We already defined them, as the completion of the Rational Numbers using linear orderings.

Lemma: for every $x\in \mathbb{R}$ and $n\in {\mathbb{N}}^{+}$, there exists $r,s\in \mathbb{Q}$ such that $r<x\le s$ and $s-r\le 1/n$

Def: Let $x,y\in \mathbb{R}$. We let $x+y=inf\{r+s\mid r,s\in \mathbb{Q},x\le r,y\le s\}$. The symbol inside the set is the sum of the rationals. We note the the infimum always exist since it is complete linear ordering.

Lemma: Let $x,y,z\in \mathbb{R}$

• $x+y=y+x$
• $(x+y)+z=x+(y+z)$
• $x+0=x$
• There exists a unique $w\in \mathbb{R}$ such that $x+w=0$. We denote $w=-x$, the opposite of $x$
• If $x<y$ then $x+z<y+z$

We denote ${\mathbb{R}}^{+}:=\{x\in \mathbb{R}\mid x>0\}$

Def: Let $x,y\in {\mathbb{R}}^{+}$. We let $x\cdot y=inf\{r\cdot s\mid r,s\in \mathbb{Q},x\le r,y\le s\}$

Lemma 1: Let $x,y,z\in {\mathbb{R}}^{+}$:

• $x\cdot (y+z)=x\cdot y+x\cdot z$
• $x\cdot y=y\cdot x$
• $(x\cdot y)\cdot z=x\cdot (y\cdot z)$
• $x\cdot 1=x$
• There exists a unique $w\in {\mathbb{R}}^{+}$ such that $x\cdot w=1$. We denote $w=1/x$, the reciprocal of $x$
• if $x<y$, then $x\cdot z<y\cdot z$

We define the absolute value for real numbers, let $x\in \mathbb{R}$, then $$ |x| = \begin{cases}x & x \ge 0 \-x & x < 0\end{cases}$$
and notice that if $x\ne 0$, the $|x|\in {\mathbb{R}}^{+}$.

Def: Let $x,y\in \mathbb{R}$, we let $$x\cdot y = \begin{cases}|x|\cdot |y|& x, y >0 \lor x, y <0 \
-(|x|\cdot|y|) & x>0, y<0 \lor x <0, y>0 \
0 & x = 0 \lor y = 0\end{cases}$$
Lemma 2: Let $x,y,z\in \mathbb{R}$:

• $x\cdot (y+z)=x\cdot y+x\cdot z$
• $x\cdot y=y\cdot x$
• $(x\cdot y)\cdot z=x\cdot (y\cdot z)$
• $x\cdot 1=x$
• For each $x\ne 0$ there exists a unique $w\in \mathbb{R}$ such that $x\cdot w=1$. We denote $w=1/x$, the reciprocal of $x$
• if $x<y$, and $z>0$ then $x\cdot z<y\cdot z$

We define division by a nonzero real number $x$: as $y÷x=y\cdot (1/x)$

A structure $\mathfrak{U}=(A,<,+,\cdot ,0,1)$ where $<$ is a linear ordering, $+$ and $\cdot$ are binary operations and $0,1$ are constants such that all properties of lemma 1 and lemma 2 are satisfied is called an ordered field in algebra. So the contents of lemma 1 and 2, can be summarized as the real numbers are an ordered field. As the ordering of the real numbers is complete, they are a complete ordered field

Th: The structure $\mathfrak{R}=(\mathbb{R},<,+,\cdot ,0,1)$ is a complete ordered field. It is in fact unique up to isomorphism. If $\mathfrak{U}$ is also a complete ordered field, then $\mathfrak{U}$ and $\mathfrak{R}$ are isomorphic.

Th (Expansion of real numbers in base $p$)
Let $p\ge 2$ be a natural number. For every $0\le a<1$ there is a unique sequence of natural numbers $⟨a_n{⟩}_{n\in {\mathbb{N}}^{+}}$ such that

• $0\le a_n<p$ for every $n\in {\mathbb{N}}^{+}$

• There is no $n_0$ such that $a_n=p-1$ for all $n>n_0$

• $a_1/p+\cdots +a_n/p^n\le a<a_1/p+\cdots +(a_n+1)/p^n$, for each $n\ge 1$
  The real number $a$ is rational iff $⟨a_n{⟩}_{n\in {\mathbb{N}}^{+}}$ is eventually periodic

  Using the Cauchy Sequence Completion

We can also consider them the completion of the rational numbers as a metric space

If we decide to do this we have that is a complete metric space by construction, the thing we get stuck on is the order.

Let's take $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ be Cauchy sequences of rational numbers, we say that $(a_n{)}_{n\in \mathbb{N}}\sim (b_n{)}_{n\in \mathbb{N}}$, iff for every $p>0$, there's $N\in \mathbb{N}$, such that $n\ge N$, $|a_n-b_n|<p$ (this feels like $lim{a}_n-b_n=0$).

We would like to consider a pre-order on the set of all Cauchy sequences of rationals $C$. The preorder is as follows: we say that $(a_n{)}_{n\in \mathbb{N}}\lesssim (b_n{)}_{n\in \mathbb{N}}$ iff for every $p>0$, there's an $N\in \mathbb{N}$ such that $n\ge N$, $a_n-b_n<p$, and according to gpt, we consider $lim sup{a}_n-b_n\le 0$, which can be a bit problematic since $\mathbb{Q}$ is not complete.

We know that $\sim$ is an equivalence relation on $C$

Also $\lesssim$ is a preorder on $C$, and the natural equivalence relation from the preorder is actually $\sim$, so $C/\sim$, will have an order induced by the preorder.

And we have the ordered set $C/\sim$ is a complete ordered field, inhereting a lot of these s properties from $\mathbb{Q}$, then $C/\sim$ is isomorphic to $\mathbb{R}$.