uncool

Well, the reason it is of order 2 is because 2a must annihilate everything...with something of order 27 you could probably find such an a.=Uncool-

Right, that's exactly what I was trying to say, but you said it much better... =Uncool-

Well, it's because this equation is one of the few that's really nice in that the solutions are "obvious." The solutions are of the form e^(2*pi*i*k/2005), let k = 0, 1, 2, ..., 2004. This works because each to the 2005 is e^(2*pi*i*k) = 1. e^-x does not work as it is -y, not + y. =Uncool-

I'm referring to non-square matrices, which have inverse on at most one side. An example: [1 0] has infinitely many right inverses (of the form vertical [1 x]), but no left inverse. =Uncool-

The methods described above are not necessary once you notice that the equation is simply x^2005 = 1 - which has as roots the 2005th roots of unity. Basically, take e^(2*pi*i/2005), and all integer powers of this number are solutions. =Uncool=

I think your proof works out fine...though there is an easier way to do it: Let A be an invertible matrix, B is its right inverse, C is its left inverse. Then: C = CI = C(AB) = (CA)B = IB = B. However, this only works for matrices that are invertible in both directions; a matrix can have a one-sided inverse. =Uncool-

I see only two ways... First, fix the 1 and 6 (one fixes the other) to be on top and bottom (we can always rotate the cube this way). Then fix 3 in front (once again, can be rotated to fit), and 4 is fixed. Now, there are 2 ways to fill the last 2, so there are a total of 2 different ways. =Uncool-

Alright...I can create a group with any number for the order of a square element, and still have a cyclic product set. I shall provide in the case for a prime. Let us take a prime p. Then let us find a prime q such that p | q-1. This necessarily does exist. Then we can create a nonabelian group of order pq. Then let a^2 = a, b^2 = e, where a is of order p and b is of order q, and everything else follows. I believe that the same occurs when you have any squarefree m. Just take a prime q such that m|q-1, and you can create a nonabelian group of order mq where there is a cyclic group of order m, and let a^2 = a where a is one element of order m, and b^2 = e where b is an element of order q. D6 is an example where m = 2, q = 3. We can also create it for D2n, for any integer n. Any other example is too large to be written here easily, though. The elements are of order 2 because those are all that can be written for such small partial rings. =Uncool-

For the 5th: Let us assume we have a "partial ring" of order [math]p_{1} p_{2}...p_{n}[/math]. Then, by Cauchy's theorem, there is an element of order [math]p_{i}[/math] for each p. Let them be known as [math]a_{1}, a_{2},...,a_{n}[/math]. Let us look at the set of squares of the as, that is [math]a_{1}^{2} = a_{1}*a_{1}, a_{2}^{2},...,a_{n}^{2}[/math]. By 2) above, we know that each either has order 1 or [math]p_{i}[/math]. Let us take the sum of these squares. We know the additive orders of each are relatively prime. We also know, by 3), that those elements are commutative. Therefore, the order of the sum is the lcm of each of the orders - the product of each of the [math]p_{i}[/math] such that [math]a_{i}^2 \neq e[/math]. However, this is also the necessary additive order to include all possible combinations of the [math]a_{i}^2[/math] as a group. Clearly, every product can be shown to be a sum of the [math]a_{i}^2[/math], as every product can be written as [math](m_{1}a_{1}+m_{2}a_{2}+...+m_{n}a_{n})*(o_{1}a_{1}+o_{2}a_{2}+...+o_{n}a_{n}) = m_{1}o_{1}a_{1}^2+m_{2}o_{2}a_{2}^2+...+m_{n}o_{n}a_{n}^2[/math] as the products of the different orders are all e, where the m and o are all integers. =Uncool-

