Mathematics desk
< December 19	<< Nov \| December \| Jan >>	December 21 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

December 20

What's the big deal with 0.999...?

Hi. I'm not sure why this seems to be a university problem, it seems rather simple to me. I hereby present two simple and possible ways to show that they equal. First, divide 1 by 3. You get 0.333... Now, multiply 0.333... by 2. You get 0.666... So, following this rule, 0.333... multiplied by 3 would equal 0.999..., correct? So, since 1 divided by 3, then multiplied by 3, is still 1, and this equation also equals 0.999..., 0.999... = 1, because 1/3*3 = 0.999... = 1. Also, another way to figure it out. You might argue that 0.999... will never equal 1 because the lead number starts with 0. So, this way you might argue that it will take a whole infinity on 9s to reach 1. Well, there you go. You have an infinity of 9s in 0.999... So, since it takes an infinity to reach 1, and there is an infinity of 9s, it does reach 1. This is somewhat related to this paradox. It states that an arrow travelling at ten times the speed of a runner running away from the arrow will never reach the runner, because the distance between the arrow and the runner will decrease tenfold every time the runner runs a specific distance, as long as the runner's speed stays constant, and since you have an infinite number of this, the arrow will never reach the runner. Well, there you go. Of course we know by logic that this is not true, that the arrow will reach the runner. There is a set limit, this time 1. So, during one period of time, the arrow will be at 0.9, then 0.99, then 0.999, and so on and so on. This is the same deal. We can prove, because there is a set limit at 1, that the arrow will reach the runner. So, there, I used simple logic to prove it. Even elementary students can understand this if you make it simple enough. I just don't get it, how can something so easy it can be resolved with logic, be so seemingly difficult to prove? Thanks. ~A H 1^(TCU) 00:10, 20 December 2007 (UTC)[reply]

Wiki has a page about this, 0.999..., which says much more than I can. Someletters<Talk> 00:28, 20 December 2007 (UTC)[reply]

The basic problem with the first 'simple' explanation is that non-termination decimals such as 0.333... and 0.999... need to be defined precisely before you can justify assertions such as 1/3 = 0.333... or 2*0.333... = 0.666... Once a suitable definition is given, the fact that 0.999... = 1 is an immediate consequence. AndrewWTaylor (talk) 00:49, 20 December 2007 (UTC)[reply]

(e/c) There are two main problems with that approach - the first is that the result initially looks counter-intuitive, and some people find it hard to get past their intuition. The second is that it takes a surprising amount of work to prove that the things you just said are mathematically sound (infinity is a tricky thing, so saying that "X takes an infinite amount of time, I have an infinite amount of time, therefore X" doesn't really work, since the infinite amount of time you need for X could be different from the infinite amount of time you actually have). If you're feeling particularly brave, try wading through the archives of Talk:0.999... and Talk:0.999.../Arguments. Confusing Manifestation(Say hi!) 00:51, 20 December 2007 (UTC)[reply]

I suppose the easy way to prove this is to show that 1/3 produces a repeating decimal via long division, showing that the remainder is the same value you started with. And thus a one number sequence is repeated ad infidum. With this one can use a scalar to get the 0.999... and then compair 3/3 to one and note that the equality is transitive thus 1=3/3=3(1/3)=3(0.333...)=0.999..... That be about as clear as the point can be made. What gets some people is that they try to fathom infinity, which is humanly impossible you can't fathom the largest possible number (ignoring advanced ideas like cardinality which is conceptually way out of reach for someone still pondering the original question). They wonder where it terminates, when of course it can't or it wouldn't be infinity. This lack of understanding and the not so inuitive nature of the result give some people a real hard time. I also note in regards to the arrow problem, it's easy to show not only that arrow catches up but WHEN TOO!! The basic approch is to model the distance traveled by both the arrow and the person and set the two equations equal to eachother (both are functions of distance with respect to time) then you can determine when the person is impailed by the arrow. A math-wiki (talk) 01:17, 20 December 2007 (UTC)[reply]

To summarize: There is no big deal with 0.999... . The equality 0.999... = 1 follows immediately from the theory of real numbers and their decimal expansions (which itself requires some work to develop). However, some people aren't willing to accept this because of misconceptions they have about decimal expansions, due to flawed teaching methods at school. Also, the intuitive arguments you give, such as analogy with a physical problem, don't quite cut it - we do want a more or less rigorous proof. -- Meni Rosenfeld (talk) 09:27, 20 December 2007 (UTC)[reply]

Uniqueness of ODE

I have an third order Ordinary Differential Equation of the form:

$\theta '''=\alpha \theta +\beta cos(\theta ),$
$\theta (0)=\theta _{0},$
$\theta '(0)=\theta _{1},$
$\theta ''(0)=\theta _{2}$

Does this always have a unique solution?
If not,
1. When does it have a unique solution?
2. How to numerically find different solutions?
3. What extra information will make the solution unique?

Thanks, deeptrivia (talk) 23:06, 20 December 2007 (UTC)[reply]

It does have a unique solution. This is just an initial value problem where the equation involves an analytic function (which is much more than enough). -- Meni Rosenfeld (talk) 23:20, 20 December 2007 (UTC)[reply]

Meni, I've added a link to the analytic function article in your reply. --Taraborn (talk) 09:06, 22 December 2007 (UTC)[reply]