# Prove that two structures are elementarily equivalent (Logic)

in an exercise I was asked to prove that

(a) The structures $(\mathbb{R}^+,1,\cdot)$ and $(\mathbb{R},0,+)$ are elementarily equivalent.

(b) The two structures $(\mathbb{N},<)$ and $(\mathbb{Q},<)$ are not elementarily equivalent.

(c) The structure $(\mathbb{R},0,1,+,\cdot )$ is not an elementary substructure of the structure $(\mathbb{C}, 0,1,+,\cdot)$.

Can I check if what I am doing is correct?

Hope it is okay that I put all three questions together, since they are sort of related.

Sincere thanks!

(a) For (a), I tried to show that the two structures are in fact isomorphic.

Let $\mathcal{N}=(\mathbb{R}^+,1,\cdot)=(\mathbb{R}^+,J)$.

Let $\mathcal{M}=(\mathbb{R},0,+)=(\mathbb{R},I)$.

Define $e:\mathbb{R}\to\mathbb{R}^+$, $e(x)=e^x$.

Then $e$ is a bijection.

Also we have, $e(I(c_0))=e(0)=1=J(c_1)$

Denote $I(F)=+$, $J(F)=\cdot$

For each $a_1,a_2,a_3\in M$, we have $I(F)(a_1,a_2)=a_3 \iff a_1+a_2=a_3 \iff e^{a_1+a_2}=e^{a_3} \iff e^{a_1}\cdot e^{a_2}=e^{a_3} \iff J(F)(e(a_1),e(a_2))=e(a_3)$

Therefore, $e$ is an isomorphism. Therefore, $\mathcal{M}$ and $\mathcal{N}$ are isomorphic, and hence elementarily equivalent.

(b) Consider $\varphi= (\forall x_1 (\exists x_2 (P_< (x_2,x_1))))$.

Then $(\mathbb{N},>)\not\models \varphi$ but $(\mathbb{Q},<)\models \varphi$.

(c) Consider $\varphi=(\forall x_1 (\exists x_2 (F_\times (x_2, x_2)=x_1)))$

Then, $(\mathbb{R},0,1,+,\cdot)\not\models\varphi$ (since the square root of $-1$ is not in $\mathbb{R}$) but $(\mathbb{C},0,1,+,\cdot)\models\varphi$.

• Looks good. But note that in $(b)$ you wrote $(\Bbb N,>)$ rather than $(\Bbb N,<)$. – Asaf Karagila Apr 20 '13 at 16:37

This is correct, except for the $(\mathbb N,>)$ in (b), as Asaf mentioned.