Real analysis using field axioms to prove

Question

Prove that $(-x)(-y)=xy$ using field axiom theorems

My attempt

I know that $-(xy)$ is the additional inverse for $xy$, and I know $xy + -(xy) = 0$ (equation $1$)

I assumed that $(-x)(-y)=xy$ and substitute into eqn $1$ and I get

$(-x)(-y)+ -(xy) =0 $

so now I just have to show that this equation is still true to justify my claim that $xy=(-x)(-y)$

But how do I move on from here??

You can't prove (-x)(-y) = xy by assuming (-x)(-y) = xy. Could you list/link to the axioms you are allowed to use? — Piotr Benedysiuk, Mar 30 '16 at 10:02
https://kevinbinz.files.wordpress.com/2014/11/group-theory-group-marriage-1.png — Wang Jia Le, Mar 30 '16 at 10:30
See also: Simple proof using only field axioms: $(-x)(-y) = xy$ How to show that $(-a)(-b)=ab$ holds in a field? — Martin Sleziak, Nov 11 '16 at 10:02

score -1 · Answer 1 · answered Mar 30 '16 at 10:07

-1

by associativity and commutativity of multiplication we get

$(-x)(-y) = (-1.x)(-1.y) = (-1.-1)(x.y)$

now remains to prove $-1.-1 = 1$

consider $-1 + -1.-1 = -1.1 + -1.-1 = -1(-1 + 1) = 0 $

answered Mar 30 '16 at 10:07

inquisitive

1

can i say that -x is equal to (-1)(x)? , i don't think you can do that since it's not in the axioms, probably requires a proof? – Wang Jia Le Mar 30 '16 at 15:16
yes you can prove it, since x + -1.x = x.1 + -1.x = x(1+ -1) = 0 – inquisitive Mar 30 '16 at 18:12

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