Proof for the properties: 1) 0*a = 0*a + 0*a + -(0*a) = (0+0)*a + -(0*a) = 0*a + -(0*a) = 0 2) ab+ab+ab+...+ab (o(a) times) = (a+a+a+a+...+a)b = eb = e, so o(ab)|o(a). Similar argument for b. 3) -ac + (a+b)(c+d) + -bd = -ac + (a+b)c + (a+b)d + -bd= -ac + ac + bc + ad + bd + -bd = bc + ad = -ac + a(c+d) + b(c+d) + -bd = -ac + ac + ad + bc + bd + -bd = ad + bc. 4) Commutative has already been shown. All we need show is closure and inverse to show it is a group. a*r1 + a*r2 = a*(r1+r2), so it is closed. a*r1 + a*-r1 = a*(r1 + -r1) = a*e = e, so we have inverse and therefore identity, so a*r is a group. 5) This one is long. It shall be proven later. =Uncool-

I'm looking more for theoretical here, as I haven't really looked at practical uses of algebraic structures (besides cryptography a bit) recently. I cannot really say much about the number of distinct products, except that it is limited by the size of the largest commutative subgroup of the original group. Some other stuff I have found out about it: We may be able to form a version of a quotient partial ring by using what I'd call a "normal partial ideal" - that is, a normal subgroup which is also ideal (any element of the original group multiplied by any element in the subgroup is in the subgroup). This follows all the normal isomorphism theorems - If you have a "partial ring" homomorphism from R onto R' with kernel N, then R/N is isomorphic to R' (I'm pretty sure of this; haven't entirely checked though), etc. A conjecture I have come up with, and checked on a few of these "partial rings," is the following: Let N be the set of all elements which always multiply to 0, that is: N = {r in R: r*b = b*r = e for all b in R}. Then N is a normal partial ideal - easy to prove - and R/N is commutative - not proven yet. I'm trying to figure out how to show this. A note: I have found the smallest partial ring with noncommutative addition and noncommutative multiplication. It is as follows: D8 under addition, with b as reflection and a as 90 degree rotation. * | e a 2a 3a b b+a b+2a b+3a ______________________________________________ e | e e e e e e e e a | e 2a e 2a e 2a e 2a 2a | e e e e e e e e 3a | e 2a e 2a e 2a e 2a b | e 2a e 2a e 2a e 2a b+a | e e e e e e e e b+2a | e 2a e 2a e 2a e 2a b+3a | e e e e e e e e I am still looking to see if I can find a partial ring such that the set of products is not a group. As shown before, a square must divide it. I know how to create it, as long as I can make sure it is associative; there is where I run into trouble. =Uncool-

Perhaps a better way to state what I did: + | e a 2a b b+a b+2a _______________________________________ e | e a 2a b b+a b+2a a | a 2a e b+2a b b+a 2a | 2a e a b+a b+2a b b | b b+a b+2a e a 2a b+a | b+a b+2a b 2a e a b+2a | b+2a b b+a a 2a e * | e a 2a b b+a b+2a ______________________________________ e | e e e e e e a | e e e e e e 2a | e e e e e e b | e e e b b b b+a | e e e b b b b+2a | e e e b b b The "partial ring" is D6 under addition, and the multiplication table is shown below addition. =Uncool-

Sorry, I forgot to specify. The "partial ring" is D6 under addition, and the table below defines multiplication. =Uncool=

Alright, I'll take the smallest such struture, the nontrivial structure on D6, the dihedral group of order 6. Let us call the elements e, a, 2a, b, b+a, and b+2a, where e is the identity, a means one rotation of the vertices, and b means a reflected. Note that addition is noncommutative, as b+a = 2a+b. * |e a 2a b b+a b+2a ------------------------ e |e e e e e e a |e e e e e e 2a |e e e e e e b |e e e b b b b+a |e e e b b b b+2a|e e e b b b This structure is associative, distributive, and has all the properties of a ring except commutativity of addition and identity of multiplication. =Uncool=

