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Spivak Example

  1. Sep 30, 2013 #1

    I am embarking to read Spivak's book on Calculus, and have come across some difficulty with something that is perhaps rather trivial. In the third edition, there is a section entitled Basic Properties of Numbers. Near the end of page 7, the author begins discussing how he will use property 9 to derive the fact that, when negative numbers are multiplied, the result is a positive number. Before this, he must show that (-a) x b = -(a x b) is true. To prove this, we must show that
    (-a) x b + a x b = [(-a) + a] x b, which I can follow. Next, he says that, because this is true, then we can add -(a x b) to both sides of the equation:

    [itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = [(-a) + a] \cdot b + [-(a \cdot b)][/itex]

    Clearly, the first term on the RHS of the equation will yield zero.

    [itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = -(a \cdot b)[/itex]

    [itex][-a + a - a] \cdot b = -(a \cdot b)[/itex]

    [itex](-a) \cdot = -(a \cdot b)[/itex]

    But are these steps truly valid; how I can know that, when factoring the b, the negative symbol isn't appended to the b; isn't this what I am trying to prove, that the negative symbol is appended to the a?
    Last edited: Sep 30, 2013
  2. jcsd
  3. Sep 30, 2013 #2
    Are those the actual steps listed in the text? The next-to-last line looks a bit suspect. It seems like it should be

    ##(-a)\cdot b+a\cdot b=(-a+a)\cdot b## by distributivity of multiplication over addition
    ##(-a)\cdot b+a\cdot b=0\cdot b## by definition of additive inverses
    ##(-a)\cdot b+a\cdot b=0## from results related to the definition of the additive identity and distributivity of multiplication over addition
    ##[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=0+(-(a\cdot b))## because addition is well-defined as a binary operation (?)
    ##[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=-(a\cdot b)## by definition of additive identity
    ##(-a)\cdot b+[a\cdot b+(-(a\cdot b))]=-(a\cdot b)## by associativity of addition
    ##(-a)\cdot b+0=-(a\cdot b)## by definition of additive inverse
    ##(-a)\cdot b=-(a\cdot b)## by definition of additive identity

    It's possible that some of those steps were left out, but that's what is going on as far as I can tell.
  4. Sep 30, 2013 #3
    Oh, yes, I see now. Thank you very much.
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