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I don't understand. The author is mixing up physics (dimensional analysis) and maths, and trying to give the multiplicand a special role (I didn't even know th

I don't understand. The author is mixing up physics (dimensional analysis) and maths, and trying to give the multiplicand a special role (I didn't even know there was a distinction between the two terms - to me they are both factors). This might be true in the physical world, but in the world of numbers, I think the distinction is irrelevant. Furthermore, being able to compute/define multiplication through repeated addition doesn't prevent you from looking at the special properties of this new operator.

drewcoo 6y ago
Agreed. Multiplier and multiplicand are different words but the commutative property says their values can swap equivalently, so . . . what was the author's point again?
- tomxor 6y ago
  This was exactly my thought, I don't understand why those two word exist since it's commutative. Perhaps all this confusion is purely semantic.
  - ThePadawan 6y ago
    They exist to distinguish the element being operated on (the LHS) and what it is operated on by (the RHS). Total technicality, but I could see myself using the term multiplicand/multiplier in my code if I had to implement e.g. a stack-based parser for arithmetic expressions.
    - tomxor 6y ago
      > to distinguish the element being operated on (the LHS) I'm not sure what you mean, they are both operated on, it is a binary operator and commutative, there is literally no difference.
      - anbende 6y ago
        If we are just looking at numbers with no units attached there is no difference. As soon as there are units, a difference arises. I will collect 12 apples 5 times. The 12 apples are the multiplicand and the 5 times is the multiplier. The answer retains the units from the multiplicand.
        np_tedious 6y ago
        Still don't see the difference. Density times volume is mass. Volume times density is mass. Distance times force is work/energy... "5 times I will connect 12 apples"
        KMag 6y ago
        You're correct. The GP in incorrect about their dimensional analysis. 5 apples 12 times yields apples because it's apples times a dimensionless scalar (count). Newtons times meters is always Newton-meters, never Newtons or meters. Units are never magically dropped in dimensional analysis.
      - simias 6y ago
        I agree that it's a generally pointless distinctions, although it might be useful in some cases such as number systems where the multiplication isn't commutative, or for a particular implementation where the distinction matters. After all if you were to code a multiplication that was implemented naively as a series of additions it'd be generally much faster todo 2x1000 than 1000x2. To return to TFA I think the author is talking from a pedagogical standpoint, that teaching that multiplication is just a bunch of additions under a trench coat is not the best way to go. I'm not sure that I agree personally. In particular this bit regarding multiplier/multiplicand makes zero sense to me: >Different names indicate a difference in function. The multiplier and the multiplicand are not conceptually interchangeable. It is true that multiplication is commutative, but (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. Of course 3 rows and 2 rows aren't the same, but what does it have to do with the order of the multiplication? Isn't 2 rows x 3 chairs the same thing as 3 chairs x 2 rows? It's a bizarre argument.
  - SAI_Peregrinus 6y ago
    Multiplication over the reals is commutative. Matrix multiplication of non-square matrices isn't. Multiplication in a Ring isn't necessarily commutative. Other algebraic structures also have non-commutative multiplication. One could argue that these things aren't "multiplication" even if they are "products" since they don't satisfy all the properties of multiplication over the reals. But it is common to call the use of the product operation "multiplication", at least in cases where there's only one product operation to use. EG Geometric Algebra has Inner, Outer, and Geometric products, so calling them "multiplication" seems less common IME.
    - qsort 6y ago
      I really don't get the point you're making. If we're going to pull out random examples, monoids aren't guaranteed to be abelian; strings and concatenation form a monoid that's not abelian. If you have enough mathematical sophistication to conceptualize a non-commutative ring, you're well past the point where naming conventions are even remotely an issue. The original article was contrasting addition and multiplication on the basis that addends are called the same while factors are supposed to be called differently, which not only makes no sense (it's just a naming convention), but it also breaks down when you have more than two factors: what is the "c" in a x b x c called? Or we're talking about non-associative operations now?
