>>11055451

<cont.>

For an infinitely long decimal expression, we define the real number that it represents by taking a sequence of increasing finite truncations of the decimal expression. In particular, we write 0.9999... to denote the real number represented by the sequence (0, 0.9, 0.99, 0.999, and so on). To make sense of the real number 1, we inject rational numbers into reals by letting the real number q' by (q,q,q,q,....) for a rational number q. Then, in order to prove that the real numbers 1=[(1,1,1,1,...)] (where [x] denotes the equivalence class of the sequence x), and 0.9999...=[(0.9, 0.99, ...)] are equal, it suffices to show that they lie in the same equivalence class (because gain, real numbers are equivalence classes, not elements in those classes themselves). The n'th term in the canonical sequence for 0.999... is 1-(10^-n). The n'th term in the sequence for 1 is 1. Hence it suffices to prove that |1-(10^-n) - 1 | = 10^(-n) -> 0 as n->infinity. Looking back at the definition of a limit, let x>0 be any positive rational number. We want to find an N such that for all n>N, 10^-n < x, or 10^n > 1/x. But that is simple, just take any N such that 10^N > 1/x. Hence, the limit is proved and 0.9999... = 1.

Note that by similar arguments, we also have [(0.97, 0.997, 0.9997,...)]= 1 etc.

Keep in mind that everything I just explained is not complicated. In fact, it's elementary mathematics that is often left as an exercise in introductory analysis books. That means you should be able to understand what I just explained and accept that 0.999..=1.

retards