6 ms·

Not defending the article, but how would you compute: \pi*\pi using repeated addition?

by midjji 6y ago

horsawlarway 6y ago
pi + pi + pi + (.141592... of pi) ~= 9.8696 You need the concept of a ratio, so arguably I'm using multiplication to define multiplication, but you're sort of cheating by asking about a fractional number.
- bidirectional 6y ago
  Pi is definitely not a fractional number... I don't think it's cheating at all, multiplication on the naturals is repeated addition, that's not the case for the reals.
  - spacedcowboy 6y ago
    Interesting how computers (which only understand ‘1’ and ‘0’ can do multiplication of reals, then. Unless you’re intel, of course... (no, I will never let it go :)
  - Cerium 6y ago
    Can we not consider that the algorithm taught for multiplication of real numbers is repeated multiplication of natural numbers which can be seen as repeated addition of natural numbers so we could define an addition only algorithm for multiplication of real numbers.
  - horsawlarway 6y ago
    Sure it is, as long as you're willing to repeat in increments of real numbers. And that's my point, basically - By the time we're discussing real numbers, we need multiplication as an operator, because we're discussing ratios already.
  - andrewprock 6y ago
    This is an interesting statement. While it's true, it obscures the fact that pi is defined as a fraction: Circumference/Diameter. That this fraction cannot be represented as a numeric fraction is one of the great insights of early mathematics.
    - DanielMcLaury 6y ago
      I wouldn't call the fact that pi is irrational an insight of "early mathematics." We knew that sqrt(2) was irrational around 500 BC if not earlier. We didn't know pi was irrational until around the time of the American revolution.
- masswerk 6y ago
  However, pi is already a ratio…
  - horsawlarway 6y ago
    Right, that's why it's cheating. He's asking about a number set that requires the concept of ratios, and ratios are multiplication.
- alentist 6y ago
  It's not cheating. It's precisely the point. There are contexts where multiplication is not simply repeated addition.
eigenket 6y ago
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.
thrower123 6y ago
3.14159 + 3.14159 + 3.14159 + (3.14159*0.14159)
- bitlax 6y ago
  red pen time
  - thrower123 6y ago
    It's fine if you're going by significant figures
scarmig 6y ago
How would you compute \pi+\pi using addition? Irrationals break all our intuitions.

Not defending the article, but how would you compute: \pi*\pi using repeated addition?

by midjji 6y ago

Not defending the article, but how would you compute: \pi*\pi using repeated addition?

pi + pi + pi + (.141592... of pi) ~= 9.8696 You need the concept of a ratio, so arguably I'm using multiplication to define multiplication, but you're sort of cheating by asking about a fractional number.

bidirectional 6y ago
Pi is definitely not a fractional number... I don't think it's cheating at all, multiplication on the naturals is repeated addition, that's not the case for the reals.
- spacedcowboy 6y ago
  Interesting how computers (which only understand ‘1’ and ‘0’ can do multiplication of reals, then. Unless you’re intel, of course... (no, I will never let it go :)
- Cerium 6y ago
  Can we not consider that the algorithm taught for multiplication of real numbers is repeated multiplication of natural numbers which can be seen as repeated addition of natural numbers so we could define an addition only algorithm for multiplication of real numbers.
- horsawlarway 6y ago
  Sure it is, as long as you're willing to repeat in increments of real numbers. And that's my point, basically - By the time we're discussing real numbers, we need multiplication as an operator, because we're discussing ratios already.
- andrewprock 6y ago
  This is an interesting statement. While it's true, it obscures the fact that pi is defined as a fraction: Circumference/Diameter. That this fraction cannot be represented as a numeric fraction is one of the great insights of early mathematics.
  - DanielMcLaury 6y ago
    I wouldn't call the fact that pi is irrational an insight of "early mathematics." We knew that sqrt(2) was irrational around 500 BC if not earlier. We didn't know pi was irrational until around the time of the American revolution.
masswerk 6y ago
However, pi is already a ratio…
- horsawlarway 6y ago
  Right, that's why it's cheating. He's asking about a number set that requires the concept of ratios, and ratios are multiplication.
alentist 6y ago
It's not cheating. It's precisely the point. There are contexts where multiplication is not simply repeated addition.

eigenket 6y ago

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.

thrower123 6y ago

3.14159 + 3.14159 + 3.14159 + (3.14159*0.14159)

bitlax 6y ago
red pen time
- thrower123 6y ago
  It's fine if you're going by significant figures

scarmig 6y ago

How would you compute \pi+\pi using addition? Irrationals break all our intuitions.