# Re: [xmca] a minus times a plus

```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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```
```
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```