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Hi 
i just wanted to add the following regarding 3):
When you look at the definition of a subgroup in wiki, "(G,*) is group. A subset H of G is called a subgroup of G if the restriction of * to H x H is group operation on H.", you dont see any mention of a 'preexisting' binary operation on H itself.
So when answering the question, is R* (under multiplication) a subgroup of (R,+), u can disregard the given operation (multiplication) totally. What you have to show is:
(i) R* subset of R?
(ii) ( R* , + ) group? (with + being the restriction of + mentioned in the definition of a subgroup.)
R* is no subgroup of ( R , +) because the identity element is missing. (And for no other reason.)
@mahrgell: ln : R+\{0} -> R is a semigroup-homomorphism between ( R+\{0} , * ) and ( R , + ). But no Monoid-homomorphism because ln(1) = e \neq 0, i.e. the identity element of ( R+\{0} , * ) is not mapped to the identity element of ( R , + ). So its definitively no group-isomorphism between ( R+\{0} , * ) and ( R,+).
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On February 28 2019 00:30 jodljodl wrote:
@mahrgell: ln : R+\{0} -> R is a semigroup-homomorphism between ( R+\{0} , * ) and ( R , + ). But no Monoid-homomorphism because ln(1) = e \neq 0, i.e. the identity element of ( R+\{0} , * ) is not mapped to the identity element of ( R , + ). So its definitively no group-isomorphism between ( R+\{0} , * ) and ( R,+).
Are you sure about the bolded part?
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Obviously its totally wrong
lol, thank you for correcting my error
even though this is quite embarrassing to me its quiiite funny to me too
So i guess, it is a group isomorphism then
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On February 27 2019 11:13 travis wrote:
ah right, the identity in 5.) is 1.. not 0. the confusion was that I was thinking 2^0 = 1 .. temporarily forgetting that the member of my group is 2^k, not k. if that confuses you don't worry about it, my thinking clearly didn't make sense. glad I posted them though!
Nope, I understand what you did. It was an easy enough mistake to make.
The purpose of these simple examples is to help you get a feel for the definitions and what they mean/represent. It can be very confusing when definitions are given without the motivation behind them, and this can happen with purely algebraic objects like groups. So it is good to play around with some examples, and why I didn't want to leave only tell you that you were wrong, because that does not help you understand
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I don't understand reflexive relationship
I have an example on my homework I missed. it asks about symmetric/transitive/reflexive
question was to classify the problem as each and if it's not then give a counterexample
the problem I missed was:
m,n in Z
m~n if m*n > 0
I said it was reflexive, symmetric, and transitive
I got the problem wrong, it says its not reflexive because x=0 does not work
but I thought the entire point was that you pick an m,n such that it fulfills the equation....
I guess maybe I am misunderstanding, and the set is not defined by that relationship...? Instead it's just like a test of whether or not the relationship passes?
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So the task is to check wether the relation R \subset Z x Z, i.e. {(m,n) \in Z x Z | m*n > 0} is reflexive, symmetric, transitive:
reflexive: R is reflexive if for all m \in Z: (m,m) \in R.
R is not reflexive because 0 \in Z and 0*0 = 0. So (0,0) \notin R.
For (0,0) to be element of R the following had to be true 0*0 = 0 > 0. Which is obviously false.
symmetric: R is symmetric if for all m,n \in Z: (m,n) \in R => (n,m) \in R.
Suppose (m,n) \in R. => m*n>0 (* is commutative) => n*m>0 => (n,m) \in R. So R is symmetric.
transitive: R is transitive if for all m,n,l \in Z: (m,n) , (n,l) \in R => (m,l) \in R.
Suppose (m,n) , (n,l) \in R. => m*n>0 and n*l>0 => m*n>0 and n*l>0 and (m,n,l > 0 or m,n,l < 0) [the cases m,n>0 n,l<0 and m,n<0 , n,l>0 are contradictory, so you can disregard them.] =>
(m*n>0 and n*l>0 and m,n,l > 0) or (m*n>0 and n*l>0 and m,n,l < 0) => m*l >0 => (m,l) \in R. So R is transitive.
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Okay I see, so yeah, I was misunderstanding how the question works. Thank you.
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I feel bad about using this thread but maths are hard for me, sorry ^.^
We start from a group of combinations, in this case 296 combinations.
I want to assign each combination a different frequency, in this case there will be three frequencies (25%, 50% and 75%).
If I go 296/3 and give 25/50/75 to each of those three groups respectively, I'll have a total frequency of 50% obviously. I'd like to have a different total frequency, in this case 66%.
66% is 2/3, 25% is 1/4, 50% is 2/4 and 75% is 3/4. So I have groups of 3/12, 6/12 and 9/12 and I'm trying to get to 8/12 for my total. There are multiple solutions. I'm trying to find the most "harmonious" one (pretty bad term to use but by that I mean, the 33%-33%-33% solution for a 50% total frequency is more "harmonious" than, say, 1% at 25, 98% at 50 and 1% at 75, even though that would also give me 50%).
