I'm relatively new to writing proofs, so perhaps you could help me vet this proof?(Its pretty basic I think, but I'm worried that I missed some details or provided excessive detail) Any feedback is appreciated, thanks!
-Let the equivalence relation $\sim$ be defined on the set $\mathbb{Z} \times \mathbb{Z}^{*}$, where $\mathbb{Z}^{*}$ = $\mathbb{Z}-\{0\}$, such that for $a,b,c,d \in \mathbb{Z}, (a,b) \sim(c,d) $ iff $ad=bc$.
-Let $\mathbb{Q}$ be the set of equivalence classes of $\mathbb{Z} \times \mathbb{Z}^{*}$. The elements $\bar{0}, \bar{1} \in \mathbb{Q}$ are defined as $\bar 0 = [(0,1)]$ and $\bar{1} = [(1,1)]$.
-Let the operation + on $\mathbb{Q}$ be defined as $[(x,y)]+[(z,w)]=[(xw+yz,yw)]$ for $[(x,y)],[(z,w)] \in \mathbb{Q}$.
-Let the operation $^{-1}$ on $\mathbb{Q}^{*}$ ,where $\mathbb{Q}^{*}=\mathbb{Q}-\{\bar{0}\}$, be defined as $[(x,y)]^{-1}=[(y,x)]$ for $[(x,y)] \in \mathbb{Q}^{*}$.
-Let the operation $\cdot$ on $\mathbb{Q}$ be defined as $[(x,y)] \cdot [(z,w)]=[(xz,yw)]$ for $[(x,y)],[(z,w)] \in \mathbb{Q}$
-Let the operation - on $\mathbb{Q}$ be defined as $-[(x,y)]=[(-x,y)]$ for $[(x,y)] \in \mathbb{Q}$.
Let $a,b,c \in \mathbb{Q}$ where $a=[(x,y)]$, $b=[(z,w)]$ and $c=[(u,v)]$. We seek to prove that the operation + and $\cdot$ are associative and commutative, and that a(b+c)=ab+ac(distributive law).
First we prove that + is associative. $a+(b+c)=[(x,y)]+[((zv+wu),wv)]=[((x(wv)+y(zv+wu)),y(wv))]=[((xwv+yzv+ywu),(yw)v]=[(((xw+yz)v+(yw)u),(yw)v]=[((xw+yz),yw)]+[(u,v)]=(a+b)+c$.
Next we prove that + is commutative. $a+b=[((xw+yz),yw)]=[((zy+wx),wy)]=b+a$.
Next we prove that $\cdot$ is associative. $a(bc)=[(x,y)]\cdot[(zu,wv)]=[(x(zu),y(wv))]=[(xzu,ywv)]=[((xz)u,(yw)v]=[(xz,yw)]\cdot[(u,v)]=(ab)c$
Next, we prove that $\cdot$ is commutative. $ab=[((xw+yz),yw)]=[((zy+wx),wy)]=ba$.
Next we prove the distributive law. First we prove a lemma, that $[(x,y)]=[(gx,gy)]$ for $g \in \mathbb{Z}$.$x(gy)=(xg)y=y(gx)$, so the lemma follows. $a(b+c)=[(x,y)] \cdot [((zv+wu),wv)]=[(x(zv+wu),y(wv)]=[((xzv+xwu),ywv)]=[y(xzv+xwu),y(ywv))]=[((xz(yu)+yw(xu)),(yw)(yu))]$. $ab+ac$=$[((xz,yw)]+[(xu,yv)]=[(xz(yu)+yw(xu),ywyv)]$. Hence $a(b+c)=ab+ac$.
QED