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Except this isn't a pedantic falsehood. Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. Un

Except this isn't a pedantic falsehood. Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. Understanding ratio is key to understanding a lot of the physical sciences and you can build a really good intuitive understanding of a lot of simple physical concepts if you can just do dimensional analysis. But if you think of multiplication and division as a kind of addition then it all breaks down and you can't reason about much at all with those tools. In fact you can't even get over the hurdle of doing dimensional analysis in chemistry or physics in the first place. It doesn't make any sense. And you'll have been set up to fail by your teachers. They're not saying "we should be teaching kindergarteners using Euclid" they're suggesting that focusing on the process of doing multiplication hides a lot of very useful insights about why someone would do multiplication in the first place. The biggest problem with math as it was taught to me in grade school is that it was focused on rote memorization of a procedure. Which is fine if you need to calculate how long to cut 2x4s or stair stringers but it doesn't help you at all if you need to understand why that works the way it does. And god help you if nobody ever shows you the exact method of calculation for those two things in the first place. You're set up to fail if you only focus on process.

dragontamer 6y ago
> Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. I just finished studying Galois Fields, which LITERALLY redefines multiplication and addition operators to study new forms of math (IE fields in particular) In all fields and rings, multiplication and addition are related by the distributive property. A(b + c) is equal to Ab + Ac, in literally every form of math you can every think of. A and b could be matrices, vectors, polynomials, prime numbers, integers, rational, complex, real, or Galois polynomials over a weird modulus ring thingy inside of a vector inside of a matrix. Doesn't matter, addition and multiplication are and always defined in that manner.
- pdonis 6y ago
  > In all fields and rings, multiplication and addition are related by the distributive property. That's a necessary condition for "repeated addition" to work for multiplication, but not a sufficient one. Try plugging in A = pi to the distributive formula and see how well "repeated addition" works.
  - dragontamer 6y ago
    pi(3 + 5) == 3pi + 5pi == (Pi + Pi + Pi) + (Pi + Pi + Pi + Pi + Pi) == 8-pi. Am I missing something here? -------- This "multiplication is repeated addition through the distributed property" thing works on freaking __matricies__. They don't even have to be numbers or even related. Pi * ([ 1 0 ; + [ 1 0 ; 0 1 ] 2 1 ]) Pi * [ 1 0 ; + Pi * [ 1 0 ; 0 1 ] 2 1 ] [ Pi 0 ; + [ Pi 0 ; 0 Pi ] 2Pi Pi ] [ 2Pi 0 ; 2Pi 2Pi ] You can even "break up" the two matricies into its component parts: Pi * ( 1 0 + 0 0 + 1 0 + 0 0 + 0 0 ) 0 0 0 1 0 0 2 0 0 1 And you'll still get the correct answer: multiplication is repeated addition. Even with matricies. Even with vectors. Even in Galois Fields. Even with rational numbers. Even with real numbers. Even with Complex numbers. This property holds through all forms of math that I'm aware of.
    - iio8999 6y ago
      I think the issue is more about: 3 * pi = pi + pi + pi but how do you represent the other distribution, where 3 is added together pi times?
      - dragontamer 6y ago
        Break-up pi in whatever means you think is reasonable. 3 * (3 + 0.1 + 0.04 + 0.001 + 0.0005...) Aka: 9.4245... You know, how we've been multiplying 3 * pi for our whole lives. We split pi up into an infinite sum of component numbers (3, 1, 4, 1, 5, 9, 2, 6...) and then combine them together by individually multiplying the parts (3 * 3 + 3 * 0.1 + 3 * 0.04...)
        6y ago
- mnl 6y ago
  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+...+1), invoke the distributive property for · wrt + and express it as: n·m=n+...+n (m-times), so it looks like "repeated addition" for integers in this case. But it's not a good idea to let ourselves get carried away, we still have two binary operations going on. At any rate we have to impose that n·0=0, which can't be writen cleverly as "repeated addition" and worked up backwards.
  - dragontamer 6y ago
    > At any rate we have to impose that n·0=0, which can't be writen cleverly as "repeated addition" and worked up backwards. n * 0 = n * (1 + (-1)) = n + (-n) = 0. ----- The 0-element in a Galois Field works identically btw. In GF(5), the 0 element is 5 (5 mod 5 == 0). n * (5) == n * 0 == n * (1 + (-1)) == n * (1 + 4) == 0 mod5. For example, if we take "n" == 2, 2 * 5 == 10 mod 5 == 0. 2 * (1 + 4) == 2 + 8 == 2 + 3 (mod 5) == 5 mod 5 == 0. Etc. etc. ------- This property literally holds in all fields and rings (but not groups).
    - mnl 6y ago
      Repeated addition that ain't... It boggles my mind that you've been studying Galois theory yet somehow try to reduce the algebraic structures associated to two binary operations to playing with one of them.
      - dragontamer 6y ago
        The algebraic structures multiplication brings is different from the algebraic structures from addition. That's the point of rings (as opposed to groups). Nonetheless, in a ring (and all fields are rings), multiplication must and always is related to addition, through the distributed property (which I argue, the distributed property IS the mathematical term for "repeated addition"). Without the distributed property, you have no ring. You at best only have a group. Therefore, all multiplication operators ever defined (or more precisely, all rings) must have multiplication related to addition: (A * (B+C) == AB+AC)
