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Dovall rational numbers under multiplication form a group.

  1. Oct 24, 2006 #1
    I know the set of positives rationals form a group under multiplication and that the negative irrationals do not form a group under multiplication because there is no identity or inverse. My question is does the set of all rational numbers under multiplication form a group.
     
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  3. Oct 24, 2006 #2

    radou

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    Hint: does there for every [tex]a \in \textbf{Q}[/tex] exist a unique element [tex]a^{-1}\in\textbf{Q}[/tex] such that [tex]a\cdot a^{-1}=1[/tex]?
     
  4. Oct 24, 2006 #3
    Yes, it would just be 1/a. That's what I put on my test and got it wrong. He wrote that the set of rational numbers under multiplication is not a group. 1 i rational, the product of any two rationals is rational, 1/a is the inverse, and it's associative. I don't understand.
     
  5. Oct 24, 2006 #4

    radou

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    Hint 2: does there exist 1/a for a = 0? :smile:
     
  6. Oct 24, 2006 #5

    JasonRox

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    What's the inverse of 0?
     
  7. Oct 24, 2006 #6
    Wow, I can't believe I didn't think about that. Thanks
     
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