[ 3 / biz / cgl / ck / diy / fa / g / ic / jp / lit / sci / tg / vr / vt ] [ index / top / reports / report a bug ] [ 4plebs / archived.moe / rbt ]

/vt/ is now archived.Become a Patron!

# /sci/ - Science & Math

[ Toggle deleted replies ]
File: 108 KB, 400x381, oh boy.png [View same] [iqdb] [saucenao] [google] [report]

why does 0.999... = 1

is any number with an infinite amount of repeating nines equal to the next whole number in the ones place?

does 314.999... = 315?

 >> Anonymous Thu Apr 6 15:02:03 2017 No.8808660 >>8808643314.999... = 314+0.999... = 314+1 = 315But proving it for all integers has eluded mathematicians for centuries and is one of the Millenium Prize problems.
 >> Anonymous Thu Apr 6 15:10:23 2017 No.8808677 >>8808660How does not even work though?How do you go from a number that's not one to a number that is?if 0.999... = 1 then 1.000... = 0.999...which would then imply that 1.000 - 0.999... = 0and that is clearly false.
 >> Anonymous Thu Apr 6 15:10:47 2017 No.8808678 >why does 0.999... = 1Because 0.999... is equal to 0.9 + 0.09 + 0.009 + ... ; that's what the notation means. And evaluating expression, through the standard algebraic techniques for solving these types of infinite sums, yields 1.>is any number with an infinite amount of repeating nines equal to the next whole number in the ones place?>does 314.999... = 315?Yes.
 >> Anonymous Thu Apr 6 15:12:03 2017 No.8808681 >>8808677>How do you go from a number that's not one to a number that is?0.999... IS one, so this question does not make sense.>if 0.999... = 1 then 1.000... = 0.999...>which would then imply that 1.000 - 0.999... = 0Indeed.>and that is clearly false.No, it is quite true.
 >> Anonymous Thu Apr 6 15:16:16 2017 No.8808693 >>8808643>why does 0.999... = 1Because it doesn't terminate. You'd agree that 0.9 is closer 1 than 0.5, right? So then you'd also agree that 0.999 is closer to 1 than 0.99. Now just continue doing that, we can get as close to 1 as we want.
 >> Anonymous Thu Apr 6 15:24:29 2017 No.8808727 look up geometric series
 >> Anonymous Thu Apr 6 15:26:09 2017 No.8808731 $\displaystyle1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}$
 >> Anonymous Thu Apr 6 15:27:00 2017 No.8808733 Basically 0.9999... sums to 9/9 at infinite terms 9/9=1
 >> Anonymous Thu Apr 6 15:29:33 2017 No.8808741 If you don't think it is true then what number would be in between 0.999... and 1? There isn't one.
 >> Anonymous Thu Apr 6 15:34:02 2017 No.8808751 File: 112 KB, 953x613, 0.999_1.jpg [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Thu Apr 6 15:34:13 2017 No.8808753 >>8808741>there's no natural number between 1 and 2, therefore 1=2nice argument brainlet
 >> Anonymous Thu Apr 6 15:57:35 2017 No.8808813 >>8808751but what if there is actually not 0.333.... in the glass
 >> Anonymous Thu Apr 6 16:00:08 2017 No.8808815 >>8808677How is it clearly false?
 >> Anonymous Thu Apr 6 16:02:30 2017 No.8808821 >>8808677the "minus" operation can be loosely translated as "what is difference between such in such". Answer me, numerically, what is the difference between 1.000 and 0.999..?
 >> Anonymous Thu Apr 6 16:02:32 2017 No.8808822 >>8808753There are real and rational numbers between them.
 >> Anonymous Thu Apr 6 16:03:37 2017 No.8808826 >>8808813Then you're a greedy faggot who doesn't want to share precious water.
 >> Anonymous Thu Apr 6 16:11:22 2017 No.8808843 >>88088211 is a whole 0.999... is not
 >> Anonymous Thu Apr 6 16:12:30 2017 No.8808846 >>8808843What's missing?
 >> Anonymous Thu Apr 6 16:15:14 2017 No.8808854 if 0.999... =1 then does 0.888... = 0.89?