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On April 08 2019 21:52 Nebuchad wrote:
I feel bad about using this thread but maths are hard for me, sorry ^.^
We start from a group of combinations, in this case 296 combinations.
I want to assign each combination a different frequency, in this case there will be three frequencies (25%, 50% and 75%).
If I go 296/3 and give 25/50/75 to each of those three groups respectively, I'll have a total frequency of 50% obviously. I'd like to have a different total frequency, in this case 66%.
66% is 2/3, 25% is 1/4, 50% is 2/4 and 75% is 3/4. So I have groups of 3/12, 6/12 and 9/12 and I'm trying to get to 8/12 for my total. There are multiple solutions. I'm trying to find the most "harmonious" one (pretty bad term to use but by that I mean, the 33%-33%-33% solution for a 50% total frequency is more "harmonious" than, say, 1% at 25, 98% at 50 and 1% at 75, even though that would also give me 50%).
Just to rephrase, you want:
0< x, y < 1
(1/3 *(1 - x))/4 + (1/3*(1 - y))/2 + 3 * 1/3(x + y - 1)/4 ) = 2/3
Or, equivalently: 2x + y = 8 with 0 < x, y < 12
And you want the solution that minimizes x, y, and 12 - x - y simultaneously. There are probably many ways of defining that harmoneousness, but lets say MSE does that, you want to minimize
x^2 + y^2 + (12 - x - y)^2 = 2x^2 + 2y^2 + 144 + 2xy -24x - 24y
Fill in:
2x^2 + 2(8 - 2x)^2 + 144 + 2x(8 - 2x) - 24x - 24(8 - 2x) =
2x^2 + 128 + 4x^2 - 64x + 144 + 16x - 4x^2 - 24x - 216 + 48x =
2x^2 - 24x + 56
This is a parabole that is 0 in two places, both within the given interval:
x = 6 +/- 2sqrt(2)
Or: x ~ 3.17 or x ~ 8.82
However if x = 8.82 then y is negative, so we can only use x= 3.17, resulting in the distribution
x = 3.17, y = 1.66
Fill in:
24.5% at 25, 28.7% at 50 and 59.8% at 75
Checking this, it results in a final frequency of 65.3, which is probably due to rounding errors.
I'm calling it a day.
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Didn't really read the post above me but looks like he doesn't think there's a solution? The problem is 3x+6y+9(1-x-y) = 8. It's clear that the LHS taken as a function is smooth and it's clear that (1,0) evaluates to 3 and (0,0) evaluates to 9, so by the IVT there is some (x,y) which gives us 8. The only question is if its obtainable with only a collection of 296 things.
This can obviously be solved in mathematica in two seconds, but I am guessing this is a homework problem so he can't do that.
First just solve for y to get y = (1-6x)/3. Now we need to choose an x so 296x is an integer, and 296y is an integer. So x = z/296 for some integer z. Thus we have y = (1-6(z/296))/3. Note that we can choose any z we like here. The only thing left to do is to find a z so that y = w/296 for some integer w.
If you follow the above logic you'll be able to figure out that we want to find an integer w = (1/3)296-2z for some integer z. In otherwords, we want a solution to the equation w+2z = (1/3)296. But this clearly has no solution, since the gcd(1,2) = 1 and (1/3)296 is not an integer, so its obviously not a multiple of 1. Therefore your problem has no solution. Why? Because if we cant choose a z that solves this equation, then there are no x which 1) solve 3x+6y+9(1-x-y) = 8 and 2) satisfy 296x is an integer.
Just a quick note: please if you don't know whether or not your problem has a solution - say so! It's kind of annoying to have to prove it for myself. I would have tried to estimate a good guess right away, but I gotta go.
edit: I fixed this post. I originally had the problem set as 3x+6y+9(1-x-y) = 4 instead of = 8.
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On April 09 2019 00:35 Rodya wrote:
Didn't really read the post above me but looks like he doesn't think there's a solution? The problem is 3x+6y+9(1-x-y) = 4. It's clear that the LHS taken as a function is smooth and it's clear that (1,0) evaluates to 3 and (0,1) evaluates to 6, so by the IVT there is some (x,y) which gives us 4. The only question is if its obtainable with only a collection of 296 things.
This can obviously be solved in mathematica in two seconds, but I am guessing this is a homework problem so he can't do that.
First just solve for y to get y = (5-6x)/3. Now we need to choose an x so 296x is an integer, and 296y is an integer. So x = z/296 for some integer z. Thus we have y = (5-6(z/296))/3. Note that we can choose any z we like here. The only thing left to do is to find a z so that y = w/296 for some integer w.
