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Math Help - Equivalence Relation Question

  1. #1
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    Equivalence Relation Question

    I've got another question.

    Problem:
    Let R be a relation on A. (a,b) \in R IFF 3a + b = 4n for some integer n. Prove that R is an equivalence relation on Z.

    I assumed the Z part was a typo, but am not sure now...

    I know I need to show that R is reflexive, symmetric, and transitive.

    For Reflexive I've got:
    Let a \in A such that a \in Z.
    if 3a + a = 4n
    then a = n and (a,a) \in R.
    I assumed that because we have IFF I have to go the other way...
    if (a,a) \in R, then 4a = 4n and a = n which checks.

    For Symmetric and Transitive...I am at a loss, don't I have to do two statements here for the IFF...or am I getting confused and using given information in the proof?
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  2. #2
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    Quote Originally Posted by Alterah View Post
    I've got another question.

    Problem:
    Let R be a relation on A. (a,b) \in R IFF 3a + b = 4n for some integer n. Prove that R is an equivalence relation on Z.

    I assumed the Z part was a typo, but am not sure now...

    I know I need to show that R is reflexive, symmetric, and transitive.

    For Reflexive I've got:
    Let a \in A such that a \in Z.
    if 3a + a = 4n
    then a = n and (a,a) \in R.
    I assumed that because we have IFF I have to go the other way...
    if (a,a) \in R, then 4a = 4n and a = n which checks.

    For Symmetric and Transitive...I am at a loss, don't I have to do two statements here for the IFF...or am I getting confused and using given information in the proof?
    Z is definately not a typo. For e.g. the relation fails for (.5,.5)

    You need to state your working more systematically. This is not a tough question I presume.
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  3. #3
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    Quote Originally Posted by Alterah View Post
    I've got another question.

    Problem:
    Let R be a relation on A. (a,b) \in R IFF 3a + b = 4n for some integer n. Prove that R is an equivalence relation on Z.

    I assumed the Z part was a typo, but am not sure now...


    Why do you think so? It really is a transitive relation on \mathbb{Z}

    I know I need to show that R is reflexive, symmetric, and transitive.

    For Reflexive I've got:
    Let a \in A such that a \in Z.
    if 3a + a = 4n
    then a = n and (a,a) \in R.
    I assumed that because we have IFF I have to go the other way...
    if (a,a) \in R, then 4a = 4n and a = n which checks.

    For Symmetric and Transitive...I am at a loss, don't I have to do two statements here for the IFF...or am I getting confused and using given information in the proof?

    Suppose (a,b)\in R\Longrightarrow 3a+b=4n for some n\in\mathbb{Z} ; we know that (b,a)\in R\iff 3b+a=4m for some m\in\mathbb{Z} , but then we can write:

    a+3b=3a+b+2(b-a)=4n+2(b-a)=4k since a,b must have the same parity (why? Check this from the assumption that 3a+b=4n...if you

    can do arithmetic modulo 4 this is pretty simple) so then b-a is even and thus 2(b-a) is a multiple of 4...!

    This proves that if (a,b)\in R then also (b,a)\in R and you have symmetry...and more: we've discovered that if (x,y)\in R then both integers a,b have

    the same parity (I honestly didn't have a clue before beginning to do the maths for symmetry).

    Now you try to do transitivity by yourself (idea: sum up both eq's for (a,b),\,(b,c)\in R ...)

    Tonio
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  4. #4
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    Thanks, that's what I needed.
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