# Do irrational number contain infinite/every patterns of sequences?

I guess the question is

"does an 'infinite' number of patterns imply 'every' number of patterns?"

For instance, if you could quickly calculate the decimal sequence of π, could you not (in theory of course) come up with an algorithm to search that sequence for some pre-determined sequence?

Then you could do this:

start = findInPi(sequence)


So "sequence" in theory could be a decimal representation of the movie "The Life of Pi". The implication is that all digital knowledge (past, present and future) is bound up in irrational numbers (not just the group of irrational numbers, but each irrational number), and we just need to know the index to pull data out.

Once you know the index and length of data, then you could simply pass this long.

playMovie(piSequence(start, length))


From an encryption standpoint, you could pass the start, length pair around, and the irrational number would be known only by the private key holder.

Am I off base here?

• this begs the question 'what is the longest known expansion of 0314159265358979323846..... inside $\pi$' – JMP Mar 26 '15 at 11:25

No, this is not the case for every irrational number. For example, the number

$$1.01001000100001000001000000100...$$

where each run of zeroes is one longer, is clearly irrational, since the decimal expansion never repeats. But it just as clearly doesn't contain every pattern of digits, because the only digits it contains are 0 and 1.

$\pi$ in particular is suspected (but not proved) to satisfy a stronger property, namely that it is normal, which means that not only does every pattern of digits occur, but every pattern occurs infinitely many times, with the frequency one would assume in a random string of digits.

In a certain technical sense, most numbers are normal, but there are very few expressions that have been proved to produce a normal number. This is a problem for your cryptography idea, because it is hard for the two parties to agree on a particular number that contains all of the messages they want to exchange.

• It's just like with humans. We believe that most people are normal, but it is very hard to prove rigorously in any particular instance. – Marc van Leeuwen Mar 21 '13 at 13:56
• As for the cryptography idea, a pair could certainly just agree $\pi$. That way, they either succeed in communicating or disprove a famous conjecture (only one of them will get the fame though because the other will have been caught and killed by the enemy after not being able to communicate with his comrade). The bigger problem with the crytography idea is that for a typical 100-digit message one will need to know more digits of $\pi$ than there are particles in the universe. – Sean Eberhard Mar 21 '13 at 14:06
• @SeanEberhard: But if I understand the suggestion right, the choice of irrational number would be the secret key! That needs a pool of many possible numbers to choose from, or it would be easy for an attacker to brute-force the possible keys. – hmakholm left over Monica Mar 21 '13 at 16:05
• "...there are very few expressions that have been proved to produce a normal number": Are there any? In base 10? In all bases simultaneously? – TonyK Mar 26 '15 at 11:22
• @TonyK: Champernowne's constant is explicitly constructed to be normal in base ten. It is not immediately clear whether one can get normal-in-all-bases by the same technique. – hmakholm left over Monica Mar 26 '15 at 11:31

You can achieve it, its not hard. for instance I assume that 0..4=LOW (0) 5..9=HIGH (1)

I start calculating π using chudnovski algorithm... and so i start π=3.1415.... (π=0.0001....)

At the same time as I calculate the infinite tail of π, I compare the data that I wish to send to somebody else, until I find a 100% match of length.

So after I find a match, I am telling my remote friend "the data that I wish to send you starts at the 9.876.543 digit of pi and its size is 1MB after the starting point. Convert it accordingly (0-4=0; 5-9=1;)

This way you could transmit data just by providing starting point and size. Also , there will be thousands of ways to optimise the algorithm.

Inside the π lies ANYTHING ! From my dna to the farest multiverse.

• Your last paragraph assumes that $\pi$ is normal, which is not proven. And from a practical point of view, your procedure doesn't seem to really achieve anything. You don't get any compression because your start index "9.876.543" will generally need to be as long as the message you want to send (except in a vanishing minority of cases), and you don't get any encryption either because your enemy can compute digits of $\pi$ just as easily as your friend can. – hmakholm left over Monica Mar 26 '15 at 11:54
• Add to this that finding an instance of a moderately long desired message among the digits of $\pi$ will require more computing resources than the known universe has room to fit within its entire lifetime, and your blithe "its not hard" will be seen to be a bit of a stretch. – hmakholm left over Monica Mar 26 '15 at 11:56