2023-06-19 20:24:33 -07:00

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\section{Equivalence (Bonus)}
\generic{Notation:}
Let $S$ be a structure and $\varphi$ a formula. \par
If $\varphi$ is true in $S$, we write $S \models \varphi$.
\definition{}
Let $S$ and $T$ be structures. \par
We say $S$ and $T$ are \textit{equivalent} and write $S \equiv T$ if for any formula $\varphi$, $S \models \varphi \Longleftrightarrow T \models \varphi$.
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{N} ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{R} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$
\vfill
\problem{}
Show that $
\Bigl(\mathbb{Z} ~\big|~ \{ +, 0 \}\Bigr)
\not\equiv
\Bigl(\mathbb{Z}^2 ~\big|~ \{ +, 0 \}\Bigr)
$
\begin{solution}
All of the above are easy, but the last one can take a while. \par
The trick is to notice that $\mathbb{Z}$ has two equivalence classes mod 2, while $\mathbb{Z}^2$ has four.
\end{solution}
\vfill