# Math: .99repeating = 1? **URL:** \ **Category:** Factual Questions\ **Created:** [April 12, 2000, 1:59pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021 "2000-04-12T13:59:00Z")\ **Posts on this page:** 20\ **Page:** 3
**Author:** ![Monty](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/monty/32/47_2.png) [@Monty](https://boards.straightdope.com/u/Monty)\ **Post date:** [April 16, 2000, 6:06pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/41 "2000-04-16T18:06:00Z")
The Ryan posted 04-14-2000 08:04PM the stuff between the quotes; my responses are in between the quotations. > [@](#): > > ## Monty posted 04-13-2000 07:35 AM quote: > > ## The very definition of “repeating” in this sense is that, in fact, an infinite number of 9s follows the decimal point. > > Have you ever seen an infinite number of nines following a decimal point? Does such a thing exist? Does it have any meaning? No,no, and no. Nobody can actually see an infinite number of things, nor can anyone see the square root of a negative number, let alone see a negative number; however, in mathematics there are the concepts of infinites (transfinites), imaginary numbers (square roots of negative numbers), and negative numbers (numbers less than zero). Since we’re discussing a purely mathematical issue and not a physical issue, your snide comment is irrelevant, not to mention that it shows you’ve no idea of what you speak. But, to answer your questions correctly: No (but I and others have postulated them, that’s part of mathematics), Yes (they exist in mathematics, the topic under consideration in this thread), and Yes (they have meaning in the realm of mathematics, the topic under consideration in this thread). > [@](#): > > ## quote: > > ## As mathematics is rife with infinites > > Actually, mathematicians are very careful about infinities; they don’t just go throwing them around. Well, apparently according to you, there’s no such thing with wich to be careful about throwing around. BTW, do you always contradict yourself in the very same posting? Anyway, discuss why there’s a whole notational system for infinities when “they don’t just go throwing them around,” please. I missed reading the funnies this morning and need a dose of humour which I’m sure your description of Aleph, Aleph-null, etc., will bring to the discussion. [My snide comment here, feel free to point it out.] > [@](#): > > ## quote: > > ## explain exactly what the little bar over the 9 means in mathematical notation? > > I thought I already did. > .(N barred)= > lim > n -\>infinity of > N_10^(-m)+N_10^(-2m)+…N\*10^(-nm) > where m=number of digits in N Nope, see \*\* below for an authorative answer on this. > [@](#): > > ## quote: > > ## Last I checked, it meant the number under the bar repeated ad infitum as in “an infinite number of 9s” in this case. > > Yeah, that sounds like a rigorous mathematical definition. \*\*From Merriam-Webster’s online dictionary: > [@](#): > > repeating decimal (noun) > > First appeared 1773 > > : a decimal in which after a certain point a particular digit or sequence of digits repeats itself indefinitely – compare TERMINATING DECIMAL & [quote] terminating decimal (noun) First appeared circa 1909 : a decimal which can be expressed in a finite number of figures or for which all figures to the right of some place are zero – compare REPEATING DECIMAL From “Number,” Microsoft ® Encarta. {bolding and description of pictures in the article brought to you by Monty} > [@](#): > > It can be shown that every rational number can be represented as a repeating or periodic decimal; that is, as a number in the decimal notation, **which after a certain point consists of the infinite repetition of a finite block of digits**. Conversely, every repeating decimal represents a rational number. Thus, 617/50 = 12.34000…, and 2317/990 = 2.34040… . The first expression is usually written as 12.34, **omitting the infinite repetition** of the block consisting of the single digit 0. The second expression is frequently written as {picture of 2.340 with a bar over the 40, picture of 2.340 with a bar under the 40, picture of 2.340 with a short dash over the 4 and a short dash over the 0} **to indicate that the block of two digits, 4 and 0, is repeated infinitely**. That rigorous enough for you? Here’s a hint or two for you: 1. The professional mathmetician (CKDext) has already shown that your take on this issue is incorrect 2. If you really don’t know what you’re talking about, keep quiet and try to learn. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [April 17, 2000, 9:48am UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/42 "2000-04-17T09:48:00Z")
Actually, I don’t see much wrong with anything that The Ryan, Monty, or Dex said. Except when you say that the other is wrong. I thought you guys were on the same side. * * * rocks ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 17, 2000, 11:36am UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/43 "2000-04-17T11:36:00Z")
