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Thread: Define a Relation on S X S* by...

  1. #1
    Sep 2011

    Question Define a Relation on S X S* by...

    I have spent the past several hours trying to wrap my mind around this question. It should be noted that this is homework and that I am not asking for anyone to do the work for me. But I need some help to figure out what I should be doing.

    I will begin by stating the question as it is written:
    "In this exercise we will be working mod 15. Notice that in mod 15 there are zero-divisors, that is, pairs of non-zero elements a and b with ab = 0; for example, 3 x 5 = 10 x 6 = 0, etc. Let S = {0, 1, ..., 14} and let S* denote the set of non-zero elements of S.
    Define a relation ~ on S X S* by: (a, b) ~ (c, d) <===> there exists a zero divisor u with u(ad - bc) = 0."

    I began this problem by creating the set S* and filling it with the non-zero elements of S. I did this by taking two elements of S (a, b) and determined whether ab mod 15 = 0. Thus I created the set S* of ordered pairs: S* = { (3, 5), (3, 10), (6,5), (9,5), (6,10), (5, 12), (10,12) } (This may or may not be correct).

    so with this framework in mind I have attempted to answer the following:
    a) Show that (2, 10) ~ (1, 5) This question threw me off right away since (1, 5) isn't in my S* set. But plugging the values into the equation I got;

    u(ad - bc) = 0
    u(10 - 10) = 0 for all u

    b) Show that (2, 10) ~ (2, 13) Plugging the values into the equation I go;

    u(ad - bc) = 0
    u(26 - 20) = 0
    u(6) = 0
    u = 10 or u = 5.

    For these first two problems I have no idea if I'm doing this properly this is just what I have gotten based on my understanding of the question. If I am incorrect in anything please help me to understand.

    c) show that ~ is reflexive and symmetric: For this question I have no idea even how to begin. The first thing I notice is that in order for it to be reflexive, for every element a in S (a, a) is in ~. But this doesn't hold because if a = 1 then (1, 1) doesn't appear in the S* set because 1x1 = 1 mod 15 = 1 and thus is not a zero-divisor. Does this then prove that this relation is NOT reflexive?

    and finally d) find all pairs (a, b) such that (a, b) ~ (1, 5). Once again on this one I'm not sure where to start.

    I would appreciate any help in wrapping my brain around this stuff. I want to understand.
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  2. #2
    MHF Contributor
    Nov 2013

    Re: Define a Relation on S X S* by...

    S* is just S-{0}, i.e. {1,2,3,.. 14}

    you sort of stumbled your way through (a) and (b)

    for (c) show that $(a,b)\sim (a,b)$ and that

    $(a,b)\sim (c,d),~(c,d)\sim (e,f) \Rightarrow (a,b)\sim (e,f)$
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