What’s Wrong with “Multiplication Is Repeated Addition”? (2008) • Hacker News Board

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#361

Post by dragontamer » Wed, Mar 03, 2021, 11:05 PM UTC

thaumasiotes wrote: post #358

dragontamer wrote: post #355

Earlier quoted context omitted.

I see what you're saying now. So to answer your original question... > You don't think it's easier to say... No. I disagree, your way of thinking is harder for me to think. :-) You're correct, but my mind didn't work like yours. But that's the beautiful thing about mathematics: we both are correct. We just had different viewpoints about how things work. Ultimately, it seems like we're both saying the same thing, alth…

Fair enough.

I do like your format though. So I'll borrow it for my side of the discussion.

  n·0 = 0               [definition of 0]
  n·(1 + -1) = 0        [definition of additive inverse]
  n·(1 + -1) = n + (-n) [multiplication is distributive over addition]
  n + (-n) = 0          [definition of additive inverse]

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#362

Post by alentist » Wed, Mar 03, 2021, 11:06 PM UTC

eigenket wrote: post #119

midjji wrote: post #101

Earlier quoted context omitted.

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

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 ;)

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

[1] https://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_ope...

[2] https://en.wikipedia.org/wiki/Limit_of_a_sequence

[3] https://en.wikipedia.org/wiki/Order_topology#Ordinal-indexed...

[4] https://en.wikipedia.org/wiki/Product_(category_theory)

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#363

Post by mnl » Wed, Mar 03, 2021, 11:07 PM UTC

thaumasiotes wrote: post #333

mnl wrote: post #193

Earlier quoted context omitted.

Yes, they are two binary operations and depending on the sets you consider and which properties you impose for those operations you have different algebraic structures. (This used to be taught at school before "modern mathematics" were considered harmful, maybe they were but at least they were correct). The thing is that as you can write m (let it be a positive integer) as m=1+...+1 (m-times), you can write n·m=n·(1+…

> At any rate we have to impose that n·0=0, which can't be writen cleverly as "repeated addition" and worked up backwards. Are you kidding? This is the exact opposite of the truth; the nature of multiplication as repeated addition is the entire reason why multiplying by 0 gives the additive identity. It's exactly the same as how exponentiating by 0 gives the multiplicative identity, since exponentiation is just repea…

Could you sketch a proof starting from some definition of the operation product of integers as "repeated addition" without using the distributive property, which would imply that we already have another binary operation besides the sum?

I was thinking about the Peano axioms when I wrote that (hence the working backwards thing). Obviously if you start with a ring, you don't have to impose it, you get that as a property. I think I mentioned that in a later post.

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#364

Post by redshirtrob » Wed, Mar 03, 2021, 11:17 PM UTC

thaumasiotes wrote: post #346

redshirtrob wrote: post #168

Earlier quoted context omitted.

I don't think so. I was on the Calculus track in high school so we derived it in...Pre-Calculus. Prior to that the quadratic formula was something that seemed to be handed down from on high. We used it in Algebra II and maybe even before that, but I had no idea where it came from. It was a mind-opening experience when we derived it in class one day. Our teacher didn't ruin the surprise. She just said, let's complete…

> Our teacher didn't ruin the surprise. She just said, let's complete the square on a general quadratic equation. How is this not ruining the surprise? The only possible outcomes of doing that are that (1) you make a mistake; or (2) you get a formula for solving quadratic equations. Quadratic equations have the same solutions regardless of your methodology, so there's only one formula you can get.

She did not tell us we were deriving the quadratic formula. I had no idea where she was going when she started the exercise.

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#365

Post by alentist » Wed, Mar 03, 2021, 11:19 PM UTC

codethief wrote: post #353

eigenket wrote: post #119

Earlier quoted context omitted.

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 ;)

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.

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.

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#366

Post by thaumasiotes » Wed, Mar 03, 2021, 11:19 PM UTC

dragontamer wrote: post #361

thaumasiotes wrote: post #358

Earlier quoted context omitted.

Fair enough.

