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Proof using ring axioms

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data

    Using only the ring axioms, prove that in a general ring (R, +,X)
    aX (x-z) = (aXx)- (aXz) where all a,x,z are elements of R

    2. Relevant equations

    Group axiom 3: G3= There is an inverse for each element g^-1 *g =e

    Ring axiom 3: R3= Two distributive laws connect the additive and multiplicatie structures.
    For any x,y,z xX(y+z) = (xXy)+ (xXz)
    and (x+y) X z= (xXz) + (yXz)

    3. The attempt at a solution
    My attempt. I thought that this would actually be straight forward; that I would just need to put -z as the addition of its inverse. I expected the rest to just fall into place.

    Here's what I did:

    aX (x-z) = (aXx)- (aXz)


    Left hand side aX (x-z)
    = aX(x + z^-1) from G3
    = (aXx) + (aXz^-1) from R3
    = (aXx)+ (aX -z) from G3
    = (aXx)- (aXz), as required


    I'm not sure whether I am allowed to just write the last line or whether I have left out some all important step!

    Thank you very much in anticipation of your assistance.
     
    Last edited: Aug 6, 2010
  2. jcsd
  3. Aug 6, 2010 #2

    Dick

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    You are asking why aX(-z)=(-aXz)? Well, aX(-z)+aXz=aX(-z+z) from your distributive axiom. What does that tell you?
     
  4. Aug 6, 2010 #3
    Thank you Dick. I got there! Thank you for your extremely quick reply. I really appreciate your help. Many thanks.
     
  5. Aug 6, 2010 #4

    Hurkyl

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    For the record, when the group operation is x+y, the inverse of x is usually written as -x.
     
  6. Aug 6, 2010 #5
    Thank you! I shall alter my notation. Much appreciated!
     
  7. Aug 6, 2010 #6

    Dick

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    True, it is kind of confusing to be mixing the '*' notation for the group operation with the ring notation of '+' for the group operation, but dndod1 seemed to be dealing with it ok.
     
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