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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

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    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

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    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

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    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

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    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?
     
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