I was trying to analyze the following statement: "Any integer can be expressed as a function of any irrational number". For which I got:

\begin{equation} a=\lfloor{10^b \alpha}\rfloor-\lfloor{10^{b-s(a)} \alpha}\rfloor10^{s(a)} \end{equation}

where a, b and $\alpha$ represents the integer, the position of the integer in the irrational and the irrational, respectively. S is size of the integer (amount of digits). As an example:

\begin{equation} 15=\lfloor{10^4 \pi}\rfloor-\lfloor{10^{2} \pi}\rfloor10^{2} \end{equation}

Expanding the floor functions I got:

\begin{equation} a=\frac{1}{2} (10^s-1)+\frac{1}{\pi} \sum_{k=1}^{\infty} \frac{1}{k} (\sin(2\pi k 10^b \alpha)-10^s\sin(2\pi k 10^{b-s} \alpha)) \end{equation}

There are infinite values for b that solve the relation. Now, it is tempting to solve for b and find out the position of every number inside an irrational. It sounds too good to be true and I probably have something conceptually wrong. Anyway, every approach of mine results with b getting cancelled out of the relation.

Could someone show how to isolate b or explain me what is the flaw in my idea?

I would like to apologize for any dumb mistake, I am just an engineer venturing in math for fun.

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  • 7
    $\begingroup$ Not following. Are you suggesting that every natural number must appear in the decimal expansion of any given irrational? This is false...take $\alpha =.101001000100001\cdots$ for instance. $\endgroup$ – lulu Jul 22 at 0:02
  • $\begingroup$ What is a function of a number? $\endgroup$ – Elliot G Jul 22 at 0:13
  • 2
    $\begingroup$ the first comment indicates that there are some irrational numbers for which not every integer appears in the decimal representation (for example 15 does not appear in .1010010001,,,). There are the so-called normal numbers en.wikipedia.org/wiki/Normal_number for which every integer does appear, and "most" numbers are normal (a statement which could be made precise using measure theory in math). But it is usually difficult to determine if a given irrational number is normal. Your question seems to suggest that π is normal, but this is not known mathoverflow.net/q/51853 Welcome! $\endgroup$ – Mirko Jul 22 at 0:18

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