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Proving A Property Of Numbers

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    The statement that is purported to be true is [itex]\frac{a/b}{c/d} = \frac{ad}{bc}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    So, I am going along with my proof, and I believe it to be going nicely. However, there is one step that I am unsure of:

    [itex]\frac{\frac{a}{b} d}{c} = \frac{\frac{a}{b} d}{c} \cdot d \cdot d^{-1} \Rightarrow \frac{\frac{a}{b} d \cdot d^{-1}}{c d^{-1}} = \frac{\frac{a}{b}}{c d^{-1}}[/itex]. Now, what I want to do is [itex]\frac{\frac{a}{b}}{c d^{-1}} = \frac{\frac{a}{b}}{\frac{c}{d}}[/itex]. But I am having trouble justifying the step [itex]\frac{1}{d^{-1}} =
    \frac{1}{\frac{1}{d}}[/itex]
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 18, 2013 #2

    tiny-tim

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    hi embphysics! :smile:

    i'm not sure what formulas you're allowed to use :confused:

    anyway, why not just multiply the RHS, ad/bc, by the bottom of the LHS, c/d ? :wink:
     
  4. Nov 18, 2013 #3
    Well, the formulas and properties I am permitted to use are given in the first chapter of Spivak's Calculus. So, tiny-tim, I am not certain that that manipulation is defined.
     
  5. Nov 18, 2013 #4

    Ray Vickson

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    Why not just use the definition? ##A/B## is that number ##X## which, when multiplied by ##B##, gives you ##A##; that is, it is the solution of the equation ##BX = A##.
     
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