      - SAI_Peregrinus 6y ago
        My point is only that the commutative property is not inherent to all multiplication operations, so there can be a distinction between the operands. It's not necessarily a useful distinction, and in the usual use of multiplication it's utterly useless and only adds confusion. But matrix multiplication is taught in high school (and usually promptly forgotten), it's not particularly advanced math. Personally I'm of the opinion that the terminology is muddled. There's no need to distinguish the operands of a multiplication over any of the usual domains (reals, rationals, integers, etc). And when you reach the point where it does become important there's generally more than one product operation and we should stop calling it multiplication. "Matrix multiplication" is a bad term. You also typically wouldn't name the operands, since as you note there can be more than two!
  - karmakaze 6y ago
    Not all multiplication is commutative, e.g. matrices.
- gugagore 6y ago
  If, for a moment, you conceptualize of multiplication on non-negative whole numbers as repeated addition, then this is the algorithm: procedure product(multiplier, multiplicand) acc := 0 for i = 1 to multiplier acc := acc + multiplicand end return acc end Swapping the arguments is a different computation. But after thinking about it, you realize that you get the same answer all the same. That's the point of > (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. The point with bringing up dimensional analysis is that the above algorithm doesn't work because what does it mean to do `for i = 1 to 3 chairs/row`? You might think of it like procedure product(multiplier, multiplicand) acc := "0" # an "absolute" zero that cooperates with any dimension each single_multiplier in multiplier acc := acc + (multiplicand * single_multiplier) end return acc end But then what is `(multiplicand * single_multiplier)` ?
- andi999 6y ago
  > (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. The author has it all backwards. Basically that you can turn this arrangement by 90 degrees (and turning the chairs, if you do not modell them by points by 90 degrees) is the reason why multiplication is commutative. It is non trivial that 5+5+5+...5 (100 times) is the same as 100+100+100+100+100
midjji 6y ago
Not defending the article, but how would you compute: \pi*\pi using repeated addition?
- 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.
spacedcowboy 6y ago
I have a PhD in physics and more maths qualifications than I can shake a stick at; to me, multiplication is repeated addition. I’m not sure what the teacher is trying to do here, but I do think the outcome of what they’re trying to do is far more complicated than the simple “multiplication is repeated addition”. I also happen to have an 8-year-old going through third grade right now, and when we were talking through his homework, it was quite clear that using simple concepts he already knew (addition & subtraction) to explain slightly more complex things that he was learning (multiplication and division) was really useful to him. As I recall it being to me. [aside] I think the maths schedule is more advanced now than it was in my day anyway - he only did multiplication and division this year, but he also did algebra and simultaneous linear equations now, as in: a + b + 8 = 24 a - b = 4 “Solve for a and b” Pretty sure I only did that in senior school (11 and up), not at age 8. No powers as yet (presumably they’ll come after the multiplication/division stuff), so no quadratic formula, but still... [/aside]
- munchbunny 6y ago
  I think the useful distinction is that, when you teach multiplication as a mechanical computation (arithmetic) it's useful to talk about it as repeated addition. As you get to negative numbers, rational/irrational numbers, complex numbers, matrices, etc. it becomes more useful to think about multiplication in more abstract ways, among which repeated addition is still often a useful way to look at it. I also think it's not particularly useful to talk about those other ways to think about multiplication until you actually need to. It's too easy once you've mastered the concepts to forget how beginners look at them and struggle to understand them. I think the author isn't remembering what it's like to try to understand multiplication as a new concept - I certainly can't remember.
  - spacedcowboy 6y ago
    Your last two paragraphs are, I believe, the crux of my own argument. Sure, matrices aren’t even commutative, and complex numbers have their own quirks because of i^2 == -1, but these concepts build on the earlier axioms the kid has learnt. Our entire education system is built on “lies-to-children”, and as you progress they point out that what you comfortably believed was a gross simplification. This is no different.

