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In recent years it has become popular to criticize string theory as out of touch with reality. Popular books have been written by scientists, some prominent and others not so prominent, arguing that string theory makes no predictions that experiment can test, that its fundamental objects cant be observed, that physicists have wasted their time on an enterprise that isnt even scientific to begin with.
Such arguments leave an impression of utter unfamiliarity with the history of science. In times past, the same kinds of aspersions were cast against quarks, neutrinos, even the very existence of atoms. Superstrings are in good company. And string theorys limit on how low viscosity can go now seems to have established that string math does indeed mirror something real in nature. This may well be the first prediction from string theory to be validated by experiment, Steinberg writes in a recent paper (arxiv.org/abs/0903.1474).
From what I have read, this is a very poor representation for the criticism string theory has received. It seems to say that string theory is receiving the criticism because it is new and uncharted territory, and that simply isn't the case. The major problem with string theory is that it may be altered in so many ways so that there is a version of string theory that matches any experimental result. This is what makes it untestable.
In times past, the same kinds of aspersions were cast against quarks, neutrinos, even the very existence of atoms. Superstrings are in good company.
"They laughed at Columbus, they laughed at Fulton, they laughed at the Wright brothers. But they also laughed at Bozo the Clown."
- Carl Sagan
At the very last second of the video, you can hear someone whistle.
From post #13:
...so using a result in Humphreys...
Can you post this result? I know of a similar result, but one must assume G is simple to use it (and hence you'd be going by contradiction).
Yo, man! I hope Ricky is appreciating my hard work!
Certainly. But perhaps it is wiser to read more on Sylow first before trying to prove more of these.
Jane, all p-groups have a nontrivial center. That means not only are all groups of order 8 not simple, but any power of 2 (or any prime for that matter) as well.
Would you like to see a proof of this?
If you know latex, then you can use that here with [math] tags. You can also use a free file hoster and post a link here.
I was wondering, do the K-automorphisms permutes the roots of ALL polynomials, or just the minimal polynomial?
All polynomials. But remember, L is generated by the roots of a polynomial: L = K[a_1, ..., a_n]. So as soon as you define where these roots go, you define where everything in L goes.
Wilson's theorem is a nice result, and gives a good necessary and sufficient condition for prime numbers. It is however computationally inefficient for primality testing.
That
makes it easy to compute the phi function for any integer n, so long as you know its prime factorization.
Perhaps I am not understanding the question. But without changing the operation this isn't possible. Let $ stand for any operation. Assume you can define $ such that P $ Q[0] = K and P $ Q[1] = L where K is not equal to L. Then if we set Q[0] = 0 and Q[1] = 1, then:
P $ Q[0] = K and so P $ 0 = K
P $ Q[1] = L and so P $ 1 = L
But now take a second Q, Q[0] = 1 and Q[1] = 0.
P $ Q[0] = K and so P $ 1 = K
But now we have P $ 1 = K and P $ 1 = L with K and L not being equal. This is not possible.
Correct, you have not proved the statement. What you want to do instead of finding the elements that conjugate, is just to count. You should know of some ways to count the number of conjugates in a group (think about centralizers). You also have the theorem:
Two elements of S_n are conjugate if and only if they have the same cycle type. Specifically, all 3 cycles are conjugate in S_5.
A more explicit hint: Compute the number of 3 cycles in S_5, then compute the centralizer of (123) in A_5.
Is this correct?
Yes.
I'm not entirely sure how much group theory you know, so if this doesn't make sense there is an easier but less general way to explain it.
Given a group G and an element g in G, we define:
C_g is called the conjugacy class of g. It is called this because for any group, the conjugacy classes form equivalence classes. So in C_g you will find all the elements which are conjugate to g. Now your question:
However, my question is, does the conjugating permutation (gamma) also have to be in A5?
Becomes a bit trivial, if you look at it upside down. Instead of asking "does their exist", let's just conjugate an element of A_5 by an element not in A_5:
(12)(123)(21) = (132)
So the answer is no. But a more interesting question is: can we always take the conjugating element to be in A5? In the previous example, if we conjugate by (12)(45) instead:
(12)(45)(123)(54)(21) = (132)
We get the same result. The answer in general is no, which can be seen by taking any group with with a nontrivial center. For example, in the integers modulo 2, the conjugacy class of 1 is just 1 itself. However, to get from 1 to 1 you need to add and then subtract 0, which is not in the conjugacy class.
To see if this is the case with A_5, the only way I know is to do the computations by hand.
Correct.
This is most likely case division. Perhaps a clever argument exists, but dividing it up into cases will do. For permutations in S_5, you have fairly limited choices. Write your permutation in a disjoint fashion and so you have:
2, 3, 4, 5 cycles
And combinations of them:
2 and 2 (i.e. (12)(34) )
2 and 3 (i.e. (12)(345) )
Remember that it is fairly easy to go from disjoint cycles to a product of transpositions. For example:
(a b c d) = (a d) (a c) (a b)
Now just eliminate all the ones except those given in the decomposition.