 >> Anonymous Thu Apr 6 16:17:44 2017 No.8808857 >>8808846the number n such that n + 0.999.. = 1
 >> Anonymous Thu Apr 6 16:19:24 2017 No.8808859 >>8808854Try subtracting them.
 >> Anonymous Thu Apr 6 16:19:52 2017 No.8808862 >>8808643The idea of real numbers is to have a set of number such that to each point on the line there would correspond exactly one point. So we can think of real numbers as points on the line. Now let's try to figure out how to represent those numbers in decimal form. Mind you, we identify numbers with points on the line. So take one point on the line and call it 0, now take another point that is to right from our point 0 and call it 1. Now we will try to assign a decimal to each point between points 0 and 1. Divide the interval between 0 and 1 into 10 equal parts, and label them from left to right with 0,1,2,3,...,9. Now for each of those interval do the same thing. So for example, you take the interval with the label 1 divide it into ten equal parts and label each of them with numbers from 0 to 9. At this stage you divided the interval into 100 equal parts and you can identify each of those parts by two numbers that range from 0 to 9. And for example the 100th interval here is identified by the pair (9,9). Now for each of the 100 interval you repeat the process of division in to ten parts and labeling them with numbers from 0 to 9. Now you clearly see you can continue this process as long as you want, and at the n-th step of this process you will have divided the interval from 0 to 1 into 10^n parts and you can identify each part with an n-tuple of numbers from 0 to 9. Now here comes the trick, if you would take any point P on the interval from 0 to 1 at each step of the construction there is only 1 interval in which that point is contained.
 >> Anonymous Thu Apr 6 16:21:13 2017 No.8808867 >>8808857It's called zero and is expressed this: 0
 >> Anonymous Thu Apr 6 16:22:03 2017 No.8808870 >>8808862(cont.) Moreover, for each other point P', no matter how close to P, there is a step in the construction and an interval from that step such that P is in that interval and P' is not. From this remarks you see that each point on the interval is defined by an infinite sequence of interval, where each step n of the sequence is the unique interval from the n-th step of the construction that contains P. But each of those intervals has an unique label - a n-th tuple of numbers from 0 to 9. Now we will identify each number on the interval as the sequence of those n-tuples. Instead of the sequence of the n tuples w can think of each points ans an infinite string of numbers fro 0 to 9. Now we will take take infinite string of numbers, put "0." and call this new string of characters is a decimal representation of the real number P. Now my question to you is, what is the decimal representation of the ponit 1?
 >> Anonymous Thu Apr 6 16:31:04 2017 No.8808891 >>8808731I guess the question here is, why is 1/3 == .333....? Are they literally the exact same?
 >> Anonymous Thu Apr 6 16:32:14 2017 No.8808896 >>8808891Try the division
 >> Anonymous Thu Apr 6 16:32:22 2017 No.8808897 >>8808867n has to be larger than zero
 >> Anonymous Thu Apr 6 16:33:33 2017 No.8808901 >>8808897Then you'd be making a false statement
 >> Anonymous Thu Apr 6 16:33:51 2017 No.8808902 >>8808677>How do you go from a number that's not one to a number that is?"The father of Ivanka Trump"and"The current president of the USA"is the same person.0.999...and 1are the same number.And so is1/2 + 1/2and1/4 + 3/42/6and1/3 and 0.333...are the same numbers too
 >> Anonymous Thu Apr 6 16:42:05 2017 No.8808934 if 0.999... = 1 then 0.999... + 0.999... should = 2but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.
 >> Anonymous Thu Apr 6 16:51:07 2017 No.8808957 >>8808934>if 0.999... = 1 then 0.999... + 0.999... should = 2Indeed it is.>but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.You can't do computable arithmetic on these representations of numbers, but that doesn't mean the algebra doesn't go through.
