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.9999999=1

polymathematics.typepad.com — Every year I get a few kids in my classes who argue with me on this. And there are arguers all over the web. And I just know I'm going to get contentious "but it just can't be true" whiners in my comments. But I feel obliged to step into this fray.

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  • thebigmagu, on 10/12/2007, -319/+38if you actually believe that then your a ***** moron and should not be teaching anyone
    that's whats wrong with the public education system in the United States dumb ass teachers like you
    • D4V1S, on 10/12/2007, -15/+56I wouldn't say that unless I could disprove his theory. Can you?

      update: without reading KineticFlow's comment....:-)
    • thebigmagu, on 10/12/2007, -190/+19i don't have to disprove anything dirtyfratboy's logic makes no sense what so ever
    • dirtyfratboy, on 10/12/2007, -23/+164"if you actually believe that then your a ***** moron and should not be teaching anyone
      that's whats wrong with the public education system in the United States dumb ass teachers like you"

      "i don't have to disprove anything dirtyfratboy's logic makes no sense what so ever"

      First off, learn about punctuation before hating on the U.S. public education system.
      Secondly, you're a stupid idiot. This isn't my blog. I just submitted a link to this theory.
    • Jakelshark, on 10/12/2007, -70/+42.9999... does not equal 1, there a basic fundamental of math that says each and every number (including .999... and 1) have their own equal and seperate placing on the mathmatical numberline
    • twollamalove, on 10/12/2007, -18/+49To be fair have not yet read the article, but I believe the headline is referring to 0.9 (repeating), which is indeed mathematically equatable to 1.
    • ComputerWiz, on 10/12/2007, -6/+49Just look at the infinite series equation. If anyone knows calculus, this will probably make sense.

      .99999... = 9/10 9/100 9/1000 ...

      just use a/(1 - r)


      Also, isn't any rational number able to be written as a fraction with a finite number as the numerator and denominator? If so, how would you write 0.9999...? Think about that.

      Question: If any number is infinitly close to another does that make it equal to that number? Now I'm not sure because isn't that the main idea behind calculus that it isn't?
    • benhocking, on 10/12/2007, -47/+21Here's another way to show it equals 1:
      Let x = 0.9999...
      Then, 10x = 9.9999...
      Thus, 10x - x = 9.
      So, 9x = 9 and x = 1. QED.

      Oh, which is exactly the first argument in the article. I guess I should have RTFA.
    • pfunked, on 10/12/2007, -23/+35> .9999... does not equal 1, there a basic fundamental of math that says each and every number (including .999... and 1) have their own equal and seperate placing on the mathmatical numberline

      Nonsense. .999... and 1 are just different ways of writing the precisely same number, with the same place on the numberline. The same way that 3/3 is another way to write 1.
    • TheRonald, on 10/12/2007, -33/+23If your postulate was correct Euler Method and all methods based on it would be wrong. So I'm going to have to say no .9999 does not equal 1.
    • rompom7, on 10/12/2007, -42/+28benhocking: thats not entirely correct: -

      Let x = 0.9999...
      Then, 10x = 9.9999...
      Thus, 10x - x = 9x = 8.9999...

      Are you going to tell me that 0.9999... = 1 = 1.1111... ?
    • Quactaur, on 10/12/2007, -27/+9It does equal one, you can deduce it very easily too.

      If 1/9 = .1^.
      and 1/9 = 1/3 = .3^.
      Then surely 8/9 =.8^.

      Then 9/9 = 1 = .9^.
    • twollamalove, on 10/12/2007, -7/+17@romhom: you're saying that 0.9... - 0.9... != 0?

      @TheRonald: Nice tactic. Why back anything up when you can just blurt claims like that out?
    • twollamalove, on 10/12/2007, -18/+60You know what, ***** it. Today I've lost my faith in digg completely.
    • takeda, on 10/12/2007, -22/+19The article is true, my math teacher also showed that using several proofs.
      This is not even a theory, it's a fact :)
    • joeshlub, on 10/12/2007, -31/+73.99999... Isn't a number. Thats like saying infinity isn't a number. Its an innacurate representation. .99999.... is a decimal approximation. Case closed. .99999 doesn't equal one because .99999 is an approximation. So all of this stuff with using fractions doesn't apply, because 8/9 DOES NOT ACTUALLY EQUAL .8888..... because it is an approximation.

      So simply put, infinity isn't an actual number, its a concept. .9 repeating isn't a number either, its just a concept, specifically an APPROXIMATION. if it could really exist, it perhaps would be equal to one. But try writing it. just like infinity, its impossible. Concepts don't equal numbers. Case Closed. Try refuting that. Seriously, don't dig me down unless you can refute it. And if you can, At least reply, I'll be happy to hear about it.

      At best, it is, yes, APPROXIMATELY one.
    • ccanni1028, on 10/12/2007, -55/+11Using the same type of logic as what is in the arcle, there is also proof that 1=2!!!

