Proof using axioms for a field


by msbell1
Tags: axioms, fields, halmos, proof
msbell1
msbell1 is offline
#1
Nov20-10, 05:05 PM
P: 22
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!
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JonF
JonF is offline
#2
Nov20-10, 05:27 PM
P: 617
Do you know what (-1)(-1) equals? If not could you prove it equals 1?
msbell1
msbell1 is offline
#3
Nov20-10, 05:55 PM
P: 22
Thanks for the reply!
I forgot to include that multiplication is distributive with respect to addition:
a(b+c) = ab + ac (which I will use later)

Would it be ok to first rewrite (-1)a(-1)b as (-1)(-1)ab and then start working separately on the (-1)(-1) part?

like continuing the proof as

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

Then forgetting about a and b and working on (-1)(-1):

(1 + (-1))(-1) = 0 (1 + -1 = 0 (-1 is unique additive inverse of 1 such that 1 + -1 = 0) and I already had to prove that 0a = a0 = 0)

(-1)(1 + (-1)) = 0 (by commutativity of multiplication)

(-1)(1) + (-1)(-1) = 0 (multiplication is distributive w.r.t. addition)

-1 + (-1)(-1) = 0 (because 1 is multiplicative identity)
1 + -1 + (-1)(-1) = 1 + 0
(1 + -1) + (-1)(-1) = 1 + 0 (because addition is associative)
0 + (-1)(-1) = 1 + 0 (because 1 + -1 = 0)
(-1)(-1) + 0 = 1 + 0 (because addition is commutative)
(-1)(-1) = 1 + 0 (because a + 0 = a)
(-1)(-1) = 1 (because a + 0 = a)

Then plugging this result into

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

Is that an ok proof? How would a mathematician prove this? Thanks again for your reply.

JonF
JonF is offline
#4
Nov20-10, 06:02 PM
P: 617

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
msbell1
msbell1 is offline
#5
Nov22-10, 09:33 AM
P: 22
Thanks a lot--especially for the outline!


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