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Partition proof

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    For each aεℝ let Aa={(x,y) ε ℝ x ℝ: y = a - x^2}


    2. Relevant equations
    Prove that the set {Aa: a ε ℝ}


    3. The attempt at a solution
    i. let X ε Aa then since X is defined for all aεℝ then X≠∅
    ii. let X ε Aa and Y ε Aa, therefore X=(x1,y1) and Y=(x2,y2) and they produce different a values their intersection will be the empty set, thus they are disjoint.
    iii. The union over of all elements a in ℝ is all sets x,y in the cross product.

    I think my main confusions is i have no idea what X is an element of Aa means. Like is X a pair of x,y or is it a distinct element where the set is defined. I don't know what to do.
     
  2. jcsd
  3. Feb 15, 2012 #2

    Deveno

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    each Aa is a curve in the plane, namely: a parabola (opening downwards).

    from elementary analytic geometry, it should be clear that each parabola has a vertex at (0,a).

    THIS should form the basis of your contention that for any a, Aa is non-empty, since we can, in fact, show that the vertex of the parabola is one of the elements of Aa.

    to show that Aa∩Ab = ∅, if a ≠ b, you need to show that:

    if (x,y) is in Aa, it is not in Ab, and vice versa.

    for part (iii), you need to show that there exists some real number a for ANY pair (x,y), with y = a - x2. for example, for (0,0), we can let a = 0.
     
  4. Feb 15, 2012 #3
    ok great i think i get it! i think i was too caught up in trying to make it abide to some formulaic thing and i wasn't thinking outside the box
     
  5. Feb 15, 2012 #4
    And is the equivalence relation for all aεℝ and (x,y) ε ℝxℝ, xRy iff y=a-x2
     
  6. Feb 15, 2012 #5

    Deveno

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    no. we're not relating two real numbers to each other, but pairs of real numbers to OTHER pairs.

    so your equivalence relation will have the form:

    (x,y) R (x',y') if.....?

    play with this a little. is (0,0) related to (2,-3)? is (0,1) related to (2,-3)?

    i'll give you a hint: you can define R in such a way as to not even mention a.
     
  7. Feb 15, 2012 #6
    think i got it:
    for all (x,y) & (x',y') ε ℝxℝ, (x,y)R(x',y') iff x+y=x'+y'
     
  8. Feb 16, 2012 #7

    Deveno

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    well, let's just see if that's true. if it is, then (0,0) should be in the same Aa

    (with a = 0) as (-2,2), because:

    0+0 = -2+2.

    now, 0 = a - 02 means a = 0 (we knew that already, right?)

    so 2 = 0 - (-2)2 = 4...wait, what?

    your formula isn't quite right.
     
  9. Feb 16, 2012 #8
    its y-x2=y'-x'2
    ...
    ah i think this is my mind telling me i cannot do math when i am sick
     
  10. Feb 16, 2012 #9

    Deveno

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    that formula still isn't right. check your signs.
     
  11. Feb 16, 2012 #10
    haha, lol i definitely copied it from my notebook incorrectly
    y+x^2=y'+x'^2
     
    Last edited: Feb 16, 2012
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