# Simple Verification of a Group

## Homework Statement

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

## 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.