> [@](#): > > Originally posted by CKDextHavn: > \*\*Hey, Jinx! Think!! > > 1 = 1. > 4 = 4. > .25 = .25 > > Therefore, 1/4 can’t equal .25, can it? > > Sheeeeeeesh. [/sarcasm] > > .333 is not equal to 1/3 > .333333333333 is not equal to 1/3 > .33333333[1 billion 3’s]33 is not equal to 1/3 > > but .3333… where the decimal 3 repeats infinitely many times IS in fact, equal to 1/3.\*\* What kind of reply is this? You’re just reiterating my point! Do you understand the concept of a fraction? Of course 1/4 = 0.25 why should it not? And yes, 1/3 = 0.3333 repeating - get it? As for 1 = 0.999 repeating… Please go back to basic math class if: a) You cannot comprehend the number 1 b) You fail to understand the meaning of the fraction 1/1 and how to set that up as a long division. I’d really like to see a long division where 1 divided into itself can yield 0.9999… * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 17, 2000, 11:38am UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/44 "2000-04-17T11:38:00Z")
I can see why Cecil feels like there is no hope for the teeming millions! * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 17, 2000, 12:32pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/45 "2000-04-17T12:32:00Z")
> [@](#): > > I’d really like to see a long division where 1 divided into itself can yield 0.9999… Oh, this is relevant. :rolleyes: * * * _It is too clear, and so it is hard to see._ ---
**Author:** ![C\_K\_Dexter\_Haven](https://avatars.discourse-cdn.com/v4/letter/c/b2d939/32.png) [@C\_K\_Dexter\_Haven](https://boards.straightdope.com/u/C_K_Dexter_Haven)\ **Post date:** [April 17, 2000, 1:06pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/46 "2000-04-17T13:06:00Z")
Proper treatment for trolls, IMHO: answer ONCE, to be sure that you are not dealing with honest ignorance, then drop it. One more time, and then I’m done with this. Several people have offered proof that .999…(repeating infinitely) = 1. The simple multiplication proof is usually sufficient: (1) x = .99999… (2) 10x = 9.999999… Subtracting (2)-(1): (3) 9x = 9, hence x = 1. OK, that’s a proof. The testimony of every professor of mathematics in the country would be another sort of proof, I suspect. You don’t buy those? OK, then, if you thinkf .999… is different from 1, then you should be able to pick a number BETWEEN them. Go ahead, tell us one. For any two distinct real numbers, there are infinitely many real numbers between them. For instance, the midpoint. Go ahead, name one. We’re not talking about theoretical numbers, now, like “infinitesimals” or whatever, we’re talking real, honest-to-god, write-downable numbers. Name one, between .999… and 1. If you can’t, then you have to concede that .99999… is just another way of writing 1, in the same way that .333333… is just another way of writing 1/3 or that .11111… is another way of writing 1/9. Oh, and BTW, Monty, I’m no longer a professional mathematician. I did that back in the 70s, but then I became an actuary (pay was better) and now I’m a more generalized consultant. Contrary to the comment made in another thread, I did not predate Riemann, although I did have a class under Zygmund. ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 17, 2000, 1:39pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/47 "2000-04-17T13:39:00Z")
> [@](#): > > Originally posted by ZenBeam: > \*\* Oh, this is relevant. Referring to: I’d really like to see a long division where 1 divided into itself can yield 0.9999… > \*\* Your sarcasm shows you’ve missed the point. Excuse me, but it is relevant! I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)? Technically, these are two distinct points on the numberline. If you receive a paycheck, would you accept anything less than 1.0 times the amount you are owed? * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 17, 2000, 1:48pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/48 "2000-04-17T13:48:00Z")
> [@](#): > > Originally posted by CKDextHavn: > One more time, and then I’m done with this. > > (1) x = .99999… > (2) 10x = 9.999999… > Subtracting (2)-(1): > (3) 9x = 9, hence x = 1. > > OK, that’s a proof…\*\* Faulty logic, prof! Excuse me, but if you’re so sharp on mathematics, you should recognize the error in your logic! Here, you are attemtping to employ methods of solving simultaneous equations! The ground rule for aplying this method is that the two equations MUST BE independent! Since formula (b) is 10 times forumla (a), these are **NOT** independent equations! Thus, you have proved nothing! Thus, it is improper to subtract (a) from (b)! Algebra cannot be applied at whim! 😉 * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![C\_K\_Dexter\_Haven](https://avatars.discourse-cdn.com/v4/letter/c/b2d939/32.png) [@C\_K\_Dexter\_Haven](https://boards.straightdope.com/u/C_K_Dexter_Haven)\ **Post date:** [April 17, 2000, 2:27pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/49 "2000-04-17T14:27:00Z")