drewcoo 6y ago

Agreed. Multiplier and multiplicand are different words but the commutative property says their values can swap equivalently, so . . . what was the author's point again?

tomxor 6y ago
This was exactly my thought, I don't understand why those two word exist since it's commutative. Perhaps all this confusion is purely semantic.
- ThePadawan 6y ago
  They exist to distinguish the element being operated on (the LHS) and what it is operated on by (the RHS). Total technicality, but I could see myself using the term multiplicand/multiplier in my code if I had to implement e.g. a stack-based parser for arithmetic expressions.
  - tomxor 6y ago
    > to distinguish the element being operated on (the LHS) I'm not sure what you mean, they are both operated on, it is a binary operator and commutative, there is literally no difference.
    - anbende 6y ago
      If we are just looking at numbers with no units attached there is no difference. As soon as there are units, a difference arises. I will collect 12 apples 5 times. The 12 apples are the multiplicand and the 5 times is the multiplier. The answer retains the units from the multiplicand.
      - np_tedious 6y ago
        Still don't see the difference. Density times volume is mass. Volume times density is mass. Distance times force is work/energy... "5 times I will connect 12 apples"
        KMag 6y ago
        You're correct. The GP in incorrect about their dimensional analysis. 5 apples 12 times yields apples because it's apples times a dimensionless scalar (count). Newtons times meters is always Newton-meters, never Newtons or meters. Units are never magically dropped in dimensional analysis.
    - simias 6y ago
      I agree that it's a generally pointless distinctions, although it might be useful in some cases such as number systems where the multiplication isn't commutative, or for a particular implementation where the distinction matters. After all if you were to code a multiplication that was implemented naively as a series of additions it'd be generally much faster todo 2x1000 than 1000x2. To return to TFA I think the author is talking from a pedagogical standpoint, that teaching that multiplication is just a bunch of additions under a trench coat is not the best way to go. I'm not sure that I agree personally. In particular this bit regarding multiplier/multiplicand makes zero sense to me: >Different names indicate a difference in function. The multiplier and the multiplicand are not conceptually interchangeable. It is true that multiplication is commutative, but (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. Of course 3 rows and 2 rows aren't the same, but what does it have to do with the order of the multiplication? Isn't 2 rows x 3 chairs the same thing as 3 chairs x 2 rows? It's a bizarre argument.
- SAI_Peregrinus 6y ago
  Multiplication over the reals is commutative. Matrix multiplication of non-square matrices isn't. Multiplication in a Ring isn't necessarily commutative. Other algebraic structures also have non-commutative multiplication. One could argue that these things aren't "multiplication" even if they are "products" since they don't satisfy all the properties of multiplication over the reals. But it is common to call the use of the product operation "multiplication", at least in cases where there's only one product operation to use. EG Geometric Algebra has Inner, Outer, and Geometric products, so calling them "multiplication" seems less common IME.
  - qsort 6y ago
    I really don't get the point you're making. If we're going to pull out random examples, monoids aren't guaranteed to be abelian; strings and concatenation form a monoid that's not abelian. If you have enough mathematical sophistication to conceptualize a non-commutative ring, you're well past the point where naming conventions are even remotely an issue. The original article was contrasting addition and multiplication on the basis that addends are called the same while factors are supposed to be called differently, which not only makes no sense (it's just a naming convention), but it also breaks down when you have more than two factors: what is the "c" in a x b x c called? Or we're talking about non-associative operations now?
    - SAI_Peregrinus 6y ago
      My point is only that the commutative property is not inherent to all multiplication operations, so there can be a distinction between the operands. It's not necessarily a useful distinction, and in the usual use of multiplication it's utterly useless and only adds confusion. But matrix multiplication is taught in high school (and usually promptly forgotten), it's not particularly advanced math. Personally I'm of the opinion that the terminology is muddled. There's no need to distinguish the operands of a multiplication over any of the usual domains (reals, rationals, integers, etc). And when you reach the point where it does become important there's generally more than one product operation and we should stop calling it multiplication. "Matrix multiplication" is a bad term. You also typically wouldn't name the operands, since as you note there can be more than two!
- karmakaze 6y ago
  Not all multiplication is commutative, e.g. matrices.
gugagore 6y ago
If, for a moment, you conceptualize of multiplication on non-negative whole numbers as repeated addition, then this is the algorithm: procedure product(multiplier, multiplicand) acc := 0 for i = 1 to multiplier acc := acc + multiplicand end return acc end Swapping the arguments is a different computation. But after thinking about it, you realize that you get the same answer all the same. That's the point of > (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. The point with bringing up dimensional analysis is that the above algorithm doesn't work because what does it mean to do `for i = 1 to 3 chairs/row`? You might think of it like procedure product(multiplier, multiplicand) acc := "0" # an "absolute" zero that cooperates with any dimension each single_multiplier in multiplier acc := acc + (multiplicand * single_multiplier) end return acc end But then what is `(multiplicand * single_multiplier)` ?
andi999 6y ago
> (2 rows × 3 chairs/row) is not the same as (3 rows × 2 chairs/row), even though both sets contain 6 chairs. The author has it all backwards. Basically that you can turn this arrangement by 90 degrees (and turning the chairs, if you do not modell them by points by 90 degrees) is the reason why multiplication is commutative. It is non trivial that 5+5+5+...5 (100 times) is the same as 100+100+100+100+100

midjji 6y ago

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

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.

spacedcowboy 6y ago

I have a PhD in physics and more maths qualifications than I can shake a stick at; to me, multiplication is repeated addition. I’m not sure what the teacher is trying to do here, but I do think the outcome of what they’re trying to do is far more complicated than the simple “multiplication is repeated addition”. I also happen to have an 8-year-old going through third grade right now, and when we were talking through his homework, it was quite clear that using simple concepts he already knew (addition & subtraction) to explain slightly more complex things that he was learning (multiplication and division) was really useful to him. As I recall it being to me. [aside] I think the maths schedule is more advanced now than it was in my day anyway - he only did multiplication and division this year, but he also did algebra and simultaneous linear equations now, as in: a + b + 8 = 24 a - b = 4 “Solve for a and b” Pretty sure I only did that in senior school (11 and up), not at age 8. No powers as yet (presumably they’ll come after the multiplication/division stuff), so no quadratic formula, but still... [/aside]

munchbunny 6y ago
I think the useful distinction is that, when you teach multiplication as a mechanical computation (arithmetic) it's useful to talk about it as repeated addition. As you get to negative numbers, rational/irrational numbers, complex numbers, matrices, etc. it becomes more useful to think about multiplication in more abstract ways, among which repeated addition is still often a useful way to look at it. I also think it's not particularly useful to talk about those other ways to think about multiplication until you actually need to. It's too easy once you've mastered the concepts to forget how beginners look at them and struggle to understand them. I think the author isn't remembering what it's like to try to understand multiplication as a new concept - I certainly can't remember.
- spacedcowboy 6y ago
  Your last two paragraphs are, I believe, the crux of my own argument. Sure, matrices aren’t even commutative, and complex numbers have their own quirks because of i^2 == -1, but these concepts build on the earlier axioms the kid has learnt. Our entire education system is built on “lies-to-children”, and as you progress they point out that what you comfortably believed was a gross simplification. This is no different.