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

- $~$[a_n] + [b_n] = [a_n+b_n]$~$
- $~$[a_n] \times [b_n] = [a_n \times b_n]$~$
- $~$[a_n] \leq [b_n]$~$ if and only if either $~$[a_n] = [b_n]$~$ or for sufficiently large $~$n$~$, $~$a_n \leq b_n$~$.

# Proof

Firstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives $~$(x_n)_{n=1}^{\infty}$~$ and $~$(y_n)_{n=1}^{\infty}$~$ of the same equivalence class $~$[x_n] = [y_n]$~$, we don't somehow get different answers.

## Well-definedness of $~$+$~$

We wish to show that $~$[x_n]+[a_n] = [y_n] + [b_n]$~$ whenever $~$[x_n] = [y_n]$~$ and $~$[a_n] = [b_n]$~$; this is an exercise. %%hidden(Show solution): Since $~$[x_n] = [y_n]$~$, it must be the case that both $~$(x_n)$~$ and $~$(y_n)$~$ are Cauchy sequences such that $~$x_n - y_n \to 0$~$ as $~$n \to \infty$~$. Similarly, $~$a_n - b_n \to 0$~$ as $~$n \to \infty$~$.

We require $~$[x_n+a_n] = [y_n+b_n]$~$; that is, we require $~$x_n+a_n - y_n-b_n \to 0$~$ as $~$n \to \infty$~$.

But this is true: if we fix rational $~$\epsilon > 0$~$, we can find $~$N_1$~$ such that for all $~$n > N_1$~$, we have $~$|x_n - y_n| < \frac{\epsilon}{2}$~$; and we can find $~$N_2$~$ such that for all $~$n > N_2$~$, we have $~$|a_n - b_n| < \frac{\epsilon}{2}$~$. Letting $~$N$~$ be the maximum of the two $~$N_1, N_2$~$, we have that for all $~$n > N$~$, $~$|x_n + a_n - y_n - b_n| \leq |x_n - y_n| + |a_n - b_n|$~$ by the [-triangle_inequality], and hence $~$\leq \epsilon$~$. %%

## Well-definedness of $~$\times$~$

We wish to show that $~$[x_n] \times [a_n] = [y_n] \times [b_n]$~$ whenever $~$[x_n] = [y_n]$~$ and $~$[a_n] = [b_n]$~$; this is also an exercise. %%hidden(Show solution): We require $~$[x_n a_n] = [y_n b_n]$~$; that is, $~$x_n a_n - y_n b_n \to 0$~$ as $~$n \to \infty$~$.

Let $~$\epsilon > 0$~$ be rational. Then $$~$|x_n a_n - y_n b_n| = |x_n (a_n - b_n) + b_n (x_n - y_n)|$~$$ using the very handy trick of adding the expression $~$x_n b_n - x_n b_n$~$.

By the triangle inequality, this is $~$\leq |x_n| |a_n - b_n| + |b_n| |x_n - y_n|$~$.

We now use the fact that [cauchy_sequences_are_bounded], to extract some $~$B$~$ such that $~$|x_n| < B$~$ and $~$|b_n| < B$~$ for all $~$n$~$; then our expression is less than $~$B (|a_n - b_n| + |x_n - y_n|)$~$.

Finally, for $~$n$~$ sufficiently large we have $~$|a_n - b_n| < \frac{\epsilon}{2 B}$~$, and similarly for $~$x_n$~$ and $~$y_n$~$, so the result follows that $~$|x_n a_n - y_n b_n| < \epsilon$~$. %%

## Well-definedness of $~$\leq$~$

We wish to show that if $~$[a_n] = [c_n]$~$ and $~$[b_n] = [d_n]$~$, then $~$[a_n] \leq [b_n]$~$ implies $~$[c_n] \leq [d_n]$~$.

Suppose $~$[a_n] \leq [b_n]$~$, but suppose for contradiction that $~$[c_n]$~$ is not $~$\leq [d_n]$~$: that is, $~$[c_n] \not = [d_n]$~$ and there are arbitrarily large $~$n$~$ such that $~$c_n > d_n$~$. Then there are two cases.

- If $~$[a_n] = [b_n]$~$ then $~$[d_n] = [b_n] = [a_n] = [c_n]$~$, so the result follows immediately. %%note:We didn't need the extra assumption that $~$[c_n] \not \leq [d_n]$~$ here.%%
- If for all sufficiently large $~$n$~$ we have $~$a_n \leq b_n$~$, then [todo: this part]

## Additive commutative group structure on $~$\mathbb{R}$~$

The additive identity is $~$[0]$~$ (formally, the equivalence class of the sequence $~$(0, 0, \dots)$~$). Indeed, $~$[a_n] + [0] = [a_n+0] = [a_n]$~$.

The additive inverse of the element $~$[a_n]$~$ is $~$[-a_n]$~$, because $~$[a_n] + [-a_n] = [a_n-a_n] = [0]$~$.

The operation is commutative: $~$[a_n] + [b_n] = [a_n+b_n] = [b_n+a_n] = [b_n] + [a_n]$~$.

The operation is closed, because the sum of two Cauchy sequences is a Cauchy sequence (exercise). %%hidden(Show solution): If $~$(a_n)$~$ and $~$(b_n)$~$ are Cauchy sequences, then let $~$\epsilon > 0$~$. We wish to show that there is $~$N$~$ such that for all $~$n, m > N$~$, we have $~$|a_n+b_n - a_m - b_m| < \epsilon$~$.

