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Boolean rings and Boolean algebras

  1. Nov 27, 2007 #1

    quasar987

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    My professor wrote that we get a Boolean algebra from a Boolean ring (R,+,-,.,0,1) by setting xANDy=xy, xORy=x+y+xy and xNOT=1+x.

    But it seems to me that xNOT is not an involution. I.e., (xNOT)NOT = 1+(1+x), which is not x.

    (xNOT=-x would do the trick though)
     
  2. jcsd
  3. Nov 28, 2007 #2
    It seems to me that

    1+(1+x) = x

    For several reasons. What else would it be equal to? Doesn't 1 + 1 = 0?
     
  4. Nov 28, 2007 #3

    quasar987

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    I forgot about that. In an idempotent ring, x+x=0.
     
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