The real numbers, when constructed as equivalence classes of Cauchy sequences of rationals, form a totally ordered field, with the inherited field structure given by
Firstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives and of the same equivalence class , we don't somehow get different answers.
We wish to show that whenever and ; this is an exercise.
Since , it must be the case that both and are Cauchy sequences such that as . Similarly, as .
We require ; that is, we require as .
But this is true: if we fix rational , we can find such that for all , we have ; and we can find such that for all , we have . Letting be the maximum of the two , we have that for all , by the triangle_inequality, and hence .
We wish to show that whenever and ; this is also an exercise.
We require ; that is, as .
Let be rational. Then using the very handy trick of adding the expression .
By the triangle inequality, this is .
We now use the fact that Cauchy_sequences_are_bounded, to extract some such that and for all ; then our expression is less than .
Finally, for sufficiently large we have , and similarly for and , so the result follows that .
We wish to show that if and , then implies .
Suppose , but suppose for contradiction that is not : that is, and there are arbitrarily large such that . Then there are two cases.
The additive identity is (formally, the equivalence class of the sequence ). Indeed, .
The additive inverse of the element is , because .
The operation is commutative: .
The operation is closed, because the sum of two Cauchy sequences is a Cauchy sequence (exercise).
If and are Cauchy sequences, then let . We wish to show that there is such that for all , we have .
But by the triangle inequality; so picking so that and for all , the result follows.
The operation is associative:
The multiplicative identity is (formally, the equivalence class of the sequence ). Indeed, .
is closed, because the product of two Cauchy sequences is a Cauchy sequence (exercise).
If and are Cauchy sequences, then let . We wish to show that there is such that for all , we have .
But by the triangle inequality.
Cauchy sequences are bounded, so there is such that and are both less than for all and .
So picking so that and for all , the result follows.
is clearly commutative: .
is associative:
distributes over : we need to show that . But this is true:
To get from a ring to a field, it is necessary and sufficient to find a multiplicative inverse for any not equal to .
Since , there is some such that for all , . Then defining the sequence for , and for , we obtain a sequence which induces an element of ; and it is easy to check that .
; but the sequence is for all , and so it lies in the same equivalence class as the sequence .
We need to show that:
We may assume that the inequalities are strict, because if equality holds in the assumption then everything is obvious.
If , then for every we have by well-definedness of addition. Therefore .
If and , then , so it is certainly true that .
For the former: suppose , and let be an arbitrary equivalence class. Then ; ; but we have for all sufficiently large , because for sufficiently large . Therefore , as required.
For the latter: suppose and . Then for sufficiently large , we have both and are positive; so for sufficiently large , we have . But that is just saying that , as required.