An order relations exercise I just did. I think it's fine, but the second proof felt a bit too wordy or discursive, instead of going straight to the point with brief and accurate statements. How could I improve that?
Over $\mathbb{N}^*$ is defined the relation of order $S$:
$$aSb \iff \exists k \in \mathbb{N} : b = ak$$
Is $S$ or total order?
No. Observe that $\lnot 2 R 3 \land \lnot 3R2$.
Demonstrate that $a = 1$ is a minimal and first element.
If we always have $a = 1$, then the relation would always check for
$$\exists k \in \mathbb{N} : b = k$$
Since $b \in \mathbb{N}^*$, then $b \in \mathbb{N}$, so there will always be a $k \in \mathbb{N}$ that fulfils $b = k$. Thus, $a = 1$ relates to all elements in $\mathbb{N}^*$, which means that $a$ is the first element.
Since it is the first element, it is also a minimal element.
Look at the definition for "minimal element". It says that $a$ is minimal if and only if whenever $bSa$ then $b=a$. So $1$ is minimal if and only if whenever $bS1$ we have $b=1$. But $bS1 \iff \exists k\in\mathbb{N}:1=kb$, and the only factor of $1$ in $\mathbb{N}^*$ is $1$.