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> You can definitely see multiplication as repeated addition. Only for rational numbers. Doesn't work for real and complex numbers. > It's even a useful thing

> You can definitely see multiplication as repeated addition. Only for rational numbers. Doesn't work for real and complex numbers. > It's even a useful thing to do, how else would you define multiplication? Axiomatically, not algorithmically.

WallWextra 6y ago
If you define the reals axiomatically, you still need an existence proof. Which will involve addition and multiplication algorithms.
- yakubin 6y ago
  Not algorithms. There will be infinite addition involved, and algorithms are finite. Thinking of multiplication as repeated addition also won't explain anything about it. It's a separate operation. Deal with it. For similar reasons, you can't calculate x-th power of a number, when x is irrational, by decomposing it into exponentiation and roots. This metaphor is just training wheels. At some point you should lose it. Together with the notion that "multiplication is repeated addition" comes the notion that numbers are quantities. Only some of them are, and this isn't really what makes them numbers. Now what exactly gets repeated, when you don't have quantities?
  - prionassembly 6y ago
    Multiplication on the naturals is repeated addition. You might see the natural numbers as training wheels for higher mathematics, but number theorists might disagree...
    - yakubin 6y ago
      Number theorists work with integers, not naturals. And they too will agree with me, as they mostly work with polynomials, which show exactly that. I also didn't say natural numbers are training wheels, just this metaphor, which to use HN lingo: doesn't scale.
      - alisonkisk 6y ago
        Number theorists work with complex numbers and irrational numbers too https://www.britannica.com/science/Riemann-zeta-function https://www.britannica.com/science/Riemann-zeta-function
  - gugagore 6y ago
    Algorithms are not finite. E.g. Newton's method.
    - yakubin 6y ago
      First sentence on Wikipedia[1]: > In mathematics and computer science, an algorithm (/ˈælɡərɪðəm/ (About this soundlisten)) is a finite sequence of well-defined, computer-implementable instructions, typically to solve a class of problems or to perform a computation. Newton's method is finite too. You perform finitely many iterations. It doesn't calculate roots. It calculates their approximations. [1]: <https://en.wikipedia.org/wiki/Algorithm https://en.wikipedia.org/wiki/Algorithm>
      - gugagore 6y ago
        If you use a termination condition that has to do with convergence of iterates instead of a number of iterations (often the case), then you generally don't know beforehand the length of the finite sequences. Maybe you know a bound, but in general you might not even have that. In an important sense, it only becomes a finite algorithm. It isn't one. You cannot write the finite sequence of instructions down. It's got loops. To your point about approximations vs not, if you have an algorithm that, for any desired approximation accuracy can compute the square root to that accuracy in a finite number of steps, then that process is as much "the square root" as anything involving the real numbers.
        yakubin 6y ago
        > To your point about approximations vs not, if you have an algorithm that, for any desired approximation accuracy can compute the square root to that accuracy in a finite number of steps, then that process is as much "the square root" as anything involving the real numbers. Not really, since approximations, no matter how accurate, don't preserve algebraic properties. You only get to know what it's bigger/smaller than.
        gugagore 6y ago
  - titzer 6y ago
    Algorithms can work on symbolic formulas, and symbols can represent anything; infinite objects, operations on infinite objects, infinite sets of operations on infinite objects, and so on.
  - bopbeepboop 6y ago
    Algorithms aren’t necessarily finite, eg “while True, print 1”. Also, an irrational exponent is the product of component factors. b = Prod(0,inf) a^[x_i * 10^(-i)] = a^x So even with irrational numbers, operations can be decomposed - such as exponentiation into multiplication of integer exponents and roots. Which makes sense, because in the sequence definition of reals you need a way to generate the resulting sequence from the two original sequences. I think you’re trying to claim more than is true.
  - eru 6y ago
    > There will be infinite addition involved, and algorithms are finite. People are just bit sloppy, and say algorithm when they mean something slightly different. See https://stackoverflow.com/questions/28841260/what-is-the-difference-between-codata-and-data https://stackoverflow.com/questions/28841260/what-is-the-dif... and https://en.wikipedia.org/wiki/Corecursion https://en.wikipedia.org/wiki/Corecursion Basically, you don't want an 'algorithm' here to produce the whole number. All you need is some scheme that will produce the next digit in finite time (and the next one and the next one etc).
    - alisonkisk 6y ago
      Which is a technical way of saying "in real life people use finite rational or algebraic approximations for reals, so uncountability of reals and infinite precision aren't a problem".
      - eru 6y ago