 >> Anonymous Thu Apr 6 16:53:10 2017 No.8808967 >>8808693Except not TO one
 >> Anonymous Thu Apr 6 16:54:26 2017 No.8808972 File: 76 KB, 325x282, image (3).png [View same] [iqdb] [saucenao] [google] [report] >>8808693That logic isn't sound in the case of equality. Its only sound in the case of equivalency. The real proof is as follows:0.99~ = 0.33~ * 30.33~ = 1/31/3 * 3 = 1Therefore 1 = 0.33~ * 3 OR 1 = 0.99~
 >> Anonymous Thu Apr 6 16:55:59 2017 No.8808977 >>8808643The proof's pretty straightforward. I think someone posted it here last time someone started a threat like this.let x = 0.99....10x = 9.99910x - x = 99x = 9x=1Alternatively,1/3 = 0.33...0.33 * 3 = 0.99...1/3 * 3 = 3/3 = 10.99... = 1
 >> Anonymous Thu Apr 6 16:57:23 2017 No.8808979 >>8808897Can you justify that?
 >> Anonymous Thu Apr 6 17:01:51 2017 No.8808992 >>8808677Okay, so, we can think about this as "distances between two numbers." PropositionIf a - b != 0, a and b are different numbersif a - b = 0, a and b are equivalent2-3 = -1, so 2 and 3 are different numbers2-2 = 0, so 2 and 2 are the same number1 - 0.99... = 0.00... so 1 and 0.99... are the same number
 >> Anonymous Thu Apr 6 17:03:35 2017 No.8808997 >>8808826but 0.333... is an infinite nnumber clearly there is only finite amount of water in the glass.
 >> Anonymous Thu Apr 6 17:05:53 2017 No.8809002 >>8808902How can I be the same person as my wife's son's father?
 >> Anonymous Thu Apr 6 17:09:49 2017 No.8809008 File: 1.83 MB, 200x200, mind_blown.gif [View same] [iqdb] [saucenao] [google] [report]
 >> Anonymous Thu Apr 6 17:11:55 2017 No.8809011 0.999... does not existprove me wrong
 >> Anonymous Thu Apr 6 17:13:54 2017 No.8809015 >>8809011It's defined as the geometric series 0.9 + 0.09 + 0.009 + ..., i.e. the sum from i=0 to infinity of 0.9 * 0.1^i. Elementary analysis clearly shows that this limit does indeed exist.
 >> Anonymous Thu Apr 6 17:14:45 2017 No.8809018 >>8809015how can something requiring """infinity""" exist
 >> Anonymous Thu Apr 6 17:15:56 2017 No.8809022 >>8809018Go study some analysis and find out. We epsilon-delta now.
 >> Anonymous Thu Apr 6 17:16:29 2017 No.8809024 >>8809022ok i actually will do this thank you
 >> Anonymous Thu Apr 6 17:20:41 2017 No.8809028 >>8809002You're not. Some other guy is.
 >> Anonymous Thu Apr 6 17:21:29 2017 No.8809031 >SCIENTISTS SAY THEY CAN ONLY SAY WITH 99.999...% CERTAINTY THAT YOU CAN NEVER GO BACK IN TIME
 >> Anonymous Thu Apr 6 17:29:25 2017 No.8809041 Figured it out ur welcome every11/3 = 0.333.... so3/3 = 0.999.... 3/3 = 10.9999..... = 1
 >> Anonymous Thu Apr 6 17:34:20 2017 No.8809051 >>8809041>1/3 = 0.333.... soprove it
 >> Anonymous Thu Apr 6 17:34:39 2017 No.8809053 >>8809024The summary (lightly simplified, but not much) of it is this:Consider the "infinite sum" 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + ...; in other words, 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... . I claim that these infinitely many values together sum to 1.What that means is the following: consider the sequence of *partial sums* of this infinite sum: the first number, the sum of the first two numbers, the sum of the first three numbers, etc. This yields 0.5, 0.75, 0.875, 0.9375, 0.96875 ...; AKA 1/2, 3/4, 7/8, 15/16, 31/32, ... . This is an infinite sequence of numbers that get ever closer to 1, but never reach it.Now it so happens that the numbers in this sequence, while all being less than 1, get *as close to 1* as you like. All numbers from 15/16 (= 0.9375) onwards are between 0.9 and 1; all numbers from 127/128 (= 0.9921875) onwards are between 0.99 and 1. More generally, for *any distance D larger than zero* that you can name -- here I named 0.1 and 0.01 -- there is a *point in the sequence* such that *all points afterwards* are no further than D removed from 1. All points in the sequence after 15/16 are AT MOST 0.1 removed from 1 (i.e. between 0.9 and 1.1); all points after 127/128 are at most 0.01 removed (between 0.99 and 1.01).Based on this, in calculus, we say that the infinite sum has the value 1. For you can make *finite approximations of the infinite sum* that are *as close to 1 as you would like*. This is called a limit, and the formal statement would be that the limit as i goes to infinity of (1 - 1/(2^i)) is 1. Moreover, formally, the SUM for i from 0 to infinity of 1/(2^i) is 1.0.999..., AKA 0.9 + 0.09 + 0.009 + ..., AKA the sum for i from 0 to infinity of 0.9 * 0.1^i, has a value of 1 by much the same reasoning.See https://en.wikipedia.org/wiki/Limit_of_a_sequence and https://en.wikipedia.org/wiki/Series_%28mathematics%29 for further introductory reading.
