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Question about natural numbers.

  1. Aug 24, 2012 #1
    Are there an [itex] \aleph_0 [/itex] # of natural numbers with an
    [itex] \aleph_0 [/itex] # of digits?
  2. jcsd
  3. Aug 24, 2012 #2


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    Every natural number has a finite number of digits.
  4. Aug 24, 2012 #3
    Adding to the above (which is correct), the set of infinite digit strings is uncountable.
  5. Aug 24, 2012 #4
    ok I understand what you guys are saying but it still seems strange to me.
    I feel like that is saying the natural numbers are not bounded but they have a finite number of digits. I mean you couldn't put a bound on the number of digits.
  6. Aug 24, 2012 #5


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    Each individual natural number has a finite number of digits.
    The entire set is unbounded.
  7. Aug 24, 2012 #6
    You can't. This doesn't change the fact that every natural number has a finite number of digits.
  8. Aug 24, 2012 #7
    For any natural number you pick, I can pick one with more digits. For example, if you picked x I could pick 10x, or 100,000,000,000,000,000x.

    However all three of those numbers have a finite number of digits.

    As the natural numbers get larger and larger so do the number of digits.

    Say you have f(x) = # of digits x has for all natural numbers.

    Then it is certainly true that as x approaches infinity, so does f(x).
  9. Aug 24, 2012 #8
    To make the above a bit more rigorous, the number of digits in a natural number [itex]n[/itex] is given by [itex]\lfloor \log_{10}(n) \rfloor[/itex] and this obviously goes to infinity.
  10. Aug 24, 2012 #9
    I could see the problem with saying that there are natural numbers with an
    [itex] \aleph_0 [/itex] of digits because then I would have 10 choices for each number in the slot and I would have [itex] 10^{\aleph_0} [/itex] numbers which would be uncountable and a contradiction because the set of naturals is countable. Could I use this as a proof by contradiction to verify it?
  11. Aug 24, 2012 #10


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    No. The contradiction does not verify that every natural number has a base 10 representation with only finitely many digits.
  12. Aug 25, 2012 #11


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    If I understood you correctly, you want to compose all strings of finite length

    with terms in {0,1,..,9} . If you write those strings as


    and let N→∞ , then(a) problem is that your sum will diverge much of the time, so that

    many of those strings are not natural numbers.
  13. Aug 25, 2012 #12
    ya thats what i am kinda saying
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