if "f" is a surjective map find domaim "f"?
820313
If f : A-->B is surjective map, then its onto but not one-one .\nso you can write domain of f as set of all values that would give values in B.\n\nYou can write as\ndom(f) = {x : f(x)=y, x is in A, y is in B}\n\nSince the function is not one-one, there can be more than one x giving the same values of y, I dont think you need more conditions in set representation.
805944
if B' is subset of A' then\n\nif x belongs to B' then x belongs to A'\n\nif x belongs to A then x cannot belong to B' because if x belongs to B' then it belongs to A then x belongs to A and A'\n\nso if  x belongs to A the x belongs to B(as it does not belong to B')\n\nso A is subset of B
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
461122
Since f(x) is a polynomial of degree two in x,\nLet f(x) = ax^2 + bx + c, where a,b,c are constants.\n\nLet's find f(x) first.\n\nf(0) = 1 implies, c = 1.\n\nf(1) = 3 implies, a + b + c = 3\nor  a + b = 2 , or b = 2 - a  ...(1)\n\nf(-1) = 5 implies, a - b + c = 5\nor a - b = 4 ...(2)\n\nSubstituiting (1) into (2), we get,\n\na - [ 2 - a  ] = 4\n2a = 6 or a = 3.\nThus, b = 2 - a = 2 - 3 = -1.\n\nHence, f(x) = 3x^2 - x + 1\n\nNote that by long division or by the identity below,\n\n3x^2 - x + 1 = 3(x + 1)^2 - 7x - 2\n             \nWe get,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - [7x + 2]/ [1 + x]^2   \n\nAnd that together with below,\n\n[7x + 2]/ [1 + x]^2  = 7 /(x + 1) - 5 /(x+1)^2\n\nGives,\n\n[ 3x^2 - x + 1 ] /[1 + x]^2 = 3 - 7 /(x + 1) + 5 /(x+1)^2\n  \nNow, we have an "integratable form".\n\n∫ [f(x) / (1 + x)^2] dx\n\n= ∫ [3x^2 - x + 1] / (1 + x)^2] dx\n\n= ∫ [ 3 - 7 /(x + 1) + 5 /(x+1)^2 ] dx\n \n= 3x - 7 log(x + 1) - (5/[x + 1]) + C\n\nwhere C is any arbitary constant.\n\nNote: Sometimes log(x + 1) can also be written as ln (x + 1)\n\nRemark. It is generally accepted that log and ln are both in base "e", the Euler number. If its in a different base, then usually its said or stated. For example logarithm of x in base 2 can be written as log_2(x).\nJust symbols :)\n\nRegards
933970
f-g(x)=(x+2)-(3x^2-5) So\nf-g(x)=x+7-3x^2
621502
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.
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
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 !
1159041
If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings.  If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones.  If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.
109954
get "x" by it's selef
576001
By the transitive property, a=c
553317
To show the order of the automorphism of even, it suffices to find an involution i.e., an automorphism of order 2.  Consider the map g -> g^{-1}.  If the group is commutative this map is a homomorphism, and it obviously has order 2.
567979
inverse of f(x) is (x-1)/2\n\nworking:\nlet inverse of f(x) be y\n       2x+1  = y\n       2x    = y-1\n        x    = (y-1)/2\nthus, inverse of f(x)= (x-1)/2
394920
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
457089
G=uys\nF=***\nE=verthing
734599
kbhit will return a integer to you, go to asciitable.com to find out what des each interger corresponds to what key.\n\ngood luck
864853
a)  f(0) = f(0-0) = f(0) - f(0) = 0.\n\nb)  Fix k in the reals, and define f(x) = kx.\n\nc)  Suppose you have f(x)=f(y) for some reals x and y.  Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0.  Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.\n\nI have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective.  There are such functions, but they are not necessarily easy to find.  Cheers!