# Number system: a basic question

1. Oct 19, 2004

### anuj

Why we have a number system where

-a x b = b x (-a)

i.e. why we dont have a system where

-a x b not equal to b x (-a)

and

-a x (-b) = - a x b
a x b = a x b

so that the multiplication/ divison of two -ve numbers results in a -ve number and that of +ve numbers in to a +ve number.

2. Oct 19, 2004

### matt grime

let's suppose this system has an additive identity 0, and a multiplicative identity 1, and. then

1*0 = 1*(0+0) => 1*0=0 (similarly 0*0=0, in fact 0*a=0 for all a)

then, 0= 1*(0) = 1*(1-1) = 1*1 + 1*(-1) if we're to have multiplication behaving reasonably (ie distributively) and thus 1*(-1) = -(1*1)

thus if you were to require certain things to fail your system cannot be a ring, which would be a shame, since it's rather nice that a number system is a ring.

there a lots of systems which aren't rings, however calling them a number system is not reasonable.

3. Oct 19, 2004

### MathematicalPhysicist

like the number systems of the user: "doron shadmi"?

4. Oct 19, 2004

### matt grime

bugger, i'd not spotted that. guess i'd got used to the lack of such posts. the 'any comments' should have given it away. (apologies if anuj is indeed not he, otherwise, lock, anyone?)

5. Oct 19, 2004

### MiGUi

That would be true if a and b are integers, reals or else. But if a and b are matrices... thats not true AB is not BA. (maybe because its mixed tensor nature :? )

6. Oct 19, 2004

### matt grime

matrices fail to commute for geometric reasons. nothing to do with mixed tensors. however matrices do not even form a division algebra, so fail one of the criteria given. besides, are they a number system? also the proofs i provide are valid in the ring of matrices anyway. (what does commutativity have to do with anything?)

7. Oct 20, 2004

### anuj

Are we sure that our number system is a ring and not an one dimensional arrow pointing towards +infinity (the imaginary numbers and its arrow not considered). The two ends i.e. -infinity and +infinity are open ends of the ring.

8. Oct 20, 2004

### Hurkyl

Staff Emeritus
Yes.

(But the mathematical word "ring" has nothing to do with the english word "ring")