      1/0=0
      2/0=0
      0=0 so 1=2


      !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
    • psiit, on 10/12/2007, -30/+10Actually, I _think_ I can prove it wrong (although I can be proved wrong just as easily)
      The math:
      x = 0.99999.....
      10x = 9.9999.....
      10x - x = 9x

      but here is where I think teh mistake is
      9x = 9
      substitute the values
      9(0.9999....) = 9

      I dunno about you guys, but for me it gives 8.999999.....1 = 9. That in my opinion is an inequality :/
    • benhocking, on 10/12/2007, -6/+7@rompom7, no, but I will tell you that 8.9999... = 9. I'm not sure what subtraction method you used to get 8.9999..., however. As for 1.1111..., that equals 10/9:
      Let x = 1.1111...
      10x = 11.1111...
      9x = 10
      x=10/9
    • benhocking, on 10/12/2007, -5/+16@psiit: Actually, you're just reinforcing its consistency. You can't stick a 1 after an _infinite_ series of 9's. And, 8.9999... is, in fact, 9.
    • aristotle1990, on 10/12/2007, -9/+21There's an extremely simple proof to show that 0.999.... = 1.

      1/9 = 0.111...
      2/9 = 0.222...
      8/9 = 0.888...
      9/9 = 0.999...

      9/9 = 1.

      0.999... is the furthest you can get to one. Precisely because it doesn't end, it equals one.
    • drwtsn32, on 10/12/2007, -4/+13psiit:

      8.999999..... does indeed = 9

      The number isn't "8.9999.........1". If the 9's are infinitely repeating, you can't all of the sudden stop and put a 1 at the end.
    • swalter, on 10/12/2007, -6/+21@thebigmagu
      "dumb ass teachers" aren't the problem with the public education system in the United States, it's ignorant parents and people like yourself. Why don't you try to sue him or something.

      If you read the article, the author gives numerous examples proving his theory.
    • CharlesDarwin, on 10/12/2007, -20/+13Good Christ diggers can't do math either! What ***** noobs!
    • psiit, on 10/12/2007, -6/+2@ ben & drw: Yup, I realized that after reading the article all the way through. But still what I meant was that 9(0.99999....) is less than 9 (as in 8.99999...; shouldn't have added the 1). But arguing that 9(0.999...) < 9 is the same as arguing that 1(0.999...) < 1, so doesn't really matter.
    • Greyhaven7, on 10/12/2007, -6/+0@twollamalove...
      you're going to lose people with that inequality symbol... ;)
    • BenjaminJoffe, on 10/12/2007, -11/+20I read digg all the time but this is the first time that I have felt compelled to comment.

      First of all please do not comment on this article unless your opinion is well formed. I see many comments by people who have obviously not studied maths to a high degree. What seems right to you is irrelevant unless you can understand the underlying mathematics.

      For those who disagree with the article and continue to believe that 0.999... and 1 and different I propose another way of thinking about it.

      For two numbers to be distinct we must be able to observe that they are distinct. If these two numbers where "plotted on the number line" as many of you state, then they would clearly be very close together. So close in fact that they would appear the same, zoom in a little, a little further, keep doing that until the world ends and the two points would still appear at the same place.

      So if there is no way of ever knowing these two numbers to be different then they are not different.
    • MonaLisa, on 10/12/2007, -8/+12It's most definitely true. It's entertaining to read the comments and flawed logic of those here who think it is not true: it reflects the general lack of rigor in this country. I teach calculus, among other things, at a university and I am amazing at how poorly students can perform these days, and it is just getting worse.
    • cmiz, on 10/12/2007, -15/+3i love arguments where everyone turns into a genius...

      as far as i understand it, the whole 1/9, 2/9 ... 9/9 argument doesn't really work because 0.1 repeating isn't _really_ 1/9 because it's a repeating decimal, but it's the best our fragile little non-infinite minds can deal with.

      as to whether they are equal or not, they are indeed mathematically convergent, so they're essentially the same thing unless you really feel like dealing with all sorts of really high level math, but 1 is an integer whereas .9 repeating is not... so they're not QUITE the same. and those are my thoughts on the matter.

      and don't forget that 2 = 3 for very large values of 2
    • subscribtion, on 10/12/2007, -14/+9If they were equal, their difference would equal 0. However, 1 - .99(repeating) = 1/10^infinite which is not equal to 0.
    • coolmos, on 10/12/2007, -17/+1Think about it for a while:

      0.9 clearly does not equal 1
      0.99 doesn't
      0.999 doesn't
      No, you need an infinite number of nines to make it equal to 1. The problem is that, to get an infinite number of nines, you need an infinite amount of time.
      So, 0.999....=1. But WHEN will it be one? Never. Since you need an infinite amount of time.
    • umrgregg, on 10/12/2007, -3/+150.99... = 1.

      Answer without gee wiz tricks or arrogance here: http://mathforum.org/dr.math/faq/faq.0.9999.html
    • glc17, on 10/12/2007, -3/+7Wow, bigmagu. You're the one who has no idea what he's talking about. This isn't just something that's taught in school. It's something that's accepted as mathematical proof, and a simple proof at that.
    • newspimp, on 10/12/2007, -7/+3@BenJ

      My simple question:

      If they cannot be proven to be different (as you propose in your model, which shows them ever closer on a numberline), how can you then prove that they are the same?
    • Dimensio, on 10/12/2007, -0/+9"It's entertaining to read the comments and flawed logic of those here who think it is not true: it reflects the general lack of rigor in this country."

