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!

Proof of the Archimedean Property

  1. Dec 22, 2014 #1
    I am reading Rudin's proof of this property, but I find one assertion he makes quite disagreeable to my understanding; I am hoping that someone could expound on this assertion. Here is the statement and proof of the archimedean property:

    (a) If ##x \in R##, ##y \in R##, and ##x > 0##, then there exists a positive integer ##n## such that ##nx > y##.

    (a) Let A be the set of all nx. If (a) were false, then y would be an upper bound of A. But then A has a least upper bound in R...

    I understand the method of proof he is employing, a proof by contradiction, which leads to y being larger than every element A (hence, it is an upper bound). However, I do not understand how this immediately implies that A must have a least upper bound. Which theorem is he invoking to come to such a conclusion?

    EDIT: Also, what is the quantification on x and y?
     
  2. jcsd
  3. Dec 22, 2014 #2

    ShayanJ

    User Avatar
    Gold Member

    I think its like this: If there is no n, such that [itex] nx > y [/itex], then y is greater than all the elements of A. But because the elements of A can be arbitrarily large, this means that [itex] y=\infty [/itex]. But this is a contradiction because we chose y in the beginning and we could have chosen a finite real number.
    And about your last question, the theorem can be written like below:
    [itex]
    \forall x,y \in \mathbb R \ \ \ (x>0 \rightarrow (\exists n\in \mathbb N \ \ \ (nx>y)))
    [/itex]
     
  4. Dec 23, 2014 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    My guess is "the completeness of the real number system". http://en.wikipedia.org/wiki/Completeness_of_the_real_numbers
     
  5. Dec 23, 2014 #4
    So, would the idea of ##A## having a least upper bound even be necessary for the proof?
     
  6. Dec 23, 2014 #5

    ShayanJ

    User Avatar
    Gold Member

    Actually it isn't clear to me. That's why I proposed my own version of it. Maybe if you give the proof completely, I can tell!

    EDIT:Well, I didn't use a least upper bound for A, so it doesn't seem necessary!
     
  7. Dec 23, 2014 #6

    jbriggs444

    User Avatar
    Science Advisor

    If A is the set of all real numbers of the form nx for some natural n and if y is an upper bound on A then by the least upper bound property, A must have a least upper bound. Call this bound B.

    B is a real number. B minus x is a real number less than B. Since B is a least upper bound on A, B-x cannot be an upper bound on A. So there must be at least one member of A between B-x and B. That is, there is a real number of the form nx between B-x and B. Pick any such number and add x to it.

    This new number is also of the form nx for some n. Thus it is a member of A and is greater than B. Contradiction.

    Edit: Note that the proof offered by Shyan invokes infinity without definition and is, thus, invalid.
     
  8. Dec 23, 2014 #7

    ShayanJ

    User Avatar
    Gold Member

    Actually I thought I have the idea right and people can fix the probable handwavy things in it. You mean it can't be fixed?
     
  9. Dec 23, 2014 #8
    Normally you do not know that to each real number there is a bigger natural number when you prove this theorem.
     
  10. Dec 23, 2014 #9

    jbriggs444

    User Avatar
    Science Advisor

    Pulling that proof back up and looking at it...

    ##\infty## is not a real number. So y cannot be equal to ##\infty##. Further, there is no predicate "is finite" defined for the real numbers.

    Edit: I missed a flaw. We didn't pick y in the beginning. The other team did. The assertion is that the relationship holds for all y. Even if there is such a thing as an infinite y, there still has to be an n large enough so that nx > y.
     
    Last edited: Dec 23, 2014
  11. Dec 24, 2014 #10
    Oops. Silly me. I thought we were on the same team. :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof of the Archimedean Property
  1. A proof. (Replies: 2)

  2. Distributive Property? (Replies: 4)

Loading...