Ordered field consequence proof

Question

Hey can anyone help me get started on this proof?

Prove OFC7: for elements x,y in an ordered field, (-x)(-y) = x*y.

I am allowed to use the following axioms:

F1. x+(y+z) = (x+y)+z; F2. x+y=y+x; F3. there is a specified element 0 such that x+0=x; F4. there is a specified element -x such that x+(-x)=0; F5. x*(y*z)=(x*y)*z; F6. x*y=y*x; F7. there is a specified element 1 such that x*1=x and 1 does not equal 0; F8. for any x not equal to 0, there is a specified element x^-1 such that x*(x^-1) = 1; F9. x*(y+z)=x*y+x*z. OFC1. If x+y=x+z, then y=z. OFC2. -(-x)=x OFC3. If x is not equal to 0 and xy=xz, then y=z. OFC4. x*0=0 OFC5. If x*y=0 then x=0 or y=0. 0FC6. (-x)*y=-(x*y)

That's true in any ring. See this math.SE answer for instance (and the question is likely a duplicate). — Zev Chonoles, Sep 07 '13 at 14:30

score 0 · Accepted Answer · answered Sep 07 '13 at 14:33

This is true in any field $F$ (not just ordered) and the proof is fairly standard.

Lemma: For all $x\in F$, $-x=(-1)x$.
Proof: $x+(-1)x=(1+(-1))x=0\cdot x=0=x+(-x)$ (by the distributive law) and then use cancellation.

It follows from the Lemma that $(-1)(-1)=-(-1)=1$ and hence $(-x)(-y)=(-1)x(-1)y=(-1)(-1)xy=xy$.

score 0 · Answer 2 · answered Sep 07 '13 at 14:37

0

It is true in any ring.

We know that $(-x)y=-xy$ since $xy+(-x)y=(x+(-x))y=0y=0.$

Now $-xy+(-x)(-y)=(-x)y+(-x)(-y)=(-x)(y+(-y))=-x*0=0.$ That means that $(-x)(-y)$ is the negative of $-xy$, thus it must be $xy$.

answered Sep 07 '13 at 14:37

Stefan Hamcke

27,733

Yeah the way to start is by using this (-x)(-y) + - (xy) = (-x)(-y) + (-x)y. Then after using the axioms (-x)(-y) = xy by F4 and F3. – Sep 07 '13 at 15:02

Ordered field consequence proof

2 Answers2