
Re: Can addition be defined in terms of multiplication?
Posted:
Aug 17, 2013 5:11 PM


On Fri, 16 Aug 2013, Shmuel (Seymour J.) Metz wrote:
> In <alpine.LNX.2.00.1308161806110.15401@badwlrzclhri01.ws.lrz.de>, on > 08/16/2013 > at 06:14 PM, Helmut Richter <hhrm@web.de> said: > > >Given a multiplication on a set (e.g. defined as a commutative > >and associative operation allowing cancellation (ab = ac implies > >b=c)), > > That would be a strange definition in general, although it's > reasonable if you're only concerned with groups. Is the "*" operation > in Z a multiplication? Certainly 0b=0c doesn't imply b=c.
I was imprecise in stating the problem. Of course, it was meant in a way so that it *can* have solutions, e.g. the ordinary multiplication of integers is indeed the multiplication in a ring. Let me try to do better:
Given a structure (M, 0, 1, ·) where
M is a set, 0 elem M is a constant, 1 elem M is a constant, · : M×M > M is an operation
with the following properties:
(1) · is commutative and associative (2) 0·x = 0 for all x (3) 1·x = x for all x (4) x·z = y·z and z ¬= 0 ==> x = y for all x, y, z
where condition (4) comes from the first intended application of the problem; the problem may be meaningful also without it. Condition (4) is not generally fulfilled in rings.
Now we want to decide the question whether there is an operation + so that (M, 0, 1, +, ·) is a ring.
Example 1: M = {0, 1, 2, 3, 4} If x, y ¬= 0, x·y is the unique z in {1, 2, 3, 4} such that x+y1 == z (mod 4)
Then there is an addition which makes it a ring, even a field: 1+1=2; 1+3=0; 1+4=3; 2+2=3; 2+3=1; 2+4=0; 3+3=4; 3+4=2; 4+4=1;
because {1, 2, 3, 4} is a group under · that is isomorphic to the multiplicative group of F5, and the isomorphism consists of swapping 3 and 4.
Example 2: M = {0} u {x elem Z: x == 1 (mod 3)} x·y is defined as ordinary multiplication
Question: Is there an addition which makes that a ring? Or why not?
 Helmut Richter

