# Is infinite sequence of irrational numbers digits mathematically observable?

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

Is infinite sequence of irrational numbers digits mathematically observable?

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 $$\left\{0,0,0\right\}$$, $$\left\{1,1,1\right\}$$ or $$\left\{2,2,2\right\}$$ in this writing?

By symbolic mathematical definition,

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

Question: Is there an exist such a $$n\in\mathbb{Z^{+}}$$, 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 ($$e,\pi\cdots$$ 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.

• Possible duplicate of Does Pi contain all possible number combinations? – Xander Henderson Mar 26 at 2:05
• arxiv.org/abs/math/0411418 – Count Iblis Mar 26 at 2:19
• I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask? – Tanner Swett Mar 26 at 2:33
• What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields. – Theo Bendit Mar 26 at 4:28
• There are certainly irrational numbers which do not contain 000 anywhere in their decimal expansion. For example, the number whose digits begin 0.10110111011110111110111111011111110111111110 and so on, with the length of the sequence of 1s increasing after each 0. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces 000 as a sequence of digits. – Daniel Wagner Mar 26 at 5:22

## 2 Answers

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...

There is a 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 (and infinitely many times that is). Unfortunately, it is currently unknown whether $$\sqrt{2}$$ has this property, same is true for other familiar irrational constants such as $$e$$, $$\pi$$, and so on (it seems that only constants for which it is known were carefully constructed for this purpose). So in your second example, $$012345$$ would be in $$\sqrt{2}$$'s digits infinitely many times (it already appears once in the first million digits referred above).

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

• Well, for $e$ is it possible? – Elvin Mar 25 at 22:35
• $e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist. – Eevee Trainer Mar 25 at 22:37
• By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago. – Paul Sinclair Mar 26 at 1:22
• You said "you might wan to check it out", I think you mean "want" instead of "wan". – numbermaniac Mar 26 at 4:15
• "By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD – Eevee Trainer Mar 26 at 6:16

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.)

• Is it known a non-normal number? – Elvin Mar 25 at 22:58
• Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p – Eevee Trainer Mar 25 at 22:59