0.999… is 1

Write a nought, a decimal point, and nines that never stop: 0.999…. Most people's first answer is that it falls just short of 1. It does not. In the ordinary real numbers, 0.999… and 1 are two ways of writing the same number, and the reason is a single line of arithmetic.

Why it is worth a look

D. O. Tall and R. L. E. Schwarzenberger, writing in 1978, report that first year university mathematics students, fresh from school, were asked whether 0.999… is equal to one or just less than one. The majority said less. Some wrote that the difference is "infinitely small"; one called it "the nearest you can get to one without actually saying it is one." That is a finding about how people learn, not about the number. The authors think it more likely that the answers are students' own attempts to resolve conflicts inherent in their earlier experience of limiting processes.

What the dots mean

Cut the nines off after n places and call the result k_n: 0.9, 0.99, 0.999, and so on. Each of these is an ordinary finite decimal, and each is less than 1. The gap is exact: 1 − k_n = 1/10ⁿ. After three nines it is a thousandth; after six, a millionth.

The symbol 0.999… does not mean any one of these. It means the number the sequence k₁, k₂, k₃, … closes in on: its limit. A sequence has limit s when, for any accuracy you name, however fine, every term from some point on is within that accuracy of s. Name any accuracy and you can pick n so that 1/10ⁿ is smaller. So the limit is 1, and 0.999… is 1.

Petr Eisenmann, writing from his experience as a college teacher, turns it into a question for anyone who still says "less": if 0.999… is below 1, what is 1 − 0.999…? It cannot be 0.0000000001, or any decimal with finitely many zeros before a 1, because adding that to 0.999… gives more than 1. What his students propose next is "infinitely many zeros and at the end 1". A decimal with infinitely many places has no end for that 1 to sit at (that reply is ours, not his).

Why it feels wrong

Tall and Schwarzenberger name several causes. Students read the symbol as a very large but finite number of nines. They imagine infinitesimals, quantities "infinitely close but not equal". And they expect every real number to have exactly one decimal expansion, so two different strings for one number looks like a contradiction.

That third expectation is nearly right. The only numbers with two expansions are those with a terminating one, and the second form ends in nines: 1.000… = 0.999…, and 2.317000… = 2.316999….

The school arguments, and their soft spots

The paper lists four arguments that might convince students at school; it flags two of them as shaky.

Thirds. 1/3 = 0.333…, so 3 × 1/3 = 0.999… = 1. This leans on accepting 0.333…, which students usually do.

Ninths. Long division gives 1/9 = 0.111… up to 8/9 = 0.888…, so 9/9 = 0.999…. Getting 0.999… out of 9 ÷ 9 needs "90 divided by 9 is 9, remainder 9", which the authors call "slightly dubious": a remainder should be less than the divisor.

Times ten. 10 × 0.999… = 9.999…; subtract 0.999… and 9 × 0.999… = 9. One student asked what happens to "the nine at infinity". Eisenmann calls the trick "a little bit of cheating": it works on infinite decimals digit by digit without asking whether that is allowed. His warning is 1 − 1 + 1 − 1 + …, which pairing one way makes 0 and another way makes 1.

Halfway. If (1 + a)/2 = a, then a = 1. Long division of 1.999… by 2 gives 0.999…, so a = 0.999… passes that test.

The limit argument is the one that needs no trick. The others are checks that it agrees with ordinary arithmetic.

In short

0.999… names the limit of 0.9, 0.99, 0.999, …, whose gaps to 1 are 1/10, 1/100, 1/1000, …. No positive gap survives all of them, so the limit is 1 exactly.