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As stated, I'm to show that $2$ is not definable in $(\mathbb{Q},+)$.

I tried proving it by contradiction by showing that if $2$ were definable, then we could define $\mathbb{N}$ and multiplication over $\mathbb{N}$, which would be impossible because the automorphism $x\mapsto 2x$ does not preserve $\times$ over $\mathbb{N}$ but my definition of multiplication was flawed. Is there an easy trick to this problem?

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I thought that is a supplementary problem! :-) – Asaf Karagila May 9 '14 at 19:47
@AsafKaragila Yeah, it is. I already turned my portfolio in though. Now I'm just working on problems for fun. – Rustyn May 9 '14 at 19:47
up vote 5 down vote accepted

Hint: Show that any definable element in a structure $M$ is fixed by all automorphisms of $M$. Can you find an automorphism of $\mathbb{Q}$ which does not fix $2$?

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Yeah so I wrote that down. Assume $2$ is definable, that is $\mathcal{A}=\{2\}$ is a definable set. All one needs is an automorphism, (say $x\mapsto 2x$) that does not fix $2$. We have that $2\in \mathcal{A}$ if and only if $2+2 \in \mathcal{A}$. Obviously the automorphism $x\mapsto 2x$ does not fix $2$, but I was thinking that this argument was too simple. I guess I'll have to look at the invariance of automorphisms proof once more. – Rustyn May 9 '14 at 19:56

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