Why is Q (the rationals) a denumerable group?
413947
Hmmm,  what you have written is correct, if you can define an injection from Q to a coutable set, or alternatively a surjection from a countable set to Q, then Q is countable (or denumerate, they mean the same thing). \n\nWe know N is denumerate, since we can label it in the obvious way. Now there is a theorem which states that any cartesian product (which is just an ordered pair: lik (1,2) or (4,29)) of denumerate sets is also countable hence NxN is also denumerate. (i'll leave the proof of that out for now, if you really need it contact me and I can give you it)\n\nSo if you can define a function from NxN to Q, which is surjective, you will then know Q is countable. If we take the function defined by f(m,n)= m/n  then this function is clearly surjective (but it wont be injective since f(1,1)=f(2,2)= ... = 1 for example). Heance Q is countable.
642364
Exactly one if you are considering groups, and none if you are considering rings. For groups, look at the image of 1 in Z/nZ.  The image will have order at most n, but everything in Z other than 0 has infinite order. Thus, the image of 1 is 0, so all images are 0 and we have the zero homomorphism. For rings, the image of 1 has to be 1 and we just showed that is impossible.
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"a congruent to b mod H" just means they both have the same remainder when divided by H.  e.g. 12 and 27 mod 5:  each of them when divided by 5 leaves remainder 2.\n\nThe word "arbitrary" means you just make it up or pick it -- it isn't derived by any rule.\n\nabelian means commutative:  i.e. when the operation which is part of the definition of the group is performed on two members, order doesn't matter.  e.g. the set of integers under addition is abelian because \nm + n = n + m, but the set of integers under subtraction (actually I don't think that's even a group!!) isn't abelian because \nm - n is not equal to n - m.\n\nI don't know.  Exactly what symbols do you mean?  Use words if you can't get the symbol on the keyboard.
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No. A rational number is one that can be written as p/q where p and q are integers. Zero can be written as 0/1. In fact,every integer is rational.
402130
I get that the irreducible monics, with the primitive ones so marked, are:\n\nx^3 + 0 x^2 + 2 x + 1 (primitive)\nx^3 + 0 x^2 + 2 x + 2\nx^3 + 1 x^2 + 0 x + 2\nx^3 + 1 x^2 + 1 x + 2\nx^3 + 1 x^2 + 2 x + 1 (primitive)\nx^3 + 2 x^2 + 0 x + 1 (primitive)\nx^3 + 2 x^2 + 1 x + 1 (primitive)\nx^3 + 2 x^2 + 2 x + 2\n\nYour mileage may vary. :-)\n\nDegree three reducibles factor in one of the two forms:\n\n(x - a)(x - b)(x - c)\n(x - a)(x^2 + bx + c) with x^2 + bx + c irreducible\n\nSo if p is a degree three reducible then at least one of p(0), p(1), and p(2) must be 0 modulo 3.  Applying that test leaves the eight monic irreducibles shown above.\n\nThe primitive ones are found by verifying that x + <p(x)> has multiplicative order 26 in (Z/3Z)[x].
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This is an unsolved problem, as of 2004; it is known that being a Hodge cycle is a necessary condition to be an algebraic cycle that is rational, and numerous particular cases of the conjecture are known.\nHopefully our generation will be bright enough to solve the problem.  Until it's solved, we can just bang the pencils on our head and which ourselves good luck!
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The problem is to show that f is identically 0 using the density of the rational numbers. I would strongly suggest to the asker that they do their own homework at this level. If you don't understand the questions or techniques, you will have a *lot* of trouble later on.
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ok, i think i have an idea for this one. \nSketch of a proof:\nSuppose that a is an element of order k. Also suppose that f is an isomorphism. Then we know that f(a*a*a...k times)=f(a)*f(a)*f(a)*...k times (homomorphism definition). We know that f(a)^k is the identity because isomorphisms take the identity to the identity. Now see that k must be the first number with the quality f(a)^k=e. This comes as a result of the isomorphism having a kernel with one element. If we suppose for the sake of contradiction that f(a)^m=e with m<k, then we would have f(a*a*a*a...m times)=e, so that means that f(a^m)=e at the same time as f(a^k)=e with a^m not equal a^k, this is a contradiction.\n\nQED...ish
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One basic idea for analyzing a mathematical object is to look at morphisms into 'simple' objects of that type.\n\nFor example, a group G operating on a set S is a homomorphism of G into the symmetric group Sym(S). We regard the symmetric groups as 'simple' objects that we can use to understand other groups.\n\nA group representation is a homomorphism of the group G into the group gl(V) of invertible linear transformations on some vector space V. The good thing about this is that you get all of linear algebra to help you understand the group gl(V) and its subgroups, and hence to understand the group G. The homomorphism is called a representation of G.\n\nAmong the questions that can be asked:\n1. Are there any invariant subspaces of V? (an invariant subspace is one that is preserved by every linear transformation in the image of G).\n2. Can the vector space V be written as a direct sum of invariant subspaces in some nice way?\n3. Are there enough representations to distinguish all elements of the group?\n4. What happens if you look at representations in vector spaces over different base fields?\n\nIf your group is a Lie group (a differentiable manifold that is also a group), the you can look at smooth homomorphisms. It turns out the there is a vast amount of geometry that appears. This geometry is closely related to problems in particle physics.\n\nThis is just the beginning of the study of representations. There is much,much more. Yes, it is fun!