      The problem is likely that it does not "feel" true to those who ojbect. The reality and truth of the situation is irrelevant; how they feel about the presentation is all that matters when evaluating the validity of the claim. Unfortunately for these individuals, neither mathematics nor reality are determined by how an individual feels about a particular point.
    • whoutz, on 10/12/2007, -1/+13I quote the recurring_decimal page on wikipedia which puts it best, "The method of calculating fractions from repeated decimals, especially the case of 1 = 0.99999..., is sometimes contested by the mathematically naive:"

      Hasn't anyone had freshman calculus? Here's the proof on wikipedia article:
      http://en.wikipedia.org/wiki/Proof_that_0.999..._equals_1

      I also suggest you read about power series and reimann sums.
    • zediker, on 10/12/2007, -15/+3It comes from a confusion of symbolism, numbers, and approximation.

      It is incorrect to say .9repeating equals 1.However it is correct to say .9repeating is approximatly 1. THAT is what slips up most people.

      And the fractions examples are also due to peoples confusion with symbols, approximation, and whole numbers. 1/3 is the symbolic representation of .3repeating. Therefore we cannot say 1/3 equals .3repeating, but it is approximatly .3repeating. As such, 3/3 is approximatly .9repeating which is approximatly 1. Thus, as to not have to wright .9repeating whenever we turn 3/3 into a decimal, we shorthand it to 1 because it is approximatly equal to 1. It makes the math easier, because we dont deal with the significant figures out to infinite decimal places, because the error .0repeating with 1 at the end is approximatly 0, and as such, we shorthand it to 0, thus we have an error of '0', which is negligible.

      So, as a recap, .9repeating DOES NOT equal 1, it is very close, but DEFINATLY not equal to 1.
    • simpleid, on 10/12/2007, -1/+11I'm sorry, it's a fact. It equals 1. You'll be ok. I didn't say it was up for debate, so don't try. It's been debated. It's been decided.

      This isn't new. It's old. It's 1. Waste all the time any of you want.

      Why don't you just go varify by e-mailing any mathematicians you can find instead of trying to argue so pointlessly. Why? Because you obviously don't 'get' it.
    • ErrandboyOfDoom, on 10/12/2007, -5/+2If we choose to let a dot and a hooked circle with a bar over it represent the same thing as a vertical bar with a serif on top, that's our business, and requires no proof.

      A smiley face can equal one if we want it to, it's just a choice of representation.
    • harr, on 10/12/2007, -4/+4"So, 0.999....=1. But WHEN will it be one? Never. Since you need an infinite amount of time."

      Time doesn't come into maths like this, you don't keep on adding nines one by one, it is a number so they are already there.

      The first proof posted in the link obviously proves it. It isn't like proofs that 1=2 where there is a flaw such as dividing by zero, it does nothing impossible. It multiplies a number by ten then subtracts a number then divides by nine. Spot the flaw in that.

      Some of the other demonstrations are just pointless because 0.333...*3 does equal 0.999..., but it does not prove that 1/3=0.333... However that does not mean that the original proof is incorrect.

      If people cannot understand multiplication, subtraction and division either they weren't taught maths or don't have the ability in maths necessary to count.
    • TheRonald, on 10/12/2007, -5/+2@ twollamalove

      Blurt out claims, if you've studied math you'd know that a core postulate of Euler's method is that the number of numbers between any two numbers is infinite. Wow I just used the words number alot. Back to Euler's method, it's that belief that allows you to calculate the area under a curve.
    • icexe, on 10/12/2007, -6/+5"if you actually believe that then your a ***** moron"

      it's YOU'RE you dumbass..maybe you should have paid more attention in school.
    • ae3145, on 10/12/2007, -1/+13Hi.

      I have a math degree. These are my conclusions of the day: Digg for the quick news, Slashdot for the commentary.

      This is high school stuff guys, don't embarrass yourselves. If you want something to think about, consider what e^(pi i) = -1 actually means. Google it, do some research, learn, enjoy.

      Take care,
      g
    • bonchbonchbonch, on 10/12/2007, -3/+4@zediker:

      It IS 1.
    • djdole, on 10/12/2007, -11/+1@takeda
      "The article is true, my math teacher also showed that using several proofs.
      This is not even a theory, it's a fact :)"

      There is a reason that teaching is one of the lowest paying careers in the industry...it's pretty much summed up by the following; "Those who can, do. Those who can't, teach." (And it's scary, because it's true for most teachers.)

      I'd recommend your teacher look up the definitions of ACCURATE and PRECISE.
      Decimals are accurate...but not precise.
      And this blogger doesn’t know the difference.
    • Apocalyps3, on 10/12/2007, -0/+7Can ANY of you give a number which lies between 1 and .9 repeating? according to one of the definitions of the sum of an infinite geometric series (and other series, too, but we won't get into those) "That smallest number that can't be exceeded by anything on the list is the definition of the sum of the geometric series." .9 repeating does in fact equal 1! how can you not perceive that? that is the easiest way to explain it...if u cannot get it, then go read the whole article thoroughly
    • Frinkiac7, on 10/12/2007, -0/+6Let me see if I can explain this a little simpler than previously described...

