Re: [xmca] a minus times a plus

[Ed Wall <ewall@umich.edu>]

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
> _______________________________________________
> xmca mailing list
> xmca@weber.ucsd.edu
> http://dss.ucsd.edu/mailman/listinfo/xmca

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