If you took calculus, believe it or not, you have the tools to prove that 0.9999… = 1 . Think back to the concept of a “limit.” OK, I really don’t expect you to remember it, so please be patient as I remind you and hopefully give you some intuition *why* it works.
To prove it to yourself that 0.9999… = 1, consider that if they weren’t equal, there would be a number E that is greater than zero such that E = (1 — 0.9999…). So now we have a game. You give me a candidate value for E, say 0.0001, and then I can give you a number D of 9’s repeating which causes (1 — 0.9999…) to be smaller than E (in this case 0.99999 (D = 5), because 1 — 0.99999 < 0.0001 ). Since we’re playing this game, you counter and make E smaller, say 10^(-10), and I turn around and say “make D = 11” (because 1 — 0.99999999999 < 10^(-10) ). Every number E that you give me, I can find a D. Specifically, if E > 10^(-X) for some positive integer X, then setting D = X will do it. It’s a proof by contradiction. There is no E that is greater than zero such that E = (1 — 0.9999…). Therefore 0.999… = 1.
This game is the definition of a limit. We define the limit as x approaches infinity of f(x) = y if for every epsion > 0 (E) there exists a delta (D) such that x > delta implies that |f(x) — y| < epsilon. Here f(x) = SUM from n=1 to x of 9 * (10^(-n)) = 0.9… [with 9 repeated x times]. As it turns out, all limits are proof by contradiction. If the two quantities we are comparing aren’t equal, then there would be a non-zero difference, and by playing this game “for every epsion, there exists a delta”, you prove that such a non-zero difference doesn’t exist, and therefore the two quantities are equal.