## Proof using axioms for a field

Hi, I am trying to work through Finite Dimensional Vector Spaces by Halmos, and I am having some difficulty with the first problem on page two (the specific problem is included below). The last class I took involving formal proofs was linear algebra about 8 years ago, and I am very rusty, but I am trying to regain some of those skills now. Here is the problem:

1. The problem statement, all variables and given/known data
Prove that (-a)(-b)=ab (assuming that F is a field and that a, b, and c belong to F)

2. Relevant equations
A reminder of the axioms:
addition is defined in the usual way, and it is commutative (a + b = b + a), associative (a + (b + c) = (a + b) + c), 0 is the unique additive identity such that a + 0 = a, and -a is the unique additive inverse such that a + -a = 0
multiplication is defined in the usual way:
it is commutative (ab = ba), associative (a(bc) = (ab)c), 1 is the unique additive identity such that a1 = a, and 1/a is the unique multiplicative inverse such that a(1/a) = 1.

3. The attempt at a solution
In the previous part of this problem I was asked to prove that (-1)a = -a, so I will use that proposition to prove the next part. I start out by using that to write:

(-a)(-b)=(-1)a(-1)b

However, once I write this I am stuck--for the past day I have been thinking about this off and on and have not been able to make any progress, and I was wondering if anyone could suggest how I should proceed.

Thank you for the help!

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Do you know what (-1)(-1) equals? If not could you prove it equals 1?

## Proof using axioms for a field

yup looks good, so an outline would look like:

lemma 1: -a = (-1)a
lemma 2: (-1)(-1)=1
proof:
(-a)(-b) =(-1)a(-1)b
(-1)a(-1)b = (-1)(-1)ab
(-1)(-1)ab = 1ab
1ab = ab

 Thanks a lot--especially for the outline!

 Tags axioms, fields, halmos, proof