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Another equivalence relation

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Decide if the following are Equivalence relations and if so describe their classes

    i) a[itex]\equiv[/itex] b if 2 divides a^2+b^2
    ii) a[itex]\equiv[/itex] b if 2b[itex]\geq[/itex] a

    2. Relevant equations

    3. The attempt at a solution

    ii) isnt an equivalence relation. it is reflexive but not symmetric. 2a [itex]\geq[/itex] b

    i) Its reflexive as [itex]a^2[/itex]+[itex]a^2[/itex] = 2[itex]a^2[/itex] which is divisable by 2.
    its symmetric
    [itex]a^2[/itex]+[itex]b^2[/itex] = 2x
    [itex]b^2[/itex]+[itex]a^2[/itex] = [itex]2a^2[/itex]+[itex]2b^2[/itex]-[itex]a^2[/itex]-[itex]b^2[/itex]
    [itex]b^2[/itex]+[itex]a^2[/itex] = [itex]2a^2[/itex]+[itex]2b^2[/itex]-2x
    [itex]b^2[/itex]+[itex]a^2[/itex] = 2[itex]a^2[/itex]+[itex]b^2[/itex]-x)
    so its divisable by 2

    its symetric
    [itex]a^2[/itex]+[itex]b^2[/itex] = 2x
    [itex]b^2[/itex]+[itex]c^2[/itex] = 2y

    we need to show [itex]a^2[/itex]+[itex]c^2[/itex] = 2w

    [itex]b^2[/itex] = 2y-[itex]c^2[/itex]
    this gives[itex]a^2[/itex]+2y-[itex]c^2[/itex] = 2x
    [itex]a^2[/itex]+[itex]c^2[/itex] = 2x-2y-[itex]2c^2[/itex]
    [itex]a^2[/itex]+[itex]c^2[/itex] = 2(x-y-[itex]c^2[/itex])

    so it is an equivalence relation. As for the classes are there infinity/2 classes? plus how do i describe them? if a & b are both even or both odd they are divisable by 2 but if a is odd and b is even or vise versa then it is not...
  2. jcsd
  3. Aug 15, 2011 #2


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    Rephrasing what you said, [itex]a \equiv b[/itex] if and only if [itex]a[/itex] and [itex]b[/itex] have the same parity (even or odd). So all the even numbers are equivalent to each other, and all the odd numbers are equivalent to each other. No even number is equivalent to any odd number. So how many equivalence classes are there?
  4. Aug 15, 2011 #3
    I think you mean
    [itex] a^2+c^2=2x-2y+2c^2 [/itex]

    Also, you don't have to put [/ itex] until you're ready to use words again. :)
  5. Aug 15, 2011 #4
    Hi, thanks for reply.

    My first parts are right?

    There are infinite classes?
  6. Aug 15, 2011 #5
    The other parts look fine. But uh, what are a and b? Integers? Naturals? Reals?
  7. Aug 15, 2011 #6
    thanks arcananoir so that gives [itex] a^2+c^2=2(x-y+c^2) [/itex] which is divisible by 2
  8. Aug 15, 2011 #7
  9. Aug 15, 2011 #8
    Okay then there are not infinite classes. Look at what jbunniii said.
  10. Aug 15, 2011 #9
    damn i'm not sure. if there are infinite odd and infinite even equivalences then are there not infinite classes {1,1}, {1,3}, {2,4}, {-2,4} .....
  11. Aug 15, 2011 #10


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    I think you are confused about what an equivalence class is. Every number lies in exactly one equivalence class, so you can't have both {2,4} and {-2,4} as equivalence classes. And what is {1,1}? Isn't it just {1}?

    It might help to look at a concrete example. Let's take the number 1. It lies in exactly one equivalence class. What is this class? It is the set of ALL numbers equivalent to 1. Can you tell me what those numbers are?
  12. Aug 15, 2011 #11
    you can also look at it this way:
    what is the class 2/R?
    it's the set of all integers such that 2Rb, or, 2 relates to it (some other integer).
    What numbers relate to 2?

    Now, since we're partitioning the integers, all sets must be disjoint, no overlap.

    What is the class created by 4/R? (What numbers relate to 4?)

    Are these the same numbers that relate to 2?

    If so, 2, and 4 create the same equivalence class.

    What other numbers create this equivalence class?

    Now, what numbers have we left out?

    In the end, how many disjoint sets do you have?

    Remember, each integer can only be in ONE set.
  13. Aug 16, 2011 #12
    Thanks for your patience!!

    So there are 2 equivalence classes, The set of all even integers and the set of all odd integers?
  14. Aug 16, 2011 #13


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