Talk:Recurring decimal

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I was taught to represent a recurring decimal not using elipses but by placing a dot over the repeating digit or, where the repeating sequence is more that digit long, over the first and last digit. I cannot work out how to do that here! Paul Beardsell 14:00, 24 Feb 2004 (UTC)

I like to do it like this:

0.5\,\overbrace{0317186}\dots\,

The overbrace indicates the repitend. Michael Hardy 18:03, 13 Jan 2005 (UTC)

A recent edit changed the text

The method of calculating fractions from repeated decimals, especially the case of 1 = .99999..., is often contested by amateur mathematicians.

to

The method of calculating fractions from repeated decimals, especially the case of 1 = .99999..., is sometimes contested.

Although I understand the motivation was to avoid slighting amateur mathematicians, I think something important has been lost here. The new text suggests that there is serious disagreement in the mathematical community about the truth of 1 = .99999...., which, of course, there is not.

I think some qualification is necessary here. Perhaps something like "The result of this method is sometimes surprising to students."

-- Dominus 14:52, 24 Feb 2004 (UTC)

You're right: I'll fix it. Paul Beardsell 02:16, 25 Feb 2004 (UTC)

It's not contested by amateur mathematicians; it's contested only by the mathematically naive. Michael Hardy 17:58, 13 Jan 2005 (UTC)

Contents

[edit] Infinity divided by infinity

Your "proof" lies on two notions. First there is that \infty + 1 = \infty. You can say XXX... =\infty But Your second notion \frac{\infty}{\infty}=1 is totally nonsense. One can divide something by very small number i.e. differential from zero but by not exact zero.

The above is nonsense, since ∞ + 1 is not mentioned in this article. Michael Hardy 21:32, 8 May 2005 (UTC)

Adapted from the article: "It follows that

\lim_{n \to \infty}{10^n-1 \over 10^n} = 0.9999..."

Previous equation has severe error. One can subtract 2 (instead of 1) from infinity and yet result infinity. So the expression of (-1) in the numerator 10n − 1 instead of just 10n, when n\rarr\infin, is simply cosmetic. I refer this as a trap of infinity. I take "the proof" as suggestion that someones (especially computerized mathematicians) tend to "think" like computers (i.e. digitally) rather than analogously. -Santa Claus 22:39 At Northpole Time.

The above is utter nonsense. No one was subtracting anything from infinity. Here's the relevant identity:
{10^n-1 \over 10^n}=0.999\dots 9\,
(only finitely many "9"s appear above).
And who are these "computerized" mathematicians? The fact that 0.9999... = 1 was appreciated LONG before there were computers. Michael Hardy 21:18, 8 May 2005 (UTC)

By stating that you truly are a dishonest one in it's very nature. In the article is written exactly as follows

  • "It follows that
\lim_{n \to \infty}{10^n-1 \over 10^n} = 0.9999...

On the other hand we can evaluate this limit easily as 1, also, by dividing top and bottom by 10n. "

You are confused. First one states that
{10^n-1 \over 10^n}=0.999\dots 9\,
there being only finitely many (i.e., n is a finite number) "9"s. Then one takes the limit as n approaches infinity on BOTH sides. This was abbreviated in the article because it was expected that the reader would understand it (or, perhaps more precisely, it never occurred to the person who wrote it that that particular misunderstanding might occur). Michael Hardy 00:12, 9 May 2005 (UTC)
Oh ... now I see that it was stated explicitly. You must not have read it. Michael Hardy 00:15, 9 May 2005 (UTC)


“there being only finitely many (i.e., n is a finite number) "9"s. "Then one takes the limit as n approaches infinity on BOTH sides"
  • First you wrote that "n” is a finite number" What is this finite? When this "n" stops approaching infinity i.e. "becomes" finite? What is the domain (of definition) of n?
  • "Then one takes the limit as n approaches infinity on BOTH sides" First you wrote that "n is a finite number", then you wrote that (it continues its journey from above-mentioned finite to infinite [true?]) "approaches infinity". Could you define your concept (definition) of "finitely many" as in an unambiguous manner, in a distinction of infinite?

