Abstract Algebra question on field extension (serious answers only please)?
344061
Since [L : K] = [L : E][E : K] and [L : K] is prime, \neither [L : E] = 1 or [E : K] = 1.\n\nThus either E = L or E = K
1057306
Here's a copy of the actual paper. Good luck understanding it:\n\nhttp://math.stanford.edu/~lekheng/flt/wiles.pdf
1013129
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
1012764
I seem to be the first with the correct solution... the confusion is obviously the notation.  If you can't type superscripts easily, you can always use ^ to mean exponentiation.  (Or Alt-253 on a PC keypad will get you a superscript ²).\n\nThe solution is to simply substitute x in your equation with whatever is put inside the parentheses.  It can be confusing if that includes another equation, but it is still simple substitution.\n\nGiven:\nf(x) = 2(x^2) -1 \n\nf(2x) = 2((2x)^2) - 1\nf(2x) = 2(4(x^2)) - 1\nf(2x) = 8(x^2) -1\nf(-x) = 2(-x^2) - 1\nf(-x) = 2(x^2) - 1\n\nf(2x) - f(-x) = 6(x^2)\n\nf(x+1) = 2((x+1)^2) - 1\nf(x+1) = 2(x^2 + 2x + 1) - 1\nf(x+1) = 2x^2 + 4x + 2 - 1\nf(x+1) = 2x^2 + 4x + 1\n\nSo your answers are:\n1) 6x^2\n2) 2x^2 + 4x + 1\n\nWritten using Alt-253 for the exponent, the answers are:\n1) 6x²\n2) 2x² + 4x + 1
769323
6 - 12x\n-or-\n-12x + 6\n\n-or-\n\n6(1-2x)\n\n...although that depends on whether you want the simplest expanded form or the simplest factored form.
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped
700951
4x + 6 is not quite right\nwatch the order of operations and distributive properties\n\nrepost your answer and I'll tell you if you got it\n\n(you didn't think I'd just give up the answer, did you ?)
773695
Look in your text book under "factoring polynomials"... or you could try this web site that explains this fairly thoroughly...\n\nhttp://faculty.ed.umuc.edu/~swalsh/Math%20Articles/Factoring.html
577215
This again is a direct theorem from the field axioms: http://mathworld.wolfram.com/FieldAxioms.html . \nSuppose a*b = 0. If a is non-zero, multiply both sides by a^(-1) to get a^(-1)ab = a^(-1)0, and thus b = 0, since the product of 0 and any number is 0.\nIf b is non-zero, multiply both sides by b^(-1) to get the similar result a = 0. Thus, either b or a must be 0. \nThe final case is that both a and b are 0, which is trivial.\nFurther discussion of this theorem can be found in the source.\nThis theorem can be used to solve equations by getting 0 on one side of the equation and factoring the other side completely, then using the theorem to imply that at least one of the factors must be zero. For example, x^2 = -3x - 2 can be rearranged to be x^2 + 3x + 2 = 0. We factor the left side to read (x + 1)(x + 2) = 0. From the theorem we know that either x + 1 = 0 or x + 2 = 0 or both. Thus we have two solutions, x = -1 or x = -2. It is obvious that in this case, only one factor could be 0.
864310
If K is a Galois extension field of F, the you know that the degree of the Galois group is [K:F]. Thus, G is a cyclic group of degree [K:F].\n\nBy the fundamental theorem of Galois theory, every subgroup of G corresponds to a field intermediate between F and K and vice versa. If H is a subgroup, then the corresponding subfield is the stabilizer of H. So the question boils down to what the subgroups of a cyclic group are. \n\nBut subgroups of cyclic groups are cyclic with order dividing the order of G. In fact, there is exactly one subgroup of order k for each k dividing the order of G. Furthermore, the index of a subgroup in G is the degree of the extension from F to the subfield corresponding to that subgroup.
978070
Try\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm\n\nYou can also try http://www.helpalgebra.com\n\nGoodLuck!
464936
Commutative operation.
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D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
783870
7x^2+16x over (x-2)(x+2)(x+3)
241036
I'm assuming you're looking for information on factoring polynomials since it's a College Algebra courese.  \nThis website is very helpful, with the steps for factoring laid out, and it also provides websites you can use if you need more information.  Good luck!\n\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm
446810
Let r be an element if R, and hence in L.  Since L is an algebraic extension of K, all elements of L, there is an irreducible polynomial:\n\n    P(x) = a0 + a1x + a2x^2 + ... anx^n\n\nwith the ai's in K such P(r) = 0.\n\nSubtracting a0 from both sides, we see:\n\n    -a0 = r ( a1 + a2r + a3r^2 .. + anr^(n-1))\n\nDividing both sides by -a0 we get (a0 is not zero because P is irreducible):\n\n      1 = r (-a1/a0 - a2/a0 r - a3/a0 r^2 ...)\n\nThe right side is in R since the the coefficients -ai/a0 are in K and r is in r.  So we have an inverse of r in R.
457166
This is a Frequently Asked Question. Following is my previous answer:\n\nThis question may be asked so frequently (along with the infamous 0.999...) because no concrete axioms or definitions are ever given for the real number system until one studies higher algebra or real analysis; it is assumed that everyone knows what a real number is until then. The proliferation of these questions show that this assumption is no good.\nFor the sake of conciseness, I will not define what a real number is here, as only the algebraic structure of the real numbers is necessary to answer this question. The algebraic structure of the real numbers is what we call a field. A field is an algebraic structure that follows a finite set of axioms. These axioms define what statements are true and false in this system based on the laws of logic. You can find the field axioms here: http://mathworld.wolfram.com/Field.html .\nEach line is preceded by "For all real numbers a, b and c, the following is true...". Note that the proper form of the additive identity and multiplicative identity is that it prescribes the existence of some number 1 and some number 0 such that the equations hold. These same numbers are then used to prescribe the existence of additive and multiplicative inverses.\nDivision a/b is just shorthand for a*(b^(-1)). So in your question, you are looking for 1*0^(-1). It is easy to show that there is no such number as 0^(-1) in any field. Suppose 0^(-1) exists.\nThen 0*0^(-1) = 1. But multiplying both sides by 0^(-1) gives 0 = 0^(-1), which means 0*0 = 0 = 1. This is obviously false.\nSuppose 0 = 1. 0*a = a for all numbers a. But 0*a = (1 + (-1))*a = a + (-a) = 0. Thus 0 is not equivalent to 1 and our sole assumption that 0^(-1) exists must be false. Thus the expression 1*0^(-1) is undefined in a field.\nProof by contradiction is very powerful for many types of statements.\nAs to the limit argument, note that you are evaluating a limit of the function f(x) = 1/x. As we approach 0 from the negative x-axis, this limit decreases without bound, but as we approach 0 along the positive x-axis, the limit increases without bound. Thus, the limit of f(x) at 0 does not exist, so f(0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nWhile 1/0 is not defined in any number system, |1/0| is defined as positive infinity in extended number systems (which are not fields by the above argument) for various purposes. See http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html .