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Multiplicative Inverse Manipulation Valid?

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Let S be a field containing elements x and y. If [tex]a \neq 0[/tex] and [tex]b \neq 0[/tex] then [tex]ab \neq 0[/tex]

    2. Relevant equations
    Field Axioms:

    Associative law for addition
    Existence of additive identity
    Existence of additive inverse
    Commutative law for addition
    Associative law for multiplication
    Existence of multiplicative identity
    Existence of multiplicative inverses
    Commutative law for multiplication
    Distributive law

    3. The attempt at a solution

    The proof I want to implement is as follows:

    Suppose [tex]ab = 0[/tex]. Then,

    (1)[tex]1 = 1 \cdot 1[/tex] (Existence of additive identity)
    (2)[tex]= (a*a^{-1})(b*b^{-1})[/tex] (Existence of multiplicative inverses)
    (3)[tex]= (a^{-1})(b^{-1})ab[/tex] (Associative law for multiplication)
    (4)[tex]= (a^{-1})(b^{-1})0 = 0[/tex] (A lemma that a*0 = 0, if a is a real)

    which is a contradiction.

    I have two primary concerns. The both involve the idea of substitution, which the professor didn't mention if we are allowed to use or not. Substitution is of course not an axiom (or we weren't given it).

    Specifically I am focused on lines (2) and (4). In line (2), it seems the substitution is more valid in a sense that the product of a real and its multiplicative identity defines 1. But on line (4), it is simply substituting 0 in place for ab. We are allowed to add and multiply both sides by the same quantity and that's really all the algebra that were explicitly mentioned for use.
    Last edited: Oct 2, 2008
  2. jcsd
  3. Oct 2, 2008 #2
    OK sorry for the bump. The focus of my question is simply this:

    We have a field that contain the 9 axioms listed above. These axioms involve binary operations. Adding or multiplying the same thing seems justified under the binary system we have. But what about a = b? Can we logically deduce that a and b are interchangeable? It's an equivalence statement so I can't think of a possible argument against using this in a proof. Nonetheless, it's not an axiom, do I have to prove it or does it follow logically.
  4. Oct 2, 2008 #3
    "=" is an equivalence relation, so it is transitive. So substitution is justified.
  5. Oct 2, 2008 #4


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    Homework Helper

    Proof looks good (classical approach) - but I would eliminate the reference to 'a is real' in your final line, as it seems to me that you are referring to fields in general, not to the reals.
  6. Oct 2, 2008 #5
    Sorry I was aware of that as well, that's what I meant but I didn't want to define the field again (laziness on my part). Thanks for the replies.
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