 >> Anonymous Thu Apr 6 17:35:52 2017 No.8809057 >>8809051So this is the stock response, but I've never quite understood it. If you won't permit .333... as 1/3's decimal representation, then what is?
 >> Anonymous Thu Apr 6 17:36:33 2017 No.8809059 >>8809053"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series
 >> Anonymous Thu Apr 6 17:38:55 2017 No.8809063 >>8809059That's not a limit though
 >> Anonymous Thu Apr 6 17:39:24 2017 No.8809064 >>8809059>"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series1+2+3+... = -1/12 is NOT true in the theory of infinite series. 1+2+3+... is just infinite, just as you'd expect it to be.
 >> Anonymous Thu Apr 6 17:41:14 2017 No.8809070 >>8809051I mean it's literally the decimal representation of 1/3. Just divide 1 by 3.
 >> Anonymous Thu Apr 6 18:14:26 2017 No.8809145 >>8808751I appreciate this picture, these threads are just boring. 1 Million "3*1/3=3/3=0.3...=1"-posts. But yellow 2 is false, there are many models with infinitely small numbers (although not $\mathbb{R}$).>>8808972>That logic isn't sound in the case of equality. Its only sound in the case of equivalency.And that is exactly the point. One way to construct $\mathbb{R}$ is by equivalence classes of cauchy series of Rationals. Unfortunately, your proof (as all the others in this thread) is not valid, it just shifts the "problem" from 0.9...=1 to 0.3...=1/3 (yes I know, I'm able to divide, but it's about representation).
 >> Anonymous Thu Apr 6 18:50:44 2017 No.8809212 >>8809145So, open the book and give a proof.
 >> Anonymous Thu Apr 6 18:57:52 2017 No.8809231 Lets say theres some infinitesmal k between 1 and .999...1 = .999... +kWe can divide both sides by any number, lets divide by 3 for familiarity and demonstration's sake. .333... = .333... + (.333...)k => .333... = .333... +kWell, thats odd, both sides are not equal. Huh, it turns out that the infinitesmal doesnt make sense for any other iteration of 1 = .999... +k. If thats the case, then 1 =/= .999... +k. Thus 1 = .999...
 >> Anonymous Thu Apr 6 19:00:29 2017 No.8809240 >>8808643>0.999... = 1>0.999... = 1 - (1/infinity)>0.999... - (1/infinity) = 0.999>[0.999... - 2(1/infinity) = 0.999>1 = 0.999 - (infinity)(1/infinity) = 1-1>1=0I maed a meem xD
 >> Anonymous Thu Apr 6 19:11:52 2017 No.8809256 File: 47 KB, 649x491, IMG_4954.jpg [View same] [iqdb] [saucenao] [google] [report] >brainlets still getting baited by trolls who contest this
 >> Anonymous Thu Apr 6 21:30:39 2017 No.8809546 >>8809145If none of the proofs satisfy you please provide one, you vapid twit
 >> Anonymous Thu Apr 6 21:49:05 2017 No.8809575 if 0.999... = 1what does 0.1/10^(infinity) +0 .999... = ?
 >> Anonymous Thu Apr 6 21:51:04 2017 No.8809581 >>8808751What we adjusted our model to include an infinitely small number? Let's call it Horse.Horse would have these properties:Horse + Horse = 2 HorseHorse * X = X HorseHowever because the Horse is infinitely small, any X Horse is also equal to Horse.So Horse = 2 HorseDivide by Horse1 = 2That doesn't make much sense until you realize there is no Horse. Or should I say, zero is also a Horse? It looks like a Horse, it walks like a Horse, gosh darn-it it must be Horse.
