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pivoxa15

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Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.

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- Thread starter pivoxa15
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pivoxa15

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Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.

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matt grime

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What? That makes no sense. Mainly because you do not ask a question at all.

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Dick

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pivoxa15

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I was looking for a confirmation that my claim is correct.

So far we know 0 and 1 is in R. Take all 4 combinations of these two elements with respect to the 2 operations. So 8 additions and multiplications all together, all of which gives 0 or 1 so R contains only two elements 0 and 1. The only non trivial one is 1+1 which has to be 0 as by the theorem in the OP.

So far we know 0 and 1 is in R. Take all 4 combinations of these two elements with respect to the 2 operations. So 8 additions and multiplications all together, all of which gives 0 or 1 so R contains only two elements 0 and 1. The only non trivial one is 1+1 which has to be 0 as by the theorem in the OP.

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Dick

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matt grime

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pivoxa15

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Remember I am talking about a boolean wring with multiplicative identity.

I am claiming there is only one unique boolean ring with multiplicative identity.

Or with even this consideration taken into account, it is not correct? If so why?

If the boolean ring dosen't have multiplicative identity than other combinations may be possible.

I am claiming there is only one unique boolean ring with multiplicative identity.

Or with even this consideration taken into account, it is not correct? If so why?

If the boolean ring dosen't have multiplicative identity than other combinations may be possible.

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HallsofIvy

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I don't see why those prevent other elements. What about a set {0, 1, a, b} with operation tables:

Then 0 is in R. Make the multiplicative identity 1 is also in it. Therefore R can only take 0 and 1 and no more because 1+1=0. 0+0=0. 1+0=1 always. So 2 or other elements can never occur.

addition:

0 1 a b

0 0 1 a b

1 1 0 b a

a a b 0 1

b b a 1 0

multiplication

0 1 a b

0 0 0 0 0

1 0 1 a b

a 0 a b 1

b 0 b 1 a

- #9

matt grime

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Remember I am talking about a boolean wring with multiplicative identity.

and?

I am claiming there is only one unique boolean ring with multiplicative identity.

but that is patently silly.

Just write down something to produce a counter example. It is trivial to produce such example. Hell, diagonal matrices with 0s and 1s on the diagonals gives infinitely many counter examples without having to think at all.

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