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Real Analysis: Relations.

  1. Jan 18, 2006 #1
    I'm doing this problem in the book - their are 2 of this kind and they have no answers in the back.. so i thought ill post one.

    Let S be the Cartesian coodinate plane R x R and define a relation R on S by (a,b)R(c,d) iff a+d=b+c. Verify that R is an equivalence relation and describe the equivalence class E(7,3)

    ....So i want to show that the 3 properties of an equivalence reltion holds: (reflective propery, symmetric propery and transitive propery.)

    1. reflective propery

    (a,b)=(a,b) and (c,d)=(c,d)......(not hard)

    2. symmetric propery

    Let (a,b) be in A and (c,d) be in B and if A R B, then (a,b) = (c,d)
    so, (c,d) = (a,b) and thus, B R A.

    How does that sound?

    3. transitive propery

    i have no idea how to start this :cry:

    i mean, the property states "if xRy and yRz, then xRz"

    so will this work:?

    if (a,b)=(c,d) and (c,d)=(e,f), then (a,b)=(e,f)

    but then i feel like i making things up when i put in (e,f)?
     
  2. jcsd
  3. Jan 18, 2006 #2

    Hurkyl

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    If you're trying to prove "=" is an equivalence relation on ordered pairs, then you've correctly stated what needs to be proved, at least for (1) and (3).

    Where did "A" and "B" come from?


    But this is all moot, since you're trying to prove "R" is an equivalence relation. :tongue2:
     
  4. Jan 18, 2006 #3
    I was trying to get fancy with the A and B.
    My prof did that in class today (but he was talking about similar triangles and stated "Let A be so and so, and Let be be so and so.. etc..)

    so i guess 2 would be something like this:

    (a,b) = (c,d) and (c,d) = (a,b)

    Just 2 more things.

    1) where does a+b = c+d come in?
    2)What does it mean "and describe the equivalence class E(7,3)"
    I suppose E(7,3) assumes a=7 and b = 3. So wont that be something like (a,b)=(c,d)=(7,3)
     
    Last edited by a moderator: Jan 18, 2006
  5. Jan 18, 2006 #4

    Hurkyl

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    Again, that's what the symmetric relation would look like, for "=".


    Incidentally, what you could say is something akin to:

    Let A, B be in S.
    We may write A = (a, b) and B = (c, d) for some a, b, c, d.
    (more stuff)


    It's the definition of (a,b)R(c,d), as you stated in your original post.


    Where did a and b come from! You're falling into a trap -- let me demonstrate it with a common mistake people make:


    What you just said is like "defining" the function f via:

    f = x²

    this is nonsense: where did x come from? Now, if you introduced x as a dummy variable, you could define f pointwise via:

    f(x) = x²

    which does make sense, and means exactly the same thing as

    f(a) = a²

    or, as one professor I knew would sometimes write to emphasize this point:

    [tex]
    f(\spadesuit) = \spadesuit^2
    [/tex]
     
    Last edited: Jan 18, 2006
  6. Jan 18, 2006 #5

    matt grime

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    None of your 'proofs' that R is an equivalence relation mention R. You might want to think why that is not a good thing. If I wished to show that (a,b)R(a,b) ie it were reflexive then I might want to consider what R actually is.
     
  7. Jan 18, 2006 #6
    Matt Grimme:

    the question said "Verify that R is an equivalence relation" so R would be "="
     
  8. Jan 18, 2006 #7

    Hurkyl

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    Why would "R" be "="?

    "=" is an equivalence relation. Not all equivalence relations are "=".
     
  9. Jan 18, 2006 #8
    okay i think im on something:

    Q) Let S be the Cartesian coodinate plane R x R and define a relation R on S by (a,b)R(c,d) iff a+d=b+c. Verify that R is an equivalence relation and describe the equivalence class E(7,3)

    check: reflective propery, symmetric propery and transitive propery.

    reflective propery:

    is (a,b)R(a,b)?
    (a,b)R(a,b) iff a+b=b+a

    since a+b = b+a, the relationship is reflective.

    symmetric propery

    is (a,b)R(c,d) the same as (c,d)R(a,b)

    (a,b)R(c,d) iff a+d=b+c
    and (c,d)R(a,b) iff c+b=d+a
    thus we have a+d=b+c and c+b=d+a which are the same, thus the symmetric property holds

    transitive propery:

    if (a,b)R(c,d) and (c,d)R(e,f) the (a,b)R(e,f)
    so, if a+d=b+c and c+f=d+e, then a+f = b+e (which can easily be shown with some arithmatic)

    thus, the tranistive property holds.

    How does this sound?
     
  10. Jan 18, 2006 #9

    Hurkyl

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    Looks good!
     
  11. Jan 18, 2006 #10
    oh boo-yeah, i feel good now!
    Thank you!

    but i still don't get what "discrie the equivelence class E(7,3)" means?
     
    Last edited by a moderator: Jan 18, 2006
  12. Jan 18, 2006 #11

    Hurkyl

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    I suppose (based on the explanatory text you gave) they're using the "E" to mean "the equivalence class of". So, EX means the equivalence class of X!
     
  13. Jan 18, 2006 #12
    the question says "verify that R is an equivalence relation and describe the equivalence class E(7,3)"

    So, based on (a,b)R(c,d) iff a+d = b+c ... i came up with this:

    (5,2)R(2,1) = (7,3) since 5+2=7 and 2+1=3

    but thats just one example... i have no idea
     
  14. Jan 19, 2006 #13
    What would happen if you wrote (7,3)R(a,b)?
     
  15. Jan 19, 2006 #14

    matt grime

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    Where did that come from? The equivalence class of something is the set of all elements related to it.
     
  16. Jan 19, 2006 #15
    The equivlence class is something like this:
    E(7,3) = { (c,d) in S: (c,d)R(7,3) }

    from what daveb wrote, What would happen if you wrote (7,3)R(c,d)?

    you would get
    7 + d = 3+ c
    4 = c - d

    so we want all the points (c,d) that satisfy 4 = c - d.

    such possibilities are (4,0) (0,-4) (6, 2) (8,4) etc... and we notice from observation (and we could so some aritimatic to show) that the points all lie on the line y = x - 4.

    Thus, "discribe the equivalence class E(7,3)" is the line y = x -4

    How does that sound?
     
  17. Jan 20, 2006 #16
    testing testing
     
  18. Jan 20, 2006 #17

    matt grime

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  19. Jan 20, 2006 #18

    HallsofIvy

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    NOW you've got it!
     
  20. Jan 20, 2006 #19
    now im happy, thanks everyone :)
     
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