\<\< The ground rule for aplying this method is that the two equations MUST BE independent! \>\> Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other. A long division? Sure. (.99999…)/1 = .99999… Happy? The way that you toss out incorrect statements and ignore the comments of any others, leads me to suggest that this conversation is finished. Fighting ignorance is one thing; fighting stubborness is something else. ---
**Author:** ![kbutcher](https://avatars.discourse-cdn.com/v4/letter/k/db5fbb/32.png) [@kbutcher](https://boards.straightdope.com/u/kbutcher)\ **Post date:** [April 17, 2000, 2:38pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/50 "2000-04-17T14:38:00Z")
What’s interesting about this thread to me, is the logic. Many have issued mathmatical proofs that .999… = 1. The detractors, on the other hand, have offered no mathmatical proof whatsoever. In fact, it would seem that the mathmaticians have won the point. The only avenue left to those who do not beleive, is using “logic” to state why the proofs are wrong, not showing any mathmatical flaws in the proofs themselves. Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999…? ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 17, 2000, 2:54pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/51 "2000-04-17T14:54:00Z")
> [@](#): > > I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)? CKDextHavn demonstrated 0.999 repeating = 1, so I won’t bother repeating him. I’d like to see you demonstrate that they are unequal. Kbutcher writes > [@](#): > > Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999…? Nope. So far, all we’ve seen is handwaving. * * * _It is too clear, and so it is hard to see._ ---
**Author:** ![Boris\_B](https://avatars.discourse-cdn.com/v4/letter/b/fbc32d/32.png) [@Boris\_B](https://boards.straightdope.com/u/Boris_B)\ **Post date:** [April 17, 2000, 3:48pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/52 "2000-04-17T15:48:00Z")
Well, I tried this prove on the other thread on this subject, but I’ll try it again. I don’t know if it’s mathematically valid or not. 1 - 0.999… = D D x 10 trillion = D Therefore, D = 0 The idea is, multiply the difference between one and point nine repeating by the highest finite number you can think of, and the product will be no different. When multiplying yields no change, one of your factors is zero. QED ---
**Author:** ![sloth](https://avatars.discourse-cdn.com/v4/letter/s/d26b3c/32.png) [@sloth](https://boards.straightdope.com/u/sloth)\ **Post date:** [April 17, 2000, 7:14pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/53 "2000-04-17T19:14:00Z")
Jinx - If 1/3= .3repeating (as you agree) Is not .3repeating + .3 repeating + .3 repeating = .9 repeating true? If so by substitution aren’t 1/3+1/3+1/3 = .9 repeating and since 1/3 \* 3 = 1, 1=.9 repeating? An ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 18, 2000, 4:01pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/54 "2000-04-18T16:01:00Z")
I know this is beating a dead horse, but like cold fusion fiasco, this is math chicanery! > [@](#): > > Originally posted by CKDextHavn: > \*\*\<\< The ground rule for aplying this method is that the two equations MUST BE independent! \>\> > > Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other. > > A long division? Sure. (.99999…)/1 = .99999… > > Happy?\*\* a1) So, like what if I used 8x instead of 10x? I could fudge “x” to equal whatever I wanted! Ever studied matrices and determinants? What’s the value of a determinant when the equations are not unique? a2) If your initial premise is let x=0.999… then how can x \<\> x? b) Your “long division” fails to prove that 1/1 = 0.999… c) Equations must be consistent? No, the term is independent…otherwise, “x” can be shown to equal whatever value you wish by yielding x\<\>x. * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 18, 2000, 4:13pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/55 "2000-04-18T16:13:00Z")
> [@](#): > > Originally posted by CKDextHavn: > \*\*\<\< The ground rule for aplying this method is that the two equations MUST BE independent! \>\> > > Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other. > > A long division? Sure. (.99999…)/1 = .99999…Happy?\*\* Sorry if this might double-post because my first posting didn’t show for some reason. Anyway, I know this is beating a dead horse, but what we have here is math chicanery! a1) The equations must be independent! What if a selected 8x instead of 10x? What would the solution be, then? I can “fudge” x to equal whatever I wish! You have simply proven x\<\>x…an old math mindbender. a2) Even used matrices? What is the value (solution) would the determinant of your pair of equations have? And, then ask yourself, what is the significance of this solution? What is trying to tell me? It’s trying to tell you that the second equation is just a scalar multiple of the first (i.e.: not independent). b) As for your example of long division, you failed to show that 1/1 = 0.999… All you have shown is that any number divided by unity is equal to itself! Hmmf! * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 18, 2000, 4:20pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/56 "2000-04-18T16:20:00Z")
