Reverse order of a ring When we think of an ordered structure with an order $\le$ we assume there is an opposite order $\le^{op}$ as well:

*

*$a \le^{op} b \iff b \le a$.

I would suggest this is a fundamental principle for all ordered structures:
If a structure is ordered one way $(\le)$ it is also ordered the opposite way $(\le^{op})$.
The reverse order principle works on the simplest algebraic structures:

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*An ordered set with a unary operation $f(a) = b$:
$a \le b \implies f(a) \le f(b)$ or
$a \le^{op} b \implies f(a) \le^{op} f(b)$ for any $a,b$;

*An ordered semigroup with a binary operation $f(a,b) = c$:
$a \le b \implies f(a,c) \le f(b,c) \land f(c,a) \le f(c,b)$ or
$a \le^{op} b \implies f(a,c) \le^{op} f(b,c) \land f(c,a) \le^{op} f(c,b)$ for any $a,b,c$;

*An ordered group (same as an ordered semigroup).

Now, let's apply the principle to an ordered ring $R(+,\cdot,0,1,\le)$.

*

*The compatibility rule for addition is fine:
$a \le b \implies a + c \le b + c \land c + a \le c + b$ or
$a \le^{op} b \implies a + c \le^{op} b + c \land c + a \le^{op} c + b$ for any $a,b,c$;


*But what about the compatibility rule for multiplication?
$0 \le a \land 0 \le b \implies 0 \le a \cdot b$ or
$0 \le^{op} a \land 0 \le^{op} b \implies 0 \le^{op} a \cdot b$.
Checking the last statement on the ring of integers with the regular order and operations:
Taking $a = -1$ and $b = -1$: $0 \le^{op} -1 \land 0 \le^{op} -1 \implies 0 \le^{op} (-1) \cdot (-1) = 1$ (false).
It looks like the reverse order principle does not work on rings:
If a ring is ordered one way $(\le)$ it may not be ordered the opposite way $(\le^{op})$.
Is this correct?
If yes, why do we ignore the reverse order principle for rings?
Are there (non-standard) definitions of an ordered ring that accept the opposite order?
 A: The reverse order isn't consistent with the compatibility axioms of the order with the ring structure.
Suppose you reverse the order in a ring with identity.
Then $0\leq^{op} -1$, and from the multiplicative axiom $0\leq^{op} (-1)(-1)=1$.
From the additive axiom and $0\leq^{op} -1$, adding $1$ to both sides would yield $1\leq^{op} 0$, and from the previous deduction $1=0$.  Presumably this isn't the case for the ring you started with, so you have a contradiction.

Note: I think what's written at the wiki link you cited causes considerable confusion. In my opinion, it's better written at opposite category where they write "$x ≤^{op} y$ if and only if $y ≤ x$."  I believe this was the intent of what's written at the partial order wiki, because partial orders can be viewed as categories, and the reverse order ought to be the opposite category.
I think the reason what's written at the order wiki is confusing is that most people will interpret $a\leq b$ and $b\geq a$ as meaning exactly the same thing, that $b$ is the bigger thing.  But the whole point of the reverse order is to make big things small and small things big. If we were to follow in the notation suggested at the order wiki, we would say this in the opposite order "$b$ used to be the bigger one, so now in the new order $a$ is bigger, and I will write $b\geq a$."
If you just view the thing in between as a separator, and always read it as "the thing on the left is smaller" then it is a correct description of the reverse order. I just think it is extremely confusing to reverse both the inputs and the direction of the symbol.
