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Let $A$ be the set of all $q$ in $\mathbb{Q}$ such that $q\leq0$ or $1\leq q$, and let $\mathcal{A}=(A,<)$ be a structure.

I have to show that 2 is not a definable element of this structure, e.g. there doesn't exist a formula $\varphi = \varphi(x)$ such that $\mathcal{A}\models\varphi[q]$ if and only if $q=2$.

I've been trying to use induction, but it doesn't seem to work..

EDIT: the first part of the exercise is: Prove that for all formulas $\varphi=\varphi(x), (A,<)\models \varphi[2]$ iff $(A,<)\models \varphi[3]$, but how would I prove such a statement?

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HINT: If $x$ is a definable element in a structure $\mathcal M$, then any automorphism of $\cal M$ must satisfy $f(x)=x$. To show that $2$ is not definable, find an automorphism of $\cal A$ such that $2\neq f(2)$.

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  • $\begingroup$ Ah, that seems logical. Could you maybe also give a hint on the edit I just made? $\endgroup$
    – konewka
    Sep 11, 2014 at 19:55
  • $\begingroup$ Recall that $\mathcal M\models\varphi(x)\iff\mathcal M\models\varphi(f(x))$. $\endgroup$
    – Asaf Karagila
    Sep 11, 2014 at 19:56
  • $\begingroup$ And then the automorphism $\psi:\mathcal{A}\rightarrow\mathcal{A}$ defined by $\psi(a)=\begin{cases} 2 && a=3 \\ 3 && a=2 \\ a && \text{else}\end{cases}$ would do the trick? $\endgroup$
    – konewka
    Sep 11, 2014 at 19:58
  • $\begingroup$ Is this an automorphism? Does it preserve the order of $\cal A$? $\endgroup$
    – Asaf Karagila
    Sep 11, 2014 at 20:12
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    $\begingroup$ Close, it's true for $\Bbb R$, but not for $\Bbb Q$ since $2$ is not in the range of $x^2$ when considering the interval defined on $\Bbb Q$. But you're getting closer. $\endgroup$
    – Asaf Karagila
    Sep 11, 2014 at 20:41

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