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Multiplicative Inverse Proof

  1. Feb 25, 2009 #1

    jgens

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    1. The problem statement, all variables and given/known data

    Prove (ab)-1 = a-1b-1 for all a,b =! 0.

    2. Relevant equations

    Multiplicative inverse property: (a)(a-1) = 1
    Commutivity: ab = ba
    Associativity: (ab)c = a(bc)
    Transitivity: If a = b and b = c then a = c

    3. The attempt at a solution

    Early today, my friend and I had a discussion regarding this statement and its proof using the principles stated above. The proof I provided is as follows:

    (ab)(ab)-1 = 1 and 1 = (a)(a-1)(b)(b-1); therefore it follows by transitivity that (ab)(ab)-1 = (a)(a-1)(b)(b-1). By commutivity and associativity it follows that (ab)(ab)-1 = (ab)(a-1)(b-1), ultimately yielding (ab)-1 = (a-1)(b-1).

    The proof my friend posted is as follows:

    (ab)-1 = 1/(ab) = (1/a)(1/b) = (a-1)(b-1)

    I argued that my friend's proof was not valid as it relies on the unproven (yet true) assumption that (ab)-1 = 1/(ab) which he would need to demonstrate was true before using it in a proof. I think my friend is right on this one and my criticisms aren't truly valid; however, I thought I would check here with people who are familiar with mathematics to ensure that any of the proofs, criticisms, etc. are valid or invalid.

    Thanks!
     
  2. jcsd
  3. Feb 25, 2009 #2

    Tom Mattson

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    Your proof is correct, and your friend's isn't. These proofs should only refer to addition, multiplication, and the axioms. Division is defined later in terms of multiplication. So things like [itex]1/(ab)[/itex] shouldn't even be showing up here.
     
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