This is the same question as Q2 of Need help with understanding the proof for Division Algorithm, from the book *Contemporary Abstract Algebra* by Joseph A. Gallian
"Q2: I don't understand how 0∉S implies a≠0. We could still have a−bk>0 even if a=0 by choosing k<0."
The moderator would not allow me to post my request there. I did not understand the answer given and would appreciate if someone could expound the answer in more detail.
This answer was: "2 If a=0, then we can take k=0 to see that 0=a−b⋅0∈S."
Below are some details on my thought process of trying to understand this answer:
I still can't see the logic on how this answer answers the question. Are we using the argument of modus tollens? I'm reading the answer logically as:
If a=0, then there exists a k such that 0=0−b⋅0∈S.
So not [there exists a k (=0) such that 0=0−b⋅0∈S] which I can't for the life of me determine the negation. Maybe? (For all k, 0=a−b⋅k∉S)? which is true because we assumed 0∉S and thus we can then conclude a≠0.
It seems Brian (in the original post answer) is saying we use the contrapositive of "If a=0, then we can take k=0 to see that 0=a−b⋅0∈S" to arrive at the solution, but I can't see how to do so with logic.
We want to prove that $0 \notin S \implies a \ne 0$.
Hence, it is equivalent to prove that if $a=0 \implies 0 \in S$.
That is we want to prove that $a=0 \implies 0 \in S=\{ a-bk|a -bk \ge 0, k \in \mathbb{Z}\}$.
That is we want to show that we can pick a value $k\in \mathbb{Z}$, such that $a-bk=0-bk=-bk = 0$.
A possible choice is to let $k=0$ and this would have shown that $0 \in S$.