8
$\begingroup$

By way of motivating example, someone came up with a completely evil user interface where the user was required to select their ten digit phone number by moving a sliding window over the digits of $\pi$:

https://static.boredpanda.com/blog/wp-content/uploads/2018/06/funny-worst-input-fields-9-5b235c0f48bf9__700.gif

Given that $\pi$ is infinite and non-repeating, my gut says that all possible combinations of ten digits must exist somewhere in the digits of $\pi$, but I don't know how to prove that.

Help appreciated.

Best,

Glenn

$\endgroup$
5
  • 2
    $\begingroup$ You need more than non-repeating. i.e. you could create a non-repeating decimal that has no nines in it. You need something stronger like $\pi$ is a "normal number." While believed to be true, that fact hasn't actually been proven. $\endgroup$
    – Doug M
    Jan 3, 2020 at 0:04
  • 2
    $\begingroup$ “Infinite and non-repeating” is not enough; the number $0.01001000100001000001\ldots$ satisfies this condition, but clearly will not satisfy your desired conclusion. If $\pi$ is normal, then this would follow, but the normality of $\pi$ is an open question, and this condition is slightly weaker than “normal number”, since you only require each combination to appear once. It may even be possible to establish it by inspection with known expansions. $\endgroup$
    – Arturo Magidin
    Jan 3, 2020 at 0:05
  • 1
    $\begingroup$ See also this $\endgroup$
    – Arturo Magidin
    Jan 3, 2020 at 0:07
  • 11
    $\begingroup$ The fact that a(10) exists in the sequence A036903 from the OEIS can be seen as a proof that all possible sequences of 10 digits appear at least once in $\pi$. In this case, if you download the first 241,641,121,048 digits (about 242 billion) you can do an exhaustive search and prove that all 10-digit sequences appear. $\endgroup$
    – DecimalTurn
    Aug 23 at 21:31
  • $\begingroup$ See also this Github repo where you can find code that reproduces theses results. $\endgroup$
    – DecimalTurn
    Aug 24 at 2:31

4 Answers 4

Reset to default
10
$\begingroup$

As DecimalTurn pointed out in the comments, this OEIS page says that the first 241641121048 digits of $\pi$ contains all ten-digit strings.

$\endgroup$
4
$\begingroup$

Independently of guts, this is a property called normality. It is unknown whether $\pi$ is normal (in any base, including base 10). We can't even show (currently) that there is no place in the decimal expansion $\pi$ after which only the digits $0$ and $1$ appear.

$\endgroup$
1
  • 10
    $\begingroup$ Normality is sufficient, but not necessary. You could have all ten-digit sequences, in order, separated by (say) $7$'s. Then the number would not be normal—it wouldn't even be simply normal—but it would still contain all ten-digit sequences. $\endgroup$
    – Brian Tung
    Sep 6 at 21:23
1
$\begingroup$

It may well be true, but you cannot prove it simply using that fact that it is infinite and non-repeating. So is the number $0.1001000010\ldots$ (it has a $1$ at the $n$th digit after the dot if $n$ is a perfect square and $0$ otherwise), but it clearly doesn't contain all phone numbers.

$\endgroup$
0
$\begingroup$

Infinite and non-repeating does not necessarily imply to contain all ten digits combinations.

Consider the infinite nonrepeating sequence $$101001000100001....$$ or $$12233344445555566666....$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .