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Proof using axioms for a field 
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#1
Nov2010, 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 stuckfor 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! 


#2
Nov2010, 05:27 PM

P: 617

Do you know what (1)(1) equals? If not could you prove it equals 1?



#3
Nov2010, 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. 


#4
Nov2010, 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 


#5
Nov2210, 09:33 AM

P: 22

Thanks a lotespecially for the outline!



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