 >> Anonymous Thu Apr 6 21:52:24 2017 No.8809584 >>8809575Sky's the limit.
 >> Anonymous Thu Apr 6 21:55:43 2017 No.8809589 >>88095750.1/10^infinity = Horse
 >> Anonymous Thu Apr 6 21:58:33 2017 No.8809595 File: 288 KB, 1399x953, .9999.jpg [View same] [iqdb] [saucenao] [google] [report] i hate you /sci/
 >> Anonymous Thu Apr 6 21:59:06 2017 No.8809596 >>88095750 + 1 = 1 ? am I missing something?
 >> Anonymous Thu Apr 6 22:02:19 2017 No.8809603 >>8809575First of all, unless you're working in an extended number system, infinity isn't a number. It's a possible cardinality for a set, but that doesn't make it a number. You can't just raise something to the power of infinity.What you CAN do is add 0.999... to the LIMIT of 0.1/10^(x) as x APPROACHES infinity (or, to use more appropriate language, grows without bound). The result is, of course, 0.999..., because the limit of 0.1/10^(x) as x grows without bound is exactly 0.So, if your point is 0.999... can't equal 1 because adding the limit of 0.1/10^(x) as x diverges would also give you 1, I agree with your premise but not your conclusion, because the limit of 0.1/10^(x) as x diverges is exactly 0, and it's absolutely expected behavior for adding 0 to a number to give you the same number back.
 >> Anonymous Thu Apr 6 22:11:12 2017 No.8809623 >>8808678It's what this anon said. If you understand how to do the sum of a geometric series, .999... can be represened by the series of 9*(1/10)^n starting at n = 1. Sure, you can argue that it will never EQUAL one, and there is no proof that would adequtely address that. What every proof does show is that the difference between .999... and 1 is absolutely and completely negligible in every possible sense; a condition that is sufficient to say that they are "equal".
 >> Anonymous Thu Apr 6 22:20:00 2017 No.8809648 >>8809623So they're not equal but we say they're equal because it's very close?Why didn't someone just post this at the beginning of the thread?why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.It's unbelievable how hard people try to shill 0.999... = 1, you'd think they were being paid.
 >> Anonymous Thu Apr 6 22:21:55 2017 No.8809653 >>8809648>So they're not equal but we say they're equal because it's very close?It's not so much that it's "very" close. It's INFINITELY close, and any two numbers that are infinitely close are equal.
 >> Anonymous Thu Apr 6 22:30:43 2017 No.8809677 >>8809002kek
 >> Anonymous Thu Apr 6 22:51:44 2017 No.8809717 Suppose $0.999 \dots \neq 1$. Using $d \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, $d(x, y)=|x-y|$, we get a metric space. Metric spaces are Hausdorff, so there are disjoint nbds $U \ni 0.999 \dots, V \ni 1$.On the other hand, let $W$ be any nbd of $0.999 \dots$. Now, since $d(0.999 \dots, 1)< \varepsilon$ for all $\varepsilon > 0$, there is an open ball $B(0.999 \dots, r) \subset W$, but then $1 \in W$, contradicting the Hausdorff property.
 >> Anonymous Thu Apr 6 22:59:17 2017 No.8809734 >>8809717>d(0.999…,1)<ε for all ε>0you can't just assert this when you're trying to prove that .999...=1
 >> Anonymous Thu Apr 6 23:12:27 2017 No.8809763 Yes I do and I will ask a question please answer me. I have been thinking about it since yesterday night.1)Take n number of things out of which p are identical to one another and other q are identical to one another.Total number of arrangement of n number of things among themselves without repitition isnPn = n!2)But total number of unique arrangement isnPn / (p! * q!)>How do i conceptually and logically infer 2 from 1?Please explain.Don't give example to prove this.Best is to give pure mathematical path of infering 2 from 1.pls halp
 >> Anonymous Thu Apr 6 23:24:20 2017 No.8809787 >>8808660Base case1 + 0.99... = 2Induction Hypothesisx_n + 0.99... = x_n + 1Induction Stepx_(n+1) + 0.99... = x_n + 1 + 0.99... = x_n + 2where's my prize
 >> Anonymous Thu Apr 6 23:29:03 2017 No.8809793 >>8809648>why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.nice strawman faggotif two numbers x and y are distinct, then there is a number between the two of themnow tell me which number is between x=0.999... and y=1 ?
