# How can you define a function with domain the whole of #RR# and range the whole of #CC#?

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(It is possible to define a bijection constructively)

(It is possible to define a bijection constructively)

##### 1 Answer

#### 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)))#