But $~$|a_n+b_n - a_m - b_m| \leq |a_n - a_m| + |b_n - b_m|$~$ by the triangle inequality; so picking $~$N$~$ so that $~$|a_n - a_m| < \frac{\epsilon}{2}$~$ and $~$|b_n - b_m| < \frac{\epsilon}{2}$~$ for all $~$n, m > N$~$, the result follows. %%

The operation is associative: $$~$[a_n] + ([b_n] + [c_n]) = [a_n] + [b_n+c_n] = [a_n+b_n+c_n] = [a_n+b_n] + [c_n] = ([a_n]+[b_n])+[c_n]$~$$

## Ring structure

The multiplicative identity is $~$[1]$~$ (formally, the equivalence class of the sequence $~$(1,1, \dots)$~$). Indeed, $~$[a_n] \times [1] = [a_n \times 1] = [a_n]$~$.

$~$\times$~$ is closed, because the product of two Cauchy sequences is a Cauchy sequence (exercise). %%hidden(Show solution): If $~$(a_n)$~$ and $~$(b_n)$~$ are Cauchy sequences, then let $~$\epsilon > 0$~$. We wish to show that there is $~$N$~$ such that for all $~$n, m > N$~$, we have $~$|a_n b_n - a_m b_m| < \epsilon$~$.

But $$~$|a_n b_n - a_m b_m| = |a_n (b_n - b_m) + b_m (a_n - a_m)| \leq |b_m| |a_n - a_m| + |a_n| |b_n - b_m|$~$$ by the triangle inequality.

Cauchy sequences are bounded, so there is $~$B$~$ such that $~$|a_n|$~$ and $~$|b_m|$~$ are both less than $~$B$~$ for all $~$n$~$ and $~$m$~$.

So picking $~$N$~$ so that $~$|a_n - a_m| < \frac{\epsilon}{2B}$~$ and $~$|b_n - b_m| < \frac{\epsilon}{2B}$~$ for all $~$n, m > N$~$, the result follows. %%

$~$\times$~$ is clearly commutative: $~$[a_n] \times [b_n] = [a_n \times b_n] = [b_n \times a_n] = [b_n] \times [a_n]$~$.

$~$\times$~$ is associative: $$~$[a_n] \times ([b_n] \times [c_n]) = [a_n] \times [b_n \times c_n] = [a_n \times b_n \times c_n] = [a_n \times b_n] \times [c_n] = ([a_n] \times [b_n]) \times [c_n]$~$$

$~$\times$~$ distributes over $~$+$~$: we need to show that $~$[x_n] \times ([a_n]+[b_n]) = [x_n] \times [a_n] + [x_n] \times [b_n]$~$. But this is true: $$~$[x_n] \times ([a_n]+[b_n]) = [x_n] \times [a_n+b_n] = [x_n \times (a_n+b_n)] = [x_n \times a_n + x_n \times b_n] = [x_n \times a_n] + [x_n \times b_n] = [x_n] \times [a_n] + [x_n] \times [b_n]$~$$

## Field structure

To get from a ring to a field, it is necessary and sufficient to find a multiplicative inverse for any $~$[a_n]$~$ not equal to $~$[0]$~$.

Since $~$[a_n] \not = 0$~$, there is some $~$N$~$ such that for all $~$n > N$~$, $~$a_n \not = 0$~$. Then defining the sequence $~$b_i = 1$~$ for $~$i \leq N$~$, and $~$b_i = \frac{1}{a_i}$~$ for $~$i > N$~$, we obtain a sequence which induces an element $~$[b_n]$~$ of $~$\mathbb{R}$~$; and it is easy to check that $~$[a_n] [b_n] = [1]$~$. %%hidden(Show solution): $~$[a_n] [b_n] = [a_n b_n]$~$; but the sequence $~$(a_n b_n)$~$ is $~$1$~$ for all $~$n > N$~$, and so it lies in the same equivalence class as the sequence $~$(1, 1, \dots)$~$. %%

## Ordering on the field

We need to show that:

- if $~$[a_n] \leq [b_n]$~$, then for every $~$[c_n]$~$ we have $~$[a_n] + [c_n] \leq [b_n] + [c_n]$~$;
- if $~$[0] \leq [a_n]$~$ and $~$[0] \leq [b_n]$~$, then $~$[0] \leq [a_n] \times[b_n]$~$.

We may assume that the inequalities are strict, because if equality holds in the assumption then everything is obvious. %%hidden(Show obvious bits): If $~$[a_n] = [b_n]$~$, then for every $~$[c_n]$~$ we have $~$[a_n] + [c_n] = [b_n] + [c_n]$~$ by well-definedness of addition. Therefore $~$[a_n] + [c_n] \leq [b_n] + [c_n]$~$.

If $~$[0] = [a_n]$~$ and $~$[0] \leq [b_n]$~$, then $~$[0] = [0] \times [b_n] = [a_n] \times [b_n]$~$, so it is certainly true that $~$[0] \leq [a_n] \times [b_n]$~$. %%

For the former: suppose $~$[a_n] < [b_n]$~$, and let $~$[c_n]$~$ be an arbitrary equivalence class. Then $~$[a_n] + [c_n] = [a_n+c_n]$~$; $~$[b_n] + [c_n] = [b_n+c_n]$~$; but we have $~$a_n + c_n \leq b_n + c_n$~$ for all sufficiently large $~$n$~$, because $~$a_n \leq b_n$~$ for sufficiently large $~$n$~$. Therefore $~$[a_n] + [c_n] \leq [b_n] + [c_n]$~$, as required.

For the latter: suppose $~$[0] < [a_n]$~$ and $~$[0] < [b_n]$~$. Then for sufficiently large $~$n$~$, we have both $~$a_n$~$ and $~$b_n$~$ are positive; so for sufficiently large $~$n$~$, we have $~$a_n b_n \geq 0$~$. But that is just saying that $~$[0] \leq [a_n] \times [b_n]$~$, as required.