There is no "after infinity". You can justify the idea by defining a decimal representation of a number x as a vector x_2, x_1, x_0, x_{-1}, x_{-2} ..., with x

There is no "after infinity". You can justify the idea by defining a decimal representation of a number x as a vector x_2, x_1, x_0, x_{-1}, x_{-2} ..., with x_n ∈ {0, 1, ..., 9}. Negative indexes are digits after the comma, positive indexes before the comma. You can recover the original number simply using x = \sum_{n=-∞}^{n=∞} 10^n x_{n} (1) (this sum always converges as long as x_n = 0 when n > N, for some big enough N ∈ ℕ). For example, the number three is represented by x_0 = 1, x_n = 0 otherwise. 0.9... is defined as x_n = 0 for n >= 0, 9 for n < 0. Now, by using limits, we could recover the original number using the formula above. However, we can do it without them. For that, we define just enough operations for the proof. 1. If z = x - y and for all x_n, y_n we have that x_n >= y_n, then z_n = x_n - y_n for all n. 2. If z = 10x, then z_n = x_{n-1}. For the first operation, in order to be rigorous, we need to ensure that if z_n = x_n - y_n in the same conditions, then z = x - y. The proof of this part just consists of plugging the recovery formula (1): z = \sum 10^n z_n = \sum 10^n (x_n - y_n) = (\sum 10^n x_n) - (\sum 10^n y_n) = x - y. We can perform all those operations as we are guaranteed that (1) always converges. Now, let y = 0.9... defined as in the example above (y_n = 0 for n >= 0, 9 otherwise), and let x = 10y (therefore x_n = 0 for n > 0, 9 otherwise). Now, define z_n = x_n - y_n, so that z_n = 9 for n = 1, 0 otherwise, which yields z = 9. As we demonstrated above, this implies that z = x - y, therefore 9 = 10y - y => y = 1, so 1 = 0.9... . PS: I don't think one can make a proof without at least using some bits of limits to be able to switch between decimal representation as a vector and the number itself. However I don't think it's a problem, because you need the same bits to be able to talk of "0.9..." as a well-defined number.