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X and Y are both metric space. Thanks a lot! 155.69.135.100 ( talk) 12:26, 8 August 2010 (UTC)
On page 13 0f "Elliptic Curves" by Dale Husemuller (1986) I read the following: "As for cubics, there is no known method for determining, in a finite number of steps, whether there is a rational point on a given cubic curve. This important question is still open."
My question is whether a method has been found, and if that is the case, where do I find such a method?
amri@vianet.ca -- 66.225.171.69 ( talk) 16:23, 8 August 2010 (UTC)
I know that we're lucky to have many talented set theorists on this page, so I'm sure my question will be answered in no time. Given a set X, take two subsets, say A and B, and consider the union A ∪ B and the intersection A ∩ B. For a set C ⊆ A, let A − C = {a ∈ A : a ∉ C}. Let P(X) denote the power set of X. Define a binary operation ƒ : P(X) × P(X) → P(X) given by ƒ(A,B) = (A ∪ B) − (A ∩ B). Notice that this is well defined since A ∩ B ⊆ A ∪ B. Let us write A⋅B in place of ƒ(A,B). I want to show that this binary operation is associative, i.e. for any subsets, A, B and C, of the set X we have (A⋅B)⋅C = A⋅(B⋅C). Explicitly this associativity condition becomes:
I've done some examples with Venn Diagrams and it seems to be true. The figure that I've included shows (A⋅B)⋅C = A⋅(B⋅C) in red, when A, B and C were solid disks. It doesn't matter which way I do it; I get the same set. I was hoping that someone might be able to help me with the following.
I thank you all in advance for what I am sure will be your helpful and well-thought replies. — Fly by Night ( talk) 19:53, 8 August 2010 (UTC)
Thanks to everyone for their answers. I now consider my question to be resolved. Thanks to Trovatore, Tango and Algebraist for their, as ever, insightful comments. I'm also going to archive the thread because I don't think it's moving in such a positive direction. Seeing as the question has been resolved in the OP's eyes, I don't see any problem in archiving. Thanks again to everyone. — Fly by Night ( talk) 22:25, 9 August 2010 (UTC)
Mathematics desk | ||
---|---|---|
< August 7 | << Jul | August | Sep >> | August 9 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
X and Y are both metric space. Thanks a lot! 155.69.135.100 ( talk) 12:26, 8 August 2010 (UTC)
On page 13 0f "Elliptic Curves" by Dale Husemuller (1986) I read the following: "As for cubics, there is no known method for determining, in a finite number of steps, whether there is a rational point on a given cubic curve. This important question is still open."
My question is whether a method has been found, and if that is the case, where do I find such a method?
amri@vianet.ca -- 66.225.171.69 ( talk) 16:23, 8 August 2010 (UTC)
I know that we're lucky to have many talented set theorists on this page, so I'm sure my question will be answered in no time. Given a set X, take two subsets, say A and B, and consider the union A ∪ B and the intersection A ∩ B. For a set C ⊆ A, let A − C = {a ∈ A : a ∉ C}. Let P(X) denote the power set of X. Define a binary operation ƒ : P(X) × P(X) → P(X) given by ƒ(A,B) = (A ∪ B) − (A ∩ B). Notice that this is well defined since A ∩ B ⊆ A ∪ B. Let us write A⋅B in place of ƒ(A,B). I want to show that this binary operation is associative, i.e. for any subsets, A, B and C, of the set X we have (A⋅B)⋅C = A⋅(B⋅C). Explicitly this associativity condition becomes:
I've done some examples with Venn Diagrams and it seems to be true. The figure that I've included shows (A⋅B)⋅C = A⋅(B⋅C) in red, when A, B and C were solid disks. It doesn't matter which way I do it; I get the same set. I was hoping that someone might be able to help me with the following.
I thank you all in advance for what I am sure will be your helpful and well-thought replies. — Fly by Night ( talk) 19:53, 8 August 2010 (UTC)
Thanks to everyone for their answers. I now consider my question to be resolved. Thanks to Trovatore, Tango and Algebraist for their, as ever, insightful comments. I'm also going to archive the thread because I don't think it's moving in such a positive direction. Seeing as the question has been resolved in the OP's eyes, I don't see any problem in archiving. Thanks again to everyone. — Fly by Night ( talk) 22:25, 9 August 2010 (UTC)