# Finite Sequence in an Infinite Non-Cyclic Sequence

My friend claims, that the digits in the decimal representation of pi contains every finite sequence of digits. For example my phone number will occur eventually. He claims that this is because there are infinite digits and they are non-cyclic.

While I agree that a decimal representation pi does have an infinite amount of non-cyclic digits, I am not convinced that these conditions are sufficient to imply that every finite sequence of digits are contained.

For instance, if we consider the decimal representation of pi: $$3.14159265358979...$$

If we (quite artificially) omit all occurrences of the digit $$1$$, we get: $$3.459265358979...$$

Or if you don't like that, we replace all occurrences of the digit $$1$$, with the digit $$0$$, we get: $$3.04059265358979...$$

I claim that (in both cases), we now have another sequence that is both infinite and non-cyclic. But clearly, the new decimal representation will not contain all finite sequences of digits. For example my phone number (which contains the digit $$1$$), will never occur.

I presume that the original claim is probably true, but the reasoning is (as demonstrated) apparently unsound.

My question is, is there some other condition of the decimal representation of pi, that is sufficient to guarantee that every finite sequence is contained?

Is this condition a property of all (natural) irrational numbers or transcendental numbers?

I don't think this related question is same, as it refers to an infinite random sequences.

## 1 Answer

It is not true that an irrational number must contain all finite digit-sequences, for example $$0.01001000100001\cdots$$ has only digits $$0$$ and $$1$$.

Neither is it true for transcendental numbers (look here https://en.wikipedia.org/wiki/Liouville_number under Lioville's constant)

It is weird that $$\pi$$ is widely accepted to be normal ( a stronger property than containing all finite digit strings ) because we cannot even rule out that eventually only the digits $$0$$ and $$1$$ occur. Neither can we prove for any digit that it occurs infinite many often in $$\pi$$.

It is conjectured that every irrational algebraic number as well as $$\pi$$ and $$e$$ are normal, but for none of those numbers , it could be proven in any base.

Even worse, noone has the slightest idea how a proof or disproof could work.

So, most mathematicians would agree to your friend, but it cannot be claimed because the problem is currently utterly open.

• Summarizing: The condition required is that pi is a normal number. This is probably true but as of yet unprovable. – Elements in Space May 17 at 6:35
• The only heuristic we have is that the awfully many digits that have been calculated , behave like "random" digits. So I would not consider it to be "probably true" as, for example, the unproven Goldbach conjecture. On the other hand, nothing speaks against the normality, and some use the argument that "almost all" real numbers are normal. Again : We have no proof of the normality of $\pi$ .What has however been shown : Every digit string of length $11$ or less occurs somewhere in $\pi$ – Peter May 17 at 10:40
• @ElementsinSpace Concerning the condition : No, it is not necessary for a number to be normal even if it contains every finite digit string. Normality is a stronger property. However, even to decide the weaker property is currently out of reach. In fact, we do not know whether , lets say, digit $1$ occurs infinite many often. – Peter May 17 at 10:43