I do like your format though. So I'll borrow it for my side of the discussion. n·0 = 0 [definition of 0] n·(1 + -1) = 0 [definition of additive inverse] n·(1 + -1) = n + (-n) [multiplication is distributive over addition] n + (-n) = 0 [definition of additive inverse]

I feel like I'm not understanding something. Isn't "n·0 = 0" the demonstrandum? I would have written something like

  n·0 = 0                [we seek to show this]
  -------
  n·0 = n·(1 + -1)       [definition of additive inverse]
  n·(1 + -1) = n + (-n)  [multiplication is distributive...]
  n + (-n) = 0           [definition of additive inverse]
  QED

(Sorry, but what you have above makes me slightly queasy, and I'm hoping the fuss about rigor can be acceptable in a math thread.)

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#367

Post by kmill » Wed, Mar 03, 2021, 11:21 PM UTC

pdonis wrote: post #330

kmill wrote: post #319

Earlier quoted context omitted.

My computer can manipulate and prove things about real numbers just fine. I think it can do at least everything I can do normally, though much more laboriously. What I was mostly responding to in your previous comment is the idea of "emulating" multiplication for whole numbers by repeated addition. What I got from that is that you were thinking of the whole numbers as being inside the real numbers. My point was that…

> My computer can manipulate and prove things about real numbers just fine. If you mean it can, with appropriate software, do symbolic manipulations of general formulas that are valid for real numbers, yes, of course. But that's not the same as doing specific concrete computations with them. > What I was mostly responding to in your previous comment is the idea of "emulating" multiplication for whole numbers by repea…

> If you mean it can, with appropriate software, do symbolic manipulations of general formulas that are valid for real numbers, yes, of course. But that's not the same as doing specific concrete computations with them.

No, I don't mean computer algebra systems, if that's what you mean. In something like Lean, you can set up an actual construction of the system of real numbers (in mathlib it's the Cauchy completion of the rationals, which in turn are a given by a numerator and positive nonzero denominator that are coprime; an integer is essentially a natural number and a sign bit, and natural numbers are inductively constructed from zero and taking successors). Maybe this will convince you that these are "the" real numbers: you can prove there are uncountably many of them, and more importantly that it is the unique complete ordered field (up to isomorphism).

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#368

Post by dash2 » Wed, Mar 03, 2021, 11:28 PM UTC

pdonis wrote: post #281

dash2 wrote: post #236

Earlier quoted context omitted.

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 b…

> 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 be…

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.)

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#369

Post by alentist » Wed, Mar 03, 2021, 11:29 PM UTC

horsawlarway wrote: post #109

midjji wrote: post #101

Earlier quoted context omitted.

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.

It's not cheating. It's precisely the point. There are contexts where multiplication is not simply repeated addition.

Re: What’s Wrong with “Multiplication Is Repeated Addition”? (2008)

#370

Post by dragontamer » Wed, Mar 03, 2021, 11:31 PM UTC

thaumasiotes wrote: post #366

dragontamer wrote: post #361

Earlier quoted context omitted.

I do like your format though. So I'll borrow it for my side of the discussion. n·0 = 0 [definition of 0] n·(1 + -1) = 0 [definition of additive inverse] n·(1 + -1) = n + (-n) [multiplication is distributive over addition] n + (-n) = 0 [definition of additive inverse]

I feel like I'm not understanding something. Isn't "n·0 = 0" the demonstrandum? I would have written something like n·0 = 0 [we seek to show this] ------- n·0 = n·(1 + -1) [definition of additive inverse] n·(1 + -1) = n + (-n) [multiplication is distributive...] n + (-n) = 0 [definition of additive inverse] QED (Sorry, but what you have above makes me slightly queasy, and I'm hoping the fuss about rigor can be accept…

I see. I'm not a real mathematician, I just dabble in it on occasion. And I fully admit: my math professors always criticized my proof constructions throughout my life, so its definitely not something I was ever that good at.

I'll happily take your advice however! Proper mathematical rigor is always something I appreciate, even if its something I'm not very much practiced at.