      We already know that 1/3 = 0.333 (repeating forever. No approximations. It just goes on like this. In fact, if you have the time and boredom, you can do the long division to find this accurate out to the infiniteth (new word of the day) place)

      Now, 1/3 1/3 1/3 = 3/3, right? Good. And we know without thinking to hard that 3/3 = 1 because the fration is a whole (yes, that sounds contradictory as it was stated, but humor me for the time being).

      Good, so let's look at it from another point of view. (remember, 1/3 = 0.333 (repeating) because you took the time to do the long division yourself)

      Substituting the fractions from the first equation with those of decimal form, we get:

      0.333 (repeating) 0.333 (repeating) 0.333 (repeating) = 0.999 (repeating) again, simple math. No approximations.

      Since we used the same number in different forms for the same equation, we can thus come to the conclusion that .999 (repeating) = 1. Again, no need for approximations, limits, or confusion. Just simple math.

      Wasn't that fun? I think it was.

      This is why I love the fact that math is universal. It has the power to confuse the hell out of you no matter what language you speak.
    • MightyGiant, on 10/12/2007, -0/+2@DirtyFratBoy

      To avoid arguments like the one right there, please remember to put clips from articles in quotes.
  • KineticFlow, on 10/12/2007, -37/+61This article tries to explain that 0.99999... = 1, but all his arguments are conversely supporting that those two rational numbers are not equal at the same time. Non-mathematical arguments made, such as "looking for patterns" and inaccurate "definitions" of words.. and he didn't learn this until he went to university?

    He also has an ignorant attitude against those who criticize his opinions, saying that ..
    "...sum of an infinite series (go talk to some math professors, and see how far you get) or to deny the very existence of the number .9999.... "
    What? SUM of an infinite SERIES? Series, by definition, is the sum of a sequence of terms.

    He's suffering in the world of real numbers (not even irrational, but rational numbers).
    I wonder what he would say to his students when he has to teach complex numbers.
    • Jakelshark, on 10/12/2007, -7/+22there is such thing as infinite geometric series, its algebra 2 stuff
    • Alegis, on 10/12/2007, -15/+30Agreed. The problem with such theories is that one calculates with "infinity" as if it's a normal number.
      The definition of 0.9 (- above the 9) is that it its a series of infinite 9. By definition, it isn't even 1.

      If you can prove otherwise, it means those calculation methods are invalid. Like dividing by 0 for example, or tangens of 90 degrees.
      • drakaan, on 10/10/2007, -0/+1here's proof...

        1/3 = .3 (imagine there's a line above the three)
        2/3 = .6 (imagine there's a line above the 6)
        1/3 + 2/3 = .3(repeating) + .6(repeating)
        1/3 + 2/3 = 1
        .3(repeating) + .6(repeating) = .9(repeating)
        1 = .9(repeating)

        That's pretty simple, really.
    • twollamalove, on 10/12/2007, -6/+24This is no different that a converging integral in calculus. No one would ever try to tell their calculus teacher that just because two lines never actually touch (they get infinitely close) the area between them cannot be determined.

      The point is that
      lim n-> infinity PI(m=0->n) (0.9 * 10^(-m)) = 1

      0.9 (repeating) is infinitely close to 1. Better put 0.9 (repeating) = 1.
    • Mohonri, on 10/12/2007, -5/+26I'll take a different tack from everyone else here. Instead of proving that .999... = 1 (which it does), I challenge you to prove that it does not.

      If .9999.... != 1, then they must be two distinct numbers.
      If they are two distinct numbers, then there *must* be a third number X such that .99999.... < X < 1.
      If there is no such number X, then .99999.... = 1.

      Let me know when you find it.
    • Momoru, on 10/12/2007, -5/+24The bottom line is that there is nothing you can subtract from 1 to get .9999999...
    • masamunecyrus, on 10/12/2007, -10/+3@Alegis:

      Agreed.

      If you can say that 0.9..9 =1, you can conversely say that 1 = 0.9..9.
      Therefore, wouldn't this principle apply to any such number? Then 0.0..1 = 0, and thus 0 = 0..1. Then, someone would have to say that 1/0 = 0, which is inherently false, as zero is ~not~ a number.

      And I would also like to see this guy teach the concept of " i ".
    • e03179, on 10/12/2007, -9/+5The author is playing with definitions. To see how the author is playing with definitions, take a look at the following example:

      (1/3) = 0.333333...

