1.4. Natural Numbers, Integers, and Rational Numbers
Theorem 1.4.2: The Integers
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, 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.