# Does .99(repeating) = 1 **URL:** \ **Category:** Factual Questions\ **Created:** [April 24, 2003, 11:03am UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821 "2003-04-24T11:03:48Z")\ **Posts on this page:** 20\ **Page:** 5
**Author:** ![FranticMad](https://avatars.discourse-cdn.com/v4/letter/f/c68b51/32.png) [@FranticMad](https://boards.straightdope.com/u/FranticMad)\ **Post date:** [April 25, 2003, 1:11pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/81 "2003-04-25T13:11:16Z")
> [@](#): > > \*Originally posted by Libertarian \* > \*\*For the record, argument from authority is a fallacy only when the authority is illegitimate. \*\* Unless the argument is demonstrably wrong for other reasons, in which case argument from authority is a diversionary, rhetorical techinique. However, since the authorities referred to are legitimate, and my valiant but misguided logic about numbers was not (as clearly demonstrated by, well, everyone), I see that 0.99999 truly is the same as 1.0. Thanks for the straight dope! ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 25, 2003, 2:46pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/82 "2003-04-25T14:46:04Z")
> [@](#): > > \*Originally posted by DrMatrix \* > \*\ ***ultrafilter** > > I disagree with your definition of closure. The reals are not closed under division. S is closed under an operation we’ll call @, if for every r, s in S, r@s is in S. > > If we haven’t yet defined negative integers, would you say the non-negative integers are closed under subtraction? \*\* I found [this page](http://mathworld.wolfram.com/SetClosure.html) over at mathworld that seems to agree with me. There’s no restriction that a binary operator be total. As for subtraction on the non-negative integers, it’s a partial function. a - b is only defined when a \> b. For such a pair (a, b), a - b is always a non-negative integer. So yes, the set is closed under subtraction. ---
**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 25, 2003, 5:19pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/83 "2003-04-25T17:19:47Z")
Did you define a binary operator back then, or did you just say division? ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [April 25, 2003, 6:07pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/84 "2003-04-25T18:07:03Z")
For the record, even though I was posting too late and it came out curtly, my question was meant sincerely. I was trying to understand where exactly you were going wrong so that we could try to explain that point. You can think of the equality of .99999… and 1 as the [pons asinorum](http://mathworld.wolfram.com/PonsAsinorum.html) of infinity. Until you fully grasp the notion of an unending series the equality seems suspect. Again, for the record, arguing from authority in a math problem is somewhat different than arguing from authority in a debate about foreign policy. This equality is a settled proof. It cannot be disputed any more than someone can insist that you can trisect a circle with straightedge and compass. There are people who do still dispute this, of course. All one can do is try to discern what is not being grasped. You say now that you understand, although you don’t say directly what it was that made you change your mind. I hope you can make it clear, because this topic comes up again and again and again and maybe it could be settled next time is less than 100 posts. 🙂 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 25, 2003, 6:35pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/85 "2003-04-25T18:35:22Z")
> [@](#): > > \*Originally posted by RM Mentock \* > \*\*Did you define a binary operator back then, or did you just say division? \*\* Division on the reals is a binary operator that’s not defined when the denominator is zero. ---
**Author:** ![Bippy\_the\_Beardless](https://avatars.discourse-cdn.com/v4/letter/b/ac8455/32.png) [@Bippy\_the\_Beardless](https://boards.straightdope.com/u/Bippy_the_Beardless)\ **Post date:** [April 25, 2003, 6:50pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/86 "2003-04-25T18:50:48Z")
> [@](#): > > \*Originally posted by The Ryan \* > \*\*Bippy the Beardless > > 😕 > If you have a set in which all of the elements are equal, how can there be an infinite number of elements? Isn’t there only one element? Aren’t you just creating an infinite number of names for the same thing? > \*\* Yes, I was making the old error of thing = means equals (a purely mathematical/logical thing) so was considering the possibility of a format of logic wher A=B and A\>B and A!\ **Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [April 25, 2003, 7:04pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/87 "2003-04-25T19:04:43Z") > [@](#): > > It cannot be disputed any more than someone can insist that you can trisect a circle with straightedge and compass. There are people who do still dispute this, of course. I’m one of those people–you **can** trisect a circle with straightedge and compass. [sup](I believe you meant _angle_, not _circle_ ;))[/sup] ---
**Author:** ![Exapno\_Mapcase](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/exapno_mapcase/32/1051_2.png) [@Exapno\_Mapcase](https://boards.straightdope.com/u/Exapno_Mapcase)\ **Post date:** [April 25, 2003, 8:18pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/88 "2003-04-25T20:18:51Z")
Trisect this, buddy! Yeah, of course I meant angle. Grumble. ---
**Author:** ![John\_Kentzel-Griffin](https://avatars.discourse-cdn.com/v4/letter/j/ccd318/32.png) [@John\_Kentzel-Griffin](https://boards.straightdope.com/u/John_Kentzel-Griffin)\ **Post date:** [April 25, 2003, 8:56pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/89 "2003-04-25T20:56:04Z")
> [@](#): > > \*Originally posted by \*\*\*ultrafilter \*\*I found [this page](http://mathworld.wolfram.com/SetClosure.html) over at mathworld that seems to agree with me. There’s no restriction that a binary operator be total. Not by my reading of that page. Their page for [Binary Operator](http://mathworld.wolfram.com/BinaryOperator.html) implies what I’d call closure. “An operator defined on a set S which takes two elements from S as inputs and returns a single element of S.” There is something unsettling (to me) about having subtraction over positive integers be closed until we define negative integers and then it’s no longer closed. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 25, 2003, 9:05pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/90 "2003-04-25T21:05:38Z")
The partial function on **N** [sup]2[/sup] does exhibit closure, but the total doesn’t. I don’t see what’s troubling about that. Mathworld’s definition is ambiguous. I expect the term is used both ways in the literature. So long as we’re clear on exactly what we mean, it’s not a huge problem, IMO. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 25, 2003, 9:43pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/91 "2003-04-25T21:43:49Z")
The partial function on **N** [sup]2[/sup] does exhibit closure, but the total doesn’t. I don’t see what’s troubling about that. Mathworld’s definition is ambiguous. I expect the term is used both ways in the literature. So long as we’re clear on exactly what we mean, it’s not a huge problem, IMO. ---
**Author:** ![Race\_Bannon](https://avatars.discourse-cdn.com/v4/letter/r/5f8ce5/32.png) [@Race\_Bannon](https://boards.straightdope.com/u/Race_Bannon)\ **Post date:** [April 25, 2003, 10:02pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/92 "2003-04-25T22:02:19Z")
> [@](#): > > \*Originally posted by ultrafilter \* > \*\*Division on the reals is a binary operator that’s not defined when the denominator is zero. \*\* Here’s what troubles me. By my schooling (and I think I was paying close attention that day), the definition of a binary operator on a set S is a map from SxS (ie. ordered pairs) into S. That means that _every_ element of SxS has to be defined, or you don’t even have a map. Closure is the property that the co-domain is also the set S that you started with (rather than some other set). By saying that division by zero is just undefined, you rule out a whole bunch of SxS from the domain of your map. So, you don’t even have a binary operator, much less closure. When you say “partial function” I’m guessing you mean that you’re not concerned that your function is undefined for some part of SxS. This is not something I’ve heard of, at least in the context of what we’re discussing. It seems to me that all of the algebraic richness of things like groups, rings, fields, and so on springs from the requirment that your operators are defined over the whole domain, as addition and multiplication certainly are (on the reals, and on the complex field). ---
**Author:** ![FranticMad](https://avatars.discourse-cdn.com/v4/letter/f/c68b51/32.png) [@FranticMad](https://boards.straightdope.com/u/FranticMad)\ **Post date:** [April 25, 2003, 10:12pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/93 "2003-04-25T22:12:52Z")
I never really doubted that .99… = 1.0, but I wanted to pose an uneducated challenge about how mathematicians define these numbers. \*\*ultrafilter[/]'s clarification of the difference between completeness and whether real numbers are _closed_ was the tipping point for me. However, it is interesting that two apparently knowledgeable people can still dispute the interpretation of Mathworld’s definition. I conclude, tentatively, that discussions like this necessarily take time to resolve. Compared to the recent fluff about GIT in Great Debates, 100 posts are not hard to bear. ---