      3*(1/3) = 1

      3*(.333333...) = 0.999999...
    • Nodren, on 10/12/2007, -9/+1the funny thing is, the guys very first algebraic example which he claims is "absolutely impeccable" was easily broken by using a number that isnt 10, 100, 1000, etc. which he even says you can plug in any number.
    • Directrix1, on 10/12/2007, -9/+0People ... 1=3/3=2/3+1/3=.6^ + .3^ right? Everyone agrees on that? .6^ + .3^ = (.6+2/30) + (.3 + 1/30) = (.66 + 2/300) + (.33 + 1/300) correct? The reason why the .6^ + .3^ != 0.9999... is because .6^ is not equal to 0.666... and .3^ is not equal to 0.333... . To say they are would be to completely disregard the fractional remainder. No matter how many places you write this out to (even infinite) you always have to add in the fractional remainder for the numbers to be equivalent. You were all tricked because of shorthand.
    • e03179, on 10/12/2007, -12/+3To say that 0.999999... is equal to 1 is to say that 0.999999...998 is equal to 0.999999...999 and that 0.999999...997 is equal to 0.999999...0998. Thus all numbers = all numbers?
    • superal1394, on 10/12/2007, -3/+7ooooh, I remember learning this this year, its really cool.

      let X=.9999999999 repeating
      10X=9.9999999999999 repeating
      10X-X=9
      9X=9
      X=1
    • lnxaddct, on 10/12/2007, -1/+8kineticflow,
      Infinite series are very important in any field of engineering or physics. They are used all of the time. The authors arguments are solid. It is funny seeing some of the comments here on digg because it brings out the ignorace. Proving that .9 repeating is 1 is baby stuff, seriously... I just assumed that everyone knew this until I read the comments in this article. If people are so damn sure about themselves that 1 and .999 repeating are not equal, than *prove* it. Hell, a proof by contradiction should be simple, right? Assume 1 and .9 repeating are equal and then find some contradiciton within. (hint: you won't be able to unless you screw up your math)
    • jejones, on 10/12/2007, -0/+8Oy vey. The sum of an infinite series is defined to be the limit of the sequence of sums of the first n terms as n goes to infinity. (Please don't go on about "infinity is not a number"; if you take calculus, you'll find that all the places where the infinity symbol is used are defined in ways that don't use it. It's just a convenient shorthand.)

      By .9999... we mean the sum of 1/10^n from n = 1 to infinity. We can all agree that .9999.... = -log10(x), the sum of 1/10^n from n = 1 to M is between 1 - x and 1.

      Sorry, folks, .999... is equal to 1.
    • jejones, on 10/12/2007, -0/+3@jejones:

      Duh. You should have written sum of 9/10^n, not sum of 1/10^n. Aside from that, it looks OK.
    • pt4117, on 10/12/2007, -5/+3@mohonri
      I don't get that. Why do they have to equal the same thing just because there is nothing in between? Why can't it just be the next number? If we are looking at whole numbers there is nothing between 1 and 2. Does that mean that they are the same thing?
  • D4V1S, on 10/12/2007, -37/+7MATH IS COOL!!!!!!!
  • shrewduser, on 10/12/2007, -34/+31.9999... is as close to one as you can get... but while the 9's go on infinately it will still never reach 1.

    from the start your defining a number that can't possibly ever be 1... so it can't = 1.
    • hessamz, on 10/12/2007, -10/+19ahhh the rounding error of life
    • picaman, on 10/12/2007, -9/+8I am so not a mathemetician, but doesn't Zeno's Paradox fit in here somewhere? The whole concept of approaching but not getting exactly to the number 1?
    • compu73rg33k, on 10/12/2007, -12/+9But if you read the article and use the algebra showed in your face, it comes out to equal exactly 1. When my math teacher told my class this theory this year, we were all saying wtf. I still think it's ***** up, but the algebra works out and it's all such SIMPLE algebra. Assigning a number to a variable, multiplying the variable by 10, subtracting, and then you get 9x = 9. Divide to get x = 1. ***** up.
    • baxtermadux, on 10/12/2007, -9/+15today's newspaper cost 25 cents. 25 cents gets you a 5 stick pack of Big Red Gum. today's Newspaper = 5 sticks of gum
    • Canthros, on 10/12/2007, -12/+20Very, very simple proof:
      1/9 = 0.111...
      9*(1/9) = 9*(0.111...) = 0.999... = 9/9 = 1
      QED.

      (Wish I could claim to have made that realization myself, but I picked it up in the last argument of this sort I saw.)

      picaman: The naysayers are basically arguing Zeno. The important thing to remember about Zeno's Paradox is that an empirical test would quite quickly prove it wrong, i.e. Achilles would win the race.
    • exipolar, on 10/12/2007, -3/+2see, it doesn't matter if you can calculate infinitaly many iterations within the inverse polynomial, it's the fact that we can express it as such that allows us to rationalize it as 1.
      e^x is semi impossible to calculate straight forward
      but it really help when you express it as:
      e^x = sum from n=0 to infinity x^n/(n!)
      calculators actually use that function to give the correct answers (probably out until the 14th digit is reliable)
      but it is true, when you do the math, both are equal, it's just that it's being expressed differently
    • Norweed, on 10/12/2007, -14/+13YES IT WILL. It will equal exactly one if the 9's go on forever.

      Have you people ever made it beyond algebra? This is NOT complicated stuff. I would have expected more from digg. Guess it's just a bunch of idiots TRYING to sound smart like the rest of the net.
    • noahhoward, on 10/12/2007, -15/+9I have taken Calculus and Calculus based Physics.

      As a matter of precision:

      1.0000000000000000000000000000000000000000000000000000000000000 = 1 not 1.1
      0.9999999999999999999999999999999999999999999999999999999999999 != 1 not 1.0

      If you are not worried about precision say in the world of baking then yes .99999 = 1

      If you are really not worried then hell yeah, even .9998 = 1.

