- #1

- 2,259

- 1

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, this conversation is about a claim that there is only one boolean ring with multiplicative identity. The claim is wrong and ridiculous.f

- #1

- 2,259

- 1

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.

Calculus and Beyond Homework Help News on Phys.org

- #2

Science Advisor

Homework Helper

- 9,426

- 6

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

- #3

Science Advisor

Homework Helper

- 26,259

- 622

- #4

- 2,259

- 1

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.

Last edited:

- #5

Science Advisor

Homework Helper

- 26,259

- 622

- #6

Science Advisor

Homework Helper

- 9,426

- 6

- #7

- 2,259

- 1

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.

Last edited:

- #8

Science Advisor

Homework Helper

- 43,008

- 974

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

Science Advisor

Homework Helper

- 9,426

- 6

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.

Share:

- Replies
- 9

- Views
- 1K

- Replies
- 0

- Views
- 420

- Replies
- 3

- Views
- 588

- Replies
- 5

- Views
- 1K

- Replies
- 1

- Views
- 718

- Replies
- 3

- Views
- 442

- Replies
- 1

- Views
- 657

- Replies
- 2

- Views
- 622

- Replies
- 5

- Views
- 795

- Replies
- 5

- Views
- 1K