A reason for my answer below: Let us define a function f(a, b) = floor(a/b) + f(a - b*floor(a/b) + floor(a/b), b) f(a, b) = 0 if 0 <= a < b I say f(a, b) = floor(a/(b-1)). Clearly, all cases do go to the base case shown above. Now, we will use WOP to show that this is the case in general. Let us assume that a is the lowest number such that f(a, b) =/= floor(a/(b-1)). Then f(a) = floor(a/b) + floor(a/(b-1) - b*floor(a/b)/(b-1) + floor(a/b)/(b-1)) = floor(a/b) + floor(a/(b-1) - floor(a/b)/(b-1) - floor(a/b) + floor(a/b)/(b-1)) = floor(a/b) + floor(a/(b-1) - floor(a/b)) = floor(a/b) - floor(a/b) + floor(a/(b-1)) = floor(a/(b-1)). This function is equal to the number of bars you get if you start out with a coupons and redeem b coupons for a bar each time. Therefore, if you start out with ab bars, you get floor(ab/(b-1)) bars at the end. =Uncool-

by partial ring I mean the algebraic structure itself, and the order is the number of elements in it. This structure is not a ring until it is given an identity element (optional, depending on who you ask) and commutativity. Basically, I've taken the definition of a ring, and removed commutativity of addition. =Uncool=

I believe the answer should eventually come out to floor(x + x/b + x/b^2 + x/b^3) = floor(x*b/(b-1)) where x is the original number you had and b is the number you need to get a free bar. In your example, it's floor(27*7/6) = floor(63/2) = 31. =Uncool=

Recently, I have come up with an algebraic structure that I would like to call "Partial rings" with some interesting properties. They are similar to rings, as will be shown later. The rules for this structure are: 1) Group under addition 2) Another function, multiplication 3) mutiplication is distributive 4) multiplication is associative 5 (optional)) Noncommutative group under addition. Some properties already found: 1) The identity under addition annihilates under multilplication (clearly) 2) The additive order of the product a*b divides the additive order of each a and b 3) The set of products is commutative 4) The set of products a*r, r is in R is a commutative group 5) If the order of the "partial ring" is squarefree, then the set of products is cyclic Some conjectures: 1) The set of products is a group. Are there any interesting applications of this "partial ring"?

The people before me have given one possible answer to the equation; now the problem is to find all answers. Let us assume for the moment that y will be positive. Let us then assume that y = e^(ax) where a is an integer, as that is the form that will work best for the equation. Then the equation reduces to: e^(ax)*a^(2005) - e^(ax) = 0. As e^(ax) =/= 0, we can remove it. Therefore, we are left with a^2005 - 1 = 0 - that is, a is one of the 2005th roots of unity. As we now have 2005 unique answers and the equation is a linear equation of order 2005 (or however you say it), we now have all the possible answers - they are linear combinations of the 2005 answers given. =Uncool-

1)I assume ' means opposite. (xy' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') = xx' + x'y' + xy + yy' = 0 + xy + x'y' + 0 = xy + x'y' Laws used here: Demorgan's, Demorgan's, Distributive, Contradiction, Exclusion (I htink the last one is right, not sure) 2)x'z + xy = x'y'z + yz + xy x'z + xy = x'yz + x'y'z + xy = x'y'z + y(x + x'z) = x'y'z + y(x + z) = x'y'z + yz + xy Laws used here: 1) Can't remember what it's called, distributive, can't remember, distributive. =Uncool=

Clearly, your teacher is wrong - as cosine can be negative, and hir answer can't be, that answer is wrong. The one question about yours is whether, when the angle is greater than 90, the dot product is positive or negative. If you get positive, then your answer is wrong too. Otherwise, your answer is right. =Uncool=

The answer works, and is entirely along the walls, actually - it's a weird answer...[hide]Didn't think that walking along 5 walls could afford a shorter path than 3...[/hide] =uncool-

Have you learned integration by parts yet? =Uncool=

Well, let us assume that there is a normal representation, that is, such that it takes the form ......ABC = 1/2. Then 1 = .....(2*A)(2*B)(2*C) with all the carry overs. But then the last digit is divisible by 2 - meaning an impossibility. This will happen in any base - so it cannot be represented *normally*. However, we could perhaps do it by putting the normal part before and the rest after. This will always have an end as anythign that wouldn't have an end can be represented beforehand. Example: 8/15 = 1/3 + 1/5 = 67 + 0.2 = 67.2 1/15 = -5/15 + 6/15 = -1/3 + 2/5 = 6 + 0.4 =6.4 =Uncool=

This can be seen as what happens when you take mod 10^n, and limit as n -> infinity. I just have to wonder, what happens for 1/5? Does it become 0.5? =Uncool=