 >> Anonymous Thu Apr 6 23:30:51 2017 No.8809799 >>8809793(1+.999...)/2
 >> Anonymous Thu Apr 6 23:30:56 2017 No.8809800 >>8808731i like this
 >> Anonymous Thu Apr 6 23:34:08 2017 No.8809806 >>88097930.99... + ((1 - 0.99...) / 2)ha I got you there is no way you'll think of a flaw in my argument
 >> Anonymous Thu Apr 6 23:34:18 2017 No.8809807 >>88097930.999..., except a 0.999... that's bigger than the one you wrote. That's which number.
 >> Anonymous Thu Apr 6 23:36:07 2017 No.8809809 >>8808751re.2: Yes they do.Example: dx
 >> Anonymous Thu Apr 6 23:39:22 2017 No.8809813 >>8809807>>8809806>>8809799let y = .0000...00009 + your number, that's closer to 1 than your xthus your x was not .999...
 >> Anonymous Thu Apr 6 23:39:42 2017 No.8809814 >>8808643In case anyone wants the real reason why it's because the sequence of numbers{.9, .99, .999, .9999, ... }converges to 1 in R. Here's what this means exactly. Let a_n be the nth term in the sequence. For any number e>0 we can find an integer N such that |a_n - 1| < e whenever n>N. This is clear since we can always make |a_n - 1| smaller by taking larger n.This limit is unique. Suppose there was some other number than 1 it converged to, say C. Then |C - 1| = |(C - a_n) + (a_n-1)| <= |C-a_n| + |a_n-1| < e + e = 2e (Here I use the triangle inequality) whenever n>N and n>N' where N' is the number associated with convergence to C. Since e is arbitrary, the inequality |C - 1| < 2e must always hold for every positive e. Therefore C=1. Since the limit of the sequence is unique and real numbers are defined by convergent sequences, we identify .99... with its limit 1.
 >> Anonymous Thu Apr 6 23:41:02 2017 No.8809816 >>8808643if two numbers differ by an arbitrarily small amount they're the same number
 >> Anonymous Thu Apr 6 23:42:56 2017 No.8809820 >>8809813you shouldn't end your arguments with ellipses...it makes you look like an edge lord........
 >> Anonymous Thu Apr 6 23:43:42 2017 No.8809824 >>8808677Your intuition wants there to be an infinitely small number, an infinitesimal between the two. Infinitesimals aren't well defined in the real number system, a whole lot of maths break if you try to shove them in. There are number systems that use infinitesimals though. Accept that in R there are multiple decimal expansions for numbers.
 >> Anonymous Thu Apr 6 23:46:03 2017 No.8809831 >>8809799False.
 >> Anonymous Thu Apr 6 23:46:04 2017 No.8809832 >>8809813Let m be a variable that changes.Let its infinitesimal unit of change dm be less than 0.0...01.Let x be 0.999... + dm.My x is now closer to 1 than 0.999... is.
 >> Anonymous Thu Apr 6 23:46:21 2017 No.8809835 >>8809820>not knowing ellipses is one way to denote an infinitely repeating decimalback to /pol/ please
 >> Anonymous Thu Apr 6 23:48:00 2017 No.8809837 >>8809832now let x = 0.999... + dm + dmstill closer.thus your x=0.999... + dm did not equal the true 0.999...
 >> Anonymous Thu Apr 6 23:49:37 2017 No.8809842 >>8809835>>not knowing all mathematicians are edgelordsno indeed i did not
 >> Anonymous Thu Apr 6 23:56:54 2017 No.8809848 >>8808677>actually 1.000... - 0.999... = 0.0...01 so it is actually true because 0.0...01 = 0
 >> Anonymous Fri Apr 7 00:02:47 2017 No.8809852 File: 2.28 MB, 1134x991, 1476667419594.png [View same] [iqdb] [saucenao] [google] [report] >>8809814i'm afraid you got meme'd, my friend. good luck in your introductory real analysis class!
