Tarski's axiomatization of the reals

Tarski's axioms imply that ${\displaystyle \mathbb {R} }$ is a totally ordered[2] abelian group under addition with distinguished element ${\displaystyle 1}$. Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete and a totally ordered abelian group, ${\displaystyle \mathbb {R} }$ is Archimedean, because the Dedekind-completion of any totally ordered abelian group with infinite elements or infinitesimals is not an abelian group, and the Dedekind-completion of any Archimedean ordered abelian group is still Archimedean.

Tarski's axioms imply that ${\displaystyle \mathbb {R} }$ is a totally ordered abelian group under addition with distinguished element ${\displaystyle 1}$. As a result, there exist maximum and minimum binary functions ${\displaystyle \max$ :\mathbb {R} \times \mathbb {R} \to \mathbb {R} } and ${\displaystyle \min$ :\mathbb {R} \times \mathbb {R} \to \mathbb {R} } , with the absolute value function defined as ${\displaystyle \vert x\vert =\max(x,-x)}$.

Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, ${\displaystyle \mathbb {R} }$ is a metric space with respect to the absolute value ${\displaystyle \vert x\vert }$ and thus a Hausdorff space, and every Cauchy net in ${\displaystyle \mathbb {R} }$ converges to a unique element of ${\displaystyle \mathbb {R} }$, and thus the absolute value ${\displaystyle \vert x\vert }$ is a complete metric on ${\displaystyle \mathbb {R} }$.

Since ${\displaystyle \mathbb {R} }$ is an abelian group, it is a ${\displaystyle \mathbb {Z} }$-module, and since ${\displaystyle \mathbb {R} }$ is totally ordered, it is a torsion-free module and thus a torsion-free abelian group, which means that the integers ${\displaystyle \mathbb {Z} }$ embed in ${\displaystyle \mathbb {R} }$, with injective group homomorphism ${\displaystyle f:\mathbb {Z} \to \mathbb {R} }$ where ${\displaystyle f(0)=0}$ and ${\displaystyle f(1)=1}$. As a result, for every integer ${\displaystyle a\in \mathbb {Z} }$ and ${\displaystyle b\in \mathbb {Z} }$ the affine functions ${\displaystyle x\mapsto ax+b}$ are well defined in ${\displaystyle \mathbb {R} }$.

Since ${\displaystyle \mathbb {R} }$ is Dedekind-complete, Archimedean, and a totally ordered abelian group, any closed interval ${\displaystyle [a,b]}$ on ${\displaystyle \mathbb {R} }$ is compact and connected. Since ${\displaystyle \mathbb {R} }$ is also a complete metric space, the intermediate value theorem is satisfied for every function from a closed interval ${\displaystyle [a,b]}$ to ${\displaystyle \mathbb {R} }$. Because ${\displaystyle x\mapsto ax+b}$ are monotonic for ${\displaystyle a>0}$, and for ${\displaystyle a<0}$ the function is just the negation of a monotonic function, ${\displaystyle x\mapsto ax+b}$ have a root for ${\displaystyle \vert a\vert >0}$. Thus ${\displaystyle \mathbb {R} }$ is a divisible group and a ${\displaystyle \mathbb {Q} }$-vector space, with an injective group homomorphism ${\displaystyle f:\mathbb {Q} \to \mathbb {R} }$ where ${\displaystyle f(0)=0}$ and ${\displaystyle f(1)=1}$.

Since every Cauchy net in ${\displaystyle \mathbb {R} }$ converges to a unique element of ${\displaystyle \mathbb {R} }$, for every directed set ${\displaystyle A}$ and Cauchy net ${\displaystyle (a_{i})_{i\in A}}$ in the rational numbers, there exists a Cauchy net of linear functions ${\displaystyle (f_{i})_{i\in A}}$ defined as ${\displaystyle f_{i}(x)=a_{i}x}$. The limit of the Cauchy net ${\displaystyle \lim _{i\in A}(f_{i})_{i}}$ exists and is a unique function ${\displaystyle g(x)=\lim _{i\in A}(a_{i})_{i}x}$. Since every real number is the limit of a Cauchy net of rational numbers, there is an ${\displaystyle \mathbb {R} }$-action ${\displaystyle \mu$ :\mathbb {R} \to (\mathbb {R} \to \mathbb {R} )} which takes a real number ${\displaystyle r}$ to the linear function ${\displaystyle x\mapsto rx}$, with ${\displaystyle \alpha (1)=x\mapsto x}$ being the idenitity function. The uncurrying of ${\displaystyle \alpha }$ leads to a bilinear function ${\displaystyle (-)(-):\mathbb {R} \times \mathbb {R} \to \mathbb {R} }$ called multiplication of the real numbers, defined on the entire domain of the binary function. Since linear functions in the function space with function composition and the identity function is a commutative monoid, ${\displaystyle \mathbb {R} }$ multiplication and the multiplicative identity element ${\displaystyle 1}$ is also commutative monoid, which means that ${\displaystyle \mathbb {R} }$ is a commutative ring.

Since ${\displaystyle \mathbb {R} }$ is a commutative ring, power series are well defined, and because all Cauchy nets converge in ${\displaystyle \mathbb {R} }$, all Cauchy sequences and all Cauchy power series converge in ${\displaystyle \mathbb {R} }$. In particular, every geometric series is a Cauchy power series and the limit of the geometric series ${\displaystyle \sum _{n=0}^{\infty }x^{n}}$ and ${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}x^{n}}$ converges in the open interval ${\displaystyle (-1,1)}$. Thus let us define functions ${\displaystyle f:(-1,1)\to \mathbb {R} }$ and ${\displaystyle g:(-1,1)\to \mathbb {R} }$ as

${\displaystyle f(x):=\sum _{n=0}^{\infty }x^{n}}$

${\displaystyle g(x):=\sum _{n=0}^{\infty }(-1)^{n}x^{n}}$

Let us define the function

${\displaystyle h(x,a):=(-a)\sum _{n=0}^{\infty }a^{n}(x+f(a+1))^{n}}$

for ${\displaystyle a<0}$ and

${\displaystyle k(x,a):=a\sum _{n=0}^{\infty }(-a)^{n}(x+g(a-1))^{n}}$

for ${\displaystyle a>0}$. These are functions which converge on the open interval ${\displaystyle (1/a,0)}$ for ${\displaystyle h(x,a)}$ and ${\displaystyle (0,1/a)}$ for ${\displaystyle k(x,a)}$, and satisfy the identity ${\displaystyle h(x,a)x=1}$ for all ${\displaystyle a<0}$ and ${\displaystyle x\in (1/a,0)}$, and ${\displaystyle k(x,a)x=1}$ for all ${\displaystyle a>0}$ and ${\displaystyle x\in (0,1/a)}$, by definition of the geometric series.

The reciprocal function is piecewise defined as ${\displaystyle {\frac {1}{x}}={\begin{cases}\lim _{a\to 0^{-}}h(x,a)&{\text{if }}x<0\\\lim _{a\to 0^{+}}k(x,a)&{\text{if }}x>0\\\end{cases}}}$

As limits preserve multiplication, ${\displaystyle {\frac {1}{x}}x=1}$. Thus, ${\displaystyle \mathbb {R} }$ is a field.