      Heres an example, lets say you are setting a ship off through space and wanted it to catch an planet in just the right place to slide right into orbit. The course vector, in a given plane of space, that you wanted was calculated to be 0.99999.

      So your computer tells you that to pick up the orbit of this planet 1 000 000 000 000 miles away you have to travel at an angle of .9999 degrees. Which means, you will hit at exactly 17 453 319 067.2531 miles above the surface of this planet.

      Lets see what happens when you sub in 1. You would reach your destination at an altitude of 17 455 064 928.21759

      17 455 064 928.21759 - 17 453 319 067.2531 = 1 745 860.96449

      You just missed your mark by approx 1.8 million miles..... how? .9999 = 1

      Lets give you the benefit of the doubt maybe you used .9999999999. You ended up about 17 455 064 926.47173 miles over the surface. This time, fully confident, you... miss your mark by 1 745 859.21863 miles.

      I don't think I need to go on .999999 does not, and will not ever = 1.

      This sounds like a result of calculator science to me.

      Just out of curiosity though.... say you are bored as you drift through space with no chance of return and you calculate using .99999999999999999999999999999999999999999999999999999999999. Yeah.. you guess it, you'd miss... by 1 745 860.96449.


      Tough luck....

      Ooo! What about if you calculate using .99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 ?

      Yeah now were thinking!!

      1745860.96449 miles off... wait... that didn't change at all. The closer you get to one... the more solid you mistake appears.





    • zediker, on 10/12/2007, -12/+5It comes from a confusion of symbolism, numbers, and approximation.

      It is incorrect to say .9repeating equals 1.However it is correct to say .9repeating is approximatly 1. THAT is what slips up most people.

      And the fractions examples are also due to peoples confusion with symbols, approximation, and whole numbers. 1/3 is the symbolic representation of .3repeating. Therefore we cannot say 1/3 equals .3repeating, but it is approximatly .3repeating. As such, 3/3 is approximatly .9repeating which is approximatly 1. Thus, as to not have to wright .9repeating whenever we turn 3/3 into a decimal, we shorthand it to 1 because it is approximatly equal to 1. It makes the math easier, because we dont deal with the significant figures out to infinite decimal places, because the error .0repeating with 1 at the end is approximatly 0, and as such, we shorthand it to 0, thus we have an error of '0', which is negligible.

      So, as a recap, .9repeating DOES NOT equal 1, it is very close, but DEFINATLY not equal to 1. But since it is so close, and has an approximate error of 0 (.0000....0001), we can shorthand it to 1.
    • noahhoward, on 10/12/2007, -9/+2I would like to add somethign that is the deciding factor on this issue.

      If you are trying to say this is a mathmatical fact you are wrong. It is not fact because the lynchpin of the problem is infinity which is a concept which is, in this case, treated like a number.

      If you carry the repeating sequence on for infinity it will NEVER = 1.

      However, if you envision the problem and take the lin 10x = .999999... when you divide by 9, it does in fact theoretically terminate the repetition and the entire number rounds to 1. It's just a little trippy.

      But anyways, you can see that outside of the realm of theoretical mathmatics and calculus concepts (read as "in the real world") .99999999999999999999999... will never = 1..


      Zediker and others have put it into clearer terms than myself though. Read their stuff.
    • superal1394, on 10/12/2007, -3/+8Simple proof, no rounding.

      let X=.9999999 repeating
      10X=9.9999999 repeating
      10X-X=9
      9X=9
      X=1
    • dorkafork, on 10/12/2007, -2/+6@zediker: "And the fractions examples are also due to peoples confusion with symbols, approximation, and whole numbers. 1/3 is the symbolic representation of .3repeating. Therefore we cannot say 1/3 equals .3repeating, but it is approximatly .3repeating."

      No, it is exactly three repeating. Your explanation makes no sense, and you've got it backwards. .3repeating is the symbolic representation of 1/3 because it is equal to 1/3. The whole point of having .3repeating is to have a numberical representation of 1/3. The only way to get .3 repeating is to do the long division of 1/3, and do it forever. If you ever stop, there would have to be a remainder, but there isn't, because 1/3 = .3repeating. By your logic, if you did the long division of 1/3, the result would be something less than 1/3. It makes 1/3 some sort of magical fraction. We can know 1/2, 1/4, 1/5 all exactly but somehow 1/3 will be forever unknown exactly.
    • lnxaddct, on 10/12/2007, -2/+6.9 repeating is equal to 1, they are the same exact number. WTH are people arguing about? This is basic math... stop showing your ignoarnce. Numbers don't get "closer" to anything... they are there, they are at some point. .9 repeating is 1. Prove me wrong.
    • rderveloy, on 10/12/2007, -9/+2"Very, very simple proof:
      1/9 = 0.111...
      9*(1/9) = 9*(0.111...) = 0.999... = 9/9 = 1
      QED."

      The statement, 1/9 = 0.111..., is incorrect since 0.111... is an approximation and not an exact value.

      This whole debate is rather pointless. Anyone who has ever taken calc 1 knows that you can have mathematical limits that will never reach 1. It will get infinitely close, but you'll never actually reach one.