 >> Anonymous Fri Apr 7 00:05:18 2017 No.8809854 >>8809852Thanks but I'm already done with it. Have to keep things simple for those not in the know.
 >> Anonymous Fri Apr 7 00:12:34 2017 No.8809861 >>8809852How do you mean?
 >> Anonymous Fri Apr 7 00:49:36 2017 No.8809916 >>8808643think of 0.9999.... as 1- lim(1/n) as n --> infinity
 >> Anonymous Fri Apr 7 06:47:08 2017 No.8810301 >>8809734Take any arbitrarily small ε. If it is between 0 and 1-0.999...9 for some number $n$ of 9's, then $|1-(0.999 \dots 9 + 9\cdot 10^{-(n+1)})|< \varepsilon$.
 >> Anonymous Fri Apr 7 11:16:00 2017 No.8810621
 >> Anonymous Fri Apr 7 11:21:07 2017 No.8810625 So you are telling me0.(9) exists but 0.(9)8 cannot, right?so much for a continious number lineCuz you know, we could pretty much deduce all numbers are equal by that logic1 = 0.(9) = 0.(9)8 = 0.(9)7 = ...
 >> Anonymous Fri Apr 7 11:24:22 2017 No.8810631 >>8809717>Now, since d(0.999…,1)<ε for all ε>0but this is equivalent to d(0.999...,1) = 0 which is equivalent to those things being equal by the axioms of metric spaces.
 >> Anonymous Fri Apr 7 11:25:57 2017 No.8810633 >>8810631>axiomsHere, found your problem>It is true, because we defined it to be true
 >> Anonymous Fri Apr 7 11:36:44 2017 No.8810652 >>8810633and that is precisely how math works, isn't it
 >> Anonymous Fri Apr 7 18:30:01 2017 No.8811315 >>8809734Alright. Let's called $u_n$ the series composed of 0.9, 0.99, 0.999 and on and on.I want to show that when n goes towards infinity (i.e, with the famous 0.9999...), I get a one.So I chose an epsilon. Any epsilon. I look at $1 - \epsilon$ and see if I can find a $u_n$ that is closer to 1 than $1 - \epsilon$.Take $\epsilon = 1 ; 1 - 1 = 0 ; u_0 = 0.9 > 0$ So if I consider a distance of 1, I can use $u_0$It works for any epsilon - it's quite easy to see. Take $\epsilon = 10^{-999}$ Then $u_1000 = 1 - 10^{1000}$ is close enough to one.It even works for an epsilon so small it can be considered 0. Therefore, 0.999... = 1.
 >> Anonymous Fri Apr 7 18:31:03 2017 No.8811318 >>8811315Shit. $u_{1000}$.
 >> Anonymous Fri Apr 7 20:09:47 2017 No.8811451 >>8808753Are you braindead?
 >> Anonymous Sat Apr 8 00:43:11 2017 No.8811872 >>88086430.9999 = 1 cannot be true because 1 - .9999 = .000...01
 >> Anonymous Sat Apr 8 03:56:46 2017 No.8812070 >>8808643>why does 0.999... = 1Because R is a Banach-space.Any more questions?
 >> Anonymous Sat Apr 8 03:57:58 2017 No.8812074 >>8811872>.000...01is a Cauchy series (roughly speaking) therefore it converges in R and is equal to 0.
 >> Anonymous Sat Apr 8 09:58:37 2017 No.8812388 Since we're having this thread, I just ran across thishttps://www.youtube.com/watch?v=wsOXvQn3JuEThis is supposed to be the top science/math female youtuber. Jesus fuck
 >> Anonymous Sat Apr 8 10:11:36 2017 No.8812406 >>8812388She'll get tons of views though as people watch it to laugh at her and then comment to prove her wrong. Meanwhile her view count grows and grows.
 >> Anonymous Sat Apr 8 10:31:53 2017 No.8812424 >>8812388>Published on 1 Apr 2012
 >> Anonymous Sat Apr 8 10:41:58 2017 No.8812431 >>8812388That's an April Fools joke my friend.
