omg new proof

Assumptions
If A = B, then
a)  A-B = 0
b) A = (A+B)/2 = B

If A > B, then
a)  A-B > 0
b) A > (A+B)/2 > B


Let A = 1, B = .999...9
a) A-B = 1-.999...9 = .000...1 =/= 0
b) (A+B)/2 = (1+.999...9)/2 = 1.999...9/2 = 1.999...95 < A

Therefore, .999...9 =/= 1

>>48
Countable, Pi is not a real number. You have two problems to deal with still, and you've basically just accepted the fact that your definition of the real numbers can not be represented in one base, contrary to the "actual" real numbers (you know, the set that every mathematician in the world is familiar with and agrees upon). Furthermore, your definition offers no advantages over the "actual" real numbers to make up for this deficiency. As you've stated it in this most recent post, the real numbers are the rational numbers.