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Defining Elements on the Real Line?

  1. Apr 10, 2010 #1
    RE-EDIT: I'm confused again, continue on reading :redface:

    Bonjourno, I'm trying to work at the lectures provided on youtube by nptelhrd but I've gotten my foot stuck in a hole in the real line only 15 minutres into it :frown:

    (In my head I say "such that" whenever the symbol : pops up!).

    We define a set;

    A : {r ∈ Q: r²<2}

    Does ∃ a largest element of A in Q?

    1: We seek to find some n ∈ N : [tex] ( r \ + \ \frac{1}{n} ) [/tex] will satisfy the conditions specified by A.

    2: [tex] (r \ + \ \frac{1}{n} )^2 \ < \ 2 [/tex]

    3: [tex] r^2 \ + \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2 [/tex]

    4: [tex] \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ 2 \ - \ r^2 [/tex]

    The R.H.S. is strictly positive due to r²<2.

    Okay, I understand up to here but then the lecturer starts to get confusing, he then says that It suffices only to find some n ∈ N :

    [tex] \frac{2r}{n} \ + \ \frac{1}{n} \ < \ 2 \ - \ r^2 [/tex]

    Notice the n and not n² on the bottom of the L.H.S. Fraction!

    He says;

    This is because;

    [tex] \frac{1}{n^2} < \frac{1}{n} \ and \ this \ implies \ \frac{2r}{n} \ + \ \frac{1}{n^2} \ < \ \frac{2r}{n} \ + \ \frac{1}{n} [/itex]

    I have no idea where this came from!

    The video is on youtube http://www.youtube.com/watch?v=0lzO...DB30C539B&playnext_from=PL&index=0&playnext=1 and I would say everything he is trying to do is described from 10:00 to 14:00.

    I would extremely appreciate it if someone could take 6 minutes to watch this and correct me as I have nobody else :redface: to explain it to me.

    What I think is going on is that he is trying to prove a least upper bound or something and that this will show that the real line can be continuously divided, or something.
    Last edited: Apr 10, 2010
  2. jcsd
  3. Apr 10, 2010 #2


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    For any integer n> 1, [itex]1> \frac{1}{n}[/itex] (divide both sides of n> 1 by the positive number n) so [itex]\frac{1}{n}> \frac{1}{n^2}[/itex] (divide both sides by n again). Now, adding any number a to both sides, [itex]a+ \frac{1}{n}> a+ \frac{1}{n^2}[/itex]. In particular, if [itex]a= \frac{2r}{n}[/itex], [itex]\frac{2r}{n}+ \frac{1}{n}> \frac{2r}{n}+ \frac{1}{n^2}[/itex].

  4. Apr 11, 2010 #3
    Thanks a lot HallsofIvy for clearing that up!
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