Let $$x=0.112123123412345123456\dots $$ Since the decimal expansion of $x$ is non-terminating and non-repeating, clearly $x$ is an irrational number.

Can it be shown whether $x$ is algebraic or transcendental over $\mathbb{Q}$ ? I think $x$ is transcendental over $\mathbb{Q}$. But I don't know how to formally prove it. Could anyone give me some help ? Any hints/ideas are much appreciated. Thanks in advance for any replies.

My Number:

$$x=0.\underbrace{1}_{1^{st}\text{ block}}\overbrace{12}^{2^{nd}\text{ block}}\underbrace{123}_{3^{rd}\text{ block}}\overbrace{1234}^{4^{th}\text{ block}}\dots \underbrace{12\dots n}_{n^{th}\text{ block}}\dots $$ where $n^{th}$ block is the first $n$ positive integers for each $n\in \mathbb{Z}^+$.

(That is the 10th block of $x $ is $12345678910$; The 11th block is $1234567891011$; ... )

  • 7
    $\begingroup$ This question is related and has some references which might be useful. $\endgroup$ – Winther Feb 16 '15 at 8:01
  • 11
    $\begingroup$ Unclear question! How about the pattern after 123456789? $\endgroup$ – Harry Peter Feb 16 '15 at 17:15
  • 6
    $\begingroup$ It would be followed by 12345678910. $\endgroup$ – dalastboss Feb 16 '15 at 17:23
  • 8
    $\begingroup$ @tomi: The "better form" may change OP's answer, for comparison Champernowne constant 0.12345678910111213... is transcendental while $\sum_{i=1}^\infty i/10^i$ = 0.123456790123... = 10/81. $\endgroup$ – kennytm Feb 28 '15 at 14:33
  • 16
    $\begingroup$ @barakmanos Actually, probability is measure theory, not cardinality, and probability zero doesn't mean much when dealing with a specific real number. $\endgroup$ – Thomas Andrews Feb 28 '15 at 14:55

Obviously, it can be outputed by Turing Machine in real time. Thus under the Hartmanis-Stearns conjecture, it is a transcendental number.


I believe your number can be written as

$$\sum _{j=1}^\infty 10^{-\sum _{m=1}^{j+\frac{1}{2}} \sum _{n=1}^m \left\lceil \log _{10}(n+1)\right\rceil } \left\lfloor c 10^{\sum _{n=0}^j \left\lceil \log _{10}(n+1)\right\rceil }\right\rfloor$$

where $c$ is the Champernowne constant.

Don't know if that helps.


Your number can be written with the following formula: $$\sum_{n=1}^{\infty} \frac{ \sum_{r=1}^n r(10)^{n-r}}{10^{\frac{n(n+1)}{2}}}$$ I don't know how to prove it is transcendental, but I hope this helps!

  • 4
    $\begingroup$ I just realized this isn't correct, I didn't account for blocks with two or three decimal digits like 1234567891011... $\endgroup$ – Rob Bland Oct 15 '15 at 2:53

protected by Community Oct 14 '15 at 18:58

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.