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

>>23
It is correct. Many mathematicians are arrogant, and later on they are rightly proved incorrected. Summation is a process, and it is not necessarily have a real number as its result. If you think it does, name the exact number. But you can't because it is proof by contradiction (no number you can provide equals ".9..." exactly).