 >> Anonymous Sat Apr 8 10:59:03 2017 No.8812454 >>88086430.999... DOES NOT = 1it's just so close that modern math can't tell the difference, just wait a couple hundred years
 >> Anonymous Sat Apr 8 11:01:34 2017 No.8812459 >>8808897That's only true if 0.999... is less than 1.000..., which is circular reasoning.
 >> Anonymous Sat Apr 8 11:02:41 2017 No.8812462 If x-y=0 then x=y1-0.999... = 0.000....=0
 >> Anonymous Sat Apr 8 12:01:26 2017 No.8812555 >>88086431=0.1×1010=1×100.9=0.1×9
 >> Anonymous Sat Apr 8 12:26:15 2017 No.8812592 I am getting so fucking sick of these thread.Just look up a few equivalent definitions for the real numbers. (inb4 Wilderberger autists). One of these definitions you can find is that as a set the reals is set of sequences of natural numbers, i.e. decimal expansions modulo an equivalence relations. This equivalence relation then has the property that two such sequences are the same if and only they converge to the same limit.So the crux of the matter is that some of these equivalence classes contain more than one element, i.e. have multiple representations as sequences. This is e.g. the case for 0.999... and 1. So some real numbers have non unique representations as sequences.To sum it all up: 0.999... = 1 is true by the definition of real numbers. The real problem is that people appearently do not define what the fuck they are talking about when posing this problem.
 >> Anonymous Sat Apr 8 12:47:32 2017 No.8812634 File: 1.09 MB, 1263x1060, 1425822523532[1].png [View same] [iqdb] [saucenao] [google] [report] >>8808751You need to improve on this. It's really ugly and hard to read. Also needs some anime bullshit.>>8812592On /g/ we made pic related and it worked pretty well.
 >> Anonymous Sat Apr 8 18:35:20 2017 No.8813166 This is retardation.0.999... = 1.000...1.000...... = 1.0010.999 = 1.0010.999 != 1
 >> Anonymous Sat Apr 8 18:45:51 2017 No.8813191 >>8808643THIS IS THE EASIEST PROOF TO UNDERSTAND:1/3 = 0.333...0.333...*3=0.999...BUT1/3*3=1THEREFORE0.999... MUST BE EQUAL TO 1
 >> Anonymous Sat Apr 8 18:51:30 2017 No.8813208 File: 56 KB, 382x358, 1491199286072.jpg [View same] [iqdb] [saucenao] [google] [report] >>8808753>unnatural numbers
 >> Anonymous Sat Apr 8 19:02:03 2017 No.8813224   >>88086430.999... = x(0.999...)2 = x20.999... = x2 = xnice fucking number faggots
 >> Anonymous Sat Apr 8 19:11:57 2017 No.8813240 0.999... = x(0.999...)^2 = x^20.999... = x^2x^2=x^11^2=1^10.333.. = y(0.333..)^2 = y^20.111.. = y^21/9 = (1/3)^2It's fun cuz the fact that it has no finite number of digits makes it so all these workThese numbers have a namethey are called nigglings dipshit math fucker
 >> Anonymous Sat Apr 8 19:30:16 2017 No.8813275 >>8809002your not, cuck
 >> Anonymous Sun Apr 9 03:53:01 2017 No.8814108 A brainlet arrives, behold.How about x+y=1x=0.99999.....y=0.000000......1In other words, you can "limes" and shit these two things together.
 >> Anonymous Sun Apr 9 04:10:30 2017 No.8814121 >>8813191so you're saying because 1 divided three times is 0.333... and 0.333...*3 is 0.999... then 0.999... is the original number that was being divided? Some how that doesn't feel right, but I guess my shitty human feels don't matter? well fuck you if that's how it is, inconsiderate dick
 >> Anonymous Sun Apr 9 04:35:39 2017 No.8814142 >>8808643$0.999...=\sum_{1}^\infty \frac{9}{10^n}=9\sum_{1}^\infty \left(\frac{1}{10}\right)^n=9\left(\frac{\frac{1}{10}}{1-\frac{1}{10}}\right)=9\left(\frac{1}{9}\right)=1$
 >> Anonymous Sun Apr 9 04:42:55 2017 No.8814146 >>8809057>cricketsevery time
 >> !niqjediPCA Sun Apr 9 06:54:36 2017 No.8814243 >>8808643>why does 0.999... = 1Because computers can't do fractions.
>>