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I have the following assignment question:

Find a first order sentence that is true in $\langle\mathbb{Q},+,\cdot ,0,1\rangle$ but not in $\langle \mathbb{R},+,\cdot,0,1\rangle$.

Most of what I can think of is second order (such as completeness, or cardinality).

I have the following candidate, though:

$$\forall x\exists y\exists z \quad \Bigg(x=\bigg(\sum_{j=0}^y 1\bigg)\cdot i\bigg(\sum_{j=0}^z 1\bigg)\Bigg).$$

Where $i(\alpha)$ denotes the multiplicative inverse of $\alpha.$

I'm just not sure that this idea of summing "$y$ times" or summing "$z$ times" is actually doable in first order logic. I'm also not sure if somehow I need to clarify that $y$ and $z$ need be integers (and if that's doable in FOL).

Any corrections or clarifications to my candidate are welcome, as well as maybe another example I'm just not thinking of.

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Is $\sigma$ a first order expression? I don't think it is. Note, since you are working first order $\exists z$ means $z$ is a rational number, so what does $\sum_{j=0}^z$ even mean? – Thomas Andrews Oct 3 '12 at 16:17
The attempted example is in principle fixable, but with a hideous amount of work, and reference to a paper by Julia Robinson on definability of the integers in the rationals. Effectively, it is not the right way to go. A very weak consequence of completeness, as in my answer, works quickly. – André Nicolas Oct 3 '12 at 16:25
@AndréNicolas Thanks, that answers a quesiton I was just wondering: if the integers were even definable in the first-order rationals. – Thomas Andrews Oct 3 '12 at 17:19
@ThomasAndrews: It was a nice result. An unsolved problem is whether the integers are existentially definable. That would settle Hilbert's 10th Problem for the rationals, a famous open problem. – André Nicolas Oct 3 '12 at 17:26
up vote 6 down vote accepted

Hint: The square root of $2$ is irrational.

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Think about roots, do they always exist in $\mathbb Q$ and do they always exist in $\mathbb R$?

Well, square roots don't have to exist, but what about other orders?

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