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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 $1234567891011$; ... )

Community
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ASB
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    [This question](http://math.stackexchange.com/questions/25205/is-this-number-transcendental?rq=1) is related and has some references which might be useful. – Winther Feb 16 '15 at 08:01
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    On which field? – Empty Feb 16 '15 at 08:10
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    @Panja.S. algebraic refers to being the root of some polynomial $p(x) \in \mathbb{Q}[x]$. – Sultan of Swing Feb 16 '15 at 08:14
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    @Winther Thanks. I'll try with that question. – ASB Feb 16 '15 at 14:50
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    Unclear question! How about the pattern after 123456789? – Harry Peter Feb 16 '15 at 17:15
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    It would be followed by 12345678910. – dalastboss Feb 16 '15 at 17:23
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    @HarryPeter my number is 0.11212312341234512345612345671234567812345678912345678910123456789101112345678910111212345678910111213... Is it clear ? :) – ASB Feb 17 '15 at 01:04
  • Proving a number to be transcendental rather than algebraic is notoriously difficult. I doubt this one is an easy case. – amcalde Feb 17 '15 at 01:46
  • The probability that it is algebraic is $\frac{|\mathbb{N}|}{|\mathbb{R}|}=0$. – barak manos Feb 22 '15 at 12:08
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    It might be better if the number was 0.1121231234123451234561234567123456781234567891234567900123456789011... Make the nth block $\sum_{i=1}^n i 10^{n-i-1}$ – tomi Feb 28 '15 at 06:38
  • oops. Make the nth block $\sum_{i=1}^n i 10^{n-i}$ – tomi Feb 28 '15 at 06:45
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    @tomi: The "better form" may change OP's answer, for comparison [Champernowne constant](http://en.wikipedia.org/wiki/Champernowne_constant) 0.12345678910111213... is transcendental while $\sum_{i=1}^\infty i/10^i$ = 0.123456790123... = 10/81. – kennytm Feb 28 '15 at 14:33
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    @barakmanos Actually, probability is measure theory, not cardinality, and probability zero doesn't mean much when dealing with a specific real number. – Thomas Andrews Feb 28 '15 at 14:55
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    Wikipedia has a [reference](http://en.wikipedia.org/wiki/Champernowne_constant#cite_note-5) to a paper proving the Champernowne has irrationality measure $10$ (which proves it is transcendental due to the the [Thue-Siegel-Roth theorem](http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem) ). It's probably helpful. I have no access to the paper. – Milo Brandt Mar 01 '15 at 02:59
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    Your number is in the form of ${\sum_{n=1}^\infty} {{\sum_{m=1}^n m10^{n-m}}\over {10^{\sum_{p=1}^n p}}}$ or if you know a little about triangular numbers $\sum_{n=1}^\infty {{\sum_{m=1}^n m10^{n-m}}\over {10^{T_p}}}$ – Renato Faraone Mar 29 '15 at 19:27
  • @barakmanos but that's true of *every* number. – fleablood Oct 08 '15 at 05:22

3 Answers3

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

XL _At_Here_There
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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.

martin
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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!

Rob Bland
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    I just realized this isn't correct, I didn't account for blocks with two or three decimal digits like 1234567891011... – Rob Bland Oct 15 '15 at 02:53
  • If I may ask : What is the procedure that you made to arrive to that expression ? – I0_0I Aug 26 '19 at 13:24