When $n>1$, there is no ordering $<$ on $\mathbb{Z_n}$ such that...

Question

When $n>1$, there is no ordering $<$ on $\mathbb{Z_n}$ such that:

1. for all $[a]_n,[b]_n\in\mathbb{Z_n}$, we have exactly one of $[a]_n<[b]_n,[a]_n=[b]_n,$ or $[b]_n<[a]_n$;
2. if $[a]_n<[b]_n$ and $[b]_n<[c]_n$, then $[a]_n<[c]_n$; and
3. if $[a]_n<[b]_n$, then $[a]_n\oplus[c]_n<[b]_n\oplus[c]_n$ for each $[c]_n\in\mathbb{Z_n}$.

Proof: Let $n\in\mathbb{Z}$ with $n>1$. Assume for contradiction that $<$ is an ordering on $\mathbb{Z_n}$ that satisfies (1), (2), (3) as listed above. Consider the two cases $[1]_n<[0]_n$ or $[0]_n<[1]_n$. These are the only cases because $[0]_n$ and $[1]_n$ are not equal to each other, so either $[1]_n<[0]_n$ or $[0]_n<[1]_n$.

• Case 1: Let $[1]_n<[0]_n$. Then by (3), we have $[k-1]_n<...<[2]_n<[1]_n<[0]_n$. Then by (2), we have $[k-1]_n<[0]_n$. Adding 1 on both sides gives $?<[1]_n$. Thus, a contradiction.
• Case 2: Let $[0]_n<[1]_n$. Then by (3), we have $[0]_n<[1]_n<[2]_n<...<[k-1]_n$. Then by (2), we have $[0]_n<[k-1]_n$. Adding 1 on both sides gives $[1]_n<?$. Thus a contradiction.

The above proof is using the hints that was given to me by my professor. What I am concerned on is how this gives a contradiction, since the question marks should be $k$s, but I am not sure how that is a contradiction? If the questions marks were 0s it would make sense to me, but I am not sure if that is correct.

Answer 1

The $k$s should be $n$s. You're proving (via repeated addition) that $[n-1]_n \lt [0]_n$ or $[0] \lt [n-1]_n$. The point is that since addition is monotone, the fact that you can use addition to cycle from $[1]$ to $[0]$ means that $[1]$ can be neither less than nor greater than $[0]$, and since they're also not equal, we have our contradiction.