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Fields Question, help needed. Basic proof.

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose F is a field and a is an element of F . Prove that the additive inverse of a is
    unique (and so we may write it as -a). Justify each line of your proof
    in terms of the eld axioms


    2. Relevant equations

    The field axioms.
    http://mathworld.wolfram.com/FieldAxioms.html

    3. The attempt at a solution

    Given: (additive inverse) for every a in F, there exists a b in F, such that a + b = b + a = 0

    Proof

    a + b = b + a = 0
    a + b = 0 (additive inverse)
    b = -a (adding (-a) to both sides)
    a - a = 0 (substitute -a in place of b)


    This proof seems incomplete to me. Have I made assumptions anywhere?
    Anything anyone would change?

    Also, this is for a first year uni maths course.

    thank you in advance
     
  2. jcsd
  3. Sep 20, 2011 #2
    Well, when you say you add "-a" you are sort of assuming that which you are trying to prove. Forget about "-a" for the proof; the book only mentioned "-a" to give the motivation for wanting to prove this is true in a field. What you need to do is start off by saying something like:

    Let a be an element of F and let b and b' be additive inverses of a. So, a + b = b + a = 0 and a + b' = b' + a = 0.


    Now, using those relations and the field axioms, you need to prove that b = b'.
     
  4. Sep 20, 2011 #3
    Thank you for your reply!

    Okay, your method makes much more sense, and would prove uniqueness. But, without subtracting a to both sides, how can I re-write b = b' as -a?
     
  5. Sep 20, 2011 #4
    Until you complete this proof, forget that "-a" is something that you are familiar with; just act as though "-a" does not exist. This uniquiness thing is essentially telling you "hey, you can write additive inverses as '-a' just as you did with addition of numbers."

    So, I'd start by writting:

    b = b + 0 (by the definition of 0)
    = what does this equal?
    ...
    = b'

    Now, use what you know about b,b' and a to get the RHS of an equation to be b' as above.

    EDIT: One last thing to point out, if you have this equation:

    a+b = a+b'

    you are NOT allowed to do something like this (though it may be tempting):

    b + a + b = b + a + b' implies b = b'

    Why can't you do this? This "cancelation" idea is not an axiom of a field. It can be deduced, but you haven't proven that yet.
     
  6. Sep 20, 2011 #5
    Thank you again. Does this work?

    (1) a + b = b + a = 0
    (2) a + b' = b' + a = 0

    Let a = 0

    (1) b + (0) = 0
    [Zero additive]
    b = 0

    (2) b' + (0) = 0
    b' = 0

    b = 0 = b'
    Therefore, b = b'


    Please reply.
     
  7. Sep 20, 2011 #6
    No, this doesn't work. First of all, at best all this does is prove that 0 is the unique additive inverse of 0 (and whether you have proven that is VERY debateable.) You are proving that FOR ALL a in F, the additive inverse of a is unique. So, you are NOT allowed to say "a=0" (or a equals anything, for that matter.) ALL you can assume about a is that it is in F.

    Let me give you another hint:

    Let a be an element of F and let b and b' be additive inverses of a, that is a+b = b+a = 0 and a+b' = b'+a = 0. Then,
    b = b + 0 = b + (a + b') = ...

    Now, keep manipulating this until the RHS of the last "=" is " b' ". Then you have proven that b = b'. Now, once you do that, you are going to have to re-write it line-by-line and explain why you can get from one step to the other from the field axioms.
     
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