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3 Responses
0.999… really is 1: if x = 0.999…, then 10x = 9.999… so 9x = 9 and x = 1 (same as the geometric series 9/10 + 9/100 + … summing to 1-no crumbs left!). A reduced fraction a/b terminates in base 10 exactly when b’s prime factors are only 2s and 5s; otherwise it repeats-hope this helps!
I used to swear there was a tiny gap, but nope-let x = 0.999…, then 10x = 9.999…, so 10x − x = 9 ⇒ 9x = 9 ⇒ x = 1. A reduced fraction has a terminating decimal iff its denominator’s prime factors are only 2s and/or 5s (e.g., 1/8 = 0.125 stops, 1/6 = 0.1666… repeats); neat explainer: https://www.mathsisfun.com/numbers/0.999….html
That “teeny crumb” feeling is super common-I had it too! Here’s the clean punchline: let x = 0.999…; then 10x = 9.999…, so 10x − x = 9, which means 9x = 9 and x = 1. Another way: 0.999… is 9/10 + 9/100 + 9/1000 + …, a geometric series whose sum is exactly 1. If there were a crumb c = 1 − 0.999…, it would have to be smaller than 1/10, 1/100, 1/1000, and so on-all at once. The only number smaller than every one of those is 0. So there’s no crumb hiding; 0.999… and 1 are the same point on the number line. Think of topping up a glass: first 90% full, then add 9%, then 0.9%, then 0.09%… those splashes land you exactly at 100%, not almost.
For which fractions “stop” and which “repeat,” there’s a neat rule. Reduce the fraction a/b to lowest terms. In base 10, the decimal terminates if and only if the denominator b has no prime factors other than 2 or 5 (because 10 = 2 × 5). So 3/8 = 0.375 and 7/20 = 0.35 terminate, while 1/6 = 0.1666…, 2/15 = 0.1333…, and 4/45 = 0.0888… repeat (their denominators bring in a factor of 3). Every rational number’s decimal either stops or eventually cycles-no middle ground.