3 ms·

You first define multiplication of natural numbers to be repeated addition, then define multiplication of rationals in terms of multiplication and addition of n

You first define multiplication of natural numbers to be repeated addition, then define multiplication of rationals in terms of multiplication and addition of naturals, then define multiplication of reals in terms of multiplication of Cauchy sequences of rationals ;)

codethief 6y ago
Exactly. I've scrolled through hundreds of comments here now and it really is beyond me how the question of whether you can define x·y in terms of addition for x and y being arbitrary reals is even a matter of debate.
- alentist 6y ago
  The question is not whether you can define it in terms of addition in some abstract way, but whether you can define it in terms of repeated addition, i.e. something of the form x + x + … + x.
  - codethief 6y ago
    > The question is not whether you can define it in terms of addition in some abstract way This is splitting hairs. The reals themselves are defined "in some abstract way". For instance, what does addition of reals even mean? How do you add two arbitrary real numbers? Exactly, by adding the elements of the Cauchy sequences. (And similarly for multiplication.) This is the definition of addition (multiplication) and the only abstraction involved here is the abstraction due to the way the reals are constructed in the first place.
    - alentist 6y ago
      > This is splitting hairs. Nope. The question addressed by the article is precisely whether multiplication is repeated addition. > adding the elements of the Cauchy sequences. (And similarly for multiplication.) This is the definition of addition (multiplication) Nope. See https://news.ycombinator.com/item?id=26336108 https://news.ycombinator.com/item?id=26336108.
alentist 6y ago
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

codethief 6y ago

Exactly. I've scrolled through hundreds of comments here now and it really is beyond me how the question of whether you can define x·y in terms of addition for x and y being arbitrary reals is even a matter of debate.

alentist 6y ago
The question is not whether you can define it in terms of addition in some abstract way, but whether you can define it in terms of repeated addition, i.e. something of the form x + x + … + x.
- codethief 6y ago
  > The question is not whether you can define it in terms of addition in some abstract way This is splitting hairs. The reals themselves are defined "in some abstract way". For instance, what does addition of reals even mean? How do you add two arbitrary real numbers? Exactly, by adding the elements of the Cauchy sequences. (And similarly for multiplication.) This is the definition of addition (multiplication) and the only abstraction involved here is the abstraction due to the way the reals are constructed in the first place.
  - alentist 6y ago
    > This is splitting hairs. Nope. The question addressed by the article is precisely whether multiplication is repeated addition. > adding the elements of the Cauchy sequences. (And similarly for multiplication.) This is the definition of addition (multiplication) Nope. See https://news.ycombinator.com/item?id=26336108 https://news.ycombinator.com/item?id=26336108.

alentist 6y ago

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