# Interactive Real Analysis

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

### Theorem 1.4.2: The Integers

Let 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:

1. [(a,b)] + [(a', b')] = [(a + a', b + b')]
2. [(a,b)] * [(a', b')] = [(a * b' + b * a', a * a' + b * b')]
then these operations are well-defined and the resulting set of all equivalence classes has all of the familiar properties of the integers (it therefore serves to define the integers based only on the natural numbers).

### 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'
Adding, we get (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'')
Hence, transitivity is proved.

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'
Adding both of those equations we get that:
• (a + a') + (d + d') = (c + c') + (b + b')
which means, in other words, that the two elements are related in turn. Hence:
• [(a + a'), (b + b')] = [(c + c'), (d + d')]
which is exactly what we wanted to show.

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'
Then, according to the definition of multiplication:
• [(a,b)] * [(a', b')] = [(a * b' + b * a', a * a' + b * b')]
• [(c, d)] * [(c', d')] = [(c * d' + d * c', c * c' + d * d')]
We need to show that these two classes are identical, i.e. the representatives are related. That means we need to show that
• (a * b' + b * a') + (c * c' + d * d') = (c * d' + d * c') + (a * a' + b * b')
It is easiest to start with what we would like to be true and work backwards. Therefore, rearranging the above equation we have:
• a * b' - a * a' + b * a' - b * b' = c * d' - c * c' + d * c' - d * d'
Factoring, we obtain:
• a * (b' - a') - b * (b' - a') = c * (d' - c') - d * (d' - c')
and factoring again:
• (a - b) * (b' - a') = (c - d) * (d' - c')
But now we can use that fact that (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.

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