Was more information that necessary given in this exercise?

Question

I had the following exercise in an exam:

Question

Let $L$ be a first order language with equality, a binary function symbol, and a binary predicate symbol.

Let $I=(\Bbb Z, +, \leq), J=(\Bbb Z^2,+, \leq_{\Bbb Z^2})$ be two interpretations of $L$, where:

$$(a,b)+(c,d)=(a+c,b+d)\\ (a,b)\leq (c,d) \iff a<c \text{ or } (a =c \wedge b\leq d) $$ Decide whether $I\simeq J$.

Attempt

If $f:\Bbb Z \to \Bbb Z^2$ is an isomorphism is must have the property: $f(n)=nf(1)\forall n$.

Proof:

If $n=0$: $$f(0)=f(0+0)=f(0)+f(0) \rightarrow f(0)=2f(0)\rightarrow f(0)=0=0\cdot f(1)$$.

$$f(n+1)=f(n)+f(1)=nf(1)+f(1)=(n+1)f(1)$$

The proving $f(-a)=-f(a)$ gives the property for all integers.

Then I said that as $f$ is bijective, $(1,0)$ must have a preimage, so there's some $k: f(k)=(1,0)$ but $f(k)=kf(1)=(ka,kb)=(1,0)\implies kb=0$ and I also have that $k\neq 0$ as $f(0)=\vec 0$. So $b= 0$ . I then did the same with the point $(0,1)$ and concluded $f(1)=(0,0)$, which is absurd because $f$ was bijective. Therefore $I\not \simeq J$.

Is this proof ok? I keep wondering if it is because I never used relation $\leq$ for anything.

Answer 1

From your comment, $\simeq$ in your question denotes isomorphism. This means that you have been given too much information:

The two group structures are not isomorphic because $\Bbb{Z}^2$ is not a cyclic group while $\Bbb{Z}$ is.

The two order structures are not isomorphic because with the lexicographic ordering there are elements $x, y \in \Bbb{Z}^2$ such that there are infinitely many $z \in \Bbb{Z}^2$ with $x < z < y$, but for $x, y \in \Bbb{Z}$ this cannot happen.

Answer 2

Yes, that looks OK. You may want to be a little more specific about how you conclude $f(0)=(0,0)$, since $0$ is not a symbol of the language.

The two structures are so different that it is okay only to use part of them to conclude that they are not isomorphic. Conversely, you could also prove this looking only at $\le$ and ignoring $+$, such as by observing that even though $\le_I$ and $\le_J$ are both total orders, $J$ has a downwards closed proper nonempty subset with no maximal element, and $I$ doesn't.