Sorry if this is pretty basic, I just want to know whether my proof works! I would love any feedback on it. What I wanted to do in this proof was try to identity the set $\mathbb{Z}$ with a subset of $\mathbb{Q}$ such that the identification would preserve the elements $0$ and $1$, and the operations + and $\cdot$ and the relation <.
-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 the relation < on $\mathbb{Q}$ be defined as $[(x,y)]<[(z,w)]$ iff either $xw<yz$ when $w<0$ and $y<0$ or when $w>0$ and $y>0$ or $xw>yz$ when $w>0$ and $y<0$ or when $w<0$ and $y>0$.
-Let $i:\mathbb{Z} \rightarrow \mathbb{Q}$ be defined by $i(x)=[(x,1)]$ for all $x \in \mathbb{Z}$.
Given that field and order properties hold for $\mathbb{Q}$, we seek to prove that :
- The function $i:\mathbb{Z} \rightarrow \mathbb{Q}$ is injective
- $i(0)=\bar{0}$ and $i(1)=\bar{1}$
- Let $x,y \in \mathbb{Z}$, then $i(x+y)=i(x)+i(y), i(-x)=-i(x), i(xy)=i(x)i(y), x<y$ iff $i(x)<i(y)$
Proofs:
Let $x,y \in \mathbb{Z}$ and let $i(x)=i(y)$. Then $[(x,1)]=[(y,1)]$, so $x\cdot1=1\cdot y$. Since $x \cdot 1=x$ and $1 \cdot y=y, x=y$.
$i(0)=[(0,1)]=\bar{0}$ and $i(1)=[(1,1)]=\bar{1}$
Let $x,y \in \mathbb{Z}$. $i(x+y)=[((x+y),1)]=[(x \cdot 1 + 1 \cdot y,1)]=[(x,1)]+[(y,1)]=i(x)+i(y)$. $i(-x)=[(-x,1)]=-[(x,1)]=-i(x)$. $i(xy)=[(xy,1)]=[(x,1)]\cdot[(y,1)]=i(x)\cdot i(y)$. Let $x>y$, then $i(x)=[(x,1)]>[(y,1)]=i(y)$ since $x\cdot 1=x>y=1\cdot y$. Now we prove the converse, let $i(x)>i(y)$. Then by hypothesis $[(x,1)]>[(y,1)]$, so $x=x \cdot 1>1 \cdot y = y$. Thus, $i(x)>i(y)$ iff $x>y$.
Thus, the function $i:\mathbb{Z} \rightarrow \mathbb{Q}$ preserves order, additive and multiplicative identities, addition and multiplication properties from $\mathbb{Z}$ and since it is injective, the subset of $\mathbb{Q}$ that it maps onto can be said to be a "copy" of $\mathbb{Z}$.