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Find feild where 1 = 0 and show that it is the only one

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Define a paddock to be a set in which A1 - A4, M1 - M4 and D holds but instead of 1[tex]\neq[/tex] 0, we have 1 = 0. Find an example of a paddock, and show that your example is the only one

    2. Relevant equations
    A1 - A4, M1 - M4 and D are all axioms

    addition axioms
    A1: a + b = b + a
    A2: (a + b) + c = a + (b + c)
    A3: there is a 0 s.t a + 0 = 0 + a = a
    A4 for every a there exsist -a s.t a + (-a) = 0

    multiplication axioms:
    M1: a.b = b.a
    M2: (a.b).c = a.(b.c)
    M3: there is a 1 s.t a.1 = 1.a = a
    M4: for every a there exsist a-1 s.t a.a-1 = 1

    D (a+b)c = ab + bc
    a, b, c, 1, 0, inverses all belong in the set

    3. The attempt at a solution
    the set {1, 0} closed under +1 and x1 (modulo multiplication of 1 and modulo addition of 1).
    1 = 0 because both 0 and 1 are the neutral element,
    if it is true how how would i go about showing its the only one. im guessing it requires proof by contradiction.
  2. jcsd
  3. Jan 14, 2009 #2


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    M3 tells you that a*1= a for all a.

    By D, a*(b+ 0)= a*b+ a*0 but by A3, b+ 0 = b so a*(b+ 0)= a*b. That is, a*(b+0) is equal to both a*b and a*b+ a*0 so they are equal: a*b= a*b+ a*0 and, adding the additive inverse of a*b to both sides, a*0= 0 for all a.

    Now you have both a*1= a and a*0= 0 as well as 0= 1. Put them together.

    ("Paddock" instead of "field". That's cute.)
  4. Jan 18, 2010 #3
    May I ask, how does putting 'a*1= a and a*0= 0 as well as 0= 1' together explicitly shows that this example of a paddock is the only one?
  5. Jan 19, 2010 #4


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    a*1= a and a*0= 0 are the same thing because 0= 1. Therefore a= 0 for all a. The "paddock" contains only a single element.
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