Date: 5/31/96 at 16:49:42 From: Anonymous Subject: Repeating Decimals Hi, I am a high school teacher who is having a running argument with some students about repeating decimals. I claim that 0.9 repeating is equal to 1. One of my proofs involves the lack of numbers between these two numbers, much like any other two identical numbers. Another good proof is the pattern that comes from 1/9 = 0.1 repeating, 2/9 = 0.2 repeating, ... 8/9 = 0.8 repeating, and therefore 9/9 = 0.9 repeating which of course = 1 (9/9). They still refuse to believe me, and have asked that I consult a higher authority. Can you help? Thanks!
Date: 6/3/96 at 10:46:14 From: Doctor Tom Subject: Re: Repeating Decimals Okay, I'm a "higher authority," and I pronounce that you're right! Is that enough? If not, there are other ways to think of it: For example, if they believe that 1/3 = .3333..., then clearly 1/3 + 1/3 + 1/3 = 1 = .99999... Or, if they believe that .00000... is zero, then if .99999... is different from 1, it must clearly be less. So what is 1 - .99999...? If it's not zero, its decimal expansion must have some non-zero term. Wherever that term is, it is easy to show that the difference between 1 and .99999... is less than that. Or, if they believe that .99999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ..., then just sum up the geometric series, where the first term is 9/10, and the factor is 1/10. If S = a + ar + ar^2 + ar^3 + ..., then S = a/(1-r). a = 9/10, (1-r) = 9/10, so S = 1. I hope this helps. -Doctor Tom, The Math Forum
One more variation to consider that has been suggested before is the following proof sent in by email@example.com (Karsten Koops): Suppose X = 0,99. periodical Multiply by 10, so you get 10X=9,99. Subtract 1X (0,99.) 9X=9 Divide by 9 and you get X=1 -Doctor Steve The Math Forum Check out our Web site: http://mathforum.org/dr.math/
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