10 ms·

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 str

by 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)
    - mnl 6y ago
      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.
      - dragontamer 6y ago
        > 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 ... ------------ 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. -------- 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.
- thaumasiotes 6y ago
  > 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"?
  - dragontamer 6y ago
    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. -------------- 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)? -------- "-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. ---------- 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).
    - thaumasiotes 6y ago
      > 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.
cannabis_sam 6y ago
> 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 :)
thaumasiotes 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. 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".
- mnl 6y ago
  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.
  - thaumasiotes 6y ago
    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.)

by mnl 6y ago

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)
  - mnl 6y ago
    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.
    - dragontamer 6y ago
      > 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 ... ------------ 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. -------- 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.
thaumasiotes 6y ago
> 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"?
- dragontamer 6y ago
  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. -------------- 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)? -------- "-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. ---------- 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).
  - thaumasiotes 6y ago
    > 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.

cannabis_sam 6y ago

> 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 :)

thaumasiotes 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. 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".

mnl 6y ago
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.
- thaumasiotes 6y ago
  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.)