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

1. Nov 30, 2003

### phoenixthoth

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
2. Nov 30, 2003

### Hurkyl

Staff Emeritus
All right, let me see if I understand you.

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

Given two natural numbers $n$ and $m$, consider the set $(\mathbb{N}_m)^n \; (\,= (\mathbb{N}_m)^{\mathbb{N}_n}\,)$, the set of all functions from $\mathbb{N}_n[/tex] to $\mathbb{N}_m$. We can then define $n/m := q$ where $q$ is a number having the property that there exists a function $f$ such that $\forall z \in \mathbb{N}_m : |f^{-1}(z)| = q$. This is well-defined; $q$ exists iff $n/m$ exists, and $q$ is unique when it exists, but I fear that you might need to know multiplication to prove the uniqueness of $q$. Edit: fixed logical mistake. Last edited: Nov 30, 2003 3. Nov 30, 2003 ### phoenixthoth yeah, that's what i meant though my $(\mathbb{N}_m)^n$ 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 4. Nov 30, 2003 ### Hurkyl Staff Emeritus There's a problem for infinite cardinals: Consider the following elements of $\mathbb{N}^\mathbb{N}$: $f(x) = x$, $g(x) = \lfloor \frac{x}{2} \rfloor$, and $h(x) = n - 1$ where $p_n$ is the smallest prime dividing $x + 2$. $f(x)$ partitions $\mathbb{N}$ into the equivalence classes $0 | 1 | 2 | 3 | \ldots$, indicating that $\frac{\mathbb{N}}{\mathbb{N}}=1$. $g(x)$ partitions $\mathbb{N}$ into the equivalence classes $0, 1 | 2, 3 | 4, 5 | 6, 7 | \ldots$, indicating that $\frac{\mathbb{N}}{\mathbb{N}}=2$. $h(x)$ partitions $\mathbb{N}$ into the equivalence classes: $$0, 2, 4, \ldots | 1, 7, 13, \ldots | 3, 23, 33, \ldots | \ldots$ indicating that indicating that $\frac{\mathbb{N}}{\mathbb{N}}=\mathbb{N}$. Notice, though, that [tex]\mathbb{N} = \mathbb{N} * 1 = \mathbb{N} * 2 = \mathbb{N} * \mathbb{N}$$.

Last edited: Nov 30, 2003
5. Nov 30, 2003

### phoenixthoth

this concept of "division" seemed to capture the statement $$\mathbb{N} = \mathbb{N} * 1 = \mathbb{N} * 2 = \mathbb{N} * \mathbb{N}$$
. i suppose no definition of division would be well defined for infinite cardinals.

6. Nov 30, 2003

### Hurkyl

Staff Emeritus
This is a superfluous condition if you do it right; I didn't do it right when I first wrote the post. If a function isn't onto, then the inverse image of some element will be the empty set (which is not $q$), so we don't have to restrict ourselves to onto functions.

7. Nov 30, 2003

### phoenixthoth

yeah i was wondering if you omitted that because you realized it was superfluous.