I'm trying to think of a simple example of a two coordinate $(a,b)\in R$ relation which is reflexive, transitive, but not symmetric and not antisymmetric over $\mathbb{N}$ (meaning $R\subseteq\mathbb{N}\times\mathbb{N}$).
I can't seem to think of one. I would be glad to see some suggestions without actually proving them. I just struggling to think of an example.
I'm not sure I can think of an intuitive mathematical example that violates both symmetry and antisymmetry, but there are certainly small artificial relations.
Consider $\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4)\}$ over $\{1,2,3,4\}$. It is not symmetric because $3\sim4$ but not $4\sim3$ and it is not antisymmetric because $1\sim2$ and $2\sim1$ but $1\neq2$.
If you want to extend that to all of $\mathbb N$, you can just do $\{(i,i)\mid i\in\mathbb N\}\cup\{(1,2),(2,1),(3,4)\}$ for the same reason.
Actually, almagest did inspire me to think of a less contrived example over $\mathbb N$: $$R=\left\{(a,b)\in\mathbb N^2\mid \left\lfloor\frac a2\right\rfloor \le \left\lfloor\frac b2\right\rfloor\right\}$$
The question asks to find a preorder on $\mathbb{N}$ that is neither an equivalence relation nor a partial order. One such relation is the relation $R$ where $(m,n) \in R$ iff $m$ and $n$ are both even, or $m$ and $n$ are both odd, or $m$ is even and $n$ is odd. This is not an equivalence relation because, assuming that the natural numbers include zero, $(0,1) \in R$, but $(1,0) \not\in R$. It is also not a partial order, because $(2,4)$ and $(4,2)$ are both in $R$, for example.
