# axioms of real numbers without multiplication

Consider the axioms of real numbers https://en.wikipedia.org/wiki/Real_number#Axiomatic_approach and suppose we remove the multiplication operation and its properties. Do we loose something?

I have the impression that multiplication can be constructed in a unique way to recover all the axioms (define multiplication by natural numbers as repeated addition, then define multiplication on rational numbers then extend by continuity on all real numbers).

The question is motivated by the fact the the axioms comprising addition, ordering and completeness are very intuitive facts about the geometric line. Multiplication, instead, is not as obvious and maybe would be nice to build it.

Yes, this has been done by Tarski.

https://en.wikipedia.org/wiki/Tarski%27s_axiomatization_of_the_reals

Here are some details that are in the spirit of Tarski's work:

Let $$M$$ be a system of magnitudes and select any element in the carrier set and call it $$1$$, so that the set $$M$$ is a pointed set and the object of study becomes $$(M,1,+)$$. We also have an injective morphism

$$\tag 1 \iota: \mathbb N^> = \mathbb N \setminus \{0\} \to M \text{ such that } 1 \mapsto 1$$

so we can view the image of the imbedding as an inclusion, $$\mathbb N^> \subset M$$.

It is not difficult to show that for any $$x \in M$$ there exist a unique element $$H(x)$$ such that $$H(x)+H(x)=x$$.

For any $$x \in M$$ there exist an $$N_x \ge 0$$ such that for all $$n \ge N_x$$ the equations $$m H^n(1) + u = x$$ with $$m \gt 0$$ have solutions ('$$m \; \text{times}$$' is shorthand for repeated addition), so we can take the maximum $$m_{(x,n)}$$ and define sets $$\{m_{(x,n)}H^{n}(1)\}$$.

For $$s, t \in M$$, we can create a set

$$\tag 2 X_{(s,t)} = \{\; (m_{(s,n)} \times m_{(t,n)})\, H^{(n+n)}(1) \;\}$$

and set $$Y_{(s,t)} = \{ m \in M \; | \; (\forall x \in X_{(s,t)}) (\exists u \in M) \,[x + u = m]\}$$.

Invoking $$\text{P-5}$$ (see above link) we can get a $$z_{(s,t)} \in M$$ that separates $$X_{(s,t)}$$ and $$Y_{(s,t)}$$. This element is clearly in $$Y_{(s,t)}$$ and is therefore unique.

We state the following two theorems without proof.

Theorem 1: The mapping $$(s,t) \mapsto z_{(s,t)}$$ is a commutative operation that distributes over addition in $$(M,1,+)$$. Moreover, $$1$$ is a multiplicative identity in $$(M,1,+,*)$$.

Theorem 2: Every element $$x \in (M,1,+,*)$$ has a multiplicative inverse.