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As far as I know the decimal numbers in any irrational appear randomly showing no pattern. Hence it may not be possible to predict the $n^{th}$ decimal point without any calculations. So I was wondering if there is a chance that somewhere down the line in the infinite list of decimal numbers in an irrational could reveal something like our date of birth in order (eg: 19901225) or a even a paragraph in binary that would reveal something meaningful.

Since this a infinite sequence of random numbers ;

  • Is it possible to calculate the probability of a birthday (say 19901225) appearing in order inside the sequence?
  • Does the probability approach to 1 since this is an infinite series.

Any discussions and debate will be welcomed.

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    $\begingroup$ For the numbers that contain every possible binary string (including this one): en.wikipedia.org/wiki/Normal_number For "non-random" irrational numbers that have some pattern, see e.g. the Liouville constant $\sum_{k=1}^{\infty}10^{-k!}$ is irrational, but quite "deterministic." $\endgroup$
    – Lord Soth
    Commented Jul 17, 2013 at 5:59
  • $\begingroup$ No. Your question is a good one nonetheless. You ask, 'are irrational numbers completely random,' but I think the question you're trying to ask is: 'is every irrational number uncomputable'? I recommend you wikipedia it, maybe look up a youtube video about the concept (I dunno if there's any good ones but its worth a look), and then maybe come back here and ask another, more precise question. $\endgroup$
    – goblin GONE
    Commented Jul 17, 2013 at 13:20

4 Answers 4

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Here are two examples of irrational numbers that are not 'completely random':

$$.101001000100001000001\ldots\\.123456789101112131415\ldots$$

Notice the string $19901225$ does not appear in the first number, and appears infinitely many times in the second.

Now, as to your question of probability, let's consider the interval $[0,1]$. Using a modified version of the argument in this question, it can be shown that given any finite string of digits, the set of all numbers containing the string in their decimal expansion is measurable, and has measure $1$.

So, as you suspect, if we choose a number at random between $0$ and $1$, the probability that it has the string $19901225$ in its decimal expansion is indeed $1$. Also, more surprisingly, and perhaps a bit creepy, the probability that we choose a number that contains the story of your life in binary is also $1$.

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    $\begingroup$ Replace "I believe it can be shown that" by "this is a classical exercise in application of Borel-Cantelli lemma to show that". // About creepiness: the number will contain the story of everybody's life, an infinite number of times. $\endgroup$
    – Did
    Commented Jul 17, 2013 at 7:34
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    $\begingroup$ But it also contains fictional versions of your life. And versions where everything is right except one crucial detail. And so on. $\endgroup$
    – GEdgar
    Commented Jul 17, 2013 at 12:17
  • $\begingroup$ How do you find a number that begins with the story of your life rather than having it bajillions of digits off into the decimal expansion? $\endgroup$
    – Darth Egregious
    Commented Jul 17, 2013 at 15:04
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    $\begingroup$ @user973810: You write the story, encode it in binary and stick it in front of the number. $\endgroup$
    – Ilmari Karonen
    Commented Jul 17, 2013 at 16:27
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Not necessarily. Consider $$0.1010010001000010000010000001....$$

Some irrational numbers do have the property you're looking for. They are called "disjunctive numbers."

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    $\begingroup$ Two people used this number to prove their points. What's its name? $\endgroup$
    – Fabinout
    Commented Jul 17, 2013 at 8:19
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    $\begingroup$ This is not a complete answer, but looking at OEIS A010054, we can see that it is an infinite sum of $x$ to the power of triangle numbers, with $x = 1/10$. The number is $r = x + x^3 + x^6 + x^{10} + x^{15} + x^{21} + x^{28} + x^{36} + x^{45} + \cdots$. It is related to the Ramanujan theta function. $\endgroup$
    – John Kemeny
    Commented Jul 17, 2013 at 10:09
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There is a notion on normal number see http://en.wikipedia.org/wiki/Normal_number which basically says that all sequence of digits of a given length are equiprobable. It is known that the set of normal number has full measure and therefore is dense. Note that normality depend on the base. So there is a notion of absolute normality which says that a number is normal in every base. $\pi$, $\sqrt(2)$, $e$ are all believed to be normal but, as far as I know, there is no proof of that. By the way, I'm pretty sure that your birth date already appeared in the known decimal of $\pi$.

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    $\begingroup$ To qualify what you said about $\pi$ and other famous irrationals being believed to be normal. No number has been proved to be absolutely normal, even though clearly many must exist. It is in fact not really known how to go about proving such a thing. So I don't think there is any strong belief about these particular numbers, rather no particular reason is known why they shouldn't be normal. $\endgroup$
    – jwg
    Commented Jul 17, 2013 at 9:22
  • $\begingroup$ I got it: here are the decimal starting from the 52392950th: 9784913659 3150321776 3512697031 4199012256 8771539127 :-) It seem that you are lucky. Mine does not shows up in the first 1000000th. $\endgroup$
    – hivert
    Commented Jul 17, 2013 at 10:05
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You may be interested in a little reading ;) http://en.wikipedia.org/wiki/Infinite_monkey_theorem

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