Could you be more precise than earlier? "0,999...." implies to me this "0.999..." is approaching one from "lower" to "upper" (i.e. 0.999....”rounds” "up" to 1?), but what is the other direction of the "0,999...."'s approach to one? Could you write to me when this "0.999..."’s arrival at one occur? And how do you define these “BOTH sides"? What is this “side”? - Santa Claus 16:17 At Northpole Time


\lim_{n \to \infty}{2^n-133 \over 2^n} = 0.9999...? - Santa Claus
Yes, this is right! Any problem with this being true? Oleg Alexandrov 00:55, 9 May 2005 (UTC)
Agreed! --BradBeattie 14:06, 9 May 2005 (UTC)

If n is finite number then {10^n-1 \over 10^n}={2^n-133 \over 2^n}. Perhaps you determine (define, indentify etc.) such a finite number n? - Santa Claus At 30/3/1426 Islamic Date


Perhaps you can write (10n, when n\rarr\infin) what "10^\infin" stands for? -Santa Claus 0:35 At Northpole Time

"Santa Claus", you are consistently showing that you're confused about basic points in calculus in just the way that I see among the duller students in courses I teach. If that's a harsh way of putting it, look in the mirror. There are only finitely many "9"s, just n of them, in the expression
0.999...9={10^n-1 \over 10^n}.\,
I.e., we have
0.9={10-1 \over 10}\,
and
0.99={10^2 -1 \over 10^2}\,
and
0.999={10^3 -1 \over 10^3}\,
and so on, each identity having just finitely many "9"s, not infinitely many.
The infinite sequence 0.9, 0.99, 0.999, has a limit. Also, the infinite sequence
{10-1 \over 10},\ {10^2-1 \over 10^2},\ {10^3-1 \over 10^3},\ \dots \,
has a limit. Since they are both the same sequence, they both have the same limit. Since the limit of the latter sequence is 1, so is that of the former.
It is certainly correct that the limit of
{10^n-6 \over 10^n}\,
as n grows, is 1. But it is NOT correct that
{10^n-6 \over 10^n}=0.99\dots 9\,
there being just n "9"s.
Before you rush in an ask to be recognized as an infallible expert, you should get beyond being so confused in really rudimentary first-year undergraduate material. Michael Hardy 20:44, 9 May 2005 (UTC)


I asked infallible professor (not Vatican's pope even he has been declared infallible one) and He replied following: There are only finitely many "9"s and "0"s, just n of them, in the expression

1.\,\overbrace{000\dots}\ - 0.\,\overbrace{999\dots}\,= \overbrace{0.00\dots}1

n, k \in N= { 1, 2, ... }


n=1, 1.\,\overbrace{0}\ - 0.\,\overbrace{9}\,= \overbrace{0.}1
n=1+1,1.\,\overbrace{00}\ - 0.\,\overbrace{99}\,= \overbrace{0.0}1
n=2+1, 1.\,\overbrace{000}\ - 0.\,\overbrace{999}\,= \overbrace{0.00}1
...n=k+1, 1.\,\overbrace{000\dots}\ - 0.\,\overbrace{999\dots}\,= \overbrace{0.00\dots}1


Thus \forall \ n \in N, 1.\,\overbrace{000\dots}\ne 0.\,\overbrace{999\dots}


Your "mathematical induction" (which it clearly isn't, as it lacks a proof for the k+1 case based on k) is flawed. The most salient reason is the fact that infinity is not a member of N. As n goes to infinity, which is a concept, a tool, not a concrete number, .0000...1 goes to .0000...0 and the two are equivalent. See, I can be smart and a wiseass without using obtrusive mathematical notation. --3rd Party

[edit] "Santa Claus"'s further confusion

If n is finite number then {10^n-1 \over 10^n}={2^n-133 \over 2^n}. Perhaps you determine (define, indentify etc.) such a finite number n? - Santa Claus At 30/3/1426 Islamic Date

