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Section B.14 The Integers as Equivalence Classes of Ordered Pairs
Define a binary relation \(\cong\) on the set \(Z=\nonnegints \times\nonnegints\) by
\begin{equation*}
(a,b)\cong (c,d)\quad\textit{iff}\quad a+d=b+c.
\end{equation*}
Lemma B.36 .
Proof. Let
\((a,b)\in Z\text{.}\) Then
\(a+b=b+a\text{,}\) so
\((a,b)\cong(b,a)\text{.}\)
Lemma B.37 .
Proof. Let
\((a,b),(c,d)\in Z\) and suppose that
\((a,b)\cong (c,d)\text{.}\) Then
\(a+d=b+c\text{,}\) so that
\(c+b=d+a\text{.}\) Thus
\((c,d)\cong (a,b)\text{.}\)
Lemma B.38 .
Proof.
Let \((a,b), (c,d), (e,f)\in Z\text{.}\) Suppose that
\begin{equation*}
(a,b)\cong(c,d)\quad\text{and} \quad (c,d)\cong (e,f).
\end{equation*}
Then \(a+d=b+c\) and \(c+f=d+e\text{.}\) Therefore,
\begin{equation*}
(a+d)+(c+f) =(b+c)+(d+e).
\end{equation*}
It follows that
\begin{equation*}
(a+f)+(c+d) =(b+e)+(c+d).
\end{equation*}
Thus \(a+f = b+e\) so that \((a,b)\cong(e,f)\text{.}\)
Now that we know that
\(\cong\) is an equivalence relation on
\(Z\text{,}\) we know that
\(\cong\) partitions
\(Z\) into equivalence classes. For an element
\((a,b)\in Z\text{,}\) we denote the equivalence class of
\((a,b)\) by
\(\langle (a,b)\rangle\text{.}\)
Let
\(\ints\) denote the set of all equivalence classes of
\(Z\) determined by the equivalence relation
\(\cong\text{.}\) The elements of
\(\ints\) are called
integers .