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Question concerning rigor of proofs

  1. Oct 10, 2011 #1
    I've just started Spivak's Calculus and I'm having a few questions concerning the validity of certain of my proofs since some of mine are not the same as the ones in the answer book.

    1. The problem statement, all variables and given/known data
    Here is one of the proof:

    I need to prove that [itex](ab)^{-1}[/itex] = [itex](a)^{-1}[/itex][itex](b)^{-1}[/itex]


    2. Relevant equations

    I must only use the basic laws of numbers :

    (PI) (Associative law for addition) a + (b + c) = (a + b) + c.
    (P2) (Existence of an additive identity) a + 0 = 0 + a = a.
    (P3) (Existence of additive inverses) a + (—a) = (—a) + a = 0.
    (P4) (Commutative law for addition) a + b = b + a.
    (P5) (Associative law for multiplication) a • (b • c) = (a • b) • c.
    (P6) (Existence of a multiplicative a identity) a • 1 = 1 • a = a; 1 ≠ 0
    (P7) (Existence of multiplicative inverses) a • [itex]a^{-1}[/itex] = [itex]a^{-1}[/itex]• a = 1, for a ≠ 0
    (P8) (Commutative law for multiplication) a • b = b • a.
    (P9) (Distributive law) a • (b + c) = a • b + a • c.

    3. The attempt at a solution

    My solution is :

    [itex](ab)^{-1}[/itex] = [itex](a)^{-1}[/itex][itex](b)^{-1}[/itex]
    [itex]\Rightarrow[/itex][itex](ab)^{-1} ab[/itex] = [itex](a)^{-1}[/itex][itex](b)^{-1} ab[/itex] (Multiply both sides by ab)
    [itex]\Rightarrow[/itex][itex](ab)^{-1} ab[/itex] = [itex](a)^{-1}a [/itex][itex](b)^{-1}b[/itex] (Commutative law for multiplication)
    [itex]\Rightarrow[/itex][itex]1 = 1[/itex] (Existence of a multiplicative inverse)

    Would this be regarded as a rigorous proof? I want to get good at this!
     
  2. jcsd
  3. Oct 10, 2011 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have more-or-less the right idea, but you can clean it up: using a^(-1)*b^(-1) = b^(-1)*a^(-1), you can see (using associativity) that [a^(-1)*b^(-1)]*(ab) = 1, so a^(-1)*b^(-1) is an inverse of ab. However, this leaves one question: how do you know that a nonzero number has just ONE inverse? (If a number had several inverses, we might not have (ab)^(-1) = a^(-1)*b^(-1).)

    RGV
     
  4. Oct 10, 2011 #3
    Hmm.. I guess I just went with the property that says that (ab)^-1 * (ab) = 1 and since they were equal, I assumed that proved it.

    I get your point that my proof doesn't necessarily prove that a^-1 * b^-1 = (ab)^-1 and nothing else (and vice-versa). So basically I should be taking one of the members and playing with it until it gives me 2nd member instead of getting them to equal the same value?
     
  5. Oct 10, 2011 #4

    Mark44

    Staff: Mentor

    The first line above is what you are trying to prove, so don't start off by assuming the two quantities are equal.

    Your last line is trivially true, but that doesn't necessarily mean that where you started from must be true.

    Here's a somewhat exaggerated example:

    Prove: 2 = 3
    ==> 0*2 = 0*3
    ==> 0 = 0

    Does the fact that 0 is obviously equal to itself then somehow imply that the first equation above is also true?

    Your work should conclude that (ab)-1 = a-1b-1, not start off assuming this is so.
     
  6. Oct 10, 2011 #5
    Thank you for your precious advice! Thanks to the school system, I've never learned to actually prove anything in mathematics. That's exactly the type of advice I was looking for :smile:
     
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