0.9999... equal to 1?
Date: 6/28/96 at 22:59:40 From: Luke Crotty Subject: 0.999999999... equal to 1? Dear Dr. Math, I am a final year high school student in Adelaide, Australia. I have two Math teachers. One insists that the value of 0.9999999999... is one. The other insists that anyone who thinks that 0.9999999... is one is an idiot! I would really like to prove one of them wrong, so any info regarding this much debated topic would be greatly appreciated. Thanks Luke Crotty
Date: 6/30/96 at 16:21:54 From: Doctor Ethan Subject: Re: 0.999999999... equal to 1? Hello! This is our standard answer for the question "Why does 0.9999... equal 1?" It is adapted from an entry in the sci.math Frequently Asked Questions file, which is Copyright (c) 1994 Hans de Vreught (firstname.lastname@example.org). Why is 0.9999... = 1 ? The first thing to realize about the system of notation that we use (decimal notation) is that things like the number 357.9 really mean "3*100 + 5*10 + 7*1 + 9/10". So whenever your write a number in decimal notation and it has more than one digit, you're really implying a sum. So in modern mathematics, the string of symbols 0.9999... is understood to mean "the infinite sum 9/10 + 9/100 + 9/1000 + ...". This in turn is shorthand for "the limit of the sequence of numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ...." One can show that this limit is 9/10 + 9/100 + 9/1000 ... using Analysis, and a proof really isn't all that hard (we all believe it intuitively anyway); a reference can be found in any of the Analysis texts referenced at the end of this message. Then all we have left to do is show that this sum really does equal 1: Proof: 0.9999... = Sum 9/10^n (n=1 -> Infinity) = lim sum 9/10^n (m -> Infinity) (n=1 -> m) = lim .9(1-10^-(m+1))/(1-1/10) (m -> Infinity) = lim .9(1-10^-(m+1))/(9/10) (m -> Infinity) = .9/(9/10) = 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 lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1 . 0.9999... = 1 Thus x = 0.9999... 10x = 9.9999... 10x - x = 9.9999... - 0.9999... 9x = 9 x = 1. Another informal argument is to notice that all periodic numbers such as 0.9999... = 9/9 = 1 are equal to the digits in the period divided by as many nines as there are in the period. Applying the same argument to 0.46464646... gives us = 46/99. 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. -Doctor Ethan, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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