Simple Verification of a Group

[Bashyboy]

Homework Statement

Determine whether the set of rational numbers with denominator equal to 1 or 2 is a group under addition.

Homework Equations

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.

 

[andrewkirk]

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.