1. ## Ordered Integral domain...

Hello,

thanks a lot!

A) prove that if a,b,c is in the ordered integral domain D and its given that a<0 and b<0 then prove that ab>0... have no idea how to approach!

B) prove that if a,b is in the ordered integral domain (D^p) and its given that a>b then prove that a^2>b^2...
--------> I know a little bit of the solution for this particular problem but not sure if this is the correct way to approach it.
we know from the ordered integral domain theorem part (c) that "Exactly one of the following is true: a=b, a>b or b>a"
a > b and a > 0 imply a^2 > ab, and a > b and b > 0 imply ab > b^2.
so by the same theorem part (g), a^2 > ab > b^2 implies a^2 > b^2.

thanks for the help!

2. It should be an axiom that if a>0 and b>0, then ab>0.

If you can prove that (-a)(-b) = ab, then you are done: Since a<0 and b<0, we have -a>0 and -b>0 (have you proven this?). By the axiom, we then have that (-a)(-b) > 0, and since (-a)(-b) = ab, we are done.

As for why (-a)(-b) = ab, you can prove first that (-a)(-b) is equal to -(a(-b)), i.e. (-a)(-b) is the additive inverse of a(-b):

a(-b) + (-a)(-b) = (a + (-a))(-b) = 0(-b) = 0 [I need only check this from "one side", since addition is commutative].

Then you can continue to prove that a(-b) is equal to -(ab), i.e. it is the inverse of ab:

ab + a(-b) = a(b + (-b)) = a0 = 0.

In conclusion, (-a)(-b) is the inverse of the inverse of ab, and is as such equal to ab. Do you know already that -(-c) = c for all elements c?