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Math Help - Proof of (a/b) + (c/d) = (ad + bc)/bd

  1. #1
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    Proof of (a/b) + (c/d) = (ad + bc)/bd

    Hi folks,

    Starting with 6 field axioms, I'm proving a sequence of 15 algebraic laws.

    Number 13 in my list is: (ab) + (cd) = (ad + bc)/(bd)

    Using the axioms and prior theorems:

    a\cdot 1/b      + c\cdot 1/d
    a\cdot 1/b\cdot d\cdot 1/d + c\cdot 1/d\cdot b\cdot 1/b
    ad\cdot 1/b\cdot 1/d + bc\cdot 1/b\cdot 1/d
    (ad +bc)\cdot 1/b\cdot 1/d

    I'm stuck here. I'm not seeing how to get from 1/b\cdot 1/d to 1/bd thus allowing the final step achieving the RHS. Unfortunately, the general case (a/b)(c/d) = (ac)/(bd) is the next theorem in the list, and so, not available.

    If you can provide a hint, I'd appreciate it.

    Thanks,
    Scott
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  2. #2
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    solved it

    a\cdot 1/b      + c\cdot 1/d
    a\cdot 1/b\cdot d\cdot 1/d + c\cdot 1/d\cdot b\cdot 1/b
    ad\cdot 1/b\cdot 1/d + bc\cdot 1/b\cdot 1/d
    (ad +bc)\cdot 1/b\cdot 1/d
    Well, while messing around with my next proof, I found I had the same dilemma and ended up solving both proofs.

    (ad +bc)\cdot 1/b\cdot 1/d = (ad + bc)/(bd)
    (ad +bc)\cdot 1/b\cdot 1/d\cdot b\cdot d = [(ad + bc)/(bd)]\cdot bd
    (ad +bc)\cdot 1\cdot 1 = [(ad + bc)\cdot 1/(bd) \cdot bd
    (ad +bc) = (ad + bc) \cdot 1
    (ad +bc) = (ad + bc)

    Scott
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  3. #3
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    Having no idea what your particular set of field axioms is, the best I can offer is what we would do.
    Maybe you can modify to suit yours.
    Note that we define \frac{a}{b} = ab^{ - 1} , so \frac{a}{b} + \frac{c}{d} = ab^{ - 1}  + cd^{ - 1} .
    Then
    \begin{gathered}  ab^{ - 1}  + cd^{ - 1}  = ab^{ - 1} \left( {dd^{ - 1} } \right) + cd^{ - 1} \left( {bb^{ - 1} } \right) \hfill \\   = \left( {b^{ - 1} d^{ - 1} } \right)\left[ {ad + cb} \right] \hfill \\   = \frac{{ad + cb}}{{bd}} \hfill \\ <br />
\end{gathered} .

    There are several gaps to be filled in. But I hope it helps.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Having no idea what your particular set of field axioms is, the best I can offer is what we would do.
    Yes, I thought of that after I posted. Not to mention the additional dozen theorems I had to work with. Sorry about that.

    Anyway, your info was helpful, and thanks for taking the time to post.

    Scott
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