Talk:Repeating decimal/Archive 1

Archive 1

Notation

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)

I was tought to pronounce repeating decimals as "0.3 repetend" as opposed to "0.3 repeating" and it appears this follows from the dictionary definition of "repetend" (not "repitend"). May want to mention this alternative vocabulary. --12.217.194.101 (talk) 05:38, 20 April 2008 (UTC)

Emphasis

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)

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 $10^{n}-1$ instead of just $10^{n}$, when $n\rightarrow \infty$, 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 ($10^{n}$, when $n\rightarrow \infty$) what "$10^{\infty }$" 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 $\forall \ n\in N,1.\,\overbrace {000\dots } \neq 0.\,\overbrace {999\dots }$

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

"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

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.

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

little bit of Geometry and divisibility (multiple)

http://wiki.riteme.site/wiki/Talk:Triangle

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)

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)

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)

The algorithm section is well and clearly written, but I think there may still a problem with the section on Why all rational numbers must have repeating or terminating decimal expansions. While it clearly demonstrates how, using long division, it's possible to express some rational numbers as repeating decimals, and why we can be certain that the block must repeat, it fails to address explicitly why this repeating series of remainders is an exclusive feature of rational numbers.

If I'm thinking correctly, this can be shown by extrapolating from the series expansion described in the previous section, but I don't think this is made explicitly clear. Rangergordon (talk) 07:56, 17 May 2008 (UTC)

If the decimal expansion repeats, the section Fraction from repeating decimal tells you how to calculate the fraction that produced it. A fraction is a rational number. Therefore a repeating decimal implies that the number is rational, and an irrational number cannot have a repeating decimal expansion.  --Lambiam 22:59, 20 May 2008 (UTC)

I don't dispute that rational numbers do have repeating or terminating decimal expansions. I have also used the algorithm you described to determine the rational form of a repeating decimal. I'm merely concerned that the section headed "Why all rational numbers must have repeating or terminating decimal expansions" does not demonstrate why all rational number must have repeating or terminating decimal expansions, but merely demonstrates how it is possible, using long division, to convert some particular rational number into its (possibly repeating) decimal expansion using long division.

At no point does the section attempt to show that, in general, a rational number must have a repeating or terminating decimal expansion. In order to do that, the section would need to show that the process of long division must always terminate, or repeat; as it is, one is left with the feeling that perhaps there may exist $p/q$ such that long division will result in a non-repeating decimal expansion. Perhaps the section should be re-titled "Using long division to convert a rational expression into its decimal expansion" or something like that. Rangergordon (talk) 04:03, 25 May 2008 (UTC)

Um, yes, that section does explain why the process of long division must terminate or repeat. The explanation is in the paragraph below the long division example. For a finite divisor there are a finite number of possible remainders. If we reach a remainder of 0 then the long division has terminated. If we never reach a remainder of 0 then we must eventually reach a remainder that has occurred before; from this point onwards the long division process will repeat itself, so the decimal expansion repeats. Gandalf61 (talk) 08:26, 25 May 2008 (UTC)

Um, nowhere does the section show that "the process of long division must terminate or repeat." It merely demonstrates that, in the case of dividing 5 by 74, we are left with a repeating decimal. Big whoop. If one were to rely on this section, one could incorrectly conclude that there may be some rational numbers which have nonrepeating or nonterminating decimal expansions. Therefore, the section is misleading. If the section wishes to demonstrate how a repeating decimal may result from long division, it succeeds. However, it fails even to demonstrate that long division may also result in a terminating decimal--and it is far from proving, as advertised, that the process of long division must always either terminate or repeat. If we were to use long division to prove irrationality, it involves some interesting number theory. Rangergordon (talk) 07:37, 4 June 2008 (UTC)

The explanation is very simple. Read the following statements carefully, and think about each one ...

• If the long division process reaches a remaider of 0 then it terminates.
• If the long division process never reaches a remaider of 0 then there are only a finite number of possible remainders (in the example in the article, the divisor is 74 and there are 73 possible non-zero remainders - the integers from 1 to 73 inclusive).
• Therefore if the long division process never reaches a remaider of 0 it must eventually reach a remainder that it has produced before.
• Once the long division process reaches a remainder that it has produced before then its outputs will start to repeat (because its outputs from some step onwards are entirely determined by the remainder at that step - it has no "memory" of previous remainders).
• Therefore the long division process must either terminate or start to repeat (and, further, we can say that for a divisor n, if it repeats, then the repeats will start within the first n steps).
• Therefore every rational numbers has a terminating or a repeating decimal expansion.