        Not quite. Consider the following real number made of binary digits: Enumerate all Turing machines and all possible inputs, iff the i-th machine/input combination holds, the i-th binary digit in our number is 0, otherwise 1. This number is well-defined (once you fix your enumeration scheme). But there's no finite algorithm to produce approximations in your sense. What I was after were what's also called Computable numbers (https://en.wikipedia.org/wiki/Computable_number https://en.wikipedia.org/wiki/Computable_number). But I used the more general term of co-recursion, that also applies to arbitrary other data-structures like infinite lists, or with some generalization, infinite event-loops where the important condition is that each run through the body of the loop only takes finite time.
- qsort 6y ago
  Well, no, not really. The standard definition of the reals is as the unique nontrivial totally-ordered, Dedekind-complete, Archimedean field up to isomorphism. So what you would really need is a uniqueness proof, with addition and multiplications "provided" by the hypothesis.
  - howling 6y ago
    And how do you prove that a totally-ordered, Dedekind-complete, Archimedean field does not lead to contradiction besides constructing it explicitly by bootstrapping from natural numbers?
    - qsort 6y ago
      Nothing in the construction requires you to show an algorithm that given x, y \in R allows you to compute x+y and xy. You would make the usual Dedekind construction and show it satisfies the axioms of such a field. (as a matter of fact, no such algorithm exists in full generality!) It's probably a tomato/tomato kind of thing, but I'm only objecting to the 'algorithm' part of parent's comment.
SAI_Peregrinus 6y ago
Multiplication of the Surreals is a recursive operation using sums (addition and subtraction on the left and right sets). Since the Reals are a strict subfield of the Surreals one can define multiplication of the reals using only the same recursive formula and restricting both operands to be Reals.
- eigenket 6y ago
  Its probably worth emphasising that the recursion you need to "construct" the Surreals is infinite, in other words this does not give a reasonable algorithm to (for example) add two real numbers, you need S_omega in order to have even all rational numbers. The construction is rather involved but if we're only interested in the reals for now you can think of it as defining a real number by a set of rational numbers, in particular define a particular "real number" to be the set of all rational numbers less than it, for example sqrt(2) is defined to be the set of all rationals p/q such that p^2/q^2 < 2. We can "recursively" define addition of these real numbers in terms of addition of rational numbers because to add two reals you "just" have to add all the rationals in their respective sets. In general there is no sensible algorithm to do anything in the real numbers, since most real numbers aren't even computable (there is no way to represent an arbitrary real number on a Turing machine).
  - marcosdumay 6y ago
    Of course, once you decide to iterate over uncountable sets, infinity starts to appear all the time. This isn't the only case of infinite calculations that can't be computed in practice but that mathematics use all the time anyway; and it reflects quite well the fact that multiplying irrational numbers isn't something that one can do practice. There is no problem with it.
    - alisonkisk 6y ago
      Multiplying algebraics is trivial. A rectangle with sides sqrt(3) and sqrt(2) has area sqrt(6), which can be approximated by a decimal if needed. There are countable/computable/constructable subsets of the reals where multiplication has a finite algorithm and is it repeated addition. One example is the algebraics, as well as extensions the including a few special constants like pi. These are the subsets of the reals most commonly used for math and science. So in a wide range of problems areas, multiplication is not just repeated addition.
xirbeosbwo1234 6y ago
>Only for rational numbers. Doesn't work for real and complex numbers. Complex numbers aren't really relevant, in my opinion, because they are usually introduced as an extension of the rules for reals and polynomials. To multiply two complex numbers, you can totally forget that i is imaginary, do the multiplication as if it's just an ordinary variable, then substitute "i" back in. But that relies on being able to multiply polynomials, which would be difficult to define in terms of repeated multiplication. To some extent all of mathematics is a lie. We can do multiplication on the reals because we have decided that it's allowed. It is reasonable to define multiplication at first as repeated addition and then define a way to extend that to the reals that is consistent with the first definition.
DanielMcLaury 6y ago
Multiplication of real and complex numbers is typically defined by starting from repeated addition, extending this notion to rationals, and then extending that notion to reals by taking limits. How exactly are you going to present multiplication of real numbers axiomatically without essentially including an axiom that bootstraps everything from repeated addition? I suppose you can try defining the reals as "the unique complete ordered field" or the complex numbers as "the unique algebraically closed field of characteristic zero with cardinality c," but I don't think either of those are pedagogically useful to someone who is still learning what multiplication is.

WallWextra 6y ago

If you define the reals axiomatically, you still need an existence proof. Which will involve addition and multiplication algorithms.

yakubin 6y ago
Not algorithms. There will be infinite addition involved, and algorithms are finite. Thinking of multiplication as repeated addition also won't explain anything about it. It's a separate operation. Deal with it. For similar reasons, you can't calculate x-th power of a number, when x is irrational, by decomposing it into exponentiation and roots. This metaphor is just training wheels. At some point you should lose it. Together with the notion that "multiplication is repeated addition" comes the notion that numbers are quantities. Only some of them are, and this isn't really what makes them numbers. Now what exactly gets repeated, when you don't have quantities?
- prionassembly 6y ago
  Multiplication on the naturals is repeated addition. You might see the natural numbers as training wheels for higher mathematics, but number theorists might disagree...
  - yakubin 6y ago
    Number theorists work with integers, not naturals. And they too will agree with me, as they mostly work with polynomials, which show exactly that. I also didn't say natural numbers are training wheels, just this metaphor, which to use HN lingo: doesn't scale.
    - alisonkisk 6y ago
      Number theorists work with complex numbers and irrational numbers too https://www.britannica.com/science/Riemann-zeta-function https://www.britannica.com/science/Riemann-zeta-function
- gugagore 6y ago
  Algorithms are not finite. E.g. Newton's method.
  - yakubin 6y ago
    First sentence on Wikipedia[1]: > In mathematics and computer science, an algorithm (/ˈælɡərɪðəm/ (About this soundlisten)) is a finite sequence of well-defined, computer-implementable instructions, typically to solve a class of problems or to perform a computation. Newton's method is finite too. You perform finitely many iterations. It doesn't calculate roots. It calculates their approximations. [1]: <https://en.wikipedia.org/wiki/Algorithm https://en.wikipedia.org/wiki/Algorithm>
    - gugagore 6y ago
      If you use a termination condition that has to do with convergence of iterates instead of a number of iterations (often the case), then you generally don't know beforehand the length of the finite sequences. Maybe you know a bound, but in general you might not even have that. In an important sense, it only becomes a finite algorithm. It isn't one. You cannot write the finite sequence of instructions down. It's got loops. To your point about approximations vs not, if you have an algorithm that, for any desired approximation accuracy can compute the square root to that accuracy in a finite number of steps, then that process is as much "the square root" as anything involving the real numbers.
      - yakubin 6y ago
        > To your point about approximations vs not, if you have an algorithm that, for any desired approximation accuracy can compute the square root to that accuracy in a finite number of steps, then that process is as much "the square root" as anything involving the real numbers. Not really, since approximations, no matter how accurate, don't preserve algebraic properties. You only get to know what it's bigger/smaller than.
        gugagore 6y ago
- titzer 6y ago
  Algorithms can work on symbolic formulas, and symbols can represent anything; infinite objects, operations on infinite objects, infinite sets of operations on infinite objects, and so on.
- bopbeepboop 6y ago
  Algorithms aren’t necessarily finite, eg “while True, print 1”. Also, an irrational exponent is the product of component factors. b = Prod(0,inf) a^[x_i * 10^(-i)] = a^x So even with irrational numbers, operations can be decomposed - such as exponentiation into multiplication of integer exponents and roots. Which makes sense, because in the sequence definition of reals you need a way to generate the resulting sequence from the two original sequences. I think you’re trying to claim more than is true.
- eru 6y ago
  > There will be infinite addition involved, and algorithms are finite. People are just bit sloppy, and say algorithm when they mean something slightly different. See https://stackoverflow.com/questions/28841260/what-is-the-difference-between-codata-and-data https://stackoverflow.com/questions/28841260/what-is-the-dif... and https://en.wikipedia.org/wiki/Corecursion https://en.wikipedia.org/wiki/Corecursion Basically, you don't want an 'algorithm' here to produce the whole number. All you need is some scheme that will produce the next digit in finite time (and the next one and the next one etc).
  - alisonkisk 6y ago
    Which is a technical way of saying "in real life people use finite rational or algebraic approximations for reals, so uncountability of reals and infinite precision aren't a problem".
    - eru 6y ago
      Not quite. Consider the following real number made of binary digits: Enumerate all Turing machines and all possible inputs, iff the i-th machine/input combination holds, the i-th binary digit in our number is 0, otherwise 1. This number is well-defined (once you fix your enumeration scheme). But there's no finite algorithm to produce approximations in your sense. What I was after were what's also called Computable numbers (https://en.wikipedia.org/wiki/Computable_number https://en.wikipedia.org/wiki/Computable_number). But I used the more general term of co-recursion, that also applies to arbitrary other data-structures like infinite lists, or with some generalization, infinite event-loops where the important condition is that each run through the body of the loop only takes finite time.
qsort 6y ago
Well, no, not really. The standard definition of the reals is as the unique nontrivial totally-ordered, Dedekind-complete, Archimedean field up to isomorphism. So what you would really need is a uniqueness proof, with addition and multiplications "provided" by the hypothesis.
- howling 6y ago
  And how do you prove that a totally-ordered, Dedekind-complete, Archimedean field does not lead to contradiction besides constructing it explicitly by bootstrapping from natural numbers?
  - qsort 6y ago
    Nothing in the construction requires you to show an algorithm that given x, y \in R allows you to compute x+y and xy. You would make the usual Dedekind construction and show it satisfies the axioms of such a field. (as a matter of fact, no such algorithm exists in full generality!) It's probably a tomato/tomato kind of thing, but I'm only objecting to the 'algorithm' part of parent's comment.

