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> Na, I'm not Yes, you are. Cauchy sequences are a model of the real numbers (which are the Dedekind-complete ordered field). > I'm explaining how to "compute

> Na, I'm not Yes, you are. Cauchy sequences are a model of the real numbers (which are the Dedekind-complete ordered field). > I'm explaining how to "compute: \pi*\pi using repeated addition" How will you define π times π as a repeated addition, i.e. something of the form x + x + … + x?

eigenket 6y ago
There is not one Dedekind-complete ordered field, for every two Dedekind-complete ordered fields there is a unique isomorphism between them. For example the nxn diagonal matrices with real entries with all the entries along the diagonal equal are a Dedekind-complete ordered field. I don't know why you are conflating the words "compute" and "define", do you understand how these are different words? I was responding to how to "compute: \pi*\pi using repeated addition", this is rather different to defining \pi*\pi as a repeated addition.
- alentist 6y ago
  > There is not one Dedekind-complete ordered field Yes there is. > For example the nxn diagonal matrices with real entries That's a model, not the theory. > I don't know why you are conflating the words "compute" and "define" This whole discussion is about definition. Your own comment started with "You first define..." And no, you are not able to compute arbitrary products of arbitrary reals in this way anyway, if by computation you mean a finitary algorithmic process.
  - eigenket 6y ago
    Ok, I think your definition of "field" is different to mine. Mine starts "A field is a set F with binary operations + and x" and then goes on to list some properties they have to have, yours seems to be doing something different. If you start with mine you get a whole bunch of different complete ordered fields, but you can easily show they're all isomorphic (e.g. Spivak does this IIRC). >This whole discussion is about definition No, it isn't this whole discussion is answering the question about computing pi * pi. Maybe I was slightly sloppy in my first answer to that question, I was only attempting to sketch the method. I don't really want to compute products of arbitrary reals in that way (or any way), but the method I sketched works for computable reals which is sufficient to cover the case of \pi. By computation I mean something that runs on a Turing machine, but I don't assume finitary (I'm happy for my Turing machine to keep producing more and more bits of precision forever).

eigenket 6y ago

There is not one Dedekind-complete ordered field, for every two Dedekind-complete ordered fields there is a unique isomorphism between them. For example the nxn diagonal matrices with real entries with all the entries along the diagonal equal are a Dedekind-complete ordered field. I don't know why you are conflating the words "compute" and "define", do you understand how these are different words? I was responding to how to "compute: \pi*\pi using repeated addition", this is rather different to defining \pi*\pi as a repeated addition.

alentist 6y ago
> There is not one Dedekind-complete ordered field Yes there is. > For example the nxn diagonal matrices with real entries That's a model, not the theory. > I don't know why you are conflating the words "compute" and "define" This whole discussion is about definition. Your own comment started with "You first define..." And no, you are not able to compute arbitrary products of arbitrary reals in this way anyway, if by computation you mean a finitary algorithmic process.
- eigenket 6y ago
  Ok, I think your definition of "field" is different to mine. Mine starts "A field is a set F with binary operations + and x" and then goes on to list some properties they have to have, yours seems to be doing something different. If you start with mine you get a whole bunch of different complete ordered fields, but you can easily show they're all isomorphic (e.g. Spivak does this IIRC). >This whole discussion is about definition No, it isn't this whole discussion is answering the question about computing pi * pi. Maybe I was slightly sloppy in my first answer to that question, I was only attempting to sketch the method. I don't really want to compute products of arbitrary reals in that way (or any way), but the method I sketched works for computable reals which is sufficient to cover the case of \pi. By computation I mean something that runs on a Turing machine, but I don't assume finitary (I'm happy for my Turing machine to keep producing more and more bits of precision forever).