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Re: [xmca] a minus times a plus

Can't you think like this---perhaps it is too much of a physicist's thinking. We can think of the following general function (operator in physics) that produces an image y of x operated upon by A. 
y = Ax
if x is from the domain of positive integers, then A = -1 would produce an image that is opposite to the one when A = +1, the identity operation.
Conceptually you would then not think in terms of a positive times a negative number, but in terms of a positive number that is projected opposite of the origin on a number line, and, if the number is unequal to 1, like -2, then it is also stretched. 

The - would then not be interpreted in the same way as the +

Cheers,
Michael




On 27-Apr-09, at 4:16 PM, Ed Wall wrote:

Mike
It is simply (of course, it isn't simple by the way) because, the negative integers (and, if you wish, zero) were added to the natural numbers in a way that preserves (in a sense) their (the natural numbers) usual arithmetical regularities. It would be unfortunate if something that was true in the natural numbers was no longer true in the integers, which is a extension that includes them. Perhaps the easiest way to the negative x positive business is as follows (and, of course, this can be made opaquely precise - smile): 

3 x 1 = 3
2 x 1 = 2
1 x 1 = 1
0 x 1 = 0
so what, given regularity in the naturals + zero) do you think happens next? This thinking works for, of course, for negative times negative. The opaque proof is more or less as follows.
Negative numbers are solutions to natural number equations of the form (I'm simplifying all this a little) 

                      x + a = 0    ('a' a natural number)
and likewise positive numbers are solutions to natural number equations of the form 

                     y = b          ('b' a natural number)
Multiplying these two equations in the usual fashion within the natural numbers gives 


             xy + ay = 0

or substituting for y


       xy + ab = 0

so, by definition, xy is a negative number.
Notice how all this hinges on the structure of the natural numbers (which I've somewhat assumed in all this). 

Ed



On Apr 27, 2009, at 6:47 PM, Mike Cole wrote:
Since we have some mathematically literate folks on xmca, could someone 
please post an explanation of why

multiplying a negative number by a positive numbers yields a negative
number? What I would really love is an explanation
that is representable in a manner understandable to old college professors 
and young high school students alike.

mike
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