Verified answer

inverse of f(x) = (1-x) / (x (x-2))

when y=0 then x=0

when y<>0 then x=+/- (1/2y) (sqrt (4^y2 + 1) -2y + 1)

since if its bijective it means there exist one...

y = p(x) / q (x) and y is defined for every x where q(x) not equal to 0

q(x) = x (x-2) is only 0 for x=0 and x=2 so for every x in (0,2) y has a value

Now prove f(x) is bijective means its one to one and onto

one to one means f(x) = f(y) gives x = y

(1-x) / (x (x-2)), = (1-y) / (y (y-2))

(x-1)/(y-1) = (x (x-2)) / (y (y-2)) (divided and multipled with -1 up and down to switch "1-x")

set t(a) = (a-1) and s(a) = (a (a-2))

What we have is t(x)/t(y) = s(x)/s(y) which means that t(a)=k * s(a) for some k (real)

x-1 = k * x * (x-2)

y-1 = k * y * (y-2)

and above holds for same k

set t=x-1 and u=y-1 gives

t = k * (t^2 - 1)

u = k * (u^2 - 1)

and above holds for same k

t / (t^2 - 1) = u / (u^2 - 1)

meaning

t = K * u ...for some other K than k :-) .....(1)

(t^2 - 1) = K * (u^2 - 1) .....(2)

so since t can be written as K times u we now only need that K must be 1

use (1) in eq (2)

(t^2 - 1) = K * t^2/K^2 - K

(t^2 - 1) = t^2/K - K

K - 1 = t^2/K - t^2

set w = t^2, meaning that w must be >=0

K - 1 = w/K - w

K^2 - K = w - Kw

K^2 - K + Kw - w = 0

K^2 + (w-1)K - w = 0

(K-1)(K+w) = 0

So K must med 1 or -w

If K = -w then since t=Ku = -w*u= -t^2 * u, gives t=-1/u , x=t+1=-1/u+1, y=u+1

x = 1 - 1/(y-1), and x,y must be in (0,2) leaves K=1 as the only solution

So now prove "onto"

x=+/- (1/2y) (sqrt (4^y2 + 1) -2y + 1)

since y is real number so is x and we can find one for every y, check eq above

Cardinality means that since they are both subsets of R they have same "amount" infinite numbers, they are not countable, but you can find a bijection (0,2) to any other subset of R for example take a subset (0,R) and define function g(x) from x in (0,2) to g in (0,R) by g(x) = x * (R/2) so for every value y in (0,R) you can find a value x in (0,2) so that y = x * (R/2)

For the 2d anticipate that F(x) is a polynomial with maximum power n. F(F(x)) will comprise power 2n and so the equation F(F(x)) = 6x - F(x) won't be in a position to be an id till n = a million. enable F(x) = ax + b ----> F(F(x)) = a(ax + b) + b = (a^2)x + ab + b F(F(x)) = 6x - F(x) ----> (a^2)x + ab + b = 6x - ax - b (a^2 + a - 6)x + (a + 2)b = 0 For this to be an id we choose for a^2 + a - 6 = 0 and (a + 2)b = 0 the 1st provides a = 2 or -3, the 2d provides a = -2 or b = 0. y = -3x does no longer be useful for useful x so a = 2, b = 0 is the only answer. So y = 2x is the only answer IF F(x) is a polynomial. i think of that via connection with power sequence all different applications are eradicated yet i'm unsure.