The [[Integers|integers]] can be formally constructed as the set of equivalence classes of ordered pairs of naturals $(a,b)$ where two pairs are equivalent if they have the same difference (i.e $(a,b)\sim(c,d)\iff a+d=b+c$). > [!NOTE] Definition (Set theoretic definition of Integers) > $\mathbb{Z}$ is the [[Equivalence relations|quotient set]] of $\sim$ by $\mathbb{N}^{2}$ (ordered pairs of [[Natural Numbers|naturals]]) where $(a,b) \sim(c,d)$ iff $a+d=b+c.$ The intuition is that $(a,b)$ stands for the result of subtracting _b_ from _a_. > [!NOTE] Lemma (Difference of ordered pairs is an equivalence relation) > The relation $\sim$ is an [[Equivalence relations|equivalence relation]] on $\mathbb{N}^{2}.$ *Proof*. Since $a+b=a+b,$ $\sim$ is reflexive. Since $a+d=b+c \implies c+b=d+a,$ $\sim$ is symmetric. Now suppose $a+d=b+c$ and $c+f=d+e$ then adding gives $a+c+d+f=b+c+e+d.$ Addition in $\mathbb{N}$ is cancellative so we may cancel terms $c$ and $d$ to obtain $a+f=b+e,$ which shows that $\sim$ is transitive. > [!NOTE] Definition/Theorem (Arithmetic Operations) > We define three operations on $\mathbb{Z}$ as follows $\begin{align} [(p,q)]_{E} +_{\mathbb{Z}} [(r,s)]_{E} &= [(p+r,q+s)]_{E} & \text{addition} \\ [(p,q)]_{E} \times_{\mathbb{Z}} [(r,s)]_{E} &= [(pr+qs, ps+qr)]_{E} &\text{multiplication} \\- [(p,q)]_{E} &= [(q,p)]_{E} & \text{negation} \end{align}$ where $p,q,r,s \in \mathbb{N}$. These operations on $\mathbb{Z}$ are [[Well-defined Function with Respect to Equivalence Relation|well-defined]]. **Proof**. Suppose that $(p,q)\,E\,(p',q')$ and $(r,s)\,E\,(r',s')$ for some $p,q,r,s \in \mathbb{N}$. Assume $p<p'$ and $r<r'$. Then $p'=p+P$ and $r'=r+R$ for some $P,R\in\mathbb{N}$. Hence $(p+P,q+P)=(p',q')$ and $(r+R,s+R)=(r',s')$. To show that addition in $\mathbb{Z}$ is well-defined we compute as follows: $\begin{align} (p'+r',q'+s') &= (p+P+r', q+P+s') \\ &= (p+P+r+R, q+P+s+R) \\ &= (p+r+P+R,q+s+P+R) \in [(p+r,q+s)]_{E} \end{align}$ To show that multiplication in $\mathbb{Z}$ is well-defined we compute as follows: To show that negation in $\mathbb{Z}$ is well-defined we compute as follows: $(q',p')=(q+P,p+P)\in[(q,p)]_{E}$