Question raised by INVARIANTOLOGY
208727
assuming that you're not talking about killing your parents or your ancestors:\n\ni am told that the cells in our bodies are constantly replacing themselves, the dead ones. the stomach lining is replaced every 48 hours (?), most everything else in your body every 4 - 6 months, and the oldest living tissue - the brain and bones - every 7 years.\n\nso at that you are NOT made of the same materials or the same cells from more than 7 years ago. you have replaced your entire body with another.\n(not counting the tips of your hair that may be older than 7 years ;-)\n\nbet you didn't even feel a thing! :-)
263546
There are two interpretations I can see here. The first is that the homology groups of the 0-dimensional sphere are needed. For this, the answer is trivial: you just get Hn(pt)+Hn(pt) (direct sum). Over a ring R, Hn(pt)=R if n=0 and is 0 if n>0. So the answer is R+R for n=0 and 0 otherwise.\n\nIf the asker actually means SO(k), the special orthogonal group, this is much trickier. It is *not true* that SO(k) is homeomorphic to S^(k-1) (this only happens if k=2; for k=3 you get a projective space). In fact, the dimension of the manifold SO(k) is k(k-1)/2. In this case, I don't know the answer off the top of my head.
149188
ok, i think one will be where you have to solve a word problem dealing with alegebra. One with number sentences that have a letter on one side of the problem and letters on both sides of the problem. i could be wrong, but i'm trying to help.
119452
you need to use another example instead of your creator. the uncreated cannot define the creator's characteristics. it only works the other way 'round. have fun though.
551700
This website looks like it's got a lot of proofs.\n\nhttp://www.cut-the-knot.org/pythagoras/index.shtml
433417
Every group has a compact normal subgroup, namely the identity.  \nAs noted, the scalar matrices form the center of your group G.  Therefore the two-element subgroup {plus/minus I} is also normal in G.\n\nAs noted, the group SL_n is normal in G.  Therefore if H is also normal in G, its intersection H' with SL_n is normal in SL_n.  It is known that if H' is a normal subgroup of SL_n, then H' is contained in the scalar matrices, which are trivial if n is odd and {plus/minus I} if n is even.\n\nIf H is not equal to H' it must contain a matrix of determinant not equal to 1.  If this determinant is not plus/minus 1, then H is not compact.  Let us note that the set of matrices with determinant plus/minus 1 is a normal subgroup of G containing SL_n.\nFor that matter, also note that the set of matrices with positive determinant is normal in G.\n\nThis discussion should enable you to classify the normal subgroups of GL_n.  If you mean *Lie* subgroups we are done.  \nIn particular, there are exactly two compact ones, already noted.
19548
if we speak of numbers and not of limits, the answer is no without exceptions: 0 is defined in group theory as the null element, so it is BY DEFINITION the element which multiplied by any other gives itself: always 0.\nInstead, what we usually assume as always true, but is not true in some finite algebraic models, is that if A*B=0 then either A=0 or B=0. This is not always true in all algebraic models (for exemple in modular algebra).
667822
Doesn't sound like you know what ID means let alone irreducible complexity or paley's watchmaker argument to back it up at all and what the hell is your question?  Seems to just be a ignorant claim without any good supporting evidence or logic to back it up.  Keep reading and maybe you'll learn something.
346861
Most introductory books on differential geometry discuss this. Looks at 'From Calculus to Cohomology' and 'Introduction to Smooth Manifolds' as references.
76562
Mathematicians have proven that, in general, 1 + 1 = 2. It took them awhile to do it, but they succeeded. Unless the time truly comes when there is a need to ignore this generalisation, this assumption made is as good as any other axioms ever declared by the wise logicians.
198209
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
534929
I have factored it please refer to.\n\nFrom this solution can be found
572989
Any chance  you could post a question that makes sense?
599251
Using the standard "construction" of the natural numbers, they ARE NOT embedded in the real numbers.  The real numbers are like a completion of a splitting field of a class of polynomials over the natural numbers. (Existance of transcendental numbers, such at pi and e, show that the real numbers themselves are not the splitting field of the set of polynomials over the naturals.  See any good textbook on abstract algebra for definitions of splitting fields, transcendental numbers, ...  )  As such, the natural numbers and real numbers are indeed disjoint sets.  HOWEVER!  Consider the function f which takes 1_N ( 1 in the natural numbers ) to 1_R ( 1 in the real numbers ), 2_N to 2_R, ... etc.  \n\nf( m_N + n_N ) = m_R + n_R\nf( m_N * n_N ) = m_R * n_R\n\nThese properties show that the map f is a homomorphism.  Moreover, the only element which maps to 0_R is 0_N showing that the map is, indeed, an embedding.\n\nWhile the sets themselves may be disjoint, the set of natural numbers is "isomorphic" to what we consider the natural numbers sitting inside the reals.  As we typically consider such sets only "up to isomorphism", we can thus assume that the natural numbers do indeed sit inside the reals.
290360
By considering a change of base point, it is clear that π\n1\n(X, {x\n0\n} depends only on the\npath component of x\n0\n. Since π\n1\n(X × Y, {(x\n0\n, y\n0\n)} = π\n1\n(X, {x\n0\n}) × π\n1\n(Y, {y\n0\n}), we can\nassume the space is path connected and omit the base point.\nSimply connected: an arcwise-connected space with trivial fundamental group. Ho-\nmotopy equivalent spaces have the same fundamental group. Thus, contractible\nspaces, which are homotopy equivalent to a point, are simply connected. Actually,\nany loop S\n1\n→ X which is homotopically trivial extends to a map S\n\n\nSorry, copy pasted it from my e-book.
482346
Denfinitely need clarification........
309863
The result is still under debate because all transformations have still not been verified. The Sobolev space can be bi variant if the gauge is not satisfied.\n\nI would suggest looking up the bulletins of the London Mathematical Society, their late 2004 editions came out with one such treatment.