A simple Intro to Real Analysis question

[Jamin2112]

Homework Statement

I'm asked to prove that

If F is an ordered field, then the following properties hold for any elements a, b, and c of F:

(a) a<b if and only if 0<b-a
(b) ...
...

Right now I'm working on (a)

Homework Equations

We're supposed to draw from the basic properties (closure, associativity, commutativity, etc.) and also from the following definition.

Definition 1.4. A field F is ordered if it has an ordering < so that:

For all a, b in F, exactly of of these holds: a<b, a=b, a>b.
For all a, b in F, if a<b, then a+c<b+c.
For all a, b in F, if a>0 and b>0, then a+b>0 and ab>0.

The Attempt at a Solution

So far I wrote:

(a) Assume a < b.
Using Definition 1.6, we can add c to both sides of the inequality. Let c = -a, the additive inverse of a, so that the left side of the inequality equals 0.
0 < b + (-a)

But I'm wondering, is it a "given" that + (-a) = -a? I'm wondering whether I can just change this to b - a to complete the proof (actually, half the proof because we're dealing with an "if and only if" proof) or if I need to draw from some property.

 

[ZioX]

What you're asking is if the additive inverse of a is the same as the product of the additive inverse of 1 and a.

0=0*a=(1-1)a=a*1+(-1)*a=a+(-1)*a

Hence, -a = (-1)*a

This is true for any ring.

 

[Jamin2112]

What you're asking is if the additive inverse of a is the same as the product of the additive inverse of 1 and a.

0=0*a=(1-1)a=a*1+(-1)*a=a+(-1)*a

Hence, -a = (-1)*a

This is true for any ring.

Yeah, but I just wondering whether it's a "given" that +(-a) = (-1)*a = -a, or whatever. I want to make my proof as concise as possible.

 

[JThompson]

The closest thing in the field properties is "for all a, there exists -a such that a+(-a)=0".

If you aren't allowed to assume anything beyond the basic properties of an ordered field, I'd say that you can't assume +(-a) = -a.