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

>>46
Of course 0.333... is not a real number, but 1/3 in base 3 is a real number because you can find a finite base in which it can be finitely represented. If you have a set S = {a, b, c} and a set T = {a, b, d}, you cannot say that any element in S v T, but not in S ^ T, has a corresponding element in the other set. Simple. Same principle here.