613style 6y ago
We can't really talk about what multiplication "is" or "isn't" independently of context. It's an operation on two objects and the context its used in is necessary for defining the operation. Though it can be helpful to think of multiplication as scaling or rotation in certain contexts, or as repeated addition in others, none of those are a universal truth. To say it's not related to addition at all is also too broad to be true. They're related in many useful ways that other commenters have pointed out, in addition to the obvious way that in some situations you can define one in terms of the other. Even at the cutting edge, our inability to prove the Goldbach Conjecture might have something to do with not fully understanding the deepest relationships between the operations.
cannabis_sam 6y ago
> Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. My apologies if this is a stupid question, but when does the intuitive (layman) understanding of multiplication as repeated addition break down (mathematically)?
- betterunix2 6y ago
  When you multiply two negative numbers.
  - cannabis_sam 6y ago
    As long as we have -1 * -1 = 1 can’t we recover the interpretation of multiplication as repeated addition? As in -x * -y = -1 * -1 * x * y = 1 * x * y = x * y Or is the issue that people are claiming a stronger relationship than an interpretation? (I guess we would need x * 0 = 0 as well)
    - betterunix2 6y ago
      Well, first, why should (-1) * (-1)= 1? Also, you take for granted that -x = (-1) * x, and why should that be the case? You had asked when the idea that multiplication is repeated addition breaks down. What does (-1) * (-1) mean if multiplication is repeated addition? Really all these statements, including x * 0 = 0, are about the relationship between addition and multiplication. 0 is the additive identity, so why should we expect anything special about it when it comes to multiplication? The answer is that addition and multiplication are related by the distributive law: 0 * x + x = 0 * x + 1 * x = (0+1) * x = x Thus (0 * x) is an identity element for addition, but the additive identity is unique (this can be proved using only the properties of addition) and therefore 0 * x = 0. From here we can show: (-1)*x + x = (-1)*x + 1* x = (-1 + 1) * x = 0 * x = 0 Thus (-1)*x is an additive inverse of x, and again, we can show that additive inverses are unique and therefore (-1)*x = -x. Now we have everything needed for your proof. Again, the idea that multiplication is repeated addition plays no role here. What actually matters is the distributive law, the meaning of "0", "1", and of "negative," and that multiplication and addition are closed (i.e. the sum or product of two numbers is a number). Those properties are part of the definition of addition and multiplication (along with the associative and commutative laws, and some form of cancellation for multiplication / multiplicative inverses) and can be taken as axioms.
aaplok 6y ago
Multiplication is literally repeated addition, as in "taking a number multiple times". It is in the name. The fact that the operation is so useful that it has been generalised to the point where that original meaning is eventually lost through more and more abstractions doesn't invalidate that, because with all the generalisation and abstraction, multiplication as repeated addition still works, and any generalisation is expected to leave that property intact. This is not unique to multiplication. Modern mathematics is all about extracting fundamental properties that make something true, to generalize results while keeping the original simpler statement true. Starting from the generalisations without explaining the intellectual process it took us to get there means just giving facts without explaining them.
- betterunix2 6y ago
  "Multiplication is literally repeated addition" How do you repeat something a negative number of times? What does it mean to multiply two negatives, if multiplication is repeated addition? This is not an overly generalized concept, we teach negative numbers to elementary school students.
  - aaplok 6y ago
    I don't know what you mean by "overly". Multiplication by negative numbers is a generalisation of the earlier concept of multiplication by positive numbers. In a sense even multiplying by 1 is a generalisation (at least if you read Euclid he doesn't really define multiplying by 1). I don't claim that we should not teach generalisations to children. What I claim is that of we teach generalisations we at least should be aware that they are generalisations, and maybe even tell them about it.
    - betterunix2 6y ago
      That sounds a bit backwards to me. If anything, negative numbers are the generalization (as is 0), and we should have a concept of multiplication that works for such generalizations. Euclid's approach to arithmetic always seemed strained to me, something motivated by a religious view of math where a compass and straightedge were the only tools anyone should need. Repeated addition is just a technique for multiplication. It is one of many techniques we teach children, and we should leave it at that -- a technique. When we get to multiplication by negative numbers or fractions, we teach children other techniques, and we do not feel any need to try to "define" multiplication in terms of those techniques. Why should we "define" multiplication in terms of repeated addition, and then do circles around ourselves trying to generalize that "definition?" We can just say that multiplication is one operation we can do on numbers; addition is another, and they are related by the distributive law (and it from the distributive law that we can derive the various techniques we use for multiplication).

dragontamer 6y ago

> Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. I just finished studying Galois Fields, which LITERALLY redefines multiplication and addition operators to study new forms of math (IE fields in particular) In all fields and rings, multiplication and addition are related by the distributive property. A(b + c) is equal to Ab + Ac, in literally every form of math you can every think of. A and b could be matrices, vectors, polynomials, prime numbers, integers, rational, complex, real, or Galois polynomials over a weird modulus ring thingy inside of a vector inside of a matrix. Doesn't matter, addition and multiplication are and always defined in that manner.

pdonis 6y ago
> In all fields and rings, multiplication and addition are related by the distributive property. That's a necessary condition for "repeated addition" to work for multiplication, but not a sufficient one. Try plugging in A = pi to the distributive formula and see how well "repeated addition" works.
- dragontamer 6y ago
  pi(3 + 5) == 3pi + 5pi == (Pi + Pi + Pi) + (Pi + Pi + Pi + Pi + Pi) == 8-pi. Am I missing something here? -------- This "multiplication is repeated addition through the distributed property" thing works on freaking __matricies__. They don't even have to be numbers or even related. Pi * ([ 1 0 ; + [ 1 0 ; 0 1 ] 2 1 ]) Pi * [ 1 0 ; + Pi * [ 1 0 ; 0 1 ] 2 1 ] [ Pi 0 ; + [ Pi 0 ; 0 Pi ] 2Pi Pi ] [ 2Pi 0 ; 2Pi 2Pi ] You can even "break up" the two matricies into its component parts: Pi * ( 1 0 + 0 0 + 1 0 + 0 0 + 0 0 ) 0 0 0 1 0 0 2 0 0 1 And you'll still get the correct answer: multiplication is repeated addition. Even with matricies. Even with vectors. Even in Galois Fields. Even with rational numbers. Even with real numbers. Even with Complex numbers. This property holds through all forms of math that I'm aware of.
  - iio8999 6y ago
    I think the issue is more about: 3 * pi = pi + pi + pi but how do you represent the other distribution, where 3 is added together pi times?
    - dragontamer 6y ago
      Break-up pi in whatever means you think is reasonable. 3 * (3 + 0.1 + 0.04 + 0.001 + 0.0005...) Aka: 9.4245... You know, how we've been multiplying 3 * pi for our whole lives. We split pi up into an infinite sum of component numbers (3, 1, 4, 1, 5, 9, 2, 6...) and then combine them together by individually multiplying the parts (3 * 3 + 3 * 0.1 + 3 * 0.04...)
      - 6y ago
mnl 6y ago
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+...+1), invoke the distributive property for · wrt + and express it as: n·m=n+...+n (m-times), so it looks like "repeated addition" for integers in this case. But it's not a good idea to let ourselves get carried away, we still have two binary operations going on. At any rate we have to impose that n·0=0, which can't be writen cleverly as "repeated addition" and worked up backwards.
- dragontamer 6y ago
  > At any rate we have to impose that n·0=0, which can't be writen cleverly as "repeated addition" and worked up backwards. n * 0 = n * (1 + (-1)) = n + (-n) = 0. ----- The 0-element in a Galois Field works identically btw. In GF(5), the 0 element is 5 (5 mod 5 == 0). n * (5) == n * 0 == n * (1 + (-1)) == n * (1 + 4) == 0 mod5. For example, if we take "n" == 2, 2 * 5 == 10 mod 5 == 0. 2 * (1 + 4) == 2 + 8 == 2 + 3 (mod 5) == 5 mod 5 == 0. Etc. etc. ------- This property literally holds in all fields and rings (but not groups).
  - mnl 6y ago
    Repeated addition that ain't... It boggles my mind that you've been studying Galois theory yet somehow try to reduce the algebraic structures associated to two binary operations to playing with one of them.
    - dragontamer 6y ago
      The algebraic structures multiplication brings is different from the algebraic structures from addition. That's the point of rings (as opposed to groups). Nonetheless, in a ring (and all fields are rings), multiplication must and always is related to addition, through the distributed property (which I argue, the distributed property IS the mathematical term for "repeated addition"). Without the distributed property, you have no ring. You at best only have a group. Therefore, all multiplication operators ever defined (or more precisely, all rings) must have multiplication related to addition: (A * (B+C) == AB+AC)

