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Math Help - Multiplicative and additive identities as successors

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    Newbie SummerMacDermott's Avatar
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    Multiplicative and additive identities as successors

    Fact: The ring of integers Z is totally ordered: for any distinct elements a and b in Z, either a>b or a<b.

    Fact: The ring of integers is discrete, in the sense that for any element a in Z, there exists an element b in Z such that there is no element c in Z with a<c<b, and the same argument holds with the greater than signs flipped. In other words, successors and predecessors exist in Z (but not in R, for example).

    Observation: In Z, the multiplicative identity 1 is the successor of the additive identity 0.

    Questions: Is this fact a coincidence? Does this have any significance? Must the multiplicative identity always succeed the additive identity in rings that have the same properties as Z, assuming there are any other?
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    Re: Multiplicative and additive identities as successors

    Try multiplying the successor of 1 with a number c where 0<c<1 and see what happens. Also try squaring the successor of 0.
    Last edited by thesmurfmaster; July 5th 2012 at 05:35 PM.
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  3. #3
    Newbie SummerMacDermott's Avatar
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    Re: Multiplicative and additive identities as successors

    Not helpful.
    Last edited by SummerMacDermott; July 6th 2012 at 12:43 AM.
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    Re: Multiplicative and additive identities as successors

    let's agree on a few things:

    we will call a ring with unity R ordered if we have a subset P such that:

    1) P is closed under multiplication and addition.
    2) for all x in R*, x2 is in P
    3) 0,-1 are not in P
    4) R = P U {0} U -P

    we will say x > y if x - y is in P.

    note rule (4) assures us that at least one of the following is true:

    a) x > y
    b) x = y
    c) y > x

    clearly rule (3) shows if (b) is true, then neither (a) nor (c) can be true; and if (a) or (c) is true, (b) is impossible. now if (a) & (c) are true, we would have (x-y) + (y-x) = 0 in P by rule (1), a contradiction. so we see exactly 1 of (a),(b), or (c) is true. this shows that ">" is irreflexive and anti-symmetric. to see that it is transitive:

    suppose x > y, and y > z. then x-y is in P, and y-z is in P, so (x-y) + (y-z) = x-z is in P (by rule (1)), so x > z. thus we have all the usual properties of a total order (trichotomy, irreflexiveness, and anti-symmetry), and this order is consistent with the ring operations in that:

    x > 0 and y > 0 means x+y > 0
    x > 0 and y > 0 means xy > 0

    in such a ring, suppose we have 0 < x < y. is it true that for 0 < z, that 0 < xz < yz?

    well clearly 0 < xz, by rule (1). if xz = yz, then (y - x)z = 0, which also contradicts rule (1) (because of rule (3)), since both y - x and z are in P. and if xz > yz, we get xz - yz > 0, so (x - y)z is in P. but y - x and z are in P, so (y - x)z is in P (by rule (1)), so (by the discussion above) (x - y)z = -(y - x)z cannot be in P. so xz < yz must be true.

    now, suppose we have such a ordered ring with unity, and that this ring is also discrete; and that the successor of 0 is e ≠ 1. that is, we have:

    0 < e < 1.

    multiplying through by e, we have:

    0 < e2 < e, contradicting that e is the successor of 0.

    thus, in a discrete ordered ring, the successor of 0 must be 1.
    Last edited by Deveno; July 6th 2012 at 02:47 AM.
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    Re: Multiplicative and additive identities as successors

    Quote Originally Posted by SummerMacDermott View Post
    Not helpful.
    I'm sorry, I thought that you wanted a nod in the right direction so you could explore the implications of 1 not being the successor of 0 for yourself, instead of being handed the answer on a platter.
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    Newbie SummerMacDermott's Avatar
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    Re: Multiplicative and additive identities as successors

    thesmurfmaster,

    Sorry, I didn't mean to be rude. What I meant was that I tried what you suggested and couldn't figure it out.
    I'm still working on Deveno's solution.
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