# Thread: proof x/(-y) = (-x)/y

1. ## proof x/(-y) = (-x)/y

I have been asked to proof this using the axioms for real numbers. Can anyone help me get on the right track?

2. ## Re: proof x/(-y) = (-x)/y

I think it's useful to write:
$\displaystyle \frac{x}{-y}=\frac{-x}{y} \Leftrightarrow (-y)\cdot (-x)=x\cdot y$
Now use the fact the multiplication is commutative and associative for real numbers.

3. ## Re: proof x/(-y) = (-x)/y

Okay I think i see the point. Using the commutative and associate law you end up with xy= xy which is of course true, so equation one holds. Thanks

4. ## Re: proof x/(-y) = (-x)/y

Though I would like to ask you one question. In the proof you are assumming to begin with that the statement is actually true and show that this leads to an equation that holds. However is this proof as valuable as if you had no assumed the statement to be true at first and then prooved it to be true with the axioms of real numbers (I have no idea how you would do this, but all algebraic rules I have seen have been prooved this way - e.g. x*0=0 prooved using x+0 = x = x *1 = x*(1+0) = x*1 + x*0 = x + x*0 => x*0 = 0)

5. ## Re: proof x/(-y) = (-x)/y

You need to reverse the steps for a formal proof. Something like the following should do:

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

So x/(-y) = x/(-y) * 1 = x/(-y) * y/y = xy/[(-y)(y)] = (-y)(-x)/[(-y)(y)] = (-y)/(-y) * (-x)/y = 1*(-x)/y = (-x)/y.