In the Principles of Mathematical Analysis, 3rd ed. by Rudin, an ordered field is defined as:
"An ordered field $\mathbb{F}$ is a field $F$ which is also an ordered set, such that for all $x, y, z \in \mathbb{F}$:
- $x+y<x+z$ whenever $y<z$
- $xy>0$ if $x>0, y>0$."
The idea of these additional axioms (apart from the field axioms for addition, multiplication and distributivity) is that it tells us what happens to the relation or ordering between two elements of the field when any of the two binary operations are applied.
I wanted to know why the second axiom is defined with respect to the additive identity. Why not any arbitrary member of the field? As in, why not define it as $xy>a$ if $x>a, y>a; a \in \mathbb{F}$?
In more abstract terms, if I create my own field with two binary operations, will ordering relative to one of the operations always be with respect to the identity element of the other binary operation? Is this a consequence of the field axioms?
Has it something to do with the following consequences:
- $0x=0$ $\forall$ $x \in \mathbb{F}$ (multiplying by the additive identity is always $0$)
- $(-x)y=-xy$ (multiplying two objects on opposite sides of the additive identity always yields an object 'lesser than' the additive identity)
- $(-x)(-y)=xy$ (multiplying two objects on the same side of the additive identity always yields an object 'greater than' the additive identity)
For some context on the kind of explanation I'm seeking, I had a similar question about why the additive identity was preferentially excluded when defining the multiplicative identity and multiplicative inverse. As in, if I wanted to create my own field (an abelian group with two binary operations and a distributive law that binds the two operations), would I always have to exclude one of the identities? This is how I resolved it.
The exclusion is because of the way distributivity is defined to place multiplication over addition. Specifically, the field axiom for distributivity demands that $x(y+z)=xy+xz$, and not $x+yz=(x+y)(x+z)$.
The consequence of our definition is that $$0x + 0x = (0+0)x,$$ or $$0x + 0x = 0x,$$ meaning that $0x = 0$, $\forall$ $x \in \mathbb{F}$.
Thus, multiplication by the additive identity always returns the additive identity, so multiplication cannot be an invertible binary operation if $0$ is included. A similar resolution would have been necessary if we had defined the distributive law otherwise, where we would have excluded $1$ from our definition of the additive identity and additive inverse.
from Hot Weekly Questions - Mathematics Stack Exchange
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