I have issues with proving the next things:
Let $P = ⟨N,\le⟩$ the poset of the natural numbers with the standard order. Consider the map $f : N → N$ defined by $f(n) = 3n$. Is $f$ residuated?
Let $P = ⟨R,\le⟩$ be the poset of the real numbers with the standard order. Consider the map $h : R → R$ defined by $h(x) = the \ greatest \ natural \ number \ \le x$ Is $h$ residuated?
I know that a map is residuated if it is monotone and for every $y \in P \ max\{x \in P: f(x) \le y \}$ exists. Now, my problem is reduced to proving that $f$ and $h$ are monotone and the max exists. For the second thing, I know that every non-empty set of natural numbers that is bounded from above does have a maximum. Somehow I think I need to prove that my sets are bounded from above, but I don't know how to procede, Im a bit lost here, so id really appreciate your input.
Fix some $y \in N$ and consider the set $A = \{x \in N \mid f(x) \leq y\}.$ We want to show that $A$ is bounded from above. Since $f$ is monotone it follows that if $x \notin A$ then $\forall z \in N : x \leq z \Rightarrow z \notin A$ (prove it by contraposition). But $y \notin A \rightarrow \forall z \in N : y \leq z \Rightarrow z \notin A$. It means that $y$ is one of the upper bounds of $A$.
Here we can't use the statement: every set of natural numbers that is bounded from above does have a maximum, because $A = \{x \in R \mid h(x) \leq y\}$ is the set of real numbers. Consider $y = 1$. In this case what is $A$? Does it have a maximum element? (Hint: $2 \notin A$, what about $x < 2$?).