It is not correct that
{10^n-1 \over 10^n}={2^n-133 \over 2^n}\,
where n is, of course finite. But it is correct that
\lim_{n\to\infty}{10^n-1 \over 10^n} =1=\lim_{n\to\infty}{2^n-133 \over 2^n}\,
the n, of course, still being finite. We don't consider any infinite n here. What this means is that the sequence
{10^1-1 \over 10},\ {10^2-1 \over 10^2},\ {10^3-1\over 10^3},\dots\,
and the sequence
{2^1-133 \over 2^1},\ {2^2-133 \over 2^2},\ {2^3-133 \over 2^3},\dots\,
both have the same limit, which is 1. But in each term of the sequence, the value of n is finite; in the first term n is 1, in the second term n is 2, and so on.
This is all very rudimentary 1st-year undergraduate-level material. You should not be claiming any kind of expertise if you don't know this stuff. Michael Hardy 21:03, 9 May 2005 (UTC)


{10^n-6 \over 10^n}=0.99\dots 9\,

I think that you think 0.99"..."9 there is forever between two nines. Apparently there isn't infinity because You ?? have symbolized "forever" (infinity) between two symbols that you called nines. 0...0, is there forever (infinity) between these two 0-symbols?

{10^n+1 \over 10^n}=0.99\dots 9\,? Santa Claus At North Pole Time

http://onlinedictionary.datasegment.com/word/symbolized

No, the dots do not mean "forever". There are finitely many "9"s; there are n of them. If, for example, n is 5, then we have 0.99999 = (105 − 1)/105. Michael Hardy 16:40, 10 May 2005 (UTC)
I'd think that in 0.999 \ldots 9 the \ldots don't mean forever as indicated by the final 9. This implies a finite number of nines between the first and the last. However, in 0.999 \ldots the \ldots does imply an infinite number of nines. --BradBeattie 13:55, 12 May 2005 (UTC)
No, the dots do not mean "forever". There are finitely many "9"s; there are n of them. If, for example, n is 5, then we have

"No, the dots do not mean "forever"". AND YET You have used dots to mean forever i.e infinite "amount" of TIME.

"Then one takes the limit as n approaches infinity on BOTH sides" First you wrote that "n is a finite number", then you wrote that (it continues its journey from above-mentioned finite to infinite [true?]) "approaches infinity". Could you define your concept (definition) of "finitely many" as in an unambiguous manner, in a distinction of infinite?"

Someone who Couldn't even answer to that question and Yet naming someones dull! "BOTH sides" -Santa Claus Perhaps I should answer to this. You have only calculated limit value when n approaches infinity from "left to right" but not opposite direction (to negative infinity). You have fallen trap of scale, Natural numbers do not have negative values thus you can calculate only ONE-SIDE LIMIT VALUE

Okay, perhaps I can explain some of the confusion you seem to be having. First, with the dots, when you write 0.9999... there are meant to be infintely many 9s, but when you write 0.9999...9 there are meant to be finitely many 9s. Second, you seem to be interpreting "take the limit of both sides" to mean a two-sided limit rather than a one-sided limit. But what is actually meant is taking the limit of both sides of an equation, just as you can, for instance, take the square root of both sides of an equation.
About the new material above that you attribute to an "infallible professor", that doesn't address the issue, because all it addresses the cases when you have finitely many 0s and 9s. But we are concerned about the case where there are infinitely many. This is formalized by limits. Eric119 23:12, 12 Jun 2005 (UTC)


"but when you write 0.9999...9 there are meant to be finitely many 9s" HAH, your "mathematical" tricks won't do the trick on me. First you mean that "..." means infinity, but paradoxusly you write later that It doesn't mean it :DDDDDDDDDDDDDD
"But what is actually meant is taking the limit of both sides of an equation, just as you can, for instance, take the square root of both sides of an equation." How low can you go? the limit value was taken of function{10^n-1 \over 10^n} NOT equation
"But we are concerned about the case where there are infinitely many. This is formalized by limits."
Can you really put the limits to unlimited (i.e. infinite)? ;) and still say it (the limited one) is unlimited.
You lack of common sense. -Santa Claus

[edit] Santa's issues

Looks to me that Santa still thinks of math as of some kind of metaphysics. Santa, you need to take a rigurous course in calculus, that will solve your problems. The issues you are worried about were solved in 19th century (they were indeed big issues, and many people voiced concerns similar to yours, but math grew out of it).

