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Math Help - addition progression in multiplication

  1. #1
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    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
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  2. #2
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    Re: addition progression in multiplication

    Quote Originally Posted by vavamustlive View Post
    multiplication is just a repeated addition.
    Mathematicians tend to rely on good definitions. This is not an example.
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  3. #3
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    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?
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  4. #4
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    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.
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