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Materials for self-study (problems and answers)

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:

- Is there a number that is palindromal in both base 2 and base 3?
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- Starting digits of $2^n$.
$$x=0.\underbrace{1}_{1^{st}\text{ block}}\overbrace{12}^{2^{nd}\text{ block}}\underbrace{123}_{3^{rd}\text{ block}}\overbrace{1234}^{4^{th}\text{ block}}\dots \underbrace{12\dots n}_{n^{th}\text{ block}}\dots $$

where $n^{th}$ block is thefirst $n$ positive integersfor each $n\in \mathbb{Z}^+$.

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

- Is the number $0.112358132134…$ rational or irrational?
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- Divisibility by $9$
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- How can I prove that all rational numbers are either terminating decimal or repeating decimal numerals?

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!

$\lim_{n\to \infty}$ of the “nth block” will be equal to the exact same digits as the Champernowne constant. The Champernowne constant is transcendental, so no matter what comes before its digits your number will be transcendental.

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