This is what the paragraph below the long division example means when it says "We eventually see a remainder that we have seen earlier because only finitely many different remainders ... can occur ... Therefore the whole sequence repeats itself, again and again". It does not spell it out as I have above, but this is a correct explanation. Gandalf61 (talk) 09:07, 4 June 2008 (UTC)

(Outdent)At this point, I would like to offer some wikilove to Gandalf61. But, just to reiterate, I never disputed the fact that the decimal expansion of a rational number always terminates or repeats, or that the decimal expansion of an irrational number never does. I'm fairly well acquainted with the concept. What I have been quibbling about all along is whether or not, from a textual reading of the article, the section claiming to prove this point actually achieved its stated goal. I wasn't even looking for a rigorous proof; I simply thought it glossed over its stated point.

Now, going by Gandalf61's thorough and meticulous outline, the step I was getting hung up on was Step 2 (there are a finite number n of possible non-zero remainders, so after at most n + 1 steps, the long-division process must repeat or terminate). This was referred to obliquely in the article, but not explicitly stated. (In fact, if the original author had been as thorough as Gandalf61 was in his talk-page explanation to me, the issue would never have come up!)

However, now that Michael Hardy (at 09:53) and Marc van Leeuwen (at 11:58) have made their edits, this problem seems to have been solved, and I think the article is much better for their efforts. Rangergordon (talk) 09:02, 6 June 2008 (UTC)

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)

The section on 0.999... in this article was POV. I have rephrased it, while retaining the mathematical content. I have also added a note referring readers to 0.999...#Skepticism in education, which has an excellent discussion on why some people make an invalid deduction in the second step of the proof. --Jtir 13:40, 26 May 2007 (UTC)

Requested move

I'd like to suggest renaming this article "repeating decimal". Here are some Google results:

• "recurring decimal": 27,700 hits
• "repeating decimal": 46,800 hits

Also, the corresponding MathWorld article uses "repeating decimal". My impression is that the "repeating" terminology is more common than "recurring" in high-school algebra classes. Jim 00:46, 24 September 2007 (UTC)

I call them repeating decimals, but I'm comfortable with either term. Michael Hardy 22:16, 2 October 2007 (UTC)

Agreed, move it. I've only ever heard 'repeating', and the other term seems odd. The way, the truth, and the light 01:26, 3 October 2007 (UTC)

I concur; seldom does "recur" occur. -GTBacchus^(talk) 05:40, 3 October 2007 (UTC)

Support. Ditto. --Eye of the Mind 23:17, 4 October 2007 (UTC)

I've moved the page to Repeating decimal, per the above discussion. Please let me know if there are any questions or concerns. -GTBacchus^(talk) 09:03, 7 October 2007 (UTC)

Clearing up some odd words

Could someone explain in reasonably simple English (i am 17) what a primitive root modulo is? I got everything else it is just that, which I dont understand. Who ever wrote the article is a a bit of a genius by the way. Thanks —Preceding unsigned comment added by 86.133.23.112 (talk) 20:01, 12 November 2007 (UTC)

Here is a reasonably simple explanation - take a look at the linked articles for more explanation:

1. The order of a number a modulo (or "with respect to") another number n is the smallest power k such that a^k is 1 more than a multiple of n. For example, the order of 10 modulo 7 is 6 because 10⁶=7x142857 + 1; the order of 10 modulo 13 is also 6 because 10⁶=13x76923 + 1.
2. A theorem called Fermat's little theorem says that if p is a prime number and a is not a multiple of p then a^p-1 is 1 more than a multiple of p. So if a is not a multiple of p then the order of a modulo p is at most p-1 - and might be smaller.
3. More than this, it can be shown that if p is a prime and a is not a multiple of p then the order of a modulo p is actually a factor of p-1. So the order of 10 modulo 7 is a factor of 7-1=6 - in fact, it is equal to 6. And the order of 10 modulo 13, which is also 6, is a factor of 13-1=12.
4. If the order of a modulo p is actually equal to p-1 then a is called a primitive root modulo p. So 10 is a primitive root modulo 7, because the order of 10 modulo 7 is 6. But 10 is not a primitve root modulo 13, because the order of 10 modulo 13 is 6, not 12.
5. If p is a prime not equal to 2 or 5, then the period of the repeating decimal representation of 1/p is equal to the order of 10 modulo p (try working out why this is so - it is not too hard). So the decimal representations of 1/7 and 1/13 both have repeating periods of 6. And in particular, if 10 is a primitive root modulo p then the order of the repeating decimal 1/p is p-1. Gandalf61 11:25, 13 November 2007 (UTC)

