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Can someone check my proof?

  1. Feb 2, 2010 #1
    [tex]\forall q \in \textbf{Q}, \exists r \in\textbf{Q}[/tex] so that [tex]q + r\in \textbf{Z}[/tex] (Q is set of all rational numbers, and Z is set of all Integers)


    let q be an arbitrary rational number
    thus, [itex]q=\frac{a}{b}[/itex] for some integers a and b, and b is not 0
    let [itex]r = \frac{b-a}{b}[/itex] where [itex]b-a,b\in\textbf{Z}[/itex], b is still not 0
    q + r = \frac{a}{b} + \frac{b-a}{b}[/tex]

    [tex] = \frac{a+b-a}{b}[/tex]

    [tex] = \frac{b}{b}[/tex]

    [tex]=1[/tex] and 1 is an integer

    End of proof

    I'm not sure if I was redundant with anything, or if I forgot to say anything. I think I only need to find one example since the second quantifier says there exists, which I think means I only need to show one algebraically for an arbitrary rational number. Also, I can take advantage of the fact that an integer is an integer, so I don't have to define it I guess...

    I'm in a first year discreet mathematics course. If there's anything wrong with my proof, please let me know.

    Thanks, appreciate it!
    Last edited: Feb 2, 2010
  2. jcsd
  3. Feb 2, 2010 #2


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    Are Q and Z restricted to > 0? If not r = -q will always work.
  4. Feb 2, 2010 #3
    Yeah, -r would work, but I was wondering if my proof works as well. I realized that -r would work afterwards, but I already wrote down my version which I think works out algebraically, but yeah I wanted to confirm before I submit this.
  5. Feb 3, 2010 #4


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    Your proof is correct, but as a mathematician I can tell you it is awful. Simple proofs are always preferred over complicated ones.

    A more interesting case is restricting Q and Z to be positive. Then something like your proof might be needed.
    (Hint: replace b-a by nb-a, where n is sufficiently large).
  6. Feb 3, 2010 #5
    Your proof is fine. I agree with mathman in spirit, but can you honestly say q+(1-q)=1 is "complicated"?
  7. Feb 4, 2010 #6


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    Your version is several lines shorter than the original.
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