addition progression in multiplication

when multiplying negatives, i find the following equivalent expressions of a series of addition a more direct explanation since multiplication is just a repeated addition.

what do mathematicians call the addition progressions from #4 to #1?

on -1:

1. -1 * -3 = **-1 + 1 + 1 + 1 + 1** = 3

2. -1 * -2 = **-1 + 1 + 1 + 1** = 2

3. -1 * -1 = **-1 + 1 + 1** = 1

4. -1 * 0 = **-1 + 1** = 0

5. -1 * 1 = **-1 + 0** = -1

6. -1 * 2 = **-1 + -1** = -2

Re: addition progression in multiplication

Quote:

Originally Posted by

**vavamustlive** multiplication is just a repeated addition.

Mathematicians tend to rely on good definitions. This is not an example.

Re: addition progression in multiplication

well, of course, definitions are better because their short, accurate, and diverse, but from a formally generalized rule their always exist a rudimentary series from which it's based upon. what is the proof that encompass -a*-b=ab without using negative numbers to prove that negative numbers is equal to a positive number since we are to prove mathematically that multiplying negative numbers results to a positive number?

Re: addition progression in multiplication

if we wish the distributive law to hold for negative, as well as positive integers:

then 0 = a*0 = a(b+(-b)) = ab + a(-b).

evidently, we will need a(-b) = -(ab) for this to be true.

we would also like for 0 = (-a)*0 = (-a)(b+(-b)) = (-a)b + (-a)(-b) = -(ab) + (-a)(-b).

this then, forces us to set (-a)(-b) = ab, unless we don't want that silly distributive law anymore.