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Evidently this is undecidable for regular spaces. It is equivalent to the question of whether herditarily separable implies hereditarily Lindelof. For hausdorff spaces, the answer is NO.\n\nThe link below is a discussion on a topology forum about this question. Very nice!
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Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
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First show that the reciprocal of every element of F[[x]] can be written as a Laurent series: this is best done by inductively defining the coefficients of the quotient. Conversely, show that the reciprocal of every Laurent series can be expressed as a Laurent series. This shows that F((x)) is a field (assuming you check the obvious properties) and contains the field of fractions of F[[x]]. Since a sufficiently high power of x can be multiplied by a Laurent series to get an element of F[[x]], this completes the proof that F((x))) is the field of fractions of F[[x]].\n\nAs for Z[[x]]. The idea is to show that the power series for e^x (which is clearly in Q[[x]]) is not of the form f(x)/g(x) for f,g in Z[[x]]. But if f/g is in lowest terms, it is easy to see that g must be a constant (otherwise there would be terms of the expansion with negative exponents), i.e. g is in Z. But the expression for e^x cannot be so written!
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I don't believe what the previous poster said makes much sense...\n\nWe can prove this by breaking things up into several cases.  First say that F is a finite field of characteristic 2.  If F has q = 2^n elements, then x^q = x for all x in F.  We can choose a = x^{2^{n-1}} and b = 0; this gives x = a^2 + b^2.  If F does not have characteristic 2 but there is an element c such that c^2 = -1, then we can factor a^2 + b^2 = (a+bc) (a-bc).  Choose a and b to satisfy a+bc = x and a-bc = 1, so that x = a^2 + b^2.\n\nIt suffices then to consider a finite field F of odd characteristic p such that -1 is not a square.  There is a norm map N: F[√-1] -> F defined as follows.  With c = √-1, define y = a+bc -> N(y) = (a+bc) (a-bc) = a^2 + b^2.  It suffices to show N is surjective i.e., given x in F there exists y in F[√-1] such that x = N(y).  The Galois group of the quadratic extension F[√-1]/F is generated by the Frobenius map y -> y^p, so N(y) = y^{1+p}.  We see that N(0) = 0, so let's consider the image of the invertible elements.  There are (p+1) elements in the kernel of N, and F[√-1]* has (p^2 - 1) elements, so the image must have (p^2 - 1) / (p+1) = (p-1) elements.  Since F* has (p-1) elements, this norm map N must be surjective.
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This hypothesis is false:\nSuppose 0/0 = 1. Since 1 is the multiplicative identity, this implies (0/0)*a = a for all real numbers a. But (0/0)*a = (0*a)/0 = 0/0 = 1. This contradicts our hypothesis, so the hypothesis must be false (proof by contradiction). QED\n0/0 is indeterminate as a limiting form. See http://mathworld.wolfram.com/Indeterminate.html . \nAs an algebraic expression over the real field, it is undefined: http://mathworld.wolfram.com/Undefined.html . \nThis is a Frequently Asked Question; the 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 0*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. Tus the expression 0*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 over a plane of the function f(x,y) = x/y. As we approach (0,0) along the x-axis, this limit is 0, but as we approach (0,0) along the y-axis, the limit diverges. Thus, the limit of f(x,y) at (0,0) does not exist, so f(0,0) remains undefined with no continuous way to complete it. The limit is not indeterminate for this function; it does not exist.\nThe expression 0/0 is said to be indeterminate as a limiting form because one can have many functions that approach the form 0/0 that have convergent limits of different values. For example, the limit as x approaches 0 of [sin(x)]/x yields the indeterminate form 0/0, but L'Hopital's rule reveals the limit to be 1. The expressions listed on this page: http://mathworld.wolfram.com/Indeterminate.html are called indeterminate because they do not have a fixed behavior when studied as limits of expressions.
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http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Dubreil.html
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1. First find the roots (or behavior/properties of the roots) of this equation.\n2. Find the smallest field that is an extension of Q and contains the roots (the splitting field).\n3. Find the group of field automorphisms over that field. That is the Galois group.\nIdentify which of the above steps you are having trouble with and we will try to help.
1091111
1, 2, 3, 4, 5,  and so on are all just labels for items.  We have defined the group of integers by 1+1=1'=2, 2+1=2'=3, and so on where for any n≠0 n•1=1+1+1+ . . . +1+1≠0 (with n 1s).  Then we have the negative integers so that for every number a, there is a solution to the equation x+a=0.\n\nOne thing that we can do instead (as is learned in slightly more advanced algebra) is basically set all of the integers equal to 0.  Then 0=1=2=3=4=5= . . .  If we do this, and keep the standard addition that we use everyday, we still have a mathematical object called a group.  It is just like the integers, except that the integers are said to be a abelian group of infinite order, and the group where all the integers are set equal is an abelian group of order 1.  In particular, 1=2.  It is perfectly valid, but since the group of order 1 isn't really that interesting.  But to answer your question, yes 1 can be equal to 2, just not in a field of characteristic 0.