(I haven’t done real math since forever)

Part 1: Proving If x,y are distinct positive real numbers and x < y, there is a z in R such that x < z < y.

Let y = x + e, where e is in R. e is also positive because:

x < y
x < x + e
0 < e

x < z < y becomes x < z < x + e. We can then choose z = x + 0.5e so that

x < z < y
x < x + 0.5e < x + e
0 < 0.5e < e
0.5e - 0.5e < 0.5e + 0 < 0.5e + 0.5e
-0.5e < 0 < 0.5e

Since we’ve shown earlier that e > 0, -0.5e < 0 < 0.5e is true no matter what e.

Part 2: Decimal Notation

Decimal notation is written as the sequence a_ma_m-1…a₁a₀.b₁b₂…b_n, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence a_ma_m-1…a₁a₀.b₁b₂…b_n itself is of the number a_m10^m + a_m-110^m-1 + … + a₀10⁰ + b₁10^-1 + b₂10^-2 + … + b_n10^-n.

Part 3: The actual proof through proof by contradiction

Let x = 0.999… and y = 1. This means there exists a z that is between 0.999… and 1. So let’s construct z = a₀.b₁b₂…b_n. a₀ has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a₀=0. z = 0.b₁b₂…b_n

0.999… = 9*10^-1 + 9*10^-2 + 9*10^-3 + … while z = b₁10^-1 + b₂10^-2 + … + b_n10^-n. If 0.999… < z, then

0.999… - z < 0
9*10^-1 + 9*10^-2 + 9*10^-3 + … - (b₁10^-1 + b₂10^-2 + … + b_n10^-n) < 0
(9*10^-1 - b₁10^-1) + (9*10^-2 - b₂10^-2) + (9*10^-3 - b₃10^-3) + … < 0
(9 - b₁)*10^-1 + (9 - b₂)*10^-2 + (9 - b₃)*10^-3 + … < 0

But as we’ve established in Part 2, b₁, b₂, b₃, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - b_n) > 0 for any n. Therefore, (9 - b₁)*10^-1 + (9 - b₂)*10^-2 + (9 - b₃)*10^-3 + … cannot be less than 0.

The theorem requires for x and y to be distinct positive real numbers and x < y, and since 0.999… and 1 can be trivially shown to be positive real numbers, this means that 0.999… and 1 are not distinct. In other words, 0.999… = 1.