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> 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.

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.

> 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.

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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.

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This property literally holds in all fields and rings (but not groups).

> n * 0 = n * (1 + (-1)) = n + (-n) = 0

You don't think it's easier to say "n·5 = n·(5+0) = n·5 + n·0"?

The (1 + (-1)) thing was very precisely written.

All rings have 0 and 1 as elements. 0 is the additive identity. 1 is the multiplicative identity. 0 and 1 are NOT necessarily numbers. In Linear Algebra of 2x2 matricies, 0 is:

    [ 0 0
      0 0 ]

And 1 is:

    [ 1 0
      0 1 ]

Because A * 1 == A, A * 0 == [0 0; 0 0]... etc. etc. In general, you cannot really assume much more than "0 exists" and "1 exists" when working with Rings (at least, if you want your proof to extend out to all possible ring algebras). 0 and 1 may look like something you wildly don't expect... they're abstract labels that are kind of undefined aside from being additive-identity or multiplicative-identity, respectively.

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As such, the concept of "5" does not necessarily exist in all possible Ring-systems. "5" exists in GF(5) for example, but not really in GF(3). Case in point, what does "5" mean in 2x2 Matrix Linear Algebra over GF(2)?

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"-A" is called the additive inverse of A, which also exists in all rings. A - A = 0.

In GF(5), -1 is 4 for example. In 2x2 Linear Algebra, -1 is [-1 0; 0 -1]. In Real Numbers, -1 is... well... -1.

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Anyway, the A * 0 == A * (1 + (-1)) == A - A == 0 thing is built up from fundamental portions of Ring theory. As such, the proof I constructed at first applies to all rings. (And then later, I did an example in the GF(5) system as a specific example).

> As such, the concept of "5" does not necessarily exist in all possible Ring-systems.

Don't think of 5 as a quantity; think of it as a variable name. The proof only depends on the concept of addition and an additive identity (and distribution of multiplication over addition, which you're using anyway); no property of 5 appeared.

> the A * 0 == A * (1 + (-1)) == A - A == 0 thing is built up from fundamental portions of Ring theory. As such, the proof I constructed at first applies to all rings.

To repeat myself:

  n·5 = n·(5+0)         [definition of 0]
  n·(5+0) = n·5 + n·0   [multiplication is distributive over addition]
  n·0 = 0               [definition of 0]

Every step in that proof is a direct application of one of the ring axioms; it doesn't matter what 5 is.

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, although we tweaked the formulas to look like the simplest ways for our own brains.

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.)

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.

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.

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)

Without the distributive property you have two magmas in general for the same set that don't see each other. I'm not sure what your point is. When you have two binary operations you need some sort of distributive property to build a structure.

All this is trivial, if you consider a ring, you get a·0=0 as a property, if your starting point is the Peano axioms for the arithmetic of natural numbers that's one of them, for the latter seeing it as "repeated addition" makes no sense, for the former, well you have a ring, you have two binary operations, not one, and of course you have some form of distributive property or else you'd be studying this set with just one binary operation at a time.

I'd like to see how "repeated addition" works in polynomial rings.

> I'd like to see how "repeated addition" works in polynomial rings.

Consider the following polynomial: x0 * b^0 + x1 * b^1 + x2 * b^2 ... xn * b^n, where "n" goes to both positive infinity and negative infinity.

When "b = 10" and when "x" can be numbers from [0-9], we have the so called base-10 set of real numbers, do we not? IIRC, if b = sqrt(-1) * 10, we then have the set of complex numbers (a non-intuitive result. I may have made a mistake somewhere, but I assure you there's a surprising property along those lines).

That's the funny thing about real numbers and complex-numbers. Real numbers and even complex-numbers ARE polynomials, and therefore a polynomial ring. 3.1415926 == 3 * 10^0 + 1 * 10 ^-1 + 4 * 10 ^-2 ...

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I'm using a lot of words here. But all I'm saying is once again: Pi * 3 == 3.14... * 3 == 3 * 3 + 0.1 * 3 + 0.04 * 3 + ... == 9.42...

We can evaluate 3 * Pi by splitting Pi up into a set of additions (3 + 0.1 + 0.04 + 0.001...), even if that set of additions is infinite. Then evaluate 3*(each component). This is possible because Pi is easily represented as a polynomial X0 * 10^0 + X1 * 10^-1 + ... Xn * 10^-n.

There's a reason why polynomial multiplication is usually called "Carry-free multiplication". Because Real-numbers are just polynomials where you have that annoying "carry the one" property to keep track of. Remove the "carry the one" property (in say: Galois extension fields), and all the math still works.

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Hmmmm... I probably could have said all that in fewer words. TL;DR: "Real numbers ARE a polynomial ring". (And complex numbers are probably a polynomial ring, I just forgot how to prove that factoid)

> I'd like to see how "repeated addition" works in polynomial rings.

EDIT: Just thought of a cute and simple retort. You ever do a CRC32 check? There ya go.

A polynomial is a polynomial, a decimal representation of a real number is a decimal representation of a real number, and your representation of complex numbers has funny properties once you begin exponentiating that.

I'd like to see how you'd show kids that: (1+x+x^2)·(1-x^3) is a "repeated addition", both belong to the ring Z[x].

> I'd like to see how you'd show kids that: (1+x+x^2)·(1-x^3) is a "repeated addition", both belong to the ring Z[x].

Is that not just (1 * (1-x^3) + x * (1-x^3) + x^2 * (1-x^3)) ??

The polynomial itself gives us the means at which we logically split up the multiplication into component parts.

Just as 3.14 * 3 == 3 * 3 + 0.1 * 3 + 0.04 * 3, when we move onto polynomials, we do the same exact thing. EDIT: remember, ALL REAL NUMBERS ARE POLYNOMIALS with a base of 10.

Or to put it another way: when x == 10, your polynomial of (1 + x + x^2) * (1-x^3) == 111 * (-999). That is to say: real numbers are simply polynomials where "x" has been defined to be a particular number, instead of an abstract entity. We call that number the radix-base.

If you instead defined the base to be x = 16 (hexadecimal numbers), you'd get 111 * (-FFF), which you'll find will satisfy similar properties. Now leave x-undefined (since it could be 10 or 16), and what do you get?

Polynomial math. Or so called "Carry-less multiplication" (https://en.wikipedia.org/wiki/Carry-less_product). We don't have a ring yet though: we still need to perform a modulus on all those polynomials to return to a proper ring (and if the modulus is irreducable, we have a Galois field). But we can already see how polynomials and the Reals are so closely related.

Precisely, you're not repeating p(x) q(x)-times, you've used that p(x) is a linear combination of monomials and then the distributive property of Z[x].

Now, you could argue that this is exactly a way to "add repeatedly", but at some point pushing analogies stops being helpful to your students.

> Now, you could argue that this is exactly a way to "add repeatedly", but at some point pushing analogies stops being helpful to your students.

That's not what this blogpost is arguing about. This blogpost is arguing that "Multiplication is Repeated Addition" is unhelpful at the elementary school level and stops being true at some point.

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My argument is otherwise. "Multiplication is Repeated Addition" is clearly helpful in grade school. Almost everybody I know has learned Multiplication through that method.

Secondly: I cannot think of a single instance where its not true. Yes, I've had to use linear-combinations to extend it out to polynomials, but clearly the property holds even in polynomial-land.

Its not useful to teach multiplication of polynomials with "Repeated Addition". But the advice is "not wrong", in fact, polynomial multiplication continues to see many similarities with Real and Complex multiplication. Especially if we consider a "Basis" to be analogs to the thing that's repeatedly-added.

I think there might be a bit of a problem with quaternions.

> 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 repeated multiplication. And this is so fundamental that 1 is frequently referred to by this property, as "the empty product".

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.

If you want to work up to it conceptually, then I'd say consider the meaning of something like

  5·3 + 2·7

We add 3 five times, and then we add 7 twice. It is then easy to extend this to

  5·3 + 2·7 + 0·4

and say, OK, add 3 five times, and then add 7 twice, and then add 4 zero times. And you get 29.

At that point you notice that when you say 5·3 means "add 3 five times", you forgot to say what you were adding it to. You're adding it to the identity, 0.

And finally we say that starting from 0 and adding something zero times leaves you where you started, at 0, so we can observe that 0·n = 0.

If you formalize that, you'll end up either deriving or postulating the distributive property, depending on what you start with. But it isn't arbitrary; it's not a coincidence that (as I mentioned above) exponentiation by 0 gives the multiplicative identity, and (as I haven't mentioned yet) exponentiation distributes over multiplication the same way multiplication distributes over addition.

(Asymmetry does start creeping in; aggregate exponentiation isn't as nice since exponentiation doesn't have the nice properties that addition and multiplication do.)

> exponentiation distributes over multiplication

things that irks me

    a**b * a**c == a**(b + c) != a**(b * c)

for exponentiation to distribute over multiplication, it must become addition, thus reinforcing the multiplication IS repeated addition?

Distributivity of exponentiation over multiplication looks like

  (ab)^x = (a^x)(b^x)

You're talking about "being exponentiated", not exponentiation. This one isn't a commutative operation.

> commutativity

yes my mistake

> 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.

Why is this not a good idea?

Can’t we just accept/postulate that the additive identity is different from the multiplicative identity?

And still define a relationship between the addition and multiplication?

Maybe I misunderstand the issue..

(I’m not trying to be pedantic, but my math background has some holes :)

> 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.

pi(3 + 5) == 3pi + 5pi == (Pi + Pi + Pi) + (Pi + Pi + Pi + Pi + Pi) == 8-pi.

Am I missing something here?

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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.

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?

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...)

> Break-up pi in whatever means you think is reasonable.

You can't. Pi is irrational.

> You can't. Pi is irrational.

You just need to break it up infinitely times. We usually call the sequence 31415926...

> You just need to break it up infinitely times.

I've responded to this elsewhere in the thread: I don't think adopting increasingly perverse interpretations of "repeated addition" as you try to include more and more numbers is a useful way to teach multiplication.

what they described is identical to long multiplication. That's not perverse, it's how most people multiply numbers.

You extend to the rationals in the natural way, and then use continuity to define what happens for irrationals. (Of course, in a discrete context that doesn't apply. But the article's author is a grade school teacher.)

> Am I missing something here?

Sorry, I should have said substitute b = pi or c = pi. Or see the sibling post by iio8999.

pi isn't a counting number like you teach to children.

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.

"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.

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.

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).

Define π times e in terms of repeated addition.

I wouldn't say multiplication is literally repeated addition. I'd say it reduces to repeated addition when the multiplier is a natural number.

I am afraid I did not make myself clear. I don't mean that multiplying by irrational numbers is repeated addition. What I mean is that multiplication by irrational (or even non-integers) was defined to generalize the earlier notion of multiplying by positive integers.

Also I mean "literally" literally :). Etymologically, Multiplication is the act of taking/creating many (multi in Latin) copies of something (as in "multiplying breads").

That these generalizations don't fit the concept of repeated addition doesn't erase the fact that the primitive concept meant precisely that. Whether we decide to hide that fact from children can be a conscious decision, but at least it should be deliberate.

Maybe it is not literally addition but it certainly reduces to addition in all cases: https://news.ycombinator.com/item?id=26329855

Nope. I'll reply there.

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.

> 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)?

When you multiply two negative numbers.

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)

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.

The original commenteer I responded to (not you), said:

>Multiplication has nothing to do with addition and at some point we have to stop teaching students that it's related.

So thank you for explaining the actual relationship!