This is similar to the question However, it is slightly different. It's probably more of a philosophy of math topic, however, I posted this on a philosophy forum and can't find many people interested in math there. I hope it hasn't been overly discussed. Thank you for reading even if you don't respond or think I'm losing a few marbles. I was talking about imaginary numbers with someone, and thinking it over. I understand how to use and apply them, but I have a problem with the operations themselves, but couldn't really express myself. After a lot of thought, the exact point where there was a disconnect came into view. Should multiplication be commutative? Granted it is for positive numbers like when the Peano axioms were written, but it was taken as a given for negative as well. It seems to me that it was forced commutative, because that's the way it had been and it worked. Fortunately, how we do it works, but unfortunately, it does change the outcome slightly. EDIT: the negative are represented by "`" in this example Say we already had real numbers, but had never touched on negatives. If we had thought of them as opposite of positives, this is what we could have ended up with: Code (Text): `3*`2 =`6 3* 2 = 6 3*`2 = ? `3* 2 = ? The problem was with the cross over of negative and positive. There are many solutions that could come from this, but the one we ended up with changed what I see as the natural outcome of multiplication. Code (Text): `3*`2 = 6 3* 2 = 6 3*`2 =`6 `3* 2 =`6 It seems to work fine because it's commutative just like the axioms that were written when negative numbers weren't even being used. I currently disagree with this outcome. If you look at where the multiplication was going you could make rules for when it crosses zero: Code (Text): 3* 2 = 6 3* 1 = 3 3* 0 = 0 So the next natural step (IMO): Code (Text): 3*`1 =`3 3*`2 =`6 And for negative: Code (Text): `3*`2 =`6 `3*`1 =`3 `3* 0 = 0 `3* 1 = 3 `3* 2 = 6 It's more complex in many ways (commutative doesn't work across the board, distributive is different, and probably associative is different also), but to me it seems cleaner (for lack of a better word). It would change many things, and no one will probably ever accept it. However, does this make sense to anyone else? FYI: the proofs I see online won't do for this, because they assume distributive property as well which is closely related (I think derived from this property actually). For instance: But that breaks on this line (for this way of thinking): "(-1)(-1) + 1(-1) = 0(-1) . distributive property". Because that already doesn't equal 0*`1 with the argument given. According to what is said above (`1)(`1) + 1(`1) would equal `1 + `1 = `2. but that would break down at (-x)*(-y) + (-(x*y)) + x*y = x*y because (`x)*y with what is said above would probably be -(x*`y) EDIT: `x*y when removing the negative would probably be closer to `x*y=`x*-`y=-(`x*`y) This breaks at (-a)(-b) + a(-b) = (-a + a)(-b) = 0 * b = 0, because you can't pull out the "-b" like that. The distributive property is different. The same with distributing the negative in the "hence". ... This is why I said things are more complex, because if you think it makes sense, then a lot of the other properties can't be assumed off the bat either. Another way it's been brought up is `1+`1 can mean a debt added by one debt, but `1*2 is a debt multiplied by 2. The problem I have with this, is that it's multiplied by what? The surplus? If you use this argument and just say it's multiplied by the quantity, then why isn't that a debt quantity or basically saying `1*`2 means a debt times 2 of debt? This might sound odd if you look at it from the debt side, but if you were, it would be 1 surplus times 2 of surplus. From the surplus side it's 1 debt times 2 of debt. The opposite multiplying a debt by a surplus or multiplying a surplus by a debt would have to be defined and explained why it would be that way, wouldn't it? Am I sounding too much like a loon with this line of thinking?