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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 alge

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 https://en.wikipedia.org/wiki/Real_closed_field [1] https://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_operations https://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_ope... [2] https://en.wikipedia.org/wiki/Limit_of_a_sequence https://en.wikipedia.org/wiki/Limit_of_a_sequence [3] https://en.wikipedia.org/wiki/Order_topology#Ordinal-indexed_sequences https://en.wikipedia.org/wiki/Order_topology#Ordinal-indexed... [4] https://en.wikipedia.org/wiki/Product_(category_theory) https://en.wikipedia.org/wiki/Product_(category_theory)

eigenket 6y ago
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.
- alentist 6y ago
  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) https://en.wikipedia.org/wiki/Theory_(mathematical_logic) [1] https://en.wikipedia.org/wiki/Model_theory https://en.wikipedia.org/wiki/Model_theory
  - eigenket 6y ago
    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.
    - alentist 6y ago
      > 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

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.

alentist 6y ago
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) https://en.wikipedia.org/wiki/Theory_(mathematical_logic) [1] https://en.wikipedia.org/wiki/Model_theory https://en.wikipedia.org/wiki/Model_theory
- eigenket 6y ago
  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.
  - alentist 6y ago
    > 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.