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I study model theory and I have questions about relations which are definable in a structure or not. I found three examples from exercises and i want to do them:

Is the relation $<$ on $\Bbb{Q}$ definable in the structure $(\Bbb{Q},+,\cdot,0,1)$ that is does there exists a formula $\phi=\phi(x_0,x_1)$ sucht that for all $p,q$ in $\Bbb{Q}$, $p if and only if $(\Bbb{Q},+,\cdot,0,1)$ realized $\phi[p,q]$ ?

Is the relation $<$ on $\Bbb{Q}$ definable in the structure $(\Bbb{Q},+,0,1)$ ?

Is the relation $+$ on $\Bbb{Q}$ definable in the structure $(\Bbb{Q},<,0,1)$ ?

I have done this already for the integers with the successor function, but I don't know how to do this in this three cases. I think the first relation is definable, but the other two not. Can someone help me? Thank you :)

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    Well, in $(\Bbb Z,+,\cdot)$ it is definable by using the fact that every positive integer can be written as the sum of $4$ squares. This can be extended to the rationals, too ($\exists$ pos.integer $b$: $\ x\cdot b$ is pos.integer) \iff x>0).2012-12-01

2 Answers 2

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For the first, use Lagrange's Four Square Theorem, as others have said.


For the second, I believe that you can show that the structure $\langle \mathbb{Q} \times \mathbb{Z} , (0,0) , (1,0) , \hat{+} \rangle$ where $\hat{+}$ is defined by $(m,i) \mathop{\hat{+}} (n,j) = (m+n,i+j)$ is an elementary extension of $\langle \mathbb{Q} , 0 , 1 , + \rangle$ (with the obvious embedding). The function $f : \mathbb{Q} \times \mathbb{Z} \to \mathbb{Q} \times \mathbb{Z}$ defined by $f ( m,i) = (m,-i)$ is an automorphism of $\langle \mathbb{Q} \times \mathbb{Z} , (0,0) , (1,0) , \hat{+} \rangle$, and should be enough to witness that $<$ is not definable. (This makes use of the fact that if $\varphi(x,y)$ defines a linear ordering in $\langle \mathbb{Q} , 0 , 1 , + \rangle$, then it defines a linear ordering in all elementary extesnions of $\langle \mathbb{Q} , 0 , 1 , + \rangle$.)


For the third, note that any strictly increasing bijection $f: \mathbb{Q} \to \mathbb{Q}$ satisfying $f(0) = 0$ and $f(1) = 1$ is an automorphism of $\langle \mathbb{Q} , 0 , 1 , < \rangle$. Just pick any of these which is not linear to witness that $+$ is not definable in the structure.

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    I got it. Thanks for your time.2016-05-10
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1) Using Lagrange's four square theorem $a

2) How can you distinguish $(\mathbb Q,+)$ from $(\mathbb Q[i],+)$?

3) Note that $x\mapsto\begin{cases}2x&x\le \frac13\\\frac12(x+1) &x\ge\frac13\end{cases}$ is an automorphism of the ordered set $(\mathbb Q,<)$.

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    @Mathbb: I should probably point out that $(\mathbb C,+)$ and $(\mathbb R,+)$ are isomorphic. The dimension argument is incorrect here.2012-12-01