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Let R be the relation on Z defined by mRn if and only if mn>0 or m=n=0 1 Prove R is an equivalence relation 2 How many distinct equivalence?

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  • 7 months ago
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    1. We're going to need to prove each of the various characteristics of an equivalence relation for two different cases:

    mn > 0

    and

    m = n = 0

    But that doesn't complicate things too much. I am assuming the conventional use of Z, denoting the set of integers.

    R is reflexive, because for any integer a, either

    a^2 > 0

    or

    a = 0

    and therefore aRa.

    Now consider any integers a and b such that aRb.

    Then either ab > 0, in which case ba > 0; or

    a = b = 0, in which case b = a = 0.

    Thus if aRb, then bRa, and R is symmetric.

    Now consider any integers a, b, c such that aRb and bRc.

    It's possible that aRb because a = b = 0,

    and in that case since b = 0,

    bRc tells us that c = 0 as well,

    so a = c = 0 and aRc.

    If a and b are nonzero, then aRb tells us that ab > 0

    and so a and b have the same sign.

    In that case, bRc tells us that b and c have the same sign,

    so a and c have the same sign and ac > 0,

    and so aRc.

    These two cases establish that R is transitive.

    We've shown R is reflexive, symmetric, and transitive,

    and so R is an equivalence relation.

    2. I assume this should have asked, "How many distinct equivalence CLASSES?"

    There are three classes: positive integers, negative integers, and zero.

    For any two positive integers m and n, mn > 0 and so mRn.

    For any two negative integers m and n, mn > 0 and so mRn.

    If m = n = 0, then mRn.

    So each of these classes has all its elements related by R.

    But if only one of integers f and g equals zero,

    then fg = 0 and f ≠ g and so fRg is not true.

    If one of f and g is positive and one negative,

    then fg < 0 and f ≠ g, and so fRg is not true..

    So no two members of two different classes, as described above,

    is related by R.

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