How is multiplying by a fraction the same as dividing by the inverse of that fraction

  1. Supposedly this proof answers my question.

    8 / 16 = .5
    8 / 8 = 1
    8 / 4 = 2
    8 / 2 = 4
    8 / 1 = 8
    8 / .5 = 16
    8 * 2/1 = 8 / .5

    I'm just wondering how this proof answers my question?
     
  2. jcsd
  3. Re: How is multiplying by a fraction the same as dividing by the inverse of that frac

    It doesn't ... and that's not a proof o__O

    It follows kind of from "definition" of division: [itex]x \over y[/itex] actually means [itex]x \cdot y^{-1}[/itex] where [itex]y^{-1}[/itex] is the number such that [itex]y \cdot y^{-1} = 1[/itex], called the "multiplicative inverse" or "reciprocal" of y. This number is unique. Obviously [itex]({a \over b})^{-1} = {b \over a}[/itex] (since [itex]{a \over b} \cdot {b \over a} = 1[/itex]) as long as we have [itex]a, b \ne 0[/itex]. (Also of note: [itex]0^{-1}[/itex] does not exist!

    So it turns out that since [itex]{x \over y} = x \cdot y^{-1}[/itex], setting [itex]y = {a \over b}[/itex] we get

    "[itex]{{x} \over {a \over b}}[/itex]" [itex]= {x} \cdot ({a \over b})^{-1} = {x} \cdot {b \over a}[/itex] (again provided [itex]a, b \ne 0[/itex]). I hope this explanation helps.
     
    Last edited: Aug 10, 2011
  4. Re: How is multiplying by a fraction the same as dividing by the inverse of that frac

    ooooooooooooooooooooooooooh awesome! :D

    Very much appreciated!
     
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