# Probability of infinite non-random subsets in an infinite set of random integers

Discussion in 'Physics & Math' started by Fafnir665, Dec 14, 2009.

1. ### Fafnir665You just got served.Registered Senior Member

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In this situation, no, no it does not matter.

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3. ### iceauraValued Senior Member

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There could be, if there were an infinite sequence of digits before it.

Why not?

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5. ### Fafnir665You just got served.Registered Senior Member

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This is a little different than the original statement, I think I would say "What is the probability that a decimal point followed by an infinite sequence of random digits would produce a consecutive subsequence which is a rational number" which would be 1, since you can take and digit or number of digits and have a rational number, so long as that number of digits is less than infinity.

This was my thinking exactly. Both methods of conceptualizing it are equally appealing. If every consecutive number is the same to infinity there must be a point where the random digits are finite, or if every part is uncountably infinite, then there is an infinite amount of digits, then the probability of it happening is 1. I think at least that the probability would be 1 or 0, there's no in between with this one.

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7. ### D HSome other guyValued Senior Member

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Yes, it does. The phrase "random integer" is meaningless nonsense. The phrase "random digit" is quite meaningful.

8. ### Fafnir665You just got served.Registered Senior Member

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Sure, if you're thinking of them in a vacuum of information, context is clearly given in the posts.

9. ### PeteIt's not rocket surgeryRegistered Senior Member

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Because an infinite sequence is unending, with no last member. Therefore there can be no members that follow afterward.

Although, I think that this should be addressed more rigorously, but that is beyond my ability. To begin, we'd probably need a more formal definition of a sequence (which I suspect would involve a mapping from the naturals to the sequence members).

10. ### Fafnir665You just got served.Registered Senior Member

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Where c is an obvious non-random consecutive subsequence, homogeneous or not, and P is the probability.

11. ### D HSome other guyValued Senior Member

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Did you just make that up?

Why don't you use the definition for a normal number instead?

12. ### Fafnir665You just got served.Registered Senior Member

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Stating a general case of what the thinking is in regards to sequences consisting of integers chosen at random rather than the digits of irrational numbers whether they are random or not.
But we're not talking about a distribution within the sequence, which is why we've started using the term "consecutive sub-sequence" since the regular definition of sub-sequence allows you to choose numbers in any manner you wish, and semantically is different from what I'm trying to discuss.

13. ### temurman of no wordsRegistered Senior Member

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That makes sense. What I really thought was that he used the word seemingly to mean obviously.

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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So, if you said "What is the probability that a decimal point followed by an infinite sequence of random digits would produce an infinite consecutive subsequence which is a rational number", then this is the same as the original statement, right?
Well, if the stream of digits in the decimal expansion of a number includes an infinite consequective sequence of a single digit (or string of digits), then that number is rational.

For example:
0.111... is rational (it equals 1/9).
Tacking on more digits after the infinite stream of 1's (if it were possible, which is questionable) does nothing to change the value. It is still equal to 1/9.

15. ### D HSome other guyValued Senior Member

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In which case the answer is of course zero.

That's not questionable. It's nonsense.

16. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's what I thought, but I'm not qualified to be authoratative

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17. ### Fafnir665You just got served.Registered Senior Member

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Yeah that works too.

If we're talking about infinity there are a couple considerations you take into account. First you take into consideration what kind of infinity you're discussing.
$\mathbb{N} = {n}_{n=1}^\infty$, or the set of natural numbers from one to infinity is a perceptively smaller space than that of all real numbers $\mathbb{R}$ when you consider that there is an infinite amount of decimal expressions between every natural number to infinity.
I think we're dealing with a space similar to that of natural numbers, or the space of decimal expression between two natural numbers. Its not really that important, but when you make a statement like "Go to infinity, and then tack on some more" it DOES make sense, even if you can't actually do it, in the same way you can say there is an infinite consecutive subsequence inside an infinite sequence.

I mean, I know the concept of infinity is hard for some people, but you don't really need to make statements calling something nonsense that is routinely discussed in that manner amongst mathematicians.

18. ### D HSome other guyValued Senior Member

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Mathematicians do not talk about adding digits to the end of an infinite sequence.

19. ### Fafnir665You just got served.Registered Senior Member

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Engineers might not.

20. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Ahh good. Thanks for the counter-example.

I wasn't sure if it was known for sure. I thought it was a conjecture.

I was waiting for this

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I do not think that this matters. We have a set of discrete digits, which happens to be infinite. We want to know the probability that that infinite set contains another infinite set which we specify---in your case, you have specified that the set contains an infinite subset of all 9's, but as has already been pointed out, it doesn't matter whether it's all 9's, or the infinitely repeating ASCII characters of Moby Dick.

I'm going to stop you here.

Another statement like this will get the thread locked, and you may be given a banning.

21. ### temurman of no wordsRegistered Senior Member

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My high school math teacher told me that you can just "concatenate" two real numbers and get another one, which kept me confused for a few years. I had asked him how to map points in a square to points on an interval, and my approach was to choose digits from the two numbers alternatively to construct a new number, but his answer was "just write one number after the other one" instead of doing a clumsy stuff like I suggested. I asked him how to do this, but he kept telling me "just write.." which made me feel stupid.

22. ### D HSome other guyValued Senior Member

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What high school math teachers teach all too often has little to do with mathematics. My high school math instructor argued with me about whether 0.999...=1 (sorry to bring this up). He took the ≠ side of the debate.

23. ### PeteIt's not rocket surgeryRegistered Senior Member

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The number of members of the sequence is question is most definitely countable.
Well, I think that it doesn't make sense, because the size of an infinite sequence is necessarily countable. That's built in to the concept of "sequence".

However...
Well, yes. As much as it pains me to point it out, the notion of tacking on more members following an infinite sequence does, in fact, have authoritative support.

Georg Cantor threw several large bricks into the works of infinity. The particular bricks of relevance here are not the uncountable infinities already mentioned, but rather the concepts of transfinite ordinals and transfinite sequences.

As interesting as this is, however, it's not directly relevant to this discussion, because the sequence under discussion in this thread question is not a transfinite sequence.