      A better way is to think of these numbers as percentages. Lets say you take a test that had 100 multiple choice questions. And let's say you get 1 wrong. Your score will be 99%, right? Now, 99% = .99 since 99/100 = .99. This isn't an approximation since there isn't an infinite series of numbers. Now, let's say you have a test with 10,000 questions, and you get 1 wrong. Your score will be 99.99% or .9999 if you divide by 100. Now, let's say you took a test with 1,000,000,000 (1 billion) questions and you got 1 wrong. Your score would be 99.9999999% or 0.999999999 if you divide by 100. You still don't have 1 (100/100).

      In all these cases, no matter how many trailing 9's you had behind the decimal point, you'll never get 100%, or 1 if you divide by 100, simply because you got 1 question wrong. Even if you were to continuously add questions to the end of a test that only had one question wrong from now to the end of time, you would never, ever, ever reach 100%. You'd get infinitely close, but you'll never reach it.

      Here's my proof that .999999... will never equal 1:

      99/100 = .99
      999/1,000 = .999
      999,999/1,000,000 = .999999
      9,999,999/10,000,000 = .9999999
      99,999,999/100,000,000 = .99999999
      999,999,999/1,000,000,000 = .999999999
      999,999,999,999/1,000,000,000,000 = .999999999999
      999,999,999,999,999/1,000,000,000,000,000 = .999999999999999
      and on and on into infinity...

      So you see, as long as you have 1 less than the total possible, you'll never reach 1 even if you were to add infinite questions to the end of the test.

      If you can disprove this, then please do so.







    • rderveloy, on 10/12/2007, -7/+4"Simple proof, no rounding.

      let X=.9999999 repeating
      10X=9.9999999 repeating
      10X-X=9
      9X=9
      X=1"

      No, there's rounding. You rounded when you said 9X=9

      First off, if you remember algebra, 10X-X does not equal 9, it equals 9X. If X = .99999999999..., then 9X = 9(.99999999999...). 9(.99999999999...) is APPROXIMATELY 8.999999999.....

      So, by saying 9X = 9, you were saying that 8.999999999... = 9 and, therefore, you rounded.
    • Azurensis, on 10/12/2007, -1/+6>Let X=.9999999 repeating
      >10X=9.9999999 repeating
      >10X-X=9
      >9X=9
      >X=1"
      >
      >No, there's rounding. You rounded when you said 9X=9

      No, that's not rounding, it's simple algebra. 10X - X = 9X, which is what is written on the left side in the next step.

      >First off, if you remember algebra, 10X-X does not equal 9, it equals 9X. If X = .99999999999..., then 9X = 9(.99999999999...). 9(.99999999999...) is APPROXIMATELY 8.999999999.....
      >
      >So, by saying 9X = 9, you were saying that 8.999999999... = 9 and, therefore, you rounded.

      That's not rounding, it's subtracting the same thing from both sides. Remember, step 1 says that "X=.9999999 repeating". Subtracting it off the right side leaves you with just 9, while removing it from the left leaves you with one fewer X's.

      Wanna really blow your own mind? Read up on Hilbert's Hotel. ;)
  • johndi, on 10/12/2007, -30/+47Many mathematicians like to play number games. Unfortunately some people buy into it and preach it like a religion, and have great faith in their beliefs. It's really sad when teachers do this, and then try to browbeat students with their dogma.

    I've seen ''proofs'' that all horses are white, 1+1=3, and that women are the root of all evil (sure it uses words, but the concept is the same). The numbers work out, but you can lie with numbers as well as words.

    Here is an interesting concept I've seen math geeks use to bend people's minds. There are an infinite number of infinities. After all if you start counting at one and go up you will never run out of numbers, the same is true if you start counting at 6, 7 or 42. If you're really adventurous you'll notice you can do this counting forwards and backwards. The truly talented will be able to do both at the same time, and it's likely a good way to throw oneself at the ground and miss ; ) Anyways, here are some math jokes to amuse.

    http://www2.cs.uregina.ca/~cowles/MathJoke.html
    • LucasOman, on 10/12/2007, -5/+18It's not a religion. It's fact. What are your qualifications to argue with these mathematical proofs? The "proof" that 1+1=3 contains an arithmetic flaw that is simply hard to identify. Those false proofs rely on such flaws to trick their readers. Can you find any flaws in this teacher's proofs? I challenge you to do so. There are none. And there are not infinite infinities, there are infinite sets that have infinite size. There are, however, only two sizes of infinity, rationals and reals.

      Honestly, I'm amazed that this article has caused such a stir on digg. I would think that only 8th graders would feel the need--and have such a lack of knowledge--to argue with such a solid fact in mathematics.
    • simpleid, on 10/12/2007, -9/+4I bet this person is religious, and actually believes there's a god.
    • rderveloy, on 10/12/2007, -9/+3"Can you find any flaws in this teacher's proofs? I challenge you to do so. There are none."

      Oh really? The first claim is a mistake:

      .9(repeating) = 1

      Why? because .99999999999... is an approximation. Ask any college calc professor.