As a side remark, I would encourange any people having this article on the watchlist put the article proof that 0.999... equals 1 on the watchlist too. Santa seems to be pushing his views on that article, making math look like some kind of pseudoscience. Oleg Alexandrov 14:34, 13 Jun 2005 (UTC)

Above is mentioned "discussion". Word should be free. I haven't changed the article. -Santa Claus.

Oh, you are more than welcome to discuss things. So it was another anon who changed proof that 0.999... equals 1. By the way, it looks that you like it here. Would you make yourself an account, so that it is easier for you to track what you write and for us too? I think that would be helpful. Oleg Alexandrov 18:43, 13 Jun 2005 (UTC)

We can talk about Quantum Physics, Special Relativity by EinStein (One Stone in English), Thermophysics and so on IF YOU keep saying that I talk about metaphysics (meaning intangible issues). We should not forget Geometry to illustrate visually mathematical functions, equations or what so ever. -Santa Claus

"there being only finitely many (i.e., n is a finite number) "9"s. Then one takes the limit as n approaches infinity on BOTH sides." Approaching is motion and is researched in Physics - Santa Claus.

[edit] What is point of this? "Fractions with prime denominators"

999,999,0 is divisible by 7 too
9,999,999,999,999,999,0 is divisible by 17 too
999,999,999,999,999,999,0 is divisible by 19 too -Santa Claus

[edit] little bit of Geometry and divisibility (multiple)

http://en.wikipedia.org/wiki/Talk:Triangle

[edit] Infinitely many times

The article included the phrase "infinitely many times" in the first paragraph. I removed "many times" because it contradicted "infinitely", and in any case "infinitely" on its own is an adequate adverb to describe what's going on. User: Eric119 reverted this explaining that I had got the word binding wrong. I would argue that if it's possible to misunderstand the meaning then the wording needs changing. Although "many times" is redundant, I am happy to leave the wording as long as "infinitely" and "many" are closely bound by a hyphen to make the meaning clear (that being one of the purposes of that particular punctuation). This solution is a result of a discussion between me and User:Michael_Hardy which we really ought not to have kept so secret :-). Bazza 15:17, 15 September 2006 (UTC)

[edit] Recurring or repeating?

Maybe it's just me, but I've always learned this concept as a "repeating" decimal. As a matter of fact, until I saw this article, I had no idea that "recurring" was even a valid name for the concept. Aside from the semantic and linguistic problems of labeling a decimal as recurring, the Google test results in ~28000 for "recurring decimal" and ~51000 for "repeating decimal". Just a little discussion about the article's title. Axem Titanium 23:51, 28 September 2006 (UTC)

[edit] Does a Recurring Decimal Always Indicate a Rational Number?

(Disclaimer: not a mathematician.) Sections of this article seem to indicate that every recurring decimal can be represented as a fraction, and therefore represents a rational number. If this is so, I think it would be good to state it explicitly in the first section, maybe by changing "real number" to "rational number." Note that the section on "Why rational numbers must have repeating..." doesn't mean that all numbers with repeating decimals are rational. Peter Delmonte 23:25, 25 October 2006 (UTC)

The section on converting a repeating decimal to a fraction does more than just "seem to indicate" that fact; it actually gives an algorithm for it. Michael Hardy 02:25, 26 October 2006 (UTC)
...OK, now I've rearranged the section headings to be more explicit about this point. Michael Hardy 02:29, 26 October 2006 (UTC)

[edit] There should be a POV tag

This article is blatantly POV, and it's of little importance that the POV may actually be correct. Stating that opponents are naive just to make a point, any point, is not an encyclopedic tool. Worse, it's not even efficient. Obviously, the trap of this situation (as of POV in general) is that it makes it harder for readers to trust what you are saying. Why not just try to simply state the facts? Luciand 13:52, 27 December 2006 (UTC)

I'm inclined to think they are little children who ought to learn to read and write before editing Wikipedia. This topic is not controversial among mathematicians. Michael Hardy 00:14, 28 December 2006 (UTC)
Sorry, this comment was intended to go in the 0.999... talk page. Moving Luciand 12:25, 28 December 2006 (UTC)
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