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Trying to define n/m when it exists in N.suppose m and n are

  1. Nov 30, 2003 #1
    trying to define n/m when it exists in N.

    suppose m and n are natural numbers and let [n,m] denote all functions from n (which is {0=Ø, 1, ..., n-1}) onto m. if [n,m] is empty, stop and say that m does not divide n.

    consider the subset of functions f in [n,m] such that ~ defines an equivalence relation on n where x~y iff f(x)=f(y) and that each equivalence class has the same size q. call this set [[n,m]]. i'm guessing that if [[n,m]] is nonempty then it will be the same q for all functions in [n,m] such that ~ defines an equivalence relation partitioning n into equal sized parts (m parts each having q elements).

    i suppose this is equivalent to saying
    [[n,m]]={f in [n,m] : for all z in m, card(f-1({z})) is constant}.

    definition: if [[n,m]] is nonempty then let n/m=q. if [[n,m]] is empty say that m does not divide n.

    question: n/m=q in this sense if and only if n/m=q in the usual sense? (i'll also have to decide if q is well defined.)

    this is a definition of division not obviously equivalent to "the inverse of multiplication." one can now define multiplication to be the inverse of division!
     
    Last edited: Nov 30, 2003
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  3. Nov 30, 2003 #2

    Hurkyl

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    All right, let me see if I understand you.

    In order to lessen potential sources of confusion, I'm going to use the notation [itex]\mathbb{N}_n = \{0, 1, \ldots, n-1\}[/itex].

    Given two natural numbers [itex]n[/itex] and [itex]m[/itex], consider the set [itex](\mathbb{N}_m)^n \; (\,= (\mathbb{N}_m)^{\mathbb{N}_n}\,)[/itex], the set of all functions from [itex]\mathbb{N}_n[/tex] to [itex]\mathbb{N}_m[/itex].


    We can then define [itex]n/m := q[/itex] where [itex]q[/itex] is a number having the property that there exists a function [itex]f[/itex] such that [itex]\forall z \in \mathbb{N}_m : |f^{-1}(z)| = q[/itex].


    This is well-defined; [itex]q[/itex] exists iff [itex]n/m[/itex] exists, and [itex]q[/itex] is unique when it exists, but I fear that you might need to know multiplication to prove the uniqueness of [itex]q[/itex].


    Edit: fixed logical mistake.
     
    Last edited: Nov 30, 2003
  4. Nov 30, 2003 #3
    yeah, that's what i meant though my [itex](\mathbb{N}_m)^n[/itex] is the set of such functions that are onto.



    that would suck.

    i'm really aiming and doing a similar thing for infinite cardinals...
     
    Last edited: Nov 30, 2003
  5. Nov 30, 2003 #4

    Hurkyl

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    There's a problem for infinite cardinals:

    Consider the following elements of [itex]\mathbb{N}^\mathbb{N}[/itex]: [itex]f(x) = x[/itex], [itex]g(x) = \lfloor \frac{x}{2} \rfloor[/itex], and [itex]h(x) = n - 1[/itex] where [itex]p_n[/itex] is the smallest prime dividing [itex]x + 2[/itex].

    [itex]f(x)[/itex] partitions [itex]\mathbb{N}[/itex] into the equivalence classes [itex]0 | 1 | 2 | 3 | \ldots[/itex], indicating that [itex]\frac{\mathbb{N}}{\mathbb{N}}=1[/itex].

    [itex]g(x)[/itex] partitions [itex]\mathbb{N}[/itex] into the equivalence classes [itex]0, 1 | 2, 3 | 4, 5 | 6, 7 | \ldots[/itex], indicating that [itex]\frac{\mathbb{N}}{\mathbb{N}}=2[/itex].

    [itex]h(x)[/itex] partitions [itex]\mathbb{N}[/itex] into the equivalence classes:

    [tex]
    0, 2, 4, \ldots | 1, 7, 13, \ldots | 3, 23, 33, \ldots | \ldots
    [/itex]

    indicating that indicating that [itex]\frac{\mathbb{N}}{\mathbb{N}}=\mathbb{N}[/itex].


    Notice, though, that

    [tex]\mathbb{N} = \mathbb{N} * 1 = \mathbb{N} * 2 = \mathbb{N} * \mathbb{N}[/tex].
     
    Last edited: Nov 30, 2003
  6. Nov 30, 2003 #5
    this concept of "division" seemed to capture the statement [tex]\mathbb{N} = \mathbb{N} * 1 = \mathbb{N} * 2 = \mathbb{N} * \mathbb{N}[/tex]
    . i suppose no definition of division would be well defined for infinite cardinals.
     
  7. Nov 30, 2003 #6

    Hurkyl

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    This is a superfluous condition if you do it right; I didn't do it right when I first wrote the post. :smile: If a function isn't onto, then the inverse image of some element will be the empty set (which is not [itex]q[/itex]), so we don't have to restrict ourselves to onto functions.
     
  8. Nov 30, 2003 #7
    yeah i was wondering if you omitted that because you realized it was superfluous.

    hmm... i wonder about 0/0

    it would seem to be the case that by my definition, 0/0=0.

    i also wonder about P(N)/N.
     
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