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Thread: stumped on this equivalence relation

  1. #1
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    stumped on this equivalence relation

    define ~ on Z by a~b iff 3a+b is a mulitple of 4.

    so far i only have reflexive...

    reflexive: 3a+a=4a

    how can i prove that its symmetric and transitive?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by glopez09 View Post
    define ~ on Z by a~b iff 3a+b is a mulitple of 4.

    so far i only have reflexive...

    reflexive: 3a+a=4a

    how can i prove that its symmetric and transitive?
    Symmetric:

    $\displaystyle a\sim b\implies 3a+b=4n$,

    But $\displaystyle 3a+b=4n\implies 9a+3b=4(3n)\implies3b+a=4(3n)-8a$ $\displaystyle \implies 3b+a=4(3n-2a)\implies b\sim a$

    Transitive:

    I'll start this, and then I'll leave it for you to finish.

    $\displaystyle a\sim b\implies 3a+b=4n$ and $\displaystyle b\sim c\implies {\color{red}3b+c}=4k$

    Now, $\displaystyle 3a+b=4n\implies 3a+(b+3b)+c-({\color{red}3b+c})=4n\implies\dots$

    Can you finish this?
    Last edited by Chris L T521; Sep 26th 2009 at 09:08 PM. Reason: fixed error.
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  3. #3
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    i am not understanding either of those...?
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  4. #4
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    Hello glopez09
    Quote Originally Posted by glopez09 View Post
    i am not understanding either of those...?
    You need to work out what it is you need to prove, in order to show that $\displaystyle \sim$ is symmetric and transitive.

    First it's symmetric if $\displaystyle a\sim b \Rightarrow b\sim a$.

    Now $\displaystyle a\sim b$ means that $\displaystyle 3a+b$ is a multiple of $\displaystyle 4$. And if we're going to show that $\displaystyle b\sim a$, we shall need to show that this will mean that $\displaystyle 3b+a$ is also multiple of $\displaystyle 4$. So, read carefully through Chris L T521's answer. He has clearly shown that if $\displaystyle 3a+b$ is a multiple of $\displaystyle 4$ then $\displaystyle 3b+a$ is also a multiple of $\displaystyle 4$.

    Then, what does it mean to prove that $\displaystyle \sim$ is transitive? It is this: if $\displaystyle a\sim b$ and $\displaystyle b\sim c$, then we must prove that $\displaystyle a\sim c$. Translating this into multiples of $\displaystyle 4$, this is: if $\displaystyle 3a+b$ is a multiple of $\displaystyle 4$, and $\displaystyle 3b+c$ is a multiple of $\displaystyle 4$, then we must prove that $\displaystyle 3a+c$ is also a multiple of $\displaystyle 4$. Chris L T521's answer has shown you how to build up an expression for $\displaystyle 3a+c$ starting with $\displaystyle 3a+b$ and $\displaystyle 3b+c$. Study it again, and make sure that you can see how this shows that $\displaystyle 3a+c$ is a multiple of $\displaystyle 4$.

    Grandad
    Last edited by Grandad; Sep 27th 2009 at 09:07 PM. Reason: Corrected typos
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  5. #5
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    OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

    this was my attempt without that.

    3a+b+3b+c = 4n+4k

    3a+4b+c = 4n+4k

    3a+c = 4n+4k-4b

    3a+c = 4(n+k-b)

    would that work?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by glopez09 View Post
    OK. i got some time to work on it. and i now understand the symmetry. but that -(3b+c) is throwing me through a loop?

    this was my attempt without that.

    3a+b+3b+c = 4n+4k

    3a+4b+c = 4n+4k

    3a+c = 4n+4k-4b

    3a+c = 4(n+k-b)

    would that work?
    Perfect!

    In the transitive case, I had to add zero:

    $\displaystyle 3a+b=4n\implies 3a+b+{\color{blue}0}=4n\implies3a+b+(3b+c){\color{ red}-(3b+c)}=4n$ $\displaystyle \implies 3a+4b+c-4k=4n$...

    Then your result follows. Does this clarify things more?
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  7. #7
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    ahhh ok thanks so much.
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