Define a relation $\propto$ on the natural numbers $\mathbb{N}$ by declaring that for $x, y \in \mathbb{N}$, $x \propto y \iff (x=y)$ or $(3x \leq y)$
a) Show that $\propto$ is a partial order on $\mathbb{N}$
b) Determine whether or not the subset $\{2, 6, 8\}$ of $\mathbb{N}$ is totally ordered with respect to $\propto$. Explain your answer.
c) Determine whether or not the poset $(\mathbb{N}, \propto)$ has a least element.
For part a), I showed that it was reflexive, transitive and antisymetric.
For part b) I determined that the subset was NOT totally ordered because 6 $\not \propto 8$ and $8 \not \propto 6$
I'm unsure about part c). My guess is that there is no least element because $1 \not \propto 2$ so $1$ cannot be the least element. However $\forall n \in \mathbb{N}$ with $n>1$, $n \not \propto 1$, so there is no number $n \not = 1$ which is the least element.
But I'm sure I'm phrasing this really badly...
You have the right idea.
Suppose $n \ne 1$ is a least element under $\propto $. Since you assume $\Bbb N = \Bbb Z^+$, then $n \ge 2$. Then $n \propto 1$ in particular. However, $n \ne 1$, and since $n \ge 2$, then $3n \ge 6$; thus $3n \not \le 1$ and $n \not \propto 1$
So the only remaining candidate is $n=1$. However, $1 \not \propto 2$ because $3 \cdot 1 \not =3\le 2$.
Thus, all possible candidates have been removed, and $\propto $ has no least element in $\Bbb N$. (Though if you were to include zero, then $0$ would be the least element, since $0 \cdot x = 0 \le y$ for all $y \in \Bbb N$.)
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.