f:R-R be a homomorphism  s.t  f^n=identy.Show  either  f or  f^2  is  identy,R=real  number?
460866
Not true. It is true for continuous homomorphisms. In that case a homomorphism is of the form f(x)=ax for some constant a. THen a^n=1, so a=+-1.
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!
933970
f-g(x)=(x+2)-(3x^2-5) So\nf-g(x)=x+7-3x^2
493069
First :\nSince f is continuous over its domain (R),\ndf/dx = 5 x^4 +2 > 0 for every x\nSo f is an increasing function over its domain\nHence for every value of y their exist a unique value of x\nHence f has an inverse and one-one function\n3- Use function graph to graph f-1 as you mentioned\n(df^-1/dx) (2)= the slope of the tangent to f-1 at x=2\n6- (1,2)\n7- 7\n8- 1/7
878633
we need to prove this by induction...\n\nfor n belonging to Z\n\nf(n)=n*(n+1)*(2n+1)/6\n\nfor n=1\n\nLHS:\n f(1)= 1^2 =1\nRHS\n1*(1+1)8(2*1+1)/6=6/6=1\n\nThus its true when n=1;\n\nAssume f(k) is true i.e. \n1^2+2^2+......+k^2=k(k+1)(2k+1).. Let this be equation 1\n\nNow we need to prove f(k+1) is true...\nie\nf(k+1)=(k+1)(k+2)(2(k+1)+1)/6\n          =(k+1)(k+2)(2k+3)/6\nL.h.s\n\nf(k+1)=1^2+2^2........+k^2+(k+1)^2\n          =k(k+1)(2K+1)/6 + (k+1)^2 from eq 1\n          =(k+1)(k(2k+1)/6 +k+1)\n          =(1/6)(k+1)(2k^2+k+6k+6)\n          =(1/6)(k+1)(2k^2+7k+6)\n          =(1/6)(k+1)(k+2)(2k+3)\n\nHence f(k+1) is true if f(k) is true....\n\nThus its true for all positive integers.......
466194
Consider K_3,3, which is the graph obtained when you draw all lines between 3 points {a,b,c} and three other points {x,y,z}. So there is a line between a and x, but not one between a and b.\n\nAn automorphism, f, will take a to one of the six points. There are two possibilities: either f(a) is in {a,b,c} or f(a) is in {x,y,z}. In the first case, {f(a), f(b), f(c)} is a permutation of {a,b,c} and {f(x),f(y),f(z)} is a permutation of {x,y,z}. In the second case, it goes the other way around. Furthermore, any two permutations in both sets will lead to an automorphism of your graph.\n\nThus, the order of the automorphism group will be 6*6*2=72.\nThere is a subgroup isomorphic to S_3 cross S_3 of index 2. Just checking commutativity tells us that the reflection between {a,b,c} and {x,y,z} is not in the center, and in fact, conjugation by the reflection reverses the order of elements in S_3xS_3. Thus, the automorphism group is the semi-direct product of S_3 cross S_3 with Z_2 where Z_2 reverses the order of the cross product.\n\nI hope this helps!
914403
i think its true\nim f is isomorphical  with G/(ker f) so\n|G| = |ker f|*|G/(ker f )| = |ker f| *|im f|.\n i applied two well known theorems.
803829
Isomorphic if you are in abstract algebra.\n\nCongruent if you are in geometry.
252734
This is how I remember it...hope it helps.\n\nA Proof\nLet a and b be any two real numbers. Consider the number x defined by\n\nx = ab + (-a)(b) + (-a)(-b).\n\nWe can write\nx = ab + (-a)[ (b) + (-b) ]       (factor out -a)\n  = ab + (-a)(0)\n  = ab + 0\n  = ab.\n\nAlso,\nx = [ a + (-a) ]b + (-a)(-b)      (factor out b)\n  = 0 * b + (-a)(-b)\n  = 0 + (-a)(-b)\n  = (-a)(-b).\n\nSo we have\n      x = ab\nand\n      x = (-a)(-b)\n\nHence, by the transitivity of equality, we have\n\n      ab = (-a)(-b).
745315
domain=any real number
359378
This is the ring of polynomials in two variables with complex coefficients. First figure out what the units are (the scalars). An irreducible will be a polynomial p such that if p=fg, then either f or g is a unit. Clearly x and y are irreducible since if x=f(x,y)g(x,y), then either f or g will be a scalar polynomial.
760867
f(x) = (x+3)(x^2 + 16)
782051
Given 2 functions f and g with the domains Df and Dg respectively.\n\nAny composite function of f and g will as domain have Dfg=Df ∩ Dg &#92;{ those interval or point where the composite functions has poles coming from the way the functions is composited}\n\nEX 1)\nf=x^2 ; Df={x in R l 0<x<20} by own choice\ng=x ..; Dg={x in Rl x belongs to [-5,5[} by own choice\n\na composite function fg=f/g+1/f-g\nfinding the domain Dfg:\nDfg={ x in R l x belongs to Df ∩ Dg &#92; {those interval or point where the composite functions has poles coming from the way the functions are composited} }\n\nDf∩Dg=\n{x in R l 0<x<20}∩{x in Rl x belongs to [-5,5[}=\n{x in R l 0<x<5}\n--------------------\n\n\nfinding: \nPole={those interval or point where the composite functions has poles coming from the way the functions are composited}\n\nlook at the composite function: fg=f/g+1/f-g.\nWe easily see the composite function has pole for g=0=x => x=0 and for f-g=0=x^2-x=x(x-1)=> x=0 or x=1=>\nPole={ x in R l x=0 or x=1}\n----------------------------------\nnow we can find the domain for the composit function fg\n\nDfg=Df ∩ Dg &#92; Pole ={x in R l 0<x<5} &#92; { x in R l x=0 or x=1} }= {x in R l 0<x<5 & x≠1}\n=======================
1149923
(2,-4) if f(x) is defined for x=2
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
186035
1) What is the difference between F(n) and f(n)? \n\n2) I cannot comprehend the second condition.\n\nEdit your question, please.
231905
I'm assuming this is a group isomorphism. I will use "e_G" for the identity element of G, and "e_H" for the identity element of H.\n\nLet G and H be groups and f:G->H be an isomorphism. Let a be an element of G of infinite order. Suppose on the contrary that f(a) does not have infinite order. Then there exists some positive integer n such that e_H=f(a)^n=f(a^n), so a^n is in the kernel of f. Since f is an isomorphism, the kernel of f is {e_G}, thus it must be that a^n=e_G, contradicting that a has infinite order.