      An approximation is an INEXACT representation of something that is still close enough to be useful.

      The teacher just uses the equals sign as if it will prove the theory for him. 1 does not equal .9999999... it's just that .999999... APPROXIMATELY equals 1.

      In the person's proofs, he/she keeps adding approximations and not whole numbers. This is why the proof is wrong. The teacher just assigns approximations to whole numbers and adds up the approximations.

      The teacher is a moron, should be fired, and is an example of how bad our schools are becoming. And, I'm sorry to say this, you are a moron for believing him.

      If I was one of his students and I got a 99% on a test in his class, by his own argument, he'd have to give me a 100%.
    • simpleid, on 10/12/2007, -2/+6An approximation is close, not exact, if it's not exact, then what's between?
    • Azurensis, on 10/12/2007, -2/+7@rderveloy

      .9... is *not* an approximation. It does represent the exact number that an infinity of nines after the decimal place is - namely 1. It is no more an approximation than all the digits of pi.
    • rderveloy, on 10/12/2007, -6/+2@Azurensis:

      "@rderveloy
      .9... is *not* an approximation. It does represent the exact number that an infinity of nines after the decimal place is - namely 1. It is no more an approximation than all the digits of pi."

      Ok... let's say you're right and that .9 repeating is EXACTLY 1.

      So, with that argument, you can make the same argument that given a test with an infinite number of questions, that even though you got one question wrong, as long as you have an infinite number of questions, you'll get a score of 100%.

      As soon as you have 1 question wrong, as long as you get all the other questions right, you'll get 99.999999999999999... percent of the questions right. And, if .9999999... is exactly one, then you'll get 100%!

      However, once you get 1 question wrong on a test with an infinite number of questions, you'll never, even until the end of time,get all the questions right because you got one wrong! Simply adding more questions doesn't change the fact that you got one question wrong!

      Now do you see how the argument that .999999... = 1 is just completely silly?

      Check out the numbers:

      (questions right / total questions) = percentage/100
      99/100 = .99
      999/1,000 = .999
      999,999/1,000,000 = .999999
      9,999,999/10,000,000 = .9999999
      99,999,999/100,000,000 = .99999999
      999,999,999/1,000,000,000 = .999999999
      999,999,999,999/1,000,000,000,000 = .999999999999
      999,999,999,999,999/1,000,000,000,000,000 = .999999999999999
      999,999,999,999,999,999/1,000,000,000,000,000,000 = .999999999999999999
      999,999,999,999,999,999,999/1,000,000,000,000,000,000,000 = .999999999999999999999
      And so on into infinity.

      If I'm wrong, then please provide proof before digging me down.
    • Azurensis, on 10/12/2007, -0/+3@rderveloy

      >So, with that argument, you can make the same argument that given a test with an infinite number of questions, that even though you got one question wrong, as long as you have an infinite number of questions, you'll get a score of 100%.

      Infinity minus 1 is still infinity, so yes, you would still have 100% of infinity.

      >As soon as you have 1 question wrong, as long as you get all the other questions right, you'll get 99.999999999999999... percent of the questions right. And, if .9999999... is exactly one, then you'll get 100%!

      Ok, since that 99.999... IS already 100%.

      >However, once you get 1 question wrong on a test with an infinite number of questions, you'll never, even until the end of time,get all the questions right because you got one wrong! Simply adding more questions doesn't change the fact that you got one question wrong!

      You don't understand the nature of infinity. Subtracting any actual number from infinity still leaves you with infinity. You can't affect infinity with addition or subtraction of anything less than another infinity.

      >Now do you see how the argument that .999999... = 1 is just completely silly?

      No, because it isn't. That number, with infinite nines out there is exactly equal to 1. For real numbers to be distinct, there has to be a difference between two of them, and there is no quantity that can be added to .9999... to make it equal 1, since it already is one.

      >Check out the numbers:
      >
      >(questions right / total questions) = percentage/100
      >99/100 = .99
      >999/1,000 = .999
      etc.
      >999,999,999,999,999,999,999/1,000,000,000,000,000,000,000 = .999999999999999999999
      >And so on into infinity.

      So? There is no infinitely large fraction that represents what you are getting at. You could write 9's on one side forever and 0's on the other, and you would still never get to 9.99...

      >If I'm wrong, then please provide proof before digging me down.

      What do you get if you subtract .999... from 1? There is no dangling 0 (or any other number) out there to save you. The 9's go on forever, and the answer is 0.
    • johndi, on 10/12/2007, -4/+1There is one significant difference between .999... and 1. You have to modify .999... to use it in calculations. Since I have an engineering background I'd definitely use the number 1 and not any form of .99999999...... It's true of any repeating decimal. With Pi most people are happy with 3.14. For fractions you can use 1/3 or 2/3 as they are, but .666... has to be rounded or truncated to use it. So they are not the same, but close enough as to make no real difference.

      Why do I equate it with a religion? Because of the fervor of it's proponents. They claim not "getting it" is proof that the education system is falling apart, and the decline of western civilization. Get a grip, it doesn't really matter. Smart people can believe either. Just like many smart people have believed in gods, and still contributed to science. No matter which side you take, it won't affect your life, and it doesn't harm anybody. