- #1

Jamin2112

- 986

- 12

## 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) ...

...

(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)

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.