Two elements in Sn are conjugate if and only if they have the same cycle types. So provided this is so, arrange their cycle types to line up, say:
(123)(57)(46) and (145)(23)(67)
Now send the first element of the first permutation to the first element of the second permutation, and so on:
1 -> 1
2 -> 4
3 -> 5
5 -> 2
7 -> 3
4 -> 6
6 -> 7
This gives (246735). This will conjugate the first into the second.
Yes Luis. They are called "File hosts", though most will typically only hold on to your file for a week. The alternate is to copy the text here, so long as there are no images and diagrams involved.
Based on that page, your calculus is wrong.
Quite fascinating Jane. That's the square root of the discriminant for a polynomial with roots a1, ..., an. Nice find.
An all too common application of this is: If I have arbitrarily many coins two different values, how many ways can I make a third value. Here you must restrict to positive solutions, but it takes very little effort to do so.
It might be worth it to point out that yuyukin has the same IP address as unnihilator.
Yuyukin, if you indeed are the same person (and I don't know if that's true), then please avoid using multiple accounts in the future. It makes things look suspicious, though you've done nothing wrong as of yet.
The fundamental theorem of algebra states that the complex field is the algebraic closure of the field of real numbers.
Im very sorry if this doesnt make you any wiser on the subject but then I have no idea what 7th Grader is.
A 7th grader is someone who is first starting to learn basic algebra. They are understanding what it means to represent and solve an algebraic equation, must of the time linear with one variable.
As for the fundamental theorem of algebra: If you have any polynomial at all, and the highest term which appears is x^n, then there are exactly n roots in the complex numbers, if you count roots that happen twice as being different. For example:
(x-1)(x-1)(x-2) is a polynomial, and we say it has three roots: there are 2 one's, and one 2. The root one is said to occur with multiplicity 2.
The fundamental theorem of algebra says that this is the case for all polynomials. x^5 - 3x + 2 has exactly 5 roots, in the complex numbers.
Very interesting! Hard to believe, but if John Conway thinks so, it must be right! He wouldn't say something without justification! Thats what I think!
That is a very dangerous way to think. Or to not think, I suppose.
I would still like to know about the set I posted in #9, George.
if you trust induction algorithm is all that entails infinity
There may be a translation problem, but that doesn't make any sense.
my inductive algorithm can produce decimals any long thus is infinity as well
No, you can't George. Induction only works when the number may be reached in a finite amount of steps. It may very well be that this finite amount is arbitrarily large, but it says nothing about what happens at infinity.
Let's say I have proven that proposition P holds for the value 1, and I know that if it holds for a number n, then it holds for n+1. Please answer me these questions:
1. Does such a proposition hold for the number 2?
2. What integer may proposition P not hold for?
I shall find the automorphism group of G = Z2 x Z4 = {(0, 0), (1, 0), (0, 1), (1, 1), (0, 2), (1, 2), (0, 3), (1, 3)}. Addition is preformed component wise. i.e. (1, 2) + (1, 3) = (1 + 1, 2 + 3) = (2, 5) = (0, 1).
This group is generated by (1, 0) and (0, 1). To see this, (x, y) = (x, 0) + (0, y). Now add (1, 0) to itself x times, and (0, 1) to itself y times.
Thus, if I have an isomorphism from G to itself, then as soon as I define where (1, 0) and (0, 1) go, I have defined the isomorphism. This is because isomorphisms preserve addition.
So now I just need to figure out where I may send (1, 0) and (0, 1). First off, isomorphisms preserve the order of elements. (1, 0) has order 2, so it must be sent to an element of order 2. Similarly with (0, 1) having order 4.
Elements of order 2: (1, 0), (1, 2)
Elements of order 4: (0, 1), (1, 1), (0, 3), (1, 3)
From this, there are a total of 8 possible maps. All that remains is to check each one to verify it is an isomorphism. While a bit tedious, there is nothing difficult going on here.
If you ignore the infinite case, there is no need for Zorn's Lemma. So we are assuming G is finite. You should be ok with the fact that G is abelian and x = -x. Do you know the fundamental theorem for finite(ly generated) abelian groups? That every finite abelian group is isomorphic to a finite number of copies of the integers modulo p (i.e. cyclic groups). Assuming you know this, consider the map that just moves around these generators. It is an automorphism and you can choose it to be non-trivial as long as G has more than one copy of the integers modulo p. We conclude from this that G is in fact the integers modulo p. But the requirement of x = -x now forces p = 2.
I would still like to know about the set I posted in #9, George.
Put it in other words, if you have got them all, you have already passed the induction phase.
Given any number k, it is either true or false that k has a property p. There is no concept of "using induction to get them". What is true was always and will always be true, whether we know it or not. Induction is just a method to know if it is true.
So what we're saying is: I know I can prove it for 1, and I can also prove that if it holds for k-1, then it must also hold for k. Every integer can be reached in a finite number of steps using this, and so we conclude that there are no integers that we can't prove it for. This of course is just saying that it isn't false for every integer, or rather it is true for every integer.