1) Let $M : =\mathbb{N} \cup \{-1,0\} $. We define the relation $R \subset M \times M$, whereas $\lfloor y \rfloor := max \{m \in \mathbb{Z} : m \leq y\}$ \
$ R :=\{ (x,y) \in M \times M: x < \lfloor \frac{y}{2} \rfloor \}$
1) Is R well-founded?
2) How does the answer of 1) change if we set $M : = \mathbb{Z}$?
To 1) I would say yes we know that $\mathbb{N}$ is well founded and $\{-1,0\}$ is also well founded so, $\mathbb{N} \cup \{-1,0\}$ should also be well-founded? To 2) I would say no, since $\mathbb{Z}$ is not well-founded.
Is this correct? Please correct me if not, any additional info would be helpful.
Here are some pointers.
To show that $R$ is well-founded on $M=\Bbb N\cup\{-1,0\}$, you need to show that every non-empty subset of $M$ has an $R$-minimal element. In other words, you need to show that if $\varnothing\ne S\subseteq M$, there is an $s_0\in S$ such that there is no $s\in S$ for which $s\,R\,s_0$, meaning that there is no $s\in S$ such that $s<\left\lfloor\frac{s_0}2\right\rfloor$. Try proving that for any $m,n\in M$, if $m\,R\,n$, then $m<n$; once you realize this fact, it’s not hard to describe an $R$-minimal element of any non-empty $S\subseteq M$.
As for the case $M=\Bbb Z$, does $M$ itself have an $R$-minimal element? Or can you prove that no matter what $m\in M$ you take, there is some $n\in M$ such that $n\,R\,m$? (Be a little careful here: $-3\,R\,-4$ even though $-3>-4$, because $-3<-2=\left\lfloor\frac{-4}2\right\rfloor$.)