The Infamous .999... = 1Date: 01/12/2002 at 08:14:36 From: Dan T Subject: The Infamous .999... = 1 Dear Dr. Math, I have seen answers to the question that .9999... = 1, but some people in my 8th grade class STILL don't agree. My geometry teacher says that it is all about giving infinity a value. A student in the class says I am trying to prove that 1 = 1/infinity. Can you straighten this out? Thanks. Dan T. Date: 01/12/2002 at 09:27:23 From: Doctor Tom Subject: Re: The Infamous .999... = 1 Hi Dan, I assume you've seen the Dr. Math FAQ: 0.9999... = 1 http://mathforum.org/dr.math/faq/faq.0.9999.html If not, take a look. Rather than saying "giving infinity a value," it's perhaps a bit clearer to say, "giving the concept of a limit of an infinite sequence of numbers a value." .9 is not 1; neither is .999, nor .9999999999. In fact if you stop the expansion of 9s at any finite point, the fraction you have (like .9999 = 9999/10000) is never equal to 1. But each time you add a 9, the error is less. In fact, with each 9, the error is ten times smaller. You can show (using calculus or other methods) that with a large enough number of 9s in the expansion, you can get arbitrarily close to 1, and here's the key: THERE IS NO OTHER NUMBER THAT THE SEQUENCE GETS ARBITRARILY CLOSE TO. Thus, if you are going to assign a value to .9999... (going on forever), the only sensible value is 1. There is nothing special about .999... The idea that 1/3 = .3333... is the same. None of .3, .33, .333333, etc. is exactly equal to 1/3, but with each 3 added, the fraction is closer than the previous approximation. In addition, 1/3 is the ONLY number that the series gets arbitrarily close to. And it doesn't limit itself to single repeated decimals. When we say: 1/7 = .142857142857142857... none of the finite parts of the decimal is equal to 1/7; it's just that the more you add, the closer you get to 1/7, and in addition, 1/7 is the UNIQUE number that they all get closer to. Finally, you can show for all such examples that doing the arithmetic on the series produces "reasonable" results: Since: 1/3 = .333333... 2/3 = .666666... 1/3 + 2/3 = .999999... = 1. By the way, there is nothing special about 1 as being a non-unique decimal expansion. Here are a couple of others: 2 = 1.9999... 3.71 = 3.709999999... 2.778 = 2.77799999999999... ...and the student who says you're trying to show that 1 = 1/infinity is wrong. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 01/12/2002 at 16:40:11 From: Dan T Subject: The Infamous .999...=1 Sorry, the equation that the student meant was: 1 = 1 - 1/infinity Date: 01/12/2002 at 16:48:09 From: Doctor Tom Subject: Re: The Infamous .999...=1 Hi Dan, Then he's basically right. As you add each new 9 to the expansion, the errors look more and more like: 1/100, 1/1000, 1/10000, ... Thus, in a sense, the error begins to look like "1/infinity," which semms as if it should be zero. The 1/infinity is meaningless, but the concept of limit is not. We can say that the limit of the sequence above is zero, and can rigorously prove it. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/