Proving (-a)(-b)=ab Using Field Axioms

[omri3012]

hall,

i need to prove by using the field axioms that:

(-a)(-b)=ab, i think i know how to this but I'm very insecure with using those

axioms cause i want to make sure I'm not using my intuition.

i tried something like: (-a)(-b)=(-1)(a)(-1)b=(-1)(-1)(a)(b)=ab and i guess it's wrong (in the formal way).

could someone show me how this mechanism works in this case?

thanks,

Omri

 

[snipez90]

Well the way you presented it, you actually used what you were trying to prove in the last equality.

Can you prove that -(ab) = (-a)(b)? The hint is that you'll need the distributive property. If you can prove this, then (-a)(-b) = ab is the exact same proof.

 

[omri3012]

so if i undrstood you correctly, i can write
-a(b+(-b))=0
-ab+(-a)(-b)=0
ab-ab+(-a)(-b)=0+ab
(-a)(-b)=ab
is that too much stepd for an answer?
thanks
Omri

 

[snipez90]

I think you are implicitly using the fact that (-a)b = -ab, but otherwise that looks fine.

 

[g_edgar]

You will have to cite which axiom you are using at each step.
You probably have them listed with some numbering, so use that.

 

[HallsofIvy]

You need to understand that, in the field axioms, "-a" does NOT mean (-1)(a). It means "the additive inverse of a". In order to prove that (-a)(-b)= ab, you need to show that "if x+ a= 0 and y+ b= 0, then xy= ab". You might start by looking at (x+a)(x+b)= 0(0)= 0.

(Yes, you can then show that if x+a= 0, x= -1(a) where "-1" is defined as the additive inverse of the multiplicative identity but I was talking about using the axioms.)