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.