(Copyright © 2000-2001 --
this item revised 26 Nov 2000)

Yes, the repeating decimal 0.999... is exactly equal to 1.

There are a number of reasons we know this is true, even though you may see a number of plausible arguments against it.

There are many proofs. All depend on understanding what a repeating
decimal means: it means you *never* stop writing the repeating
digits.

Simple division: Divide 1 by 9 and get the repeating decimal 0.111... .

1/9 = 0.111... Now multiply by 9: 9*(1/9) = 9*0.111... And simplify: 1 = 0.999...

Summation of series: 0.999... is a geometric series:

0.999... = 0.9 + 0.09 + 0.009 ... First term: a = 0.9 Ratio: r = 0.1

Such a series must converge, since |r| < 1. Any algebra book will prove that the sum of a infinite series that converges isS = a/(1-r)

Plug and chug:S = 0.9/(1-0.1) = 0.9/0.9 = 1

Q.E.D. (mathematician's Latin for "nyah, nyah! nyah! told ya so!")

(The "real" proof) The infinite decimal notation is a way of writing a real number by defining a Cauchy series which converges to a real number, and then "calling" that real number by its decimal expansion. Defining

a(n) = 0.9999..9 [9 repeated n times]

we have1 - a(n) = 10^(-n)

which converges to zero as n increases without bound. In other words, the limit of this series is 1, so by definition the number equals 1.

- 0.999... < 1 because you can put a number between them.
People who say this forget what a repeating decimal means: you keep writing those 9's forever. If you claim some number x comes between 0.999... and 1, if I write enough 9's I can give you a number that is larger than x but still smaller than 1.

- 0.999... approaches 1 but never gets there.
No, a repeating decimal is not an approximation that gradually moves toward a goal. It is a definite number. 0.999... is a constant, just another way of writing 1 -- the same as the repeating decimal 1.000... .

- Dr. Math FAQ item

- Stan Brown
- Haran Pilpel

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