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 Subject: ".999...=1?" Previous Topic | Next Topic
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David (Guest) guest
Apr-22-01, 08:45 PM (EST)

".999...=1?"

 please provide me with some evidence backing up the statement.999...=1 or the statement that it doesnt equal 1.thanx

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Subject     Author     Message Date     ID .999...=1? David (Guest) Apr-22-01 TOP RE: .999...=1? alexb Apr-22-01 1 RE: .999...=1? Have_Blue Apr-10-02 2 RE: .999...=1? alexb Jun-28-02 5 RE: .999...=1? Michael Klipper Jul-27-02 8 RE: .999...=1? stapel Apr-12-02 3 RE: .999...=1? math_spect Jun-28-02 4 RE: .999...=1? pradeep Jul-24-02 6 RE: .999...=1? dc Jul-26-02 7

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alexb
Charter Member
780 posts
Apr-22-01, 08:59 PM (EST)    1. "RE: .999...=1?"
In response to message #0

 Please have a look at/arithmetic/999999.html

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Have_Blue
Member since Apr-10-02
Apr-10-02, 06:59 AM (EST)    2. "RE: .999...=1?"
In response to message #0

 Here's a proof that I like and is quite simple. Note 0.9_ means 0.999999999999...Let N = 0.9_10N = 9.9_10N-N = 99N = 9N = 1Therefore, N = 0.999999... = 1 Q.E.D.

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Charter Member
780 posts
Jun-28-02, 01:05 PM (EST)    5. "RE: .999...=1?"
In response to message #2

 That's the proof of the following statement:Assume 0.9... is a number N such that 10N - N = 9, then N = 1.Which is obvious. What is not obvious isthat N is a number in the first place.that N satisfies 10N - N = 9.A number is a member of a group of other numbers so that some operations (like addition, subtraction, multiplication, etc.) are defined between them. It's not at all obvious that an infinite decimal is a number in any sense and it's not at all obvious that it may be subject to some arithmetic operations.

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Jul-27-02, 02:45 PM (EST)

8. "RE: .999...=1?"
In response to message #5

 Sorry if this topic has already been brought up, but I might be able to provide a good answer to the question of "Is 0.999... = 1?". This will be a long message, since I don't have the patience to read through all the cut-the-knot stuff already published.-----------------First of all, the definition of a decimal expansion is important. For finite decimals, there is no problem. For infinite decimals,we have this rule (which I write only for nonnegative decimals <= 1): 0.a1a2a3... is an infinite decimal expansion of x if and only if for any positive integer n, we have 0.a1a2...an <= x <= 0.a1a2...an + 10^(-n). This definition has two <= signs, and the second one is in fact crucial. Note that by this definition, 0.999... is a perfectly valid decimal expansion for 1, since adding 10^(-n) produces 1 exactly, and the <= holds. If it were strictly <, then this would not work.However, could it be a decimal expansion for something else? Let's say there is some z < 1 which is equal to 0.999...Then after n decimal places, where n is any integer, we have .99 (n times) <= z <= .99 + 10^(-n).Let d = 1 - z > 0. Since .99... is a convergent series converging to 1, we know that there exists an n after which all decimal expansions with n places is less than d away from 1. Let m represent this difference. Then our above inequalities become 1 - m <= 1 - d <= 1 - m + 10^(-n). However, this implies d >= m, which is false. Thus, the decimal expansion for 0.999... must be 1.From this, a corollary can be established: every infinite decimal expansion converges to exactly one number. However, the converse is not true; a number may have two decimal expansions. For example, 1/4 = 0.250000... = 0.249999....-----------------The above stuff is a perfectly valid proof. However, I want to answer the previous question about: Is it fair to say that 0.999... = N such that 10N - N = 9? Well, look at it in terms of the expansions.First of all, this is a rational number (3 / 3), so it has a repeating decimal with some period. In this case, the period is 1, because ai = a(i+1) for all positive i. So,if N = 0.a1a2a3..., then 10N = a1.a2a3a4...By the periodicity relation above, a2 = a1, a3 = a2, etc., and thus we obtain 10N = a1.a1a2a3...It's obvious that 10N - a1 = N in this case (since the repeating decimals of 10N - a1 and N are exactly equal), thus 10N - N = a1 = 9, and that proves the validity of the reasoning that you questioned.-----------------I hope that helped. I'm a computer science student at Carnegie Mellon University who just took a couple courses on this type of material.

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stapel
Member since Mar-5-02
Apr-12-02, 12:11 PM (EST)    3. "RE: .999...=1?"
In response to message #0

 Another discussion of this topic is archived at The Math Forum:

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math_spect
Member since Jun-28-02
Jun-28-02, 12:51 PM (EST)    4. "RE: .999...=1?"
In response to message #0

 Now remember, this is only true if 0.9... is a repeating decimal. If 0.9_ went out to 1000 places and stopped, it would not be equal to 1.

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pradeep guest
Jul-24-02, 08:14 PM (EST)

6. "RE: .999...=1?"
In response to message #0

 1/9 = .1111111111...2/9 = .2222222222... (2 * 1/9)7/9 = .7777777777... (7 * 1/9)8/9 = .8888888888... so 9/9 = .999999999... = 1

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dc guest
Jul-26-02, 02:45 PM (EST)

7. "RE: .999...=1?"
In response to message #0

 First we must be recognize that a real can be written in any number of ways. The notations are obviously not equal, but it may be that the represented numbers are the same.The next step is to get clear about the meaning of 0.999..., and that will ultimately require some understanding of the construction of the reals. Put a(i)= 9/10^i, then 0.999... = lim n->inf (a(1)+ a(2)+ ... + a(n)) = lim n ->inf (1-1/10^n)by definition. This limit can be shown to exist. Moreover the limit is clearly greater than any real less than 1 and clearly less than or equal to 1. The conclusion? The limit must be equal to 1.

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