How can you define a function with domain the whole of #RR# and range the whole of #CC#?
(It is possible to define a bijection constructively)
(It is possible to define a bijection constructively)
1 Answer
See explanation...
Explanation:
Define
#h(x) = (1-2x)/(x(x-1))#
graph{(sqrt(1/4-(x-1/2)^2))/(sqrt(1/4-(x-1/2)^2))(1-2x)/(x(x-1)) [-1, 2, -10, 10]}
Then:
#h^(-1)(y) = ((y-2)+sqrt(y^2+4))/(2y)#
So
Let
We can use
#k(a+bi) = h(a)+h(b)i#
#k^(-1)(a+bi) = h^(-1)(a) + h^(-1)(b)i#
Having found these bijections, next consider decimal representations of numbers in
Some numbers have two possible decimal representations; one with a tail of repeating
Then there is a bijection between
Given a number
For example:
#0.0230040291...#
splits as:
#0." "02" "3" "004" "02" "9" "1" "...#
Then recombine alternate subsequences to form two Real numbers in
#0.020049...#
#0.3021...#
These are the Real and imaginary parts of a corresponding Complex number in
This process defines a function:
#s(x): (0, 1) -> S#
Conversely, given a Complex number in
So given:
#0.020049... + 0.3021... i#
Split into subsequences:
#0." "02" "004" "9" "...#
#0." "3" "02" "1" "...#
Interlace:
#0.0230040291...#
This defines the inverse function
Hence we can define a function with domain
#f(x) = k^(-1)(s(h(x)))#
This is a bijection, with inverse:
#f^(-1)(y) = h^(-1)(s^(-1)(k(y)))#