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sci.math FAQ: Why is 0.9999... = 1?


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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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