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!

Homework Help: Sequence of integers

  1. Jun 21, 2010 #1
    if Xn is a sequence of integers and Xn--->x as n----> infinity and x is an element of the reals. show that x must be an integer.

    i know that since the sequence is convergent it will be bounded. i dont however see how i can prove the above. thank you very much for any help.
     
  2. jcsd
  3. Jun 21, 2010 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A convergent sequence also has to be Cauchy. See what you can get out of that
     
  4. Jun 21, 2010 #3
    im not exactly sure what you mean by ''bounded away'', is this the same as will not converge? it seems to me that even if x were a real value that |Xn-x|<e given that e>0. is this wrong?
     
  5. Jun 21, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    If [itex]x_n\to x[/itex], then, given any [itex]\epsilon[/itex], there exist N such that if n> N, [itex]|x- x_n|< \epsilon[/itex]. Take [itex]\epsilon[/itex] to be less than 1 and remember that [itex]x_n[/itex] is an integer.
     
  6. Jun 21, 2010 #5
    using what you said in that a convergent sequence has a cauchy sequence, and from hallsofly, could the argument go as follows:
    the sequence Xn converges and therefore is a cauchy sequence, by definition for any value n,m>N where N is a natural number |Xn-Xm|<e, given e>0 must be true. each element of Xn,m however is an integer and by the commutative ring of integers Xn-Xm will also produce an integer. if x were a real number it would then be between to integer values. The nearest value Xn can assume are these two integers, therefore taking e<1 |Xn-Xm|<e will not hold
     
  7. Jun 21, 2010 #6

    HallsofIvy

    User Avatar
    Science Advisor

    No, that's not true and not my point. The difference of two integers can be 0!
     
  8. Jun 21, 2010 #7
    it seems like i need to show that x must be an integer in order for the |Xn-x| to equal zero, but i do not see why |Xn-x|<e would not hold if Xn was an integer and x Real. is it because there is a set distance between a integer and a real value, and if epsilon was set to be this value, |Xn-x|could not be less then epsilon?
     
  9. Jun 21, 2010 #8

    Mark44

    Staff: Mentor

    Integers are real numbers! The converse is not necessarily true, though.

    If you have a sequence of integers, it either doesn't converge or converges to an integer. Examples of each kind:
    {1, 2, 3, 4, ..., n, ...}
    {4, 3, 2, 1, 0, 0, 0, ..., 0, ...}
     
  10. Jun 21, 2010 #9
    assuming x is a value between two successive integers, there will be a finite distance E between x and the nearest integer. if e<E then |Xn-x|<e will not hold. therefore by contradiction x must be an integer. This also shows that |Xn-x|=0, and therefore Xn will eventually be a constant value.
    How does that sound?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook