4 ms·

Multiplication by rational fractions p/q is just multiplication by whole p, followed by multiplication by 1/q. 1/q is just the number that when multiplied by q

by dash2 6y ago

Multiplication by rational fractions p/q is just multiplication by whole p, followed by multiplication by 1/q. 1/q is just the number that when multiplied by q gives 1. So you can cash that all out as repeated addition. Multiplication by irrationals is just an infinite sum of multiplication by rationals, no? x times pi = x times 3, plus x times 1 / 10, plus.... I don't see there's any conceptual issue here. Mathematicians feel free to correct me.

pdonis 6y ago
> multiplication by 1/q Is not repeated addition. You can't add a number to itself 1/q times. At least, not unless you're willing to adopt increasingly perverse interpretations of "repeated addition" as you try to cover more and more numbers. See my response to wruza upthread.
- dash2 6y ago
  Hmm, so just go about it differently: x * p/q with p,q integers is (x*p) / q. The first bit is repeated addition. The second bit means, find the number r such that r * q = x * p. Even if r is not integer, q is integer, so we can try different numbers, add them to themselves q times, and close in on the answer. So I still think conceptually it's fine to think of it as repeated addition? It might be algorithmically a bad way to do it ("increasingly perverse", taking limits when the number is irrational etc.). I don't know how computers actually implement multiplication - though wikipedia (https://en.wikipedia.org/wiki/Binary_multiplier https://en.wikipedia.org/wiki/Binary_multiplier) says it works via shifts and adds. Or maybe, it's bad to teach it to kids this way? But then I think we need evidence from educationalists.
  - pdonis 6y ago
    > The first bit is repeated addition So now your definition of "repeated addition" is "repeated addition, plus a version of multiplication". Division is the inverse of multiplication, so your definition is circular: you're "defining" multiplication in terms of repeated addition and multiplication. Similar objections apply to another poster's contention upthread that the "repeated addition" definition is justified because of the distributive law. The distributive law defines A times (b plus c) in terms of A times b plus A times c. So it's useless as a definition of multiplication in terms of addition.
    - dash2 6y ago
      I don't think so. Look more closely. (x * p) is repeated addition with p integer - we can do this no multiplication involved. Then dividing by q involves finding r such that r * q = x * p. Here, again, we can do this by finding r such that r, added to itself q times, equals x * p. Since q is integer, again, this again involves repeated addition. (It does need an algorithm to find the right value of r. I suppose that trying an arbitrary value, then increasing or decreasing it, would work.)
      - pdonis 6y ago
        > (x * p) is repeated addition with p integer Yes, that's the "first bit" that I was not raising an issue with. > dividing by q involves finding r such that r * q = x * p Which, as you note, requires a trial and error algorithm to find r. But in any case, r is the answer being sought, not the thing we're supposed to be adding to itself some number of times to get the answer.
        dash2 6y ago
        OK. It sounds like we agree that my definition doesn't secretly involve multiplication and isn't circular?
- im3w1l 6y ago
  You can do it like this: First define a*b for integers by addition. Define the the rational numbers as ordered pairs of integers where the last number may not be 0. Consider the rational number (a,b) to be equivalent to (c, d) if a*d = b*c (Exercise for the reader: Prove that this is an equivalence relation) Now define (a, b) + (c, d) = (ad + bc, bd), (a, b) * (c, d) = (ac, bd) (Exercise for the reader: Prove that the subset of rationals (n, 1) behaves just like the integers)

by dash2 6y ago

pdonis 6y ago

> multiplication by 1/q Is not repeated addition. You can't add a number to itself 1/q times. At least, not unless you're willing to adopt increasingly perverse interpretations of "repeated addition" as you try to cover more and more numbers. See my response to wruza upthread.

dash2 6y ago
Hmm, so just go about it differently: x * p/q with p,q integers is (x*p) / q. The first bit is repeated addition. The second bit means, find the number r such that r * q = x * p. Even if r is not integer, q is integer, so we can try different numbers, add them to themselves q times, and close in on the answer. So I still think conceptually it's fine to think of it as repeated addition? It might be algorithmically a bad way to do it ("increasingly perverse", taking limits when the number is irrational etc.). I don't know how computers actually implement multiplication - though wikipedia (https://en.wikipedia.org/wiki/Binary_multiplier https://en.wikipedia.org/wiki/Binary_multiplier) says it works via shifts and adds. Or maybe, it's bad to teach it to kids this way? But then I think we need evidence from educationalists.
- pdonis 6y ago
  > The first bit is repeated addition So now your definition of "repeated addition" is "repeated addition, plus a version of multiplication". Division is the inverse of multiplication, so your definition is circular: you're "defining" multiplication in terms of repeated addition and multiplication. Similar objections apply to another poster's contention upthread that the "repeated addition" definition is justified because of the distributive law. The distributive law defines A times (b plus c) in terms of A times b plus A times c. So it's useless as a definition of multiplication in terms of addition.
  - dash2 6y ago
    I don't think so. Look more closely. (x * p) is repeated addition with p integer - we can do this no multiplication involved. Then dividing by q involves finding r such that r * q = x * p. Here, again, we can do this by finding r such that r, added to itself q times, equals x * p. Since q is integer, again, this again involves repeated addition. (It does need an algorithm to find the right value of r. I suppose that trying an arbitrary value, then increasing or decreasing it, would work.)
    - pdonis 6y ago
      > (x * p) is repeated addition with p integer Yes, that's the "first bit" that I was not raising an issue with. > dividing by q involves finding r such that r * q = x * p Which, as you note, requires a trial and error algorithm to find r. But in any case, r is the answer being sought, not the thing we're supposed to be adding to itself some number of times to get the answer.
      - dash2 6y ago
        OK. It sounds like we agree that my definition doesn't secretly involve multiplication and isn't circular?
im3w1l 6y ago
You can do it like this: First define a*b for integers by addition. Define the the rational numbers as ordered pairs of integers where the last number may not be 0. Consider the rational number (a,b) to be equivalent to (c, d) if a*d = b*c (Exercise for the reader: Prove that this is an equivalence relation) Now define (a, b) + (c, d) = (ad + bc, bd), (a, b) * (c, d) = (ac, bd) (Exercise for the reader: Prove that the subset of rationals (n, 1) behaves just like the integers)