1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    pwsnafu

    User Avatar
    Science Advisor

    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

    pwsnafu

    User Avatar
    Science Advisor

    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

    jgens

    User Avatar
    Gold Member

    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

    Bacle2

    User Avatar
    Science Advisor

    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

    Ʃi=0Nai10i

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about natural numbers.
Loading...