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April 13[edit]

.9 repeating = 1???[edit]

This question is subject to a bit of a feud between a highschool math teacher friend and myself (a student) Obviously at first glance its obvious, of course they are different values. The math teacher( who i will call BoB)has shown me two ways to prove it to be true including this way (which was less technical) start with 1/9 = .1111-> 2/9= .2222-> .... and so on until reaching .99999-> = 9/9 which is of course = 1.0 I kind of just wanted some opinions on this and any other ways to prove disprove this, because to me just this problem seems like it makes parts of math either incorrect or wrongly labeled (fractions, irrational numbers ect..) I try to explain that the number line does not jump from .9999->8 to 1.0 to no avail. THANKS —The preceding unsigned comment was added by 66.220.213.33 (talk) 14:55, 13 April 2007 (UTC).[reply]

We actually have an article talking about this - 0.999.... --LarryMac 15:02, 13 April 2007 (UTC)[reply]

Much of the contention about this question IMO is that the quantity on the LHS is not the same type of animal as either LHS or RHS of say 2+2=4. It is not a number per se but a limit, as the number of 9s increases to +infinity. The ellipses after the more common notation "0.999..." make it explicit that this is a limit. Note the representations of the various number of ninths are also limits. Thus even though the tenths decimal place, the hundredths, etc., do not match in the expression "0.999... = 1", it is generally accepted that the equality is true (although for some modern extensions of the real line, it may not be. John Conway I think did some work like that...). I do not understand your last sentence at all, however. Baccyak4H (Yak!) 15:17, 13 April 2007 (UTC)[reply]

This is not a proof, but if 0.999 < 1, then what is 1 − 0.999...? --Lambiam ^Talk 17:12, 13 April 2007 (UTC)[reply]

One of Conway's surreal infinitesimals? Baccyak4H (Yak!) 17:22, 13 April 2007 (UTC)[reply]

Thanks for the help- my last sentence was just me trying to illustrate that IMO it made no sense to have the number .9999... if 1 is equal to it, laws of substitution mean you can simply trade them out- what i meant was that if you had a number with an infinite number of .9s and a 8 on the end then you wouldn't "skip" .999... repeating to get to one- although i know that logic is flawed as there is no end to a repeating set of .99s it kinda made sense to me .9999997 .9999998 1.0000000 ------- just trying to show what i meant--Thanks again, sorry about not finding the article first

And as for the question: what is 1 - .999... if .999 does not equal 1... it seems to me that a question like this is in itself irrational to ask as it is impossible to ever get an "accurate" number for the answer- as .999... is not a value with a limit as Baccyak4H had said it makes no sense (to me at least) to attempt to subtract it from a rational number with an end and a limit such as 1 -------- just some more babble66.220.213.33 (talk) 13 April 2007 (UTC).

If both are real numbers, then it makes sense to ask what the difference is. If 0.999... is not a real number, then of course it is not equal to 1. --Lambiam ^Talk 14:02, 14 April 2007 (UTC)[reply]

As you say, there is no number which is .9 followed by an infinite string of nines and then an 8 at the end since there is no room at the end for the 8. On the other hand .999... is a number (equal to 1) and not a limit. Finally, perhaps the thing which most resembles .9 followed by an infinite string of nines and then an 8 at the end is the limit of the sequence .98, .998, .9998, .99998, etc. The limit of this sequence is 1, so we can hardly say we are skipping over anything by saying that .99999 =1. Stefán 22:36, 13 April 2007 (UTC)[reply]

I have to speak up here: The "nothing after infinity" thing is a common misconception. It's common, and useful, in mathematics, to consider doing infinitely many things, and then doing something else. There is nothing wrong per se with considering infinitely many 9s and then an 8 at the end.

That string of digits just doesn't happen to be the decimal representation of any real number.

The thing to keep in mind is that the real numbers are primary, not their decimal representations. So you might ask, well then, why not simply ban all representations ending in infintely many nines, so that every real number would have a unique decimal representation? Well, you could do that; the biggest downside would be that the description of arithmetic would get slightly more complicated (we definitely want 1/3 to have a decimal representation, and there isn't any choice but 0.333..., and it's harder to explain why when you triple that you get 1.000... than it is to explain why you get 0.999...). But the main thing is that there isn't really much of an upside; having a unique decimal representation isn't all that useful in most cases, so why bother? --Trovatore 03:07, 14 April 2007 (UTC)[reply]

.999... is not strictly a number. It is a limit.

$.999...=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}$

It'll all make sense in Calculus, trust me.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 22:27, 13 April 2007 (UTC)[reply]

.9999... is a number, not a limit. It has the same value as the limit

\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}

, but it has also the same value as the limit

\lim _{n\rightarrow \infty }1+\sum _{i=1}^{n}{\frac {0}{10^{i}}}

, which is 1 (or 1.000... if you prefer). Stefán 22:36, 13 April 2007 (UTC)[reply]

Every real number is a limit of some sequence; in fact, of uncountably many sequences. 0.999... is a real number, and also the limit of uncountably many sequences. This false "limit/number" distinction is only going to cause confusion. Maelin (Talk | Contribs) 02:55, 14 April 2007 (UTC)[reply]

0.999... is not merely a number. How's that?--Ķĩřβȳ ♥♥♥Ťįɱé Ø 23:37, 14 April 2007 (UTC)[reply]

Series_(mathematics). I don't want to fry your brain, but consider that:

$1+1+1+...=-{\frac {1}{2}}$

But that's for a totally different reason. Riemann zeta function. Do not try to argue. You will fail. Math is all-powerful.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 22:34, 13 April 2007 (UTC)[reply]

grabs head and screams* cant wait for college now —The preceding unsigned comment was added by 69.24.162.51 (talk) 23:03, 13 April 2007 (UTC).[reply]

Here is the article on that series, by the way:

1_+_1_+_1_+_1_+_·_·_·

Enjoy. =)--Ķĩřβȳ ♥♥♥Ťįɱé Ø 23:10, 13 April 2007 (UTC)[reply]

1 + 2 + 3 + ... = - 1/12, as well. —The preceding unsigned comment was added by 203.49.240.43 (talk) 02:10, 14 April 2007 (UTC).[reply]

What would $\lim _{n\rightarrow +\infty }\left(\sum _{i=1}^{n}\left({\frac {9}{10^{i}}}\right)+{\frac {8}{10^{n+1}}}\right)$ be, if anything ? I suppose it would still be 1, because $\lim _{n\rightarrow +\infty }{\frac {8}{10^{n+1}}}=0$ , is that so ? --Xedi 14:19, 14 April 2007 (UTC)[reply]

Yes, unless that second term is part of the summation.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 23:37, 14 April 2007 (UTC)[reply]

Regarding 1 - 0.999..., the value is 0.000..., which any reasonable person would take to be 0→81.154.106.153 20:18, 14 April 2007 (UTC)[reply]

And using the zero property of subtraction, a - b = 0 iff a=b.--Ķĩřβȳ ♥♥♥Ťįɱé Ø 23:37, 14 April 2007 (UTC)[reply]

0.99999... recurring does not exist - it is futile to consider whether it equals one

Recurring decimals only exist because in the base system we have chosen (denary - base 10), because two integers have been divided and the result is not a terminating decimal. For example, in base 10, 1/3=0.33333... (recurring) but 1/4=0.25 (terminating).

In base 3, the number we call a third is not recurring - it is 0.1 (in base 3) and it is a terminating decimal.

So, is there a number which is 0.9 followed by an infinite number of '9's? If there is, then it must be the result of dividing two integers. There are no two integers which when divided give 0.9999...

It is therefore meaningless to consider whether 0.9 recurring equals one. The number simply does not exist.

Bat400 12:10, 17 August 2007 (UTC)[reply]

Lottery number 1 to 6[edit]

Is it really true that in a lottery, no sequence of numbers is any more or less likely to be drawn than other? I'm thinking specifically of the National Lottery (Lotto) in the UK - where 6 balls are selected from 49.

I understand that, statistically any group of 6 numbers has a roughly 14million to one chance of being drawn. However, it seems that drawing the numbers 1 to 6 (in any order) would be far less likely than any others. The significance of this sequence of numbers just makes it seem incredibly unlikely. In fact I'd be willing to bet my house that the numbers 1 to 6 will never be drawn in my lifetime in any lottery in world, even if I live to be 1000. --Ukdan999 15:36, 13 April 2007 (UTC)[reply]

Yes, it is true. Suppose balls 1 to 5 have already been drawn. The chances that the final ball drawn will be number 6 are 1 in 44 - exactly the same as for each of the other 43 balls still in play. Nothing "tells" the number 6 ball that it should avoid being drawn because it will complete a "significant" sequence. The inhabitants of the planet Zorg venerate the sacred numbers 4, 13, 17, 25, 33 and 42, but this does not make this sequence any more or less likely to be drawn in their National Lottery draw than in ours. Gandalf61 15:59, 13 April 2007 (UTC)[reply]