Periodic decimals

Some people call it periodic decimals so it should be added to the article. —Preceding unsigned comment added by 71.249.100.242 (talk) 19:06, 9 April 2008 (UTC)

This is not mathematical rigour

"This is because an extra 1 added to the last (i.e. infinitieth) digit would add a value of 10^−∞ = 0."

I am quite positive that what is said here is pure nonsense. The 'last' digit is undefined here. The limit of 'approaching' infinity is a particularly unhandy way of saying it. You are approaching nothing. You are simply becoming greater and greater. Some other ways which are mathematical which could in easy terms explain why this is so are: "This is because it can be proven that there is no real number between 0.999... and 1. Formally meaning that they are the same real number." or "This is because 0.999... can be seen as an infinite series which converges to 1." What is standing in those quotes above me is pure nonsense. Is exponentiation defined with the arguments 10, negative infinity even, I am not sure of that. Also, 'infinitieth' returns too few hits on Google for me to classify as an English word.Niarch (talk) 19:37, 21 April 2008 (UTC)

I agree - the "infinitieth digit" sentence is nonsense. I have removed it. Gandalf61 (talk) 19:48, 21 April 2008 (UTC)

Thank you, my dearest man, I dared not show such boldness myself. But do notice my fancy British spelling of 'rigour'. I have observed a correlation between being a female mathematician and a liking for a man who talks in British English. A male one even attested: 'I am of course far from a lady with class, but I must acknowledge to be far more enchanted by British English over American.' (Bob Planqué) http://www.few.vu.nl/~rplanque/ He also likes birds and Euler's Identity. I could recommend you to have a lecture from him where he states: 'And this is attested by many to be the most mathematically beautiful and elegant equation of the whole analysis.' while he becomes particularly sentimental. But relating the five constants with the three binary operators and the relation. I can understand. Niarch (talk) 13:00, 23 April 2008 (UTC)

Other bases approach?

"The period of the repeating decimal of 1⁄p is equal to the order of 10 modulo p. The period is equal to p-1 if 10 is a primitive root modulo p."

I know this is an article on 'decimals' (base 10) but what about other bases? The 'primitive roots' article makes plain that there are some numbers 'p' which don't possess any primitive roots - whichever 'coprime' base apart from '10' is selected - so could not reach length 'p-1' as a recurring 'digital' representation of '1/p'. I'd have thought that by exploring other bases that any number would 'succumb' but it doesn't appear to be so.

More generally, how do you calculate rational numbers as non base-10 'decimals'? —Preceding unsigned comment added by 172.143.1.114 (talk) 18:49, 15 June 2008 (UTC)

Basically, for another base b, replace each occurrence of 10^k by b^k (which, expressed in base b, is written as a digit 1 followed by k digits 0). For example, the period of the repeating hexadecimal of 1/11, 0.1745D1745D..., is equal to the order of 12 modulo 7, which equals 5. To compute the expansion of p/q in base b by long division, where 0 ≤ p < q, repeatedly execute:

next digit := floor(bp/q); p := (bp) mod q;

For example, for 1/11 in base 12, starting with p = 1, while q = 11 and b = 16:

next digit := floor(16×1/11) = floor(16/11) = 1; p := 16 mod 11 = 5;

next digit := floor(16×5/11) = floor(80/11) = 7; p := 80 mod 11 = 3;

next digit := floor(16×3/11) = floor(48/11) = 4; p := 48 mod 11 = 4;

next digit := floor(16×4/11) = floor(64/11) = 5; p := 64 mod 11 = 9;

next digit := floor(16×9/11) = floor(144/11) = 13 = D; p := 144 mod 11 = 1;

and since we had p = 1 before, the hexadecimal repeats from here on.  --Lambiam 07:24, 17 June 2008 (UTC)