Prove if ##x<0## and ##y<z## then ##xy>xz## (Rudin)

[zenterix]

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.

 

[PeroK]

Your proof looks correct.

 

[PeroK]

The way I would do this is simply to see c) as a corollary of b). If $x < 0$ then $-x > 0$ (a), hence $(-x)y < (-x)z$ (b), hence $-xy < -xz$ (1.16c), hence $xz < xy$. For this last step, it feels like you need another proposition: $x < y$ iff $-x > -y$