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Field Proofs (just needs revision)

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Thanks to everyone who has helped me so far - I'm very grateful.

    (1) Prove that the multiplicative inverse in any field is unique
    (2) Prove the cancellation law | ab = ac => b=c
    (3) Prove (-1)a = -a

    2. Relevant equations

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

    3. The attempt at a solution

    (#1)
    Let: a is an element of a field
    Let: b and c are multiplicative inverses of a, such that, (1) ab = 1, and (2) ac = 1

    Proof.
    Start with (1)

    ab = 1
    (ab)/b = (1/b)
    a (b/b)= (1/b)
    a 1 = (1/b)
    a = (1/b)

    (2)

    ac = 1
    (1/b)c = 1
    b((1/b)c)) = b(1)
    (b/b)c = b
    1c = b
    c = b

    QED

    (#2)

    Let: a, b are elements of F

    Let: z is the mutliplicative inverse of x

    Proof.

    ax = bx
    z(ax) = z(bx)
    z(xa) = z (xb)
    (zx)a = (zx)b
    1a = 1b
    a = b

    QED

    (#3)
    Let: a is in a field
    Let: -1 is the additive inverse of 1

    Proof.

    (-1)a = (-1 + 0) a
    (-1)a = (-1 + (-1 + 1)) a
    (-1)a = -1a + (-1a + 1a)
    (-1)a = -1a + (1(-a + a))
    (-1)a = -1a + (-a + a)
    (-1)a = 1(-a) + (-a + a)
    (-1)a = -a + (-a + a)
    (-1)a = -a + (a + (-a))
    (-1)a = (-a + a) + (-a)
    (-1)a = 0 + (-a)
    (-1)a = -a

    QED



    Thank you very much!
    Please be as critical as possible, I really want to learn.
    This is for a first year algebra class.

    J.Anthony
     
  2. jcsd
  3. Oct 2, 2011 #2
    for your part 2, I thought you were trying to show b=c. so why do you show a=b?

    For proof 1 (and any other) it's slicker to show the proof using this=this=this=this=conclusion than to have it set up as individual lines as you did. try starting with:
    let: b and c are multiplicative inverses of a with e the identity.
    b=b*e=b*(a*c)=......=c

    Try doing two similarly starting with b= , the crucial step will be when you get a*b as part of a line and you replace it with a*c since it's part of your assumption, then decompose back to c.
     
  4. Oct 3, 2011 #3

    Deveno

    User Avatar
    Science Advisor

    suppose ab = 1 and ac = 1.

    since fields are commutative, ba = 1.

    thus b = b1 = b(ac) = (ba)c = 1c = c.

    see how much shorter this is?

    2) is fine, but you should start with ab = ac,

    instead of introducing some variable x. so it should read:

    ab = ac
    z(ab) = z(ac)
    (za)b = (za)c
    1b = 1c
    b = c (this also saves one line, since you don't need to "switch factors").

    3) isn't very good. in these lines:

    (-1)a = -1a + (-1a + 1a)
    (-1)a = -1a + (1(-a + a))

    you assert that -1a = 1(-a), which is what you are trying to prove.


    first we show that if x + x = x, x = 0.

    x+x = x
    (x+x) + -x = x + -x = 0
    x + (x + -x) = 0
    x + 0 = 0
    x = 0.

    next we show that 0a = 0:

    0a = (0 + 0)a = 0a + 0a, so by the above, 0a = 0.

    finally, the proof that (-1)a = -a:

    (-1)a + a = (-1)a + 1a = (-1 + 1)a = 0a = 0. so
    ((-1)a + a) + -a = 0 + -a = -a
    (-1)a + (a + -a) = -a
    (-1)a + 0 = -a
    (-1)a = -a.
     
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