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Thread: Fields

  1. #1
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    Fields

    I'm terrible at these proofs, so I was wondering if someone could please check this proof for me:

    Question: Suppose a and b are non-zero elements of a field F. Using only the field axioms, prove that $\displaystyle a^{-1}b^{-1} $ is a multiplicative inverse of ab.

    Proof:

    $\displaystyle (ab)(a^{-1}b^{-1}) = (aba^{-1})(b^{-1}) = $ by associativity.

    $\displaystyle (aba^{-1})(b^{-1}) = (aa^{-1}b)(b^{-1}) $ by commutativity.

    $\displaystyle (aa^{-1}b)(b^{-1}) = (aa^{-1})(bb^{-1}) $ by associativity.

    $\displaystyle (aa^{-1})(bb^{-1}) = 1*1 $ By the axiom that every non-zero real number has a multiplicative inverse.

    $\displaystyle 1*1 = 1 $ By the axiom that 1 is the multiplicative identity.

    Since the multiplicative inverse of an element of a field is unique, this implies $\displaystyle a^{-1}b^{-1} $ is the multiplicative inverse of ab.

    How does it look? Do I have to mention at the end that the multiplicative inverse of an element of a field is unique, or would it be good enough if I left that out?
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  2. #2
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    Quote Originally Posted by JG89 View Post
    I'm terrible at these proofs, so I was wondering if someone could please check this proof for me:

    Question: Suppose a and b are non-zero elements of a field F. Using only the field axioms, prove that $\displaystyle a^{-1}b^{-1} $ is a multiplicative inverse of ab.

    Proof:

    $\displaystyle (ab)(a^{-1}b^{-1}) = (aba^{-1})(b^{-1}) = $ by associativity.

    $\displaystyle (aba^{-1})(b^{-1}) = (aa^{-1}b)(b^{-1}) $ by commutativity.

    $\displaystyle (aa^{-1}b)(b^{-1}) = (aa^{-1})(bb^{-1}) $ by associativity.

    $\displaystyle (aa^{-1})(bb^{-1}) = 1*1 $ By the axiom that every non-zero real number has a multiplicative inverse.

    $\displaystyle 1*1 = 1 $ By the axiom that 1 is the multiplicative identity.

    Since the multiplicative inverse of an element of a field is unique, this implies $\displaystyle a^{-1}b^{-1} $ is the multiplicative inverse of ab.

    How does it look? Do I have to mention at the end that the multiplicative inverse of an element of a field is unique, or would it be good enough if I left that out?
    The proof looks fine. I don't think you need to include that the inverse is unique. It is of course unique, but you don't need it directly in your proof. For instance, you don't say that 1 is the unique multiplicative identity, even though it is. But I am just talking about presentation and your proof is exactly right.
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  3. #3
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    Thanks for checking
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