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Archive-name: sci-math-faq/0.999999
Last-modified: February 20, 1998
Version: 7.5
Why is 0.9999... = 1?
In modern mathematics, the string of symbols 0.9999... is understood
to be a shorthand for ``the infinite sum 9/10 + 9/100 + 9/1000 +
...''. This in turn is shorthand for ``the limit of the sequence of
real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...''. Using
the well-known epsilon-delta definition of the limit (you can find it
in any of the given references on analysis), one can easily show that
this limit is 1. The statement that 0.9999... = 1 is simply an
abbreviation of this fact.
0.9999... = sum_(n = 1)^(oo) (9)/(10^n) = lim_(m --> oo) sum_(n = 1)^m
(9)/(10^n)
Choose varepsilon > 0. Suppose delta = 1/- log_(10) varepsilon , thus
varepsilon = 10^(-1/delta). For every m > 1/delta we have that
sum_(n = 1)^m (9)/(10^n) - 1 = (1)/(10^m) < (1)/(10^(1/delta)) =
varepsilon
So by the varepsilon - delta definition of the limit we have
lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
Not formal enough? In that case you need to go back to the
construction of the number system. After you have constructed the
reals (Cauchy sequences are well suited for this case, see
[Shapiro75]), you can indeed verify that the preceding proof correctly
shows 0.9999... = 1.
An informal argument could be given by noticing that the following
sequence of ``natural'' operations has as a consequence 0.9999... = 1.
Therefore it's ``natural'' to assume 0.9999... = 1.
x = 0.9999....
10 x = 10 * 0.9999...
10 x = 9.9999...
10 x - x = 9.99999... - 0.9999...
Thus 0.9999... = 1.
An even easier argument multiplies both sides of 0.3333... = 1/3 by 3.
The result is 0.9999... = 3/3 = 1.
Another informal argument is to notice that all periodic numbers such
as 0.46464646... are equal to the period divided over the same number
of 9s. Thus 0.46464646... = 46/99. Applying the same argument to
0.9999... = 9/9 = 1.
Although the three informal arguments might convince you that
0.9999... = 1, they are not complete proofs. Basically, you need to
prove that each step on the way is allowed and is correct. They are
also ``clumsy'' ways to prove the equality since they go around the
bush: proving 0.9999... = 1 directly is much easier.
You can even have that while you are proving it the ``clumsy'' way,
you get proof of the result in another way. For instance, in the first
argument the first step is showing that 0.9999... is real indeed. You
can do this by giving the formal proof stated in the beginning of this
FAQ question. But then you have 0.9999... = 1 as corollary. So the
rest of the argument is irrelevant: you already proved what you wanted
to prove.
References
R.V. Churchill and J.W. Brown. Complex Variables and Applications.
5^(th) ed., McGraw-Hill, 1990.
E. Hewitt and K. Stromberg. Real and Abstract Analysis.
Springer-Verlag, Berlin, 1965.
W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.
L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.
_________________________________________________________________
--
Alex Lopez-Ortiz alopez-o@unb.ca
http://www.cs.unb.ca/~alopez-o Assistant Professor
Faculty of Computer Science University of New Brunswick
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Last Update March 27 2014 @ 02:12 PM
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