Let's do some math. If the chance of an event happening each day is 1 in 14 million, and the chances of it happening every day are independent of every other day, then it should take 14 million days on average before that event occurs. 14 million days is over 38,000 years, but there are probably more than 38 daily lotteries in the world, so it probably will happen within the next 1000 years. It may have already happened; if it was in some obscure lottery on the other side of the world it probably wouldn't be much of a news story here. I believe there have been cases where that day's date or some other significant number sequence hit, resulting in thousands of winners. (It seems stupid to bet that day's date to me, as it's no more likely to win, but you are quite likely to have to share your winnings with thousands of people if it does hit.) StuRat 17:01, 13 April 2007 (UTC)[reply]

Agreed. You can't normally improve your chances of winning a lottery, but you can reduce the expected number of people who picked the exact same numbers as you. For example, since large numbers of people pick numbers based on birth dates or sequences, if you pick seemingly random numbers that include some numbers over 31 (that won't appear in birthdates and anniversaries) then if you did happen to win you'll have improved the odds that you are the sole winner. Dugwiki 17:11, 13 April 2007 (UTC)[reply]

Sure the chances of the numbers being in X ={1,2,3,4,5,6} are not likely. But the chances of the lottery numbers being in some "random" set are no different, if you chose that random set set prior to the lottery (say Y = {20,27,34,41,48}). Root⁴(one) 19:06, 13 April 2007 (UTC)[reply]

You are forgetting that, although the numbers 1 through 6 form a meaningful sequence to you, the Lottery Machine has no such emotions. So 1 through 6 are exactly as likely to be drawn as any other sequence. However, the 1 through 6 sequence will win less because lots of people will probably chose that sequence. If it comes up, the payout per winner will be less than the payout on a more random-looking sequence that few or no people have chosen. So don't use 1 through 6 sequence on the Lottery - Adrian Pingstone 21:23, 13 April 2007 (UTC)[reply]

As an aside, I just noticed that Root4's example "random" sequence is (20,27,34,41,48). Ironically, though, all the numbers in that sequence differ by exactly seven making it not look quite as random in an aesthetic sense. Dugwiki 22:07, 13 April 2007 (UTC)[reply]

That was kinda my point ;-) Root⁴(one) 02:09, 15 April 2007 (UTC)[reply]

You are falling for the gambler's fallacy. That example is actually on there. — Daniel 19:10, 14 April 2007 (UTC)[reply]

This question stumped me once, until a friend pointed out that if today's draw was, say, 3, 9, 19, 26, 31, 38, and if you looked back over the entire history of the National Lottery, it would be extremely unlikely that this sequence has ever been drawn before, or ever will be again. One particular set of numbers may be easier to remember than another, but this has nothing to do with its probability of being drawn in a lottery. No matter which sequence you choose, it is extremely unlikely to come up. The only certainty is that one set of numbers will be drawn, out of 14 million-odd equally unlikely possibilities. JackofOz 02:52, 15 April 2007 (UTC)[reply]

Thanks for all the replies. It just confirms what I thought, statistically but the whole thing still seems totally counter-intuitive. Do the odds differ if you ask the question of the numbers 1 to 6 being drawn in numerical order (rather than being drawn 6,2,1,4,3,5, for example)? Also, does this mean that the chances of 1 to 6 being drawn 10 weeks in a row are just the same as drawing a different set of numbers every week? --Ukdan999 14:33, 16 April 2007 (UTC)[reply]

To the 2nd question, I believe the answer is yes. Just as there's no mathematical reason to suppose that any specified set of 6 numbers is more or less likely be drawn than any other set of 6 numbers in any one draw, there's no mathematical reason to suppose that any specified set of 6 numbers is more or less likely to be drawn 10 draws in a row than any other set of 6 numbers would be drawn 10 times in a row. They're all equally exceedingly unlikely possibilities which, collectively, cover the entire spectrum of possibilities. JackofOz 02:31, 17 April 2007 (UTC)[reply]

Predictor -corrector stuffq[edit]

hey guys, could someone please explain how you can derive the 2 step Adams Bashforth Moulton predictor method by integration? I've been looking everywhere for an explanation but haven't come up with anything. (my lecturer "hints" that we start with y(t_j+1)= y(t_j) + the integral between t_j and t_j+1 of f(t,y(t))dt. (sorry, dont have latex)

anyway, as i said, i'm truly stuck so if you could throw a few crumbs my way then that would be great. 130.88.52.49 16:48, 13 April 2007 (UTC)[reply]

$y(t_{j+1})=y(t_{j})+\int _{t_{j}}^{t_{j+1}}f(t,y(t))dt$

Like that?--Ķĩřβȳ ♥♥♥Ťįɱé Ø 22:15, 13 April 2007 (UTC)[reply]

Yeah thats the one! thanks! any pointers?87.194.21.177 19:48, 14 April 2007 (UTC)[reply]