How can I prove that $(ℤ, ≺)$ is not isomorphic to $(ℕ, ≤)$

Question

We define the relation $≺$ between pairs of integers like this: $n≺m$ is true if and only if one of the following conditions holds:

a) $0≤n≤m$

b) $0≤n$ and $m<0$

c) $n<0 , m<0$ and $|n|≤|m|$

where $n,m ∈ ℤ$.

How can I prove that $(ℤ, ≺)$ is not isomorphic to $(ℕ, ≤)$? (I assume that $≺$ is a partial order).

My thoughts: I need to prove that no function $f: ℤ →ℕ$ is an order isomorphism for these orders.

Answer 1

Hint for an easy way out: Does each ordered set have a greatest element?

Edit: Sorry, I did misread part (c) of the definition. There's a related tactic that will work, though. The ordered set $(\mathbb Z, \prec)$ has two elements without immediate predecessors: $0$ and $-1$. The ordered set $(\mathbb N, \leq)$ has only one such element.