Let f: A->B and g:B->C.  If the composition gof and f are bijective (one to one and onto), then is g necessarily bijective?
212445
I think so, yes.\n\nBy contradiction, if g is not injective then exists x0,x1 member of B such that g(x0)=g(x1)\nAs f is bijective, there exists y0,y1 member of A st f(y0)=x0, f(y1)=x1)\n\nSo the composition f0g maps y0,y1 onto the same point in C\nThis contradicts the assumption of f0g being bijective\n\nI'm pretty sure you can do a similar thing for surjectivity
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.
545212
For each pair (A, A') of elements of B such that A is a subset of A', pick, if possible, an element C is B' so that A is a subset of B is  a subset of A'. If it isn't possible to find a C, don't pick anything. Since there are countable many pairs, (A,A'), there are only countable many C chosen. Let B'' be the collection of these C. Then elements of B'' are open with property P.\n\nTo show that B'' is a base, suppose that a point x is in an open set U. Choose A' in B with p in A' a subset of U. Then choose C' in B' with x in C' a subset of A'. Finally pick A in B with x in A a subset of C'. Now, the pair (A,A') has elements of B' of the appropriate form, so the C associated with (A, A'), satisfies\nx in C a subset of U. Notice that C and C' may be different!
532909
Composition of functions means that you evaluate one function, then take that answer and use it to evaluate the next function.  For example, if you are trying to find (f o g)(1), plug 1 into the formula for g(x).  Then take that answer and plug it into f(x).  If you are trying to find (g o f)(-2), plug -2 into the formula for f(x) and then use that answer in g(x).  To find (g o g)(-1), plug -1 into the formula for g(x), then use that answer in g(x) again.\n\nsome websites for math:\nwww.algebrahelp.com\nwww.askdrmath.com\nwww.purplemath.com
513250
say g(x)=2x^2+3x+8\nh(x)=x-8\ng(x)+h(x)=2x^2+4x=f(x)\n\n2.let f(x)=x^2\nf(-x)=(-x)^2=x^2 we say f(x)=x^2 is an even function\n\n3.let f(x)=x^3\nf(-x)=(-x)^3=-x^3\nf(-x)=-f(x)\nwe say f(x)=x^3 is an odd function
526512
If A is a subset of C and A has the same cardinality as C, then the sets A and C are the same.\n\nI guess as a proof, you could assume that there is an element in C that is not in A...in which case the cardinality of C would be greater than the cardinality of A, which is a contradiction.\n\nThen B gets trapped in the middle.  A = B = C.
412216
A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
548515
Here is an outline:\n\nf is an ONTO function. So for EVERY element b in B there is an\nelement a in A so that f(a) = b.  That is what onto means.  All the\nelements in B have some element in A that f maps it into.\n\nSince g is not one to one there must be two elements  in B that get mapped into the same element in C.  Call them b1 and b2. Show that b1 and b2 are part of the codomain of f.  f(x) maps into all B, so there must be a a1 and a2 that get mapped into b1 and b2.  So then we have g(f(a1)) and g(f(a2) that map into the same element, and hence g(f(x)) is not one to one.
183821
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.
433926
Let g(x) = f(x) - f'(x). Then the problem can be restated as saying that g(x) has at least one real root. If f(x) is a polynomial of degree n, where n≠0, then f'(x) is a polynomial of degree n-1, and so the difference of the two polynomials will be a polynomial of degree n. We consider two cases:\n\nSuppose n is odd. Then g(x) has odd degree, and all polynomials of odd degree have a real root, so we are done.\n\nSuppose n is even. Then f(x) has either an absolute minimum or an absolute maximum. Suppose without loss of generality that it has an absolute maximum at c (the arguments for polynomials with an absolute minimum are essentially the same). Then at that point f'(c) = 0, so g(c) = f(c). Since f(x) has a real root, f(c) ≥ 0. If f(c) = 0 then g(c) = 0 and we are done. Otherwise, consider that the x^n term of f(x) dominates the terms of f'(x), so [x→∞]lim g(x) = -∞. Thus, g(x) is negative at at least one point, and since g(x) is everywhere continuous, then by the intermediate value theorem it must have at least one zero. Q.E.D.\n\nNote that in the special case where n=0, f(x) is a constant function, and thus does not have a root, so it does not meet the conditions of the theorem. Finally, in the case where f(x) = 0 (which is usually considered to be a polynomial of indeterminate degree, rather than degree zero, primarily to avoid exceptions to the fundamental theorem of algebra), f(x) is everywhere equal to f'(x).
160082
Let H be a subgroup of order p. Then G acts via translationon the cosets of H. This induces a map of G into the symmetric group on n letters since there are [G:H]=n cosets. But the order of this symmetric group is n! and p does not divide n!. Hence, the kernel of this action contains H and is not all of G, so is a non-trivial normal subgroup of G.
553493
D_4 is NOT abelian.\nnow if f:G ->D_4 was onto, then:\ntake a, b in D_4, then there exist g1 and g2 in G such that\nf(g1) = a and f(g2) = b\nnow\nab =f(g1)f(g2) =f(g1g2) =f(g2g1) = f(g2) f(g1) = ba,\nwhich would imply that D_4 is also abelian!!!! contradiction, therefore f cannot be onto\n`
497325
1) yes, by definition\n2) no, example: sin x on [0,2 pi]\n3) no , example |x|, absolute value of x is continuous in 0  but not differentiable in 0.\n4) yes, since (F-G)' = f-f = 0, then F-G is a constant.\n5) yes, known theorem
393069
1. f(g(x))\n2. g(f(x))
153427
Im not entirely sure if this is correct but, I know that a relation is a partial ordering if it is reflexive, transitive and anti-symmetric.  \n\nSo to satisfy the reflexive part we need that x>=x, which is trivially satisfied by f(x)>=f(x).  For the transitive part we need that if x>=y and y>=z then x>=z, which will be true if for f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).\nFor the anti-symmetric part we need that if x>=y and y>=x then x=y, so when f(x)>=f(y) and f(y)>=f(x) then x=y\n\nSo this would suggest that the necessary and sufficient conditions on f are that: (1) when f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).  (2) if f(x)>=f(y) and f(y)>=f(x) then x=y.  So f would appear to be one to one.\n\nI hope this helps
518349
Well, let's look at a few of them to see if there is a pattern.\nf(x) = 2x + 1 = (2^1)x + (2^1 -1)\n(fof)(x) = 2(2x+1) + 1 = 4x + 3 = (2^2)x + (2^2 -1)\n(fofof)(x) = 4(2x+1) + 3 = 8x + 7 = (2^3)x + (2^3 -1)\n(fofofof)(x) = 8(2x+1) + 7 = 16x + 15 = (2^4)x + (2^4 -1)\nDo you see the pattern now?\n(fofofo...of)(x) (n functions) = (2^n)x + (2^n -1)
504762
Don't you see the pattern?\nIt's right in front of your eyes.  Let g(x) be what is in the parentheses.  So g(x) = x+2\nNow replace x+2 with just x, and that will be f(x).  \nSo f(x) = x^3 - 5x^2 + 3x - 1\nAgain, this is assuming that h(x) = f(g(x)) (based on your other problems)