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!

Homework Help: 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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    that formula still isn't right. check your signs.
  11. Feb 16, 2012 #10
    haha, lol i definitely copied it from my notebook incorrectly
    Last edited: Feb 16, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook