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.
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.
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]
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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.
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.
> 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.)
> 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 :)
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.