number theory is MR2077395 (2005f:11145) Reviewed. Burger, Edward B.(1WLMS); Tubbs, Robert(1CO), Making transcendence transparent. An intuitive approach to classical transcendental number theory. SpringerVerlag, New York, 2004. x+263 pp. ISBN: 0387214445. $\endgroup$ – Andrés E. Caicedo Apr 30 '17 at 0:57

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. Caicedo Apr 30 '17 at 1:03

4$\begingroup$ For a proof of Roth's theorem, I suggest for instance Chapter 6 of MR2216774 (2007a:11092) Reviewed. Bombieri, Enrico(1IASP); Gubler, Walter(DDORT), Heights in Diophantine geometry. New Mathematical Monographs, 4. Cambridge University Press, Cambridge, 2006. xvi+652 pp. ISBN: 9780521846158; 0521846153. $\endgroup$ – Andrés E. Caicedo Apr 30 '17 at 1:03

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 Myerson Apr 30 '17 at 6:56
In point of fact, K. Mahler proved in this paper that, if $p(x)$ in a nonconstant 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 nonLiouville number.

2$\begingroup$ NonLiouville! That's wonderful!! I'm glad to know these are transcendental. $\endgroup$ – user10290 Apr 30 '17 at 7:40

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. $\endgroup$ – Patrick Stevens Apr 30 '17 at 8:35

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/195204.pdf $\endgroup$ – José Hdz. Stgo. Apr 30 '17 at 21:17

4$\begingroup$ In this case, the underlying polynomial may be taken as p(x) = 42... $\endgroup$ – José Hdz. Stgo. Apr 30 '17 at 22:50

5$\begingroup$ @MateenUlhaq: What polynomial? There is no polynomial whose values at natural numbers are 4,2,4,2,... $\endgroup$ – Burak May 1 '17 at 19:46