SAI_Peregrinus 6y ago

Multiplication of the Surreals is a recursive operation using sums (addition and subtraction on the left and right sets). Since the Reals are a strict subfield of the Surreals one can define multiplication of the reals using only the same recursive formula and restricting both operands to be Reals.

eigenket 6y ago
Its probably worth emphasising that the recursion you need to "construct" the Surreals is infinite, in other words this does not give a reasonable algorithm to (for example) add two real numbers, you need S_omega in order to have even all rational numbers. The construction is rather involved but if we're only interested in the reals for now you can think of it as defining a real number by a set of rational numbers, in particular define a particular "real number" to be the set of all rational numbers less than it, for example sqrt(2) is defined to be the set of all rationals p/q such that p^2/q^2 < 2. We can "recursively" define addition of these real numbers in terms of addition of rational numbers because to add two reals you "just" have to add all the rationals in their respective sets. In general there is no sensible algorithm to do anything in the real numbers, since most real numbers aren't even computable (there is no way to represent an arbitrary real number on a Turing machine).
- marcosdumay 6y ago
  Of course, once you decide to iterate over uncountable sets, infinity starts to appear all the time. This isn't the only case of infinite calculations that can't be computed in practice but that mathematics use all the time anyway; and it reflects quite well the fact that multiplying irrational numbers isn't something that one can do practice. There is no problem with it.
  - alisonkisk 6y ago
    Multiplying algebraics is trivial. A rectangle with sides sqrt(3) and sqrt(2) has area sqrt(6), which can be approximated by a decimal if needed. There are countable/computable/constructable subsets of the reals where multiplication has a finite algorithm and is it repeated addition. One example is the algebraics, as well as extensions the including a few special constants like pi. These are the subsets of the reals most commonly used for math and science. So in a wide range of problems areas, multiplication is not just repeated addition.

xirbeosbwo1234 6y ago

>Only for rational numbers. Doesn't work for real and complex numbers. Complex numbers aren't really relevant, in my opinion, because they are usually introduced as an extension of the rules for reals and polynomials. To multiply two complex numbers, you can totally forget that i is imaginary, do the multiplication as if it's just an ordinary variable, then substitute "i" back in. But that relies on being able to multiply polynomials, which would be difficult to define in terms of repeated multiplication. To some extent all of mathematics is a lie. We can do multiplication on the reals because we have decided that it's allowed. It is reasonable to define multiplication at first as repeated addition and then define a way to extend that to the reals that is consistent with the first definition.

DanielMcLaury 6y ago

Multiplication of real and complex numbers is typically defined by starting from repeated addition, extending this notion to rationals, and then extending that notion to reals by taking limits. How exactly are you going to present multiplication of real numbers axiomatically without essentially including an axiom that bootstraps everything from repeated addition? I suppose you can try defining the reals as "the unique complete ordered field" or the complex numbers as "the unique algebraically closed field of characteristic zero with cardinality c," but I don't think either of those are pedagogically useful to someone who is still learning what multiplication is.