Kdtyjb

I have a little question. In fact, is too short.

Is infinity observable? (Can infinity be observed?)

I would like to explain it by example because the question seems unclear in this way.

A simple example:

$sqrt 2=1,41421356237309504880168872420969\807856967187537694807317667973799073247\846210703885038753432764157273501384623\091229702492483605585073721264412149709\993583141322266592750559275579995050115\278206057147010955997160597027453459686\201472851741864088 cdots$

Is it possible to prove that there is no combination of $left{0,0,0right}$, $left{1,1,1right}$ or $left{2,2,2right}$ in this writing?

By mathematical definition,

Let, $phi_{sqrt 2}(n)$ is n'th digit function of $sqrt 2.$

Question: Is there an exist such a $ninmathbb{Z^{+}}$, then $phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=0, phi_{sqrt 2}(n+2)=0$ ?

Or other combinations can be equal,

$$phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=1,phi_{sqrt 2}(n+2)=2, phi_{sqrt 2}(n+3)=3, phi_{sqrt 2}(n+4)=4, phi_{sqrt 2}(n+5)=5$$

Here, $sqrt 2$ is an only simple example. The question is not just
$sqrt 2$.

Generalization of the question is :

For function $phi _alpha (n)$, is it possible to find any integer sequence ? where $alpha$ is an any irrational number or constant ($e,picdots$ and etc).

I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.

Sorry about the grammar and translation errors in my English.

Thank you very much.

Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...

Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).

Also, there is one popular question here on MSE about whether $pi$ has this property, you might wan to check it out: Does Pi contain all possible number combinations? .

Less an answer than an extended comment:

This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.

Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:

$$0.12345678910111213141516171819202122232425...$$

It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.

Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)