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Equivalence relations problem (algebra)

  1. Jan 24, 2006 #1
    Z = all integers

    A = Z; m is related to n, means that [tex] m^2 - n^2 [/tex] is even;
    B = {0,1}

    I already proved that this is a equivalence relation, but i just dont know how to;

    I need to find a well defined bejection
    sigma : [tex] A_= -> B [/tex]


    I hope this makes sense.. i wrote it up as well as I can.
     
  2. jcsd
  3. Jan 24, 2006 #2

    JasonRox

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    You want a bijection from A to B?

    Is that was you are looking for?

    Note: If that is the case, that isn't possible.
     
  4. Jan 24, 2006 #3
    I'm confused, why wouldnt that be possible??
     
  5. Jan 24, 2006 #4
    A is an equivalence class,
    so yes i am looking for
    [tex] A_= -> B [/tex]
     
  6. Jan 24, 2006 #5

    JasonRox

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    A has more than two elements, where B has two elements.

    If I got the definition of bijection, which I think I do, then it's not possible.
     
  7. Jan 24, 2006 #6

    JasonRox

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    I could be reading the definition of the set A incorrectly.
     
  8. Jan 24, 2006 #7
    no, i think A is really Z/~ where ~ is the equivalence relation given. as with any equivalence relation, the underlying set (in this case the integers) gets partitioned into cosets, in this case the one where m^2 - n^2 is even & one where it's odd, so A:=Z/~ has two elements. make sigma(one coset) = 0 or 1 & sigma(other coset) = 0 or 1 (so sigma is bijective). i don't know what the actual bijection is but that's probably how it's done. might take some fiddling.
     
  9. Jan 24, 2006 #8

    JasonRox

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    Oh, I thought you said all elements (m and n) that satisfy that and that are in Z, which would be written as a pair is in A.

    You know what. I'm really confused.
     
  10. Jan 24, 2006 #9
    yes fourier jr, thats the idea..
    thanks guys for giving me a push in the right direction
     
  11. Jan 24, 2006 #10
    another way to write it is

    m~n <==> [tex]m^2 \equiv n^2 (mod2)[/tex] ie [tex]m^2 - n^2 \equiv 0 (mod2)[/tex]

    so you've got 2 possibilities: m~n or not, in which case [tex]m^2 - n^2 \equiv 1 (mod2)[/tex]

    (OOPS i just gave the solution away i think :blushing: )

    edit: only jasonrox is allowed to read this!
     
    Last edited: Jan 24, 2006
  12. Jan 24, 2006 #11
    im actually gonna post another question similar to this..
    thanks
     
  13. Jan 24, 2006 #12

    JasonRox

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    I see what's going on now. Thanks. :biggrin:
     
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