If you follow the above logic you'll be able to figure out that we want to find an integer w = (5/3)296-2z for some integer z. In otherwords, we want a solution to the equation w+2z = (5/3)296. But this clearly has no solution, since the gcd(1,2) = 1 and (5/3)296 is not an integer, so its obviously not a multiple of 1. Therefore your problem has no solution. Why? Because if we cant choose a z that solves this equation, then there are no x which 1) solve 3x+6y+9(1-x-y) = 4 and 2) satisfy 296x is an integer.
Just a quick note: please if you don't know whether or not your problem has a solution - say so! It's kind of annoying to have to prove it for myself. I would have tried to estimate a good guess right away, but I gotta go.
What? I gave a solution. It maximizes "harmoniousness" if you define that as the MSE of the difference for each of frequencies from the "ideal" 33.3%. If you define harmoniousness differently, then you'll get a different solution.
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Thanks for the answers. As per the meme, I... understood some of those words ^^'
Acro do you mind if I annoy you a little bit more in PM with this?
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Acro, the key to his problem is that there are only 296 things to split the frequencies among. That's like the whole point. Can you say exactly how many elements should have each frequency (of the 296 elements) according to your solution?
If we remove the restriction that we only have 296 discrete units, then obviously there are lots of elegant results like a 00/33/66 split.
I actually made a mistake in my post but I don't want to fix it right now.
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On April 09 2019 07:53 Rodya wrote:
Acro, the key to his problem is that there are only 296 things to split the frequencies among. That's like the whole point. Can you say exactly how many elements should have each frequency (of the 296 elements) according to your solution?
If we remove the restriction that we only have 296 discrete units, then obviously there are lots of elegant results like a 00/33/66 split.
I actually made a mistake in my post but I don't want to fix it right now.
But why? He doesn't restrict it to integers, so why should you?
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He wants to assign EACH "combination" a value. There are 296 combinations, so we need to assign values to 296 things. That's the problem. If that part of the problem is removed, then your whole thing with MSE isn't necessary anyway. A 00/33/66 split is 1) way more elegant and 2) actually gives you 8/12.
He DOES restrict the problem in this way, it's in his post. If he doesn't mean to restrict it this way then the question is odd since it's so easy to solve.
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Okay so; it's not a maths problem or something I was ordered to do. I'm doing it for myself. It could be that it's easy to solve but I can't because I suck, that's entirely possible (or likely).
Having integers isn't necessary, I'll have to do rounding on the results anyway even if it's integers. I can see how the way I worded the problem made you think that, sorry - again, I don't claim to not suck at this ^.^
That being said I do want a value for each group, I can't have one group at 0, so your easier solution wouldn't work for what I'm trying to get.
As per PM I guess I just have to choose one way to organize the numbers and then stick with it. I was sort of hoping that there would be one that would jump out, like 33-33-33 for a 50% frequency, but it doesn't seem to be the case with other frequencies.
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I sadly still don't understand your problem. If you have 296 groups, three values, (1/4, 1/2, 3/4), and you want to assign each group one of these three values such that the average value is as close to 2/3 as possible, then I read you correctly the first time.
If I'm correct that this is your problem then I can tell you that it is impossible to get the average value to be exactly 2/3. I can try to figure out how close you can possibly get if you would like. What Acro did would be an upper bound on the accuracy of my answer though.
If you want help again post in the thread - good luck though if not.
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Hm. I made a mistake above and am having the damndest time fixing it. I made a mistake in the constraints, and so the total percentage is above 100% :/
But adding that constraint makes the whole thing harder to compute because it makes it 2 variable instead of 1 (you can no longer use that the third variable is (1 - x - y), because you need to enforce that the total of x + y + z = 1. So you have a quadratic equation in two variables, which you can optimize using the ellipsoid method, but I don't have a tool handy here to do that.
Talking to Nebuchad in PM, however, it seems like his general problem has more than 3 "combinations", so the general solution requires something like that anyway.
Basically, figure out the formula you want to minimize, and formulize the constraints correctly. Then plug these into a constrained optimization tool (if you don't want the square error, but rather the simple error, you can even use simplex for this). As long as the formulas are convex (for MSE they definitely seem to be), you can use the ellipsoid method, otherwise you have a real cracker of a problem (same if you don't have linear or quadratic problems to optimize in the first place, or have non-linear constraints). Check that the solution does indeed work and you didn't make a mistake somewhere.
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had a homework question
H = cyclic subgroup <2> of R* (reals nonzero) with multiplication
G = Z
give an isomorphism G--->H
I said:
for all k > 0
0 --> 2
0-k ---> 2^(-k)
0+k --> 2^(k+1)
is this correct?
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By Z you mean the set of integers?
Sounds strange to me because Z is countable while R* is uncountable. Or are you looking "just" for an (injective?) homomorphism instead of an isomorphism?
Edit: Nevermind. I somehow missed that H isnt actually R*.
Your map isnt correct because an homomorphism must map the identity element of G to that of H, i.e. 0 ->1.
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EPT
