- #1

zenterix

- 693

- 83

- Homework Statement
- Using field axioms, definition of ordered field, and some properties that ensue, prove the property that if ##x<0## and ##y<z## then ##xy>xz##.

- Relevant Equations
- A field is a set F with two operations (called addition and multiplication) which satisfy so-called "field axioms". These axioms are properties the operations satisfy.

These axioms lead to certain properties

The properties above apply to all fields.

We can define a more specific type of field, the

**ordered field**

And the following properties follow from this definition

My question is about the proof of (c).

**My initial proof was**

Using b) with ##z=0## we have that if ##x>0## and ##y<0## then ##xy<0##.

Now assume ##x,y,z\in F## with ##x>0## and ##y<z## for a general ##z## in ##F##.

Then, ##(-y)+z>(-y)+y=0## by property (i) of ordered fields (1.17). Thus ##z-y>0##.

Then, ##x(z-y)<0## and thus

##xz=x(z-y)+xy<0+xy=xy##

where again we used property (i) of ordered fields.

**Rudin uses the following proof**

By (a), (b) and Proposition 1.16(c),

##-\left [ x(z-y)\right ]=(-x)(z-y)>0##

so that ##x(z-y)<0##, hence ##xz<xy##.

In more steps,

We start with ##-\left [x(z-y)\right ]## and by 1.16c this equals ##(-x)(z-y)##. This is larger than zero because of property (ii) of ordered fields.

But then ##x(z-y)<0## by part (a) and so ##xz=x(z-y)+xy<0+xy=xy##, where again we have used property (i) of ordered field.

Thus, ##xz<xy##.

**I find that though these proofs are all simple they aren't completely trivial because I think it is easy to use assumptions that have not been proved yet.**

My question is if my initial proof is correct.

My question is if my initial proof is correct.