613style 6y ago

We can't really talk about what multiplication "is" or "isn't" independently of context. It's an operation on two objects and the context its used in is necessary for defining the operation. Though it can be helpful to think of multiplication as scaling or rotation in certain contexts, or as repeated addition in others, none of those are a universal truth. To say it's not related to addition at all is also too broad to be true. They're related in many useful ways that other commenters have pointed out, in addition to the obvious way that in some situations you can define one in terms of the other. Even at the cutting edge, our inability to prove the Goldbach Conjecture might have something to do with not fully understanding the deepest relationships between the operations.

cannabis_sam 6y ago

> Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related. My apologies if this is a stupid question, but when does the intuitive (layman) understanding of multiplication as repeated addition break down (mathematically)?

betterunix2 6y ago
When you multiply two negative numbers.
- cannabis_sam 6y ago
  As long as we have -1 * -1 = 1 can’t we recover the interpretation of multiplication as repeated addition? As in -x * -y = -1 * -1 * x * y = 1 * x * y = x * y Or is the issue that people are claiming a stronger relationship than an interpretation? (I guess we would need x * 0 = 0 as well)
  - betterunix2 6y ago
    Well, first, why should (-1) * (-1)= 1? Also, you take for granted that -x = (-1) * x, and why should that be the case? You had asked when the idea that multiplication is repeated addition breaks down. What does (-1) * (-1) mean if multiplication is repeated addition? Really all these statements, including x * 0 = 0, are about the relationship between addition and multiplication. 0 is the additive identity, so why should we expect anything special about it when it comes to multiplication? The answer is that addition and multiplication are related by the distributive law: 0 * x + x = 0 * x + 1 * x = (0+1) * x = x Thus (0 * x) is an identity element for addition, but the additive identity is unique (this can be proved using only the properties of addition) and therefore 0 * x = 0. From here we can show: (-1)*x + x = (-1)*x + 1* x = (-1 + 1) * x = 0 * x = 0 Thus (-1)*x is an additive inverse of x, and again, we can show that additive inverses are unique and therefore (-1)*x = -x. Now we have everything needed for your proof. Again, the idea that multiplication is repeated addition plays no role here. What actually matters is the distributive law, the meaning of "0", "1", and of "negative," and that multiplication and addition are closed (i.e. the sum or product of two numbers is a number). Those properties are part of the definition of addition and multiplication (along with the associative and commutative laws, and some form of cancellation for multiplication / multiplicative inverses) and can be taken as axioms.

aaplok 6y ago

Multiplication is literally repeated addition, as in "taking a number multiple times". It is in the name. The fact that the operation is so useful that it has been generalised to the point where that original meaning is eventually lost through more and more abstractions doesn't invalidate that, because with all the generalisation and abstraction, multiplication as repeated addition still works, and any generalisation is expected to leave that property intact. This is not unique to multiplication. Modern mathematics is all about extracting fundamental properties that make something true, to generalize results while keeping the original simpler statement true. Starting from the generalisations without explaining the intellectual process it took us to get there means just giving facts without explaining them.

betterunix2 6y ago
"Multiplication is literally repeated addition" How do you repeat something a negative number of times? What does it mean to multiply two negatives, if multiplication is repeated addition? This is not an overly generalized concept, we teach negative numbers to elementary school students.
- aaplok 6y ago
  I don't know what you mean by "overly". Multiplication by negative numbers is a generalisation of the earlier concept of multiplication by positive numbers. In a sense even multiplying by 1 is a generalisation (at least if you read Euclid he doesn't really define multiplying by 1). I don't claim that we should not teach generalisations to children. What I claim is that of we teach generalisations we at least should be aware that they are generalisations, and maybe even tell them about it.
  - betterunix2 6y ago
    That sounds a bit backwards to me. If anything, negative numbers are the generalization (as is 0), and we should have a concept of multiplication that works for such generalizations. Euclid's approach to arithmetic always seemed strained to me, something motivated by a religious view of math where a compass and straightedge were the only tools anyone should need. Repeated addition is just a technique for multiplication. It is one of many techniques we teach children, and we should leave it at that -- a technique. When we get to multiplication by negative numbers or fractions, we teach children other techniques, and we do not feel any need to try to "define" multiplication in terms of those techniques. Why should we "define" multiplication in terms of repeated addition, and then do circles around ourselves trying to generalize that "definition?" We can just say that multiplication is one operation we can do on numbers; addition is another, and they are related by the distributive law (and it from the distributive law that we can derive the various techniques we use for multiplication).