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Posted

From what I can gather this was (is?) a real problem in mathematics and I honestly don't know the full resolution to this question.

 

However, by thinking of multiplication as repeated addition you can convince yourself that positive x positive = positive, negative x positive = positive x negative = negative and negative x negative = positive. At least for natural numbers. As for extending this to all real numbers I don't know.

Posted

I would imagine it being more or less defined. For example, you can easily show that if reals > 0 form a group under multiplication, then for the extension including the negative reals to form a group under multiplication the product of two negative numbers (particularly a negative number and its inverse) must be positive (=1 in the special case). It's not clear to me to what extent you can see this example I made up as an axiom (is there a reason to assume either set forms a group under multiplication?) or a proof (if you assume they should form a group under multiplication then the property you wanted follows). You probably can make up quite a lot of examples like the one I just presented. I would be interesting to hear a canonic viewpoint (or whether one exists).

Posted

Really we are thinking of the reals as a ring. I imagine you can show that minus x minus is required by the axioms of a ring.

 

But this is "chicken and egg" situation as reals are the prototype ring. That is why I don't really know the solution to this question. And for some reason it is not something I have thought about.

Posted

It may be a convention. If we take an x,y axis and shade the area (0,2) to (2,0) we 2 times 2 = 4. If we plot (0,-2) to (-2,0) this also gives an area of 4 but in the double minus quadrant. If we plot (0,-2) to (2,0) and get that area we also get 4, but half is in a minus quadrant and half in a positive quadrant. This would be called minus even of it is half and half. The double negative quadrant creates a problem since if we call this just minus, are the half breeds half minus? Or if we call the half breeds a full minus, what would we use a double minus for? Also, does that make the double + quadrant double plus?

 

There is a physical way to settle this. If we substituted charges, like electrons and protons into our grid, the double positive quadrant would have four protons or 2 time 2 protons. It ends up with four + and not 4 ++. The double negative quadrant would have 2 time 2 electrons, which would not equal four positive charges but four negative charges. The half breeds would end up with 4 charges for a net zero charge. This would have made the convention, based on physical evidence, 2X 2=+4, 2 X (-2) = 4 (o), and (-2) X (-2) = -4. This is a realty way to see how plus and minus act within multiplication tables using a physical example. But the convention has been used for such a long that we have tailored the math around it, so there is no good reason to change at this point in the game. It was defined before we could use charges to proves how these would multiply or add up with a physical experiment. I am just being playful, so takes this with a grain of salt.

Posted

It follows from the definition of multiplication of the natural numbers as repeated addition. The only way to extend multiplication to the integers and retain distributivity of multiplication over addition is to have [math](-m)\times(-n) = mn[/math]

 

In other words, it is the only way to define multiplication that makes sense.

Posted

Axioms :

1. a + (-a) = 0 for all a

2. a*(b+c) = a*b + a*c for all a,b,c

3. 1*a = a for all a

 

We have that (1+(-1)) = 0 by axiom 1 therefore

(1+(-1))*(1+(-1)) = 0

1*(1+(-1)) + (-1)*(1+(-1)) = 0 by axiom 2

1+(-1) + (-1)*1 + (-1)*(-1) = 0 by axiom 2 and 3

(0) + (-1) + (-1)*(-1) = 0 by axioms 1 and 3

-1 + (-1)*(-1) = 0

+1 + (-1) +(-1)*(-1) = +1

0 + (-1)*(-1) = 1

(-1)*(-1) = 1

 

Therefore minus * minus = plus , from the general axioms you get in things like rings and fields.

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