Why Does Multiplication of Negative Numbers Yield Positive Results?

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The discussion explores the properties of multiplication in number systems, specifically why multiplying two negative numbers results in a positive number. It emphasizes the necessity of maintaining consistency in mathematical operations to preserve the structure of a ring, which includes additive and multiplicative identities. The conversation touches on the implications of altering multiplication rules, suggesting that such changes would undermine the integrity of the number system. Additionally, it briefly addresses the non-commutative nature of matrices and their classification as a different kind of mathematical structure. Ultimately, the participants affirm that the standard number system is indeed a ring, despite some confusion over terminology.
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Why we have a number system where

-a x b = b x (-a)

i.e. why we don't 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 into a +ve number.

Any comments
 
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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.
 
matt grime said:
there a lots of systems which aren't rings, however calling them a number system is not reasonable.
like the number systems of the user: "doron shadmi"?
 
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?)
 
That would be true if a and b are integers, reals or else. But if a and b are matrices... that's not true AB is not BA. (maybe because its mixed tensor nature :? )
 
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?)
 
matt grime said:
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.

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.
 
Are we sure that our number system is a ring

Yes.

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

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