## Theorem 1.4.2: The Integers |

Let |

- Take
*(a,b) ~ (a', b’)*, i.e.*a + b’ = a’ + b* - Take
*(a', b’) ~ (a’’, b’’)*, i.e.*a’ + b’’ = a’’ + b’*

*a + b’’ = a’’ + b, i.e. (a, b) ~ (a’’, b’’)*

Next, we will have to show that the way that the definition of addition and multiplication is well-defined. In particular, we need to show that the definition of these operations does not depend on the particular representative of the equivalence classes that we chose. The idea of that proof is clear: pick different members of a class, and show that their sum or product results in the same class. Thus, suppose:

*(a, b)*and*(c, d)*are related*(a’, b’)*and*(c’, d’)*are related.- Then
*[(a,b)] + [(a’, b’)] = [(a + a’, b + b’)]* - and
*[(c, d)] + [(c’, d’)] = [(c + c’, d + d’)]* - Because
*(a, b) ~ (c, d)*we know:*a + d = c + b* - Because
*(a’, b’) ~ (c’, d’)*we know that*a’ + d’ = c’ + b’*

*(a + a’) + (d + d’) = (c + c’) + (b + b’)*

*[(a + a’), (b + b’)] = [(c + c’), (d + d’)]*

Finally, we need to show that multiplication is also well-defined. Therefore, suppose:

*(a, b)*and*(c, d)*are related, i.e.*a + d = c + b**(a’, b’)*and*(c’, d’)*are related. i.e.*a’ + d’ = c’ + b’*

*[(a,b)] * [(a’, b’)] = [(a * b’ + b * a’, a * a’ + b * b’)]**[(c, d)] * [(c’, d’)] = [(c * d’ + d * c’, c * c’ + d * d’)]*

*(a * b’ + b * a’) + (c * c’ + d * d’) = (c * d’ + d * c’) + (a * a’ + b * b’)*

*a * b' - a * a' + b * a' - b * b' = c * d' - c * c' + d * c' - d * d'*

*a * (b' - a') - b * (b' - a') = c * (d' - c') - d * (d' - c')*

*(a - b) * (b' - a') = (c - d) * (d' - c')*

As for the actual proof, we only have to read the last few lines backwards to have a perfectly good proof. This will show that the two resulting classes are the same, proving that multiplication is indeed well-defined.

As to why this definition yields equivalence classes with properties similar to
the integers, consider a few examples on your own. We do not actually have to
prove anything, because we simply **define** the integers to be the set of
equivalence classes with respect to the above equivalence relation and definition
of addition and multiplication.

Before we finish: it seems like a real coincidence that this nice factorization
worked in the proof of the well-definition of the multiplication. If you work
out some examples with actual numbers, however, it might become clear that this
is of course **not** a coincidence after all.