(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 is
S = 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 have
1 - 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.
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.
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... .