Order axioms stated in this book earlier:
First example:
Question 1
I can't get the statement: $$ \text{Since } 1_n = 1_n \times 1_n = (-1_n)\times(-1_n), \text{ either possibility implies that }1_n ∈ \mathbb{Z}_n^+ \text{ using }(O_1) $$
By ($O_2$) either a($1_n$) or b($-1_n$) ∈ $\mathbb{Z}_n^+$ by definition, but not both.
So what's meant by 'either possibility' and why this magic with $1_n = 1_n \times 1_n = (-1_n)\times(-1_n)$ is needed? What is it for?
Question 2
Words: "But this gives the same contradiction as before"
If I get it right, the contradiction was:
$\text{for } 1_n > 0_n, (n-1)_n > 0_n \text{ would then imply } 0_n = 1_n + (n-1)_n > 0_n$, $0_n > 0_n$.
But this is a logical contradiction, presumed '>' is already defined and works as order relation for $\mathbb{Z}_n$
But where is contradiction in $2_n=1_n+1_n ∈ \mathbb{Z}_n^+,3_n ∈ \mathbb{Z}_n^+, ..., (n-1)_n ∈ \mathbb{Z}_n^+$ ?
Consequences of these examples are used in the next example:
Sentence:
"As we argued for $\mathbb{Z}_n^+$, we always have 1 > 0 because $1=1^2=(-1)^2$
Question 3
How '1 > 0' follows from '$1=1^2=(-1)^2$'?
Question 1
He is trying to determine whether $1_n \in \Bbb{Z}_n^+$.
Are you happy that $1_n \times 1_n = 1_n$ and $(-1_n) \times (-1_n) = 1_n$?
If $1_n \in \Bbb{Z}_n^+$ then $1_n \times 1_n = 1_n \in \Bbb{Z}_n^+$ but that is not very exciting.
If $1_n \notin \Bbb{Z}_n^+$ then $-1_n \in \Bbb{Z}_n^+$ so $(-1_n) \times (-1_n) \in \Bbb{Z}_n^+$ but since $(-1_n) \times (-1_n) = 1_n$ this means that $1_n \in \Bbb{Z}_n^+$.
So, as he says, in either case we have proved that $1_n \in \Bbb{Z}_n^+$.
Question 2
The contradiction is that you have proved that $(n-1)_n \in \Bbb{Z}_n$ but $(n-1)_n = -1_n$ and $-1_n$ cannot be in $\Bbb{Z}_n^+$.
Question 3
See my answer to Question 1