Fesshole 🧻10h ago
My phone number is a 7-digit prime that shows up in the digits of pi. My husband's phone number is a 7-digit prime that shows up in the digits of e. We met when training students for the Maths Olympiad. The fess is how nerdy we both are.
Show threadNeal Walfield10h ago

@fesshole That the numbers are both prime is a cool coincidence, but all numbers show up in irrational numbers like pi and e.

Edit: We don't know that all numbers show up in irrational numbers; it's just a conjecture. Thanks to those who replied for clarifying (and see below for more details).

Show threadChaucerburnt9h ago
@nwalfield @fesshole that's actually not confirmed! Being irrational is not a guarantee that every digit sequence occurs; you can make an irrational with only 0s and 1s. It's generally *believed* that every finite digit sequence shows up somewhere in pi and e, but not proven.
Show threadJdeBP

@nwalfield

For the technically minded, the name for such numbers is 'normal numbers'. Not much is known about the relationship between the transcendentals and the normals, or even whether some common algebraics are normal.

So the thing about π at the end of Carl Sagan's Contact, if true, may be either profound or trivial.

@chaucerburnt @fesshole
#maths #mathematics #CarlSagan #pi

11:26amWeb
Show threadChaucerburnt8h ago
@JdeBP @nwalfield @fesshole "normal" is a stricter condition - it requires not only that every finite sequence appears, but that any two sequences of length n occur at the same frequency, for all positive integer n.
Show threadJdeBP8h ago

@chaucerburnt

The interesting question is whether it actually is stricter. It would certainly seem so on its face. But I don't know whether it is conjecture or proven that there's a difference; and I have long since learned to be wary about making claims about infinitely big things based upon intuition alone. (-:

@nwalfield @fesshole
#maths #mathematics

Show thread2e Asteroid5h ago

@JdeBP @chaucerburnt @nwalfield @fesshole Normality is definitely stricter than the condition that all digit sequences show up -- it's not too hard to build a bespoke irrational number where all digit sequences show up, but not at the same frequency.

(For example you could start with Champernowne's constant 0.12345678910111213141516171819... which is normal by construction, and every so often insert a large block of 1s, so the digit sequences 1, 11, 111... show up disproportionately often.)

Show threadChaucerburnt16m ago

@2easteroid @JdeBP @nwalfield @fesshole ha, I think we came up with almost exactly the same counterexample.

You could also make a number where the digit frequencies don't even converge: write all the n-digit sequences, let k be the total number of digits so far, then write (n mod 10) k! times.

I think we can even prove that the cardinality of all-sequence but non-normal numbers is at least as large as that of normal numbers. Take any normal number x, write out the integer part and the decimal point, then for n in N:

- enumerate the first n! digits of x after the decimal place
- enumerate (2n)! zeroes

e.g. π -> 3.1014001410000001415000...

Any two different x will generate different y (since every digit of x appears in known places in y, and those places are the same for all x). Further, since every substring of x appears in y, we haven't broken the all-sequence property with our insertions.

So we have a one-to-one mapping from normals to all-sequence non-normals.

Show threadChaucerburnt54m ago
@JdeBP @nwalfield @fesshole write a decimal point, then for n = 1 to infinity: enumerate all n-digit sequences (takes a total of (10^n) n digits), then write a sequence of n! zeroes. Every sequence appears, but density of zeroes asymptotes to 1.