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Math Help - Relation Question 2

  1. #1
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    Relation Question 2

    Let R be a relation on \mathbb{N}, where xRy \Leftrightarrow x\equiv y mod 8

    I have an answer saying this is an equivalence relationship; however, I can't seem to derive it myself. Maybe it is because my understanding of what mod 8 means... the way I understand the mod 8 term is that x = y with a remainder of 8 (or multiple of 8?), but why has \equiv been used instead of =? Does it make a difference?

    To check if the relation is reflexive would I say x = x + 8k for some integer k? It seems like that would not be true or only for k = 0. I'm am stuck for symmetry as well.

    I think I can see transitive... If

    x = y + 8k

    and

    y = z + 8m

    then

    x = (z + 8m) + 8k

    = z + 8(m + k)

    So there is a remainder that is a multiple of 8.
    Last edited by terrorsquid; September 4th 2011 at 04:48 AM.
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  2. #2
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    Re: Relation Question 2

    Quote Originally Posted by terrorsquid View Post
    Let R be a relation on \mathbb{N}, where xRy \leftrightarrow x\equiv y mod 8.
    Look at this web page..
    You can always to some research on the web.

    For this question, I would use a \equiv b\,\bmod (c) if and only if a~\&~b have the same remainder when divided by c.

    So 49 \equiv 25\,\bmod (8).
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  3. #3
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    Re: Relation Question 2

    Your relation is equivalent to x-y=8k for some integer k.

    So for example, x-x=8*0 shows that the relation is reflexive.
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  4. #4
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    Re: Relation Question 2

    Quote Originally Posted by DrSteve View Post
    Your relation is equivalent to x-y=8k for some integer k.

    So for example, x-x=8*0 shows that the relation is reflexive.
    Furthermore, we conclude that R is symmetric, since

    xRy <=> x = y (mod8) <=> y - x = 8k for some integer k <=> x - y = 8(-k), where (-k) is also/still an integer <=> yRx

    Transitivity is simple too. I'll let you fill in the middle, after I write the hypothesis and conclusion:

    To prove (xRy and yRz) => (xRz)

    (xRy and yRz) <=> (y - x = 8j and z - y = 8k for integers j, k) => ....... <=> xRz
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  5. #5
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    Re: Relation Question 2

    To prove (xRy and yRz) => (xRz)

    (xRy and yRz) <=> (y - x = 8j and z - y = 8k for integers j, k) => ...

    \therefore x - z = (y-8j) - (8k +y) = 8(-j-k)

    <=> xRz
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  6. #6
    Super Member TheChaz's Avatar
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    Re: Relation Question 2

    Yes. FYI, there should be an option to "mark question as solved" if/when you feel that is appropriate
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