In each case, I am asked to decide whether the indicated pair is a group or not. If so, prove it; if not, show which group axiom fails.
(a) $(\dfrac{1}{2}\mathbb{Z}, +)$ where $\dfrac{1}{2} \mathbb{Z} = \{\dfrac{n}{2} | n \in \mathbb{Z}\}$.
(b) $(\mathbb{R_{\geq 0}}, *)$ where $x * y = max\{x,y\}$.
As for (a), it appears that $\dfrac{1}{2}\mathbb{Z}$ is closed with respect to addition since given $x,y \in \dfrac{1}{2}\mathbb{Z}$, we have $x = \dfrac{n_1}{2}, y = \dfrac{n_2}{2}$, for some $n_1,n_2 \in \mathbb{Z}$ and so $x + y = \dfrac{n_1 + n_2}{2}$ and clearly $n_1 + n_2 \in \mathbb{Z}$ so $x+y \in \dfrac{1}{2} \mathbb{Z}$.
It's associative as well since $x,y,z \in \dfrac{1}{2}\mathbb{Z}$ is such that $x = \dfrac{n_1}{2}, y = \dfrac{n_2}{2}, z = \dfrac{n_3}{2} = \dfrac{n_1 + n_2 + n_3}{2}$ in either case.
There's an identity element since $0 \in \dfrac{1}{2}\mathbb{Z}$ because $ 0 = \dfrac{0}{n}$.
And there exists inverses since for any $x = \dfrac{n_0}{2} \in \dfrac{1}{2}\mathbb{Z}$ we have $x^{-1} = \dfrac{-n_0}{2}$ and $\dfrac{n_0 - n_0}{2} = 0 = e_{\dfrac{1}{2}\mathbb{Z}}$. So $(\dfrac{1}{2}\mathbb{Z},+)$ is a group.
(b) This is not a group since we do not have a $unique$ identity element since while $\forall x \in \mathbb{R_{\geq 0}}$ we have $x*x = x$, there is no unique $e$ such that $e*x = x$ that will not depend on $x$.
Is this reasoning correct?