> [@](#): > > Originally posted by kbutcher: > \*\*Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999…? \*\* It all depends on who taught you math. If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other. If you understand the concept of why two equations are not unique if one is a mulitple of the other, then you would agree that x\<\>x is no proof. Again, I’ll ask you, what if I used 8x instead of 10x in the example “proof” given by the Administrator on page 1 of this thread? * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [April 18, 2000, 4:25pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/57 "2000-04-18T16:25:00Z")
Sloth, the limit of 0.9999… is 1.0; I’d agree to that. In short, it will continually approach, but never reach, the value of 1.0 * * * “They’re coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time… :)” - Napoleon IV ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [April 18, 2000, 4:40pm UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/58 "2000-04-18T16:40:00Z")
> [@](#): > > Originally posted by Jinx: > If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other. Untrue; I’m afraid you’re extrapolating the requirement that matrices be non-singular in order to solve the system of equations beyond its applicable range. This requirement doesn’t mean you can’t subtract two dependent equations, it just means you can’t solve a system of n equations in n variables if any of the equations are dependant. Not applicable to the current problem. > [@](#): > > Originally posted by Jinx: > Again, I’ll ask you, what if I used 8x instead of 10x in the example “proof” given by the Administrator on page 1 of this thread? OK, easy enough, (1) x = .99999… (2) 8x = 7.999999… Subtracting (2)-(1): (3) 7x = 7, hence x = 1. ---
**Author:** ![JonF](https://avatars.discourse-cdn.com/v4/letter/j/cab0a1/32.png) [@JonF](https://boards.straightdope.com/u/JonF)\ **Post date:** [April 19, 2000, 5:03am UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/59 "2000-04-19T05:03:00Z")
> [@](#): > > If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other. No, N linear equations in N unknowns must all be _independent_ for a unique solution to exist. They need not be independent in order to validly combine the equations arithmetically. If you have N independent linear equations in N unknowns then you may multiply any equation by a constant and/or add any equation to itself or any other equation without changing the solution (except you can’t multiply an equation by -1 and then add it to itself). This applies equally to the case N = 1, which is the case in the previous messages. Demonstration for N = 1: Any linear equation in one unknown “X” may be written as: ``` aX = b ``` The matrix representation of this system is: ``` {a}{x} = {b} ``` The determinant of the coefficient matrix is: ``` a ``` and is non-zero if “a” is non-zero. The determinant of the coefficient matrix with the coefficients of X replaced by the constant vector from the right side of the equation is: ``` b ``` Therefore, by Cramer’s rule, if a \<\> 0 then: ``` X = b/a ``` (gee, we really needed matrix algebra for that one, didn’t we?) {grin} Now multiply the original equation by a constant “C” and add the result to the original equation: ``` aX = b + CaX = Cb --------------- aX + CaX = b + Cb ``` Simplifying: ``` (1+C)aX = (1+C)b ``` (can you see where this is going?) By similar reasoning, the determinant of the coefficients is (1+C)a and the determinant with the coefficients of X replaced by the constant vector from the right side of the equals sign is (1+C)b, so the solution to the “system” of “equations” is, by Cramer’s rule: ``` (1+C)b -------- (1+C)a ``` or, provided C \<\> -1, ``` b/a ``` which is the same as the solution of the original problem. Therefore, a linear equation in one unknown may be multiplied by any constant (except -1) and added to itself without changing the solution. The proof for larger values of N is left as an excercise for the student {grin}. * * * jrf ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 19, 2000, 5:14am UTC](https://boards.straightdope.com/t/math-99repeating-1/11021/60 "2000-04-19T05:14:00Z")
> [@](#): > > two equations MUST BE unique (independent) in order to subtract one equation from the other. If A = B and C = D, then A-C = B-D. Always. I’m still waiting for your proof that 1 and 0.999repeating are not equal. * * * _It is too clear, and so it is hard to see._ [Previous page](https://boards.straightdope.com/t/math-99repeating-1/11021.md?page=2) [Next page](https://boards.straightdope.com/t/math-99repeating-1/11021.md?page=4)