Boolean rings and Boolean algebras

Thread starter quasar987
Start date Nov 27, 2007
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A Boolean algebra can be derived from a Boolean ring by defining operations such as xANDy as xy, xORy as x+y+xy, and xNOT as 1+x. However, there is a concern that xNOT does not satisfy the involution property, as (xNOT)NOT results in 1+(1+x), which does not equal x. The discussion suggests that if xNOT were defined as -x, it would resolve the issue of involution. Additionally, it is noted that in an idempotent ring, the equation x+x=0 holds true, reinforcing the complexities of these definitions. The conversation highlights the nuances in the relationship between Boolean rings and algebras.

Nov 27, 2007

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)

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Nov 28, 2007

Coin

It seems to me that

1+(1+x) = x

For several reasons. What else would it be equal to? Doesn't 1 + 1 = 0?

Nov 28, 2007

quasar987

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

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