**Author:** ![KidCharlemagne](https://avatars.discourse-cdn.com/v4/letter/k/ed8c4c/32.png) [@KidCharlemagne](https://boards.straightdope.com/u/KidCharlemagne)\ **Post date:** [April 25, 2003, 10:26pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/94 "2003-04-25T22:26:28Z")
The whole idea behind calculus is that .999…=1 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [April 25, 2003, 10:47pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/95 "2003-04-25T22:47:35Z")
> [@](#): > > \*Originally posted by Race Bannon \* > \*\*Here’s what troubles me. By my schooling (and I think I was paying close attention that day), the definition of a binary operator on a set S is a map from SxS (ie. ordered pairs) into S. That means that _every_ element of SxS has to be defined, or you don’t even have a map. > Closure is the property that the co-domain is also the set S that you started with (rather than some other set). > By saying that division by zero is just undefined, you rule out a whole bunch of SxS from the domain of your map. So, you don’t even have a binary operator, much less closure. > > When you say “partial function” I’m guessing you mean that you’re not concerned that your function is undefined for some part of SxS. This is not something I’ve heard of, at least in the context of what we’re discussing. > > It seems to me that all of the algebraic richness of things like groups, rings, fields, and so on springs from the requirment that your operators are defined over the whole domain, as addition and multiplication certainly are (on the reals, and on the complex field). \*\* You tend to deal with total functions in general because they’re simpler to work with than partial functions, but there’s no logical reason why we can’t consider partial functions on any algebraic structure. ---
**Author:** ![Race\_Bannon](https://avatars.discourse-cdn.com/v4/letter/r/5f8ce5/32.png) [@Race\_Bannon](https://boards.straightdope.com/u/Race_Bannon)\ **Post date:** [April 25, 2003, 11:03pm UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/96 "2003-04-25T23:03:38Z")
> [@](#): > > \*Originally posted by ultrafilter \* > \*\*You tend to deal with total functions in general because they’re simpler to work with than partial functions, but there’s no logical reason why we can’t consider partial functions on any algebraic structure. \*\* Of course not, you can consider anything you want. I was just challenging the orthodoxy of your claim that division is closed on the reals. (since you kind of went out of your way to make that statement). The textbooks I learned from would not agree with that. ---
**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 26, 2003, 2:29am UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/97 "2003-04-26T02:29:26Z")
> [@](#): > > \*Originally posted by Exapno Mapcase \* > **Trisect this, buddy!** [Here ya go, one straightedge and compass _angle_trisection](http://mentock.home.mindspring.com/trisect.htm). 🙂 ---
**Author:** ![Desmostylus](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)\ **Post date:** [April 26, 2003, 2:47am UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/98 "2003-04-26T02:47:08Z")
Gosh, why didn’t Archimedes think of that? ---
**Author:** ![ComeToTheDarkSideWeHaveCookies](https://avatars.discourse-cdn.com/v4/letter/c/e9bcb4/32.png) [@ComeToTheDarkSideWeHaveCookies](https://boards.straightdope.com/u/ComeToTheDarkSideWeHaveCookies)\ **Post date:** [April 26, 2003, 2:56am UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/99 "2003-04-26T02:56:14Z")
[Relevant (?) Hijack]My apologies if someone else has mentioned this already, but a different spin on the same general cause for head-scratching…[Gabriel’s Horn](http://mathworld.wolfram.com/GabrielsHorn.html) illustrates how a rotated curve creates a surface of revolution that has finite volume and infinite surface area. It was the last thing that fucked with my head in 3rd semester Calculus. Thankfully I was done with all of that stuff young, young enough to consciously decide to re-allocate those brain cells to remebering the important things in life…like…what was I saying?[/Hijack] ---
**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 26, 2003, 3:18am UTC](https://boards.straightdope.com/t/does-99-repeating-1/170821/100 "2003-04-26T03:18:07Z")
Too busy inventing sarcasm, got himself in trouble with it too. [Previous page](https://boards.straightdope.com/t/does-99-repeating-1/170821.md?page=4) [Next page](https://boards.straightdope.com/t/does-99-repeating-1/170821.md?page=6)