## 1.4. Natural Numbers, Integers, and Rational Numbers

### Theorem 1.4.2: The Integers

**A**be the set

**N**x

**N**and define a relation

*r*on

**N**x

**N**by saying that

*(a, b)*is related to

*(a', b')*if

*a + b' = a' + b*. Then this relation is an equivalence relation.

If *[(a,b)]* and *[(a', b')]* denote the equivalence classes
containing *(a, b)* and *(a', b')*, respectively, and if we define
addition and multiplication of those equivalence classes as:

*[(a,b)] + [(a', b')] = [(a + a', b + b')]**[(a,b)] * [(a', b')] = [(a * b' + b * a', a * a' + b * b')]*

### Proof:

We first have to prove reflexivity, symmetry, and transitivity to show that the relation is an equivalence relation. The first two properties are easy, and are left as an exercise. As for transitivity:- 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'') = (a' + b) + (a'' + b')*. Canceling

*a'*and

*b'*on both sides yields

*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')*

*(a, b) ~ (c, d)*and

*(a', b') ~ (c', d')*to see that this equation is indeed true.

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.