1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.

    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


    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Fields Question, help needed. Basic proof.
  1. Help with a basic proof (Replies: 15)