Not only that, I have found precisely how many natural numbers there are, and how many rational numbers between 0 and 1 there are ( 0 < x < 1)

I dont know how to use latex so follow my reasoning with paper and pen

q_N = how many natural numbers in the Natural number set

q_R(0-1) = how many real numbers there're between 0 and 1

so, fundamental principle of counting, basic combinatorics, how many unique 1 digit natural numbers you can form? 10 right?

q_N(1d) = (0-9) = 10

and 2 digit natural numbers (e.g. 20)? how about 3 digits (e.g 200)?

q_N(2d) = (1-9)(0-9) = 9 × 10
q_N(3d) = (1-9)(0-9)(0-9) = 9 × 10 × 10

so
q_N = q_N(1d) + q_N(2d) + q_N(3d) + q_N(4d) + ... = 10 + 9×10^1 + 9×10^2 + 9×10^3 +... = 10 + 9×(10^1+10^2+10^3+...)

the possible ammount of unique rational numbers between 0 and 1 can be determined the same way

q_R(0-1)(1d) = ( ) = 0 (the real number is between 0 and 1, 0 < x < 1)
q_R(0-1)(2d) = (0).(1-9) = 1×9
q_R(0-1)(3d) = (0).(0-9)(1-9) = 1×10×9
q_R(0-1)(4d) = (0).(0-9)(0-9)(1-9) = 1×10×10×9

q_R(0-1) = q_R(0-1)(1d) + q_R(0-1)(2d) + q_R(0-1)(3d) + q_R(0-1)(4d) = 0 + 1×9 + 1×10×9 + 1×10×10×9 + ... = 0 + 9×(1+1×10+1×10×10+ ... ) = 9 + 9×(10^1+10^2+10^3+...)

as
q_N = 10 + 9×(10^1+10^2+10^3+...)

and
q_R(0-1) = 9 + 9×(10^1+10^2+10^3+...)

you should be able to see that
q_N = q_R(0-1) + 1

incredible right? but what is the number q_N?

q_N = 10 + 9×(10^1+10^2+10^3+...)

10^1 = 10
10^2 = 100
10^3 = 1000

notice, if you add all the powers of the the result would be 111...110

so
q_N = 10 + 9×111...110

9 × 10 = 90
9 × 110 = 990
9 × 1110 = 9990
...

hence
q_N = 10 + 999...990

but
90 + 10 = 100
990 + 10 = 1000
9990 + 10 = 10000
...

so
q_N = 100...000

and as such
q_R(0-1) = 100...000 - 1

so the real numbers greater than 0 and less than 1 contain all of the natural numbers,.except for one unknown number (my guess is either 0 or 1).

If you include 0 and 1 in the real number set though q_R[0-1] = q_N + 1