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-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 $\mathbb{Z} \times \mathbb{Z}^{*}$ with respect to $\sim$. 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 $[(x,y)] \in \mathbb{Q}^{*}$.
We seek to prove that + and $^{-1}$ are well defined. We begin with proving that + is well defined.
Let $[(x,y)]=[(a,b)]$ and $[(z,w)]=[(c,d)]$, where $[(x,y)],[(z,w)],[(a,b)],[(c,d)] \in \mathbb{Q}$.
From hypothesis, $zd=wc$. Thus, $(aw)d+(bz)d=(ad)w+b(zd)=(ad)w+b(wc)=(ad)w+(bc)w$. Multiplying both sides of the equation by $b$, we obtain $(aw)bd+(bz)bd=(ad)bw+(bc)bw$. Rearranging and simplifying, the equation becomes $(aw+bz)bd=bw(ad+bc)$. This implies that $[(a,b)]+[(z,w)]=[((aw+bz),bw)]=[((ad+bc),bd)]=[(a,b)]+[(c,d)]$.
From hypothesis, $xb=ya$. Thus, $(xw)b+(yz)b=(xb)w+y(bz)=y(aw)+y(bz)$. Multiplying $w$ on both sides of the equation yields $(xw)bw+(yz)bw=yw(aw)+yw(bz)$. Rearranging and simplifying gives $(xw+yz)bw=yw(aw+bz)$. This implies that $[(x,y)]+[(z,w)]=[((xw+yz),yw)]=[((aw+bz),bw)]=[(a,b)]+[(z,d)]$.
Putting this two results together, $[(x,y)]+[(z,w)]=[(a,b)]+[(z,w)]=[(a,b)]+[(c,d)]$. Thus + is well defined.
Next we prove that the operation $^{-1}$ is well defined. Let $[(x,y)]=[(a,b)]$, where $[(x,y)],[(a,b)] \in \mathbb{Q}^{*}$.
By hypothesis, we know that $xb=ya$. $[(x,y)]^{-1}=[(y,x)]$ and $[(a,b)]^{-1}=[(b,a)]$.Hence,$[(x,y)]^{-1}=[(a,b)]^{-1}$ if $ya=xb$, which is exactly what hypothesis implies. Hence, the operation $^{-1}$ is well defined.