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Homework Help: Linear Order on RxR. ?

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Define a relation on the plane by setting:

    [tex] (x_o,y_o)<(x_1,y_1) \mbox{ if either } y_0-x_0^2<y_1-x_1^2, \mbox{ or } y_o-x_o^2=y_1-x_1^2 \mbox{ and } x_0<x_1.[/tex]

    I have easily showed that Nonreflexivity and Transitivity hold. The only dilemma i am facing is to show that Comparability holds as well.
    That is to show that for any two elements
    [tex](x_0,y_0),(x_1,y_1)\in R^2[/tex]
    such that :
    [tex](x_0,y_0)\not=(x_1,y_1)[/tex]

    then either
    [tex] (x_0,y_0)<(x_1,y_1) \mbox{ or } (x_1,y_1)<(x_0,y_0)[/tex]

    To be more specific, is proof by cases the only way to go about it, or is there any way around it? There seem to be too many cases, and i don't really want to pursue this route. I am also thinking about proving its contrapositive, but still, it looks like i would have to treat a few cases separately.

    Any suggestions would be appreciated!
     
  2. jcsd
  3. Feb 1, 2010 #2
    Anyone?
     
  4. Feb 1, 2010 #3

    Hurkyl

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    Too many cases?

    How many do you think there are? And how involved is working through each one?

    I haven't worked the problem, but my instinct says there shouldn't be much more than 4, and each one very quick. So nowhere near "too many".
     
  5. Feb 1, 2010 #4
    Ok then, let:

    [tex](x_o,y_o),(x_1,y_1)\in R^2, \mbox{ such that } (x_o,y_o)\not=(x_1,y_1).[/tex]

    [tex]\mbox{ Then } x_o\not=x_1 \mbox { or } y_o\not=y_1.[/tex]

    Without loss of generality(?) assume that

    [tex]x_o<y_o \mbox{ or } y_o<y_1.----(1)[/tex]

    Then we would have to consider the following cases:

    [tex](i) x_o>0,x_1>0 \mbox{ or } y_o>0,y_1>0 .[/tex]

    [tex](ii) x_0<0,x_1<0 \mbox{ or } y_o<0,y_1<0.[/tex]

    [tex](iii) x_o<0,x_1>0 \mbox{ or } y_o<0,y_1>0.[/tex]

    [tex] (iv) x_o>0, x_1<0 \mbox{ or } y_o>0,y_1<0.[/tex]

    First, can i make the assumption on (1), and second are these all the cases or?
     
    Last edited: Feb 1, 2010
  6. Feb 1, 2010 #5
    ...??
     
  7. Feb 2, 2010 #6

    HallsofIvy

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    It looks pretty straight forward to me. Assume that [itex](x_0,y_0)\ne (x_1,y_1)[/itex] and [itex](x_0, y_0) is not < (x1, y1)[/itex]. Then it is NOT the case that [itex]y_0- x_0^2< y_1- x_1^2[/itex] so either [itex]y_0- x_0^2> y_1- x_1^2[/itex] or [itex]y_0- x_0^2= y_1- x_1^2[/itex].
    If the former, then [itex]y_1- x_1^2< y_0- x_0^2[/itex] so [itex](x_0, y_0)< (x_1, y_1)[/itex].

    If the latter, then [itex]y_0- x_0^2= y_1- x_1^2[/itex] but [itex]x_0[/itex] is not less than [itex]x_1[/itex].

    If [itex]x_0= x_1[/itex] then [itex]x_0^2= x_1^2[/itex] so from [itex]y_0- x_0^2= y_1- x_1^2[/itex] we get [itex]y_0= x_0[/itex] which contradicts [itex](x_0, y_0)\ne (x_1, y_1)[/itex].

    If [itex]x_1< x0[/itex] then that, together with [itex]y_1- x_1^2= y_0- x_0^2[/itex] gives [itex](x_1,y_1)< (x_0,y_0)[/itex].

    You still need to show that only one of those three can hold but that should be fairly easy.


    We want to prove that, in that case, [itex](x_1, y_1)< (x_0, y_0)[/itex]. Then
     
  8. Feb 3, 2010 #7
    Thanks! I appreciate your help!
     
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