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Simple Verification of a Group

  1. Apr 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine whether the set of rational numbers with denominator equal to 1 or 2 is a group under addition.

    2. Relevant equations


    3. The attempt at a solution

    Please have a look at the closure proof of part 5. I don't quite understand how ##q/2## implies that the denominator of ##p/q## will be 1 or 2, so I tried to conceive of my own.

    Either ##2a+d## is even or is odd. If even, then ##\frac{2a+d}{2} = \frac{2k}{2} = \frac{k}{1}## which is therefore in reduced form and puts it in ##G##. Now, if odd, then the numerator and denominator of ##\frac{2a+d}{2}## have no common factors and is therefore already in reduced form. Because the denominator is ##2##, it is by definition in ##G##.

    Is there anything wrong with this proof? Although I gave my own, I would like to understand the linked solution.
     
  2. jcsd
  3. Apr 21, 2016 #2

    andrewkirk

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    It looks mostly fine. But by using ##2a+d## you have not covered the case where both numerators are odd. You need to cover that case as well to complete the proof.
     
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