Definable relations

model-theory

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 $\Bbb{Q}$, $p<q$ 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 :)

Solution 1:

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.