```On Mon, Jul 30, 2001 at 08:10:04AM +0900, MikkelFJ wrote:
> > This seems bad, or do you have a weird definition of multiplication
> > and division that makes it work out?  I always thought even the
> > integers mod something were a ring and not a field.  Rational support
> > seems like the only sane option.

integers mod x are a field when x is a prime number, otherwise its a ring.

when you take N (the integers) mod a prime number, the multiplicative
inverse of all (resulting) numbers exists, and therefore it is a field.

You than define the divison as the inverse of the multiplication - thus the
results are different from "normal" divison.

e.g:

Integers mod 5 gives the numbers
0
1
2
3
4

and:

2*3 = 1 (because 2*3 would give 6, and 6 mod 5 = 1)
therefore 2 is the multiplicative inverse of 3 (and vice versa).

This makes 2 = 1/3, or 3 = 1/2

All in the field of integers mod 5 of course.

On the other hand, of you take the integers mod 4, this gives you the
numbers
0
1
2
3

now 2 doesn't have a multiplicative inverse, because there is no element of
the set which would give the result "1" when multiplied with 2.
0*2 = 0
1*2 = 2
2*2 = 0
3*2 = 2

This (and that there can be two non-zero numbers which have the product "0")
mean that the integers mod 4 are not a field ('cause every element of a
field must have a multiplicative inverse).

greetings, Florian Pflug

PS: I hope you understand what I mean with "multiplicative inverse". I have
_absolutly_ no idea how this is called in english - it's "multiplikatives
inverses" in german, and I tried to translate it ;-))
```