Prove there are no hidden messages in Pi

Question

Assume that a proof that pi is normal existed.

Is it then possible that starting at some finite position x in pi, from there on every p(n)'th digit is 0, where p(n) is the n'th prime?

I know probability arguments say no, but do they also prove that it is impossible?

Is there a way to disprove the statement?

Also, the digits of Pi can tell us certain mathematical truths, for instance that the circumference of a circle is less then any other shape with same area.

The question is, is there a limit to the information one can extract from the digits of Pi? Is it possible to reconstruct all theorems from the digits of Pi? What sort of truths can be extracted from the digits of Pi? Is entire math and all truths somehow encoded within the digits of Pi ?

Answer 1

If $\pi$ is normal, then in fact not only does $\pi$ have secret messages, but it will contain in its digits every possible finite message infinitely often! This is simply because it is part of the meaning of normal number that every finite sequence appears with the expected density. And so in particular, every normal number will contain long blocks of digits that exactly express the collected works of William Shakespeare, and also versions with the plays translated (and mis-translated) into every other language, in all possible ways, and so on.

Indeed, any normal number will contain within it all the theorems of mathematics, with fully correct proofs (as well as with false proofs, in all possible ways).

But of course, this is why you asked about coding on an infinite set, asking for a kind of infinitary regularity property. Leaving aside the particular question about primes, let me consider your latter question: is there a limit to the amount of information that we can extract from $\pi$?

For this, the answer is yes, there is a limit, if we make the problem precise with a particular meaning of what it should mean to extract information. The reason is that $\pi$ is a computable number; there is a computable procedure that will tell us the $n$-th digit. Thus, any procedure that extracts information from $\pi$ by means of a computable procedure inspecting the digits will necessarily be altogether computable. So for example, there can be no computable procedure that extracts information from $\pi$ so as to answer yes-or-no the question of whether a given Turing machine program halts, since the halting problem is not decidable.

More generally, there are only countably many sets that are computable from any given real, no matter the complexity of that real (since there are only countably many programs), and so there is a robust sense in which most sets of natural numbers are not reducible to $\pi$ or any other fixed real in this way.

Lastly, apart from $\pi$, it seems that there are normal numbers $z$ that have the property that the $p(n)$-th digit of $z$ is $0$, where $p(n)$ is the $n$-th prime. Since the primes have asymptotic density $0$, then unless I am mistaken, it appears that we could simply place $0$'s in the prime digits, and choose the rest of the digits randomly, to arrive at a normal number exhibiting your property.

Answer 2

(this is a too-long comment)

You're describing a set $S$ of numbers, those for which there is a $k$ with the $(k+p)$-th digit equal to 0 when p is a prime, and asking whether $\pi\in S$. (The problem doesn't essentially change whether you fix a base b or allow any base.)

We don't know much about set $S$. Finite methods are useless, since by assumption $\pi$ already contains arbitrarily long finite segments that match this description (as well as the Declaration of Independence and this post).

Unfortunately, $S$ is a tricky character. It is dense in the reals so we can't separate $\pi$ from its members by an $\varepsilon$. It is uncountable so we can't use something like Roth's theorem to separate it from $\pi$. We don't really have any methods for working on these sorts of problems.

Answer 3

This is more of an informational comment than a direct answer to your question. (Later) Just before sending this answer in, I noticed the question and activity are from almost a year ago!

The property you're describing is shared by almost all real numbers in both the Baire category sense and the Lebesgue measure sense. In fact, its complement is even smaller than first-category-and-measure-zero, being a countable union of lower porous sets. Note there is a huge difference between the notion of a normal number (to base $10$) and the property you're talking about, since the set of normal numbers is large in one way (Lebesgue measure) but small in another way (Baire category), whereas the set you're talking about (sometimes called the "absolutely disjunctive" real numbers, for which you can google the phrase I put in quotes) is large in both the Lebesgue measure sense and in the Baire category sense (and, in fact, even larger than what the conjunction of these two notions could allow for). For more details, see my 19 February 2003 sci.math post at

http://groups.google.com/group/sci.math/msg/4ec315328c1afdb8

An excerpt from the above post:

Disjunctive to base $b$ ($b$ being some integer greater than $1$) means that every finite $b$-word appears infinitely often in the $b$-ary expansion of the number (note this is equivalent to every finite $b$-word appearing at least once), and the adjective "absolutely" means that this property holds for each $b = 2,\; 3,\;...$ The result is virtually immediate [all but a $\sigma$-porous set of real numbers are absolutely disjunctive], since for each of the countably many ways of choosing a fixed $b$ and a fixed $b$-word, the collection of numbers whose $b$-ary expansions don't contain that $b$-word infinitely often is a uniform Cantor set, and hence is porous (even uniformly porous; in fact, even uniformly for the stronger lim-inf type of porosity that Julia set theorists use). Recall that the larger collection of numbers which fail to be absolutely normal (or which fail to be simply normal relative to a specific base) forms a measure zero (but with Hausdorff dimension $1$) co-meager set.

Answer 4

The statement is not clear. You say "from there on every p(n)'th digit is 0, where p(n) is the n'th prime?", but what is n? If you mean that this holds for every n, then it's easy to disprove: if every second digit after that, and also every third, fifth, seventh etc. digit after that is zero, then all digits except for the first one after that are zero. (You apply the sieve of Eratosthenes but also strike out the first multiple of each prime, i.e., not even the primes which usually survive this process, would survive yours.) That would mean that π is rational, which we know to be wrong.