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What are real numbers called that have the following property: Given a non-terminating real number r and for any digit D, P(D is in the tail of r) > 0. I do not believe transcendental numbers have this properly but I could be wrong.

For example, one can easily prove interesting things about real numbers if one can show that any digit always exists.

For example if pi or e has these properties one can show that any subsequence of pi always exists in pi or e. (and hence a also as a sub-subsequence, etc)

The argument is quite simple and interesting. I think it demonstrates to the lay person how interesting numbers can be in that an infinite sequence can contain itself as a sub-sequence... it seems counter intuitive to the lay person but is quite natural for the type of numbers I'm describing.

for example, 0.01234567890123456790123456789....

has pi "embedded" in it as a subsequence and, in fact, as any sequence embedded in it. (pi, e, 2, or any other value)

Of course, the above value is somewhat contrived but it does demonstrate that we can, instead of thinking of "real numbers" and calculations on them, use sequences an and calculations on the sequences instead.

AbstractDissonance
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    What does "P(D is in the tail of r) > 0" mean? – Misha Lavrov Nov 15 '17 at 02:23
  • @MishaLavrov: The tail of a real number is simply the decimal expansion so it is saying that any digit D(0 through 9 for base 10) can be found in the expansion. This excludes terminating numbers. It sort of is a statistical randomness argument that any digit can be found in the decimal expansion. If you know any digit can be found in the expansion and it is infinite then it is easy to construct any string of digits as a "sub-sequence). for the example gave, pi is embedded because 3 is the 3rd digit, 1 is the 9th digit, 4 is the 14th digit, 1 is the 19th digit, etc. – AbstractDissonance Nov 15 '17 at 23:25
  • this is counter intuitive to most people because pi is has an infinitely long expansion and so one would think it could not contain, as a "sub-sequence" itself but it's related to the concept of infinity itself and it also relates all reals to these special numbers. Not all numbers have this properly. The louville constant, for example, only contains 0's and one's so it is impossible to find pi base 10 in it. – AbstractDissonance Nov 15 '17 at 23:28
  • How does this exclude terminating numbers? Any digit $0$ through $9$ can be found in the expansion of $0.0123456789$. – Misha Lavrov Nov 15 '17 at 23:28
  • Misha: because a a terminating number doesn't have an infinite sub sequence to chose. We are talking about sequences where digits are simply concatenations of the sequence. If D is a digit and n_k is the kth digit in a some sequence n sequence(so it is either 0,1,...,9) then n = D_1 cat D_2 cat D_3 ... is the decimal representation of a number that has the same sequence. Another way to see it is that if n is the number > 0 & < 1 who's k digit is D_k, then n = sum(D_k*10^(-k),k=1..infinity). – AbstractDissonance Nov 16 '17 at 00:48
  • Now, we can then write pi/10 as such a sequence. D_1 = 3, D_2 = 1, D_3 = 4, D_4 = 1, D_5 = 6 and, in general, D_k = floor(pi*10^k) mod 10. So, we can see that pi can be seen as a sequence. This should all be elementary and if you don't understand it you should spend a few days learning this stuff. – AbstractDissonance Nov 16 '17 at 00:50
  • Now, if the number you have given, which is finite, then we cannot achieve a sub-sequence from it because it is a finite sequence. n_k = (k mod 10) - 1. Because k must be monotonically increasing. But we would run out of digits from your finite sequence very quickly. – AbstractDissonance Nov 16 '17 at 00:52
  • @MishaLavrov: Your sequence can only provide the sub sequence 012345689, 123456789, 23456789, 3456789, 456789, 56789, 6789, 789, 89, 9, and removing sequential digits from those above. e.g., 459 works. (in this case, all sequences digits that satisfy for j < k, D_j < D_k). E.g., to represent pi/10 using your number n, we have to have pi_1 = 3 = n_4(the 4th digit in your sequence is 3, the 1st digit of pi/10). pi_2 = 1 = n_2... but 4 > 2 so we are in a bind. our sequence should be monotonically ascending because it's actually an index in to the original sequence. – AbstractDissonance Nov 16 '17 at 00:55
  • While it works, in a sense, in this case, what if you have a number of 0.948390483724000000590000359000035900000299...? How do you know that the number has a 6? You don't unless you know it has the property that I'm talking about. (that a 6 will occur in the infinite sequence that the decimal represents) So it is a bit more complicated than it seems and if you just spend a few days thinking about it you will understand what is going on. – AbstractDissonance Nov 16 '17 at 00:59
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    I don't appreciate your condescension. I can't read your mind about what you're trying to define if you don't define it properly. – Misha Lavrov Nov 16 '17 at 01:00

2 Answers2

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The property that you are describing sounds like a variation on the concept of "Normal Numbers". In a normal number every sequence of digits occurs with equal probability. A slightly broader concept is a disjunctive sequence which is a sequence that contains all possible subsqeuences (likewise a number who's decimal expansion is called a disjunctive number).

You might be able to say "The number is normal for strings of length n" or "The number is disjunctive for strings of length n".

Q the Platypus
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  • Normal numbers does seem to be similar although I think my requirement is weaker. The digits do not have to be uniformly random. Just that any digit occurs infinitely often. e.g., the repeated sequence 00123456789 has 0 occur more often but what is important is that all the digits are "represented"(or, more accurately, that each digit occurs infinitely often in the decimal expansion). – AbstractDissonance Nov 16 '17 at 01:06
  • I think disjunctive numbers might be what I'm looking for, although i came from it from the other end Thanks. – AbstractDissonance Nov 16 '17 at 01:11
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Although Liouville's Constant is transcendental, it is also non-repeating and non-terminating, but it (by construction) only contain's 1s and 0s in it's tail, and so the statement is false.

More about Liouville's Constant here: http://mathworld.wolfram.com/LiouvillesConstant.html

Sultan of Swing
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  • What? Is English not your first language? I specifically said that I didn't think transcendental numbers would work. You have done nothing to answer my question. – AbstractDissonance Nov 15 '17 at 02:04
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    Relax; you said you weren't sure if your conjecture was true or false for transcendental numbers, and so I found a counterexample (and you had said given any real number). What you should have said was "for any algebraic irrational number..." And it probably would have made more sense. And to answer that question, I believe it is currently unproven. – Sultan of Swing Nov 15 '17 at 03:20
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    Also you asked about $\pi$ and $e$, which are transcendental. As for algebraic irrationals, very little is known about the properties of their digits. – Robert Israel Nov 15 '17 at 04:02
  • Sultan: I said that because I wasn't 100% sure and could not remember if the liuville constant was in binary or not since it is only 0's and 1's so I didn't want to make up an absolute statement nor look it up... but it's pointless because it is irrelevant does you still did NOTHING to answer the question, and irrelevant. I am looking for the class of numbers that have the property I stated... and you have not supplied that answer but basically created noise. – AbstractDissonance Nov 15 '17 at 22:10
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    Well, like I said--the statement is unproven and Robert mentioned not much is known about those kinds of numbers/properties. And as for the transcendental part...now you are 100% sure ;) – Sultan of Swing Nov 16 '17 at 00:49