Well-founded relation, well-order.

Question

On the set $\mathbb{N}\cup \{0\}$ we define the relation of strict divisibility with $$ a \text{ strictly divides } b \Leftrightarrow a | b \text{ and } a \neq b.$$ Do we get a well-founded relation? Do we get a well-order?

I'm stuck with proving that there doens't exist an infinite descending chain or a cycle. Is there any other type of way to prove that?

Thanks for the help.

There was a slightly different question asked in Well-founded ordering on natural numbers, but it didn't really help me.

Answer 1

Note that $a\text{ strictly divides }b\implies a<b$ (for $a, b\neq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.

Answer 2

For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $\mathbb N\cup\{0\}$ then can we always find an element $n\in A$ such that $n$ has no $R$-predecessors?

Here $m$ is a predecessor of $n$ if $mRn$.

Things are clear if $0\notin A$ (you can just take the smallest element of $A$).

If $A=\{0\}$ then you can take $n=0$.

If $0\in A$ and $A$ contains more elements then you can take the smallest element of $A-\{0\}$.

We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.