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Archive-Name: sci-math-faq/specialnumbers/0.999eq1
Last-modified: December 8, 1994
Version: 6.2

Why is 0.9999... = 1 ?

In modern mathematics, the string of symbols 0.9999... = 1 is
understood to be a shorthand for the infinite sum 0.9999... ''. This
in turn is shorthand for the limit of the sequence of real numbers
9/10 + 9/100 + 9/1000 + ... , 9/10 , 9/10 + 9/100 ''. 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 9/10 + 9/100 + 9/1000, ... . The statement that 1 is
simply an abbreviation of this fact.

0.9999... = 1

Choose 0.9999... = sum_(n = 1)^(oo) (9)/(10^n) = lim_(m --> oo) sum_(n
= 1)^m (9)/(10^n) . Suppose varepsilon > 0 , thus delta = 1/- log_(10)
varepsilon . For every varepsilon = 10^(-1/delta) we have that

m > 1/delta

So by the \left| sum_(n = 1)^m (9)/(10^n) - 1 \right| = (1)/(10^m) <
(1)/(10^(1/delta)) = varepsilon definition of the limit we have

varepsilon - delta

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 lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 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 .

0.9999... = 1

Thus x = 0.9999... ; 10x = 10 o 0.9999... ; 10x = 9.9999... ; 10x - x
= 9.9999... - 0.9999... ; 9x = 9 ; x = 1 ; .

An even easier argument multiplies both sides of 0.9999... = 1 by
0.3333... = 1/3 . The result is 3 .

Another informal argument is to notice that all periodic numbers such
as 0.9999... = 3/3 = 1 are equal to the period divided over the same
number of 0.46464646... s. Thus 9 . Applying the same argument to
0.46464646... = 46/99 .

Although the three informal arguments might convince you that
0.9999... = 9/9 = 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... = 1 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... 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.
0.9999... = 1 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.

This subsection of the FAQ is Copyright (c) 1994 Hans de Vreught. Send
comments and or corrections relating to this part to
hdev@cp.tn.tudelft.nl.

_________________________________________________________________

alopez-o@barrow.uwaterloo.ca
Tue Apr 04 17:26:57 EDT 1995