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Multiplicative Inverses

  1. Nov 8, 2007 #1
    In class today, my teacher said,

    "the definition for multiplicative inverses in the R (real numbers) does not give an inverse for
    the real number [1/n] element of R."

    What does this mean? It seems counterintuitive, and I can't figure out why it's true.
     
    Last edited: Nov 8, 2007
  2. jcsd
  3. Nov 8, 2007 #2
    What is the real number [1/n]? The reals with usual multiplication and adddition form a field, so every real number does have a multiplicative inverse except for 0.
     
  4. Nov 8, 2007 #3

    mathman

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    [x] is often used to mean the greatest integer in x. Using this notation [1/n]=0 for n>1.
     
  5. Nov 8, 2007 #4
    You do have to admit, though, that this is a rather odd way of referring to 0.
     
  6. Nov 8, 2007 #5
    Oops

    I figured it out actually. It has to do specifically with how we defined multiplicative inverses on the real numbers

    i.e.

    If (an) not equivalent to (0) is a cauchy sequence then (an)^-1 is equal to (bn) where bn = { 1/an when (an) not equal to 0, 1 when (an)=0.

    So, if you take the sequence (1/n), in particular, bn = n which is not a cauchy sequence. And because the real numbers are defined as the equivalence classes of cauchy sequences, this inverse does not hold.
     
  7. Nov 9, 2007 #6

    HallsofIvy

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    Okay, so your [1/n] mean "the set of all Cauchy sequences equivalent to the sequence {1/n}". You really should have said that- especially since there are many other ways of defining the real numbers! That equivalence class also includes the sequence {0} consisting of all 0s- in fact it includes every Cauchy sequence that converges to 0. Now, my question is, what is the definition, in terms of Cauchy sequences, for "multiplicative inverses in R"?
     
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