Comment by alentist | Hacker News Reader

Comment by alentist | Hacker News Reader

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You are making a mistake that many students unfamiliar with abstract algebra make, which is to confuse a particular "encoding" or "implementation" with the algebraic structure.

The concept of a real closed field [0] (and its categorical second-order version, the Dedekind-complete ordered field) stands on its own, without multiplication being defined in terms of repeated addition. It is completely independent of whether you happen to encode the reals as Cauchy sequences, Dedekind cuts, or something else. The (equivalence classes of) Cauchy sequences are not the same thing as the real numbers, even if we sometimes abuse terminology in this way for expediency. The distinction becomes increasingly important as you delve into more exotic algebraic structures.

Another illustrative example is the Hessenberg product [1]. Even the ordinary product cannot really be reduced to "repeated addition", because you have to use the infinitary concept of a limit. And not just your everyday limit [2], but a limit on a proper-class sized domain [3]!

See also the category-theoretic product [4].

[0] https://en.wikipedia.org/wiki/Real_closed_field

[1] https://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_ope...

[2] https://en.wikipedia.org/wiki/Limit_of_a_sequence

[3] https://en.wikipedia.org/wiki/Order_topology#Ordinal-indexed...

[4] https://en.wikipedia.org/wiki/Product_(category_theory)

I think its easy to get too hung up on particular definitions, you're free to define the reals to be the isomophism class of Dedekind-complete ordered fields if you like, I'm free to define the reals by some construction (Cauchy sequences, Dedekind cuts or whatever). As long as the chosen definitions are isomorphic it doesn't matter at all.

In general in a field I have addition and a multiplicative identity 1, as well as the distributive property, which means that for all elements x satisfying x = 1 + 1 + .... 1 for some number of ones I can write a.x = a.(1 + 1 + .... 1) = a + a + .... a, so there is always a subset for which multiplication works like repeated addition.

You are confusing a theory [0] with a model [1] of that theory. These are two different things.

> there is always a subset for which multiplication works like repeated addition

A subset, yes.

[0] https://en.wikipedia.org/wiki/Theory_(mathematical_logic)

[1] https://en.wikipedia.org/wiki/Model_theory

Na, I'm not - I'm explaining how to "compute: \pi*\pi using repeated addition". Step 1 - find a convenient Cauchy sequence that converges to \pi.

> 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?

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.

> 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.

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).

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