Newsgroups: sci.math From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz) Subject: sci.math FAQ: Why is 0.9999... = 1? Summary: Part 16 of many, New version, Message-ID: <DI76L5.nz@undergrad.math.uwaterloo.ca> Sender: news@undergrad.math.uwaterloo.ca (news spool owner) Date: Fri, 17 Nov 1995 17:15:05 GMT Reply-To: alopez-o@neumann.uwaterloo.ca 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

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