incredible and confusing but true

Two sets are considered equinumerous if they have the same number of elements. for

example, the set {2 5 6} is equinumerous with the set {-512 5 345}. Also, the set of all

natural numbers {1 2 3 4 5...} has the same number of elements as the set of all even

positive integers {2 4 6 8 10...}; they both have a cardinality of aleph null, meaning each

has aleph null elements. To prove that sets with a transfinite number of elements have

the same number of elements, a bijective, or one-to-one function from one set to the

other must exist. So, we can prove that there are as many even postive integers as there

are natural numbers by finding the bijective function between the two, and that function

is f(x) = 2x.

So, to prove the statement why is the total number of real numbers equal to the

number of numbers between 0 n 1?? , we have to find a bijective function mapping

the numbers in the interval (0,1) onto the set of all real numbers, from negative infinity to

positive infinity. The first such function I could find was a piecewise function defined as

f(x) = (ln(x)-ln(1/2)) when x is less than 1/2 and f(x) = -ln(1-x) + ln(1/2) when x is

greater than 1/2. Since a bijective function between (0,1) and the set of all real numbers

exists, the two sets are equinumerous, and each contains the same number of elements.