# Given that you only have multiplication and addition, how can you divide two numbers?

I have a signature given with the real numbers as its universe and addition and multiplication as functions. I need to write the following expression in First Order Logic.

• $x$ is a rational number

My idea: $\varphi(x) = \exists a \, \exists b \, x = \frac{a}{b}$.

The problem is I don't have division.

Extra question: how can I write

• $x \geq 0$?
• How about $bx=a$ instead of $x=\frac{a}{b}$. – Lee Mosher Jan 9 '17 at 16:06
• The real question IMO is, how do you denote the fact that $a$ and $b$ are integers? With only $\times,+$ and $\mathbb{R}$ it seems somewhat infeasible. – barak manos Jan 9 '17 at 16:07
• Regarding the extra question: $x\ge0\iff \exists y\colon x=y^2$ – Hagen von Eitzen Jan 9 '17 at 16:09
• @HagenvonEitzen Thank you very much. Why didn't I think that? – MessitÖzil Jan 9 '17 at 16:11
• You can't express the predicate "is an integer" in first order logic over $({\mathbb{R}},+,\cdot)$. You need second order logic to define that predicate, for instance as the smallest set containing $0$, $1$ and being closed under addition. – Magdiragdag Jan 9 '17 at 16:19

Let me strengthen what others have already said:

The structure $\mathcal{R}=(\mathbb{R}; +, \times)$ is decidable (this is due to Tarski). This immediately rules out the possibility of defining $\mathbb{Z}$ in $\mathcal{R}$, since the theory of the integers is undecidable (by Goedel). (Incidentally, since $\mathbb{Z}$ is (nontrivially) definable in $\mathbb{Q}$, this also rules out the possibility of defining $\mathbb{Q}$ in $\mathcal{R}$.)

But in fact more is true: Tarski showed that it is o-minimal, that is, every definable set is a finite union of intervals. So nothing remotely like $\mathbb{Z}$ or $\mathbb{Q}$ can be a definable subset of $\mathcal{R}$.

EDIT: I just noticed you said first order arithmetic. This answer uses second order arithmetic. It's a theorem that it is impossible to define $\mathbb{Z}$ in $(\mathbb{R},+,\cdot)$

As mentioned in the comments, division is easy: define $a/b$ to be the number $c$ such that $cb=a$. It's identifying integers that is hard. Notably, $\mathbb{Z}$ and $\mathbb{Z}[\pi]$ have pretty much the same arithmatic structure. However, you can identify $0$ as the only number that satisfies $$\varphi(x):=\forall a(ax=x)$$ and then you can identify $1$ as the only number that satisfies $$\varphi'(x)=\forall a(\varphi(a)\lor a=ax)$$

Given these two constants, we can then recursively define the integers by using the fact that they are generated by $1$ as a group under addition, a la the Peano Axioms.

• but for x to be rational, a and b must not be Integers, right? so $ax=b$ works. – MessitÖzil Jan 9 '17 at 16:28
• and it's NOT possible to say a and b are integers in first order logic over $(R,+,⋅)$ ? – MessitÖzil Jan 9 '17 at 16:35
• @StellaBiderman That's not really relevant here. The second-order Peano axioms are not first-order expressible, but that doesn't a priori mean that $\mathbb{N}$ isn't a definable subset of the field of reals. Indeed, it is a definable subset of the field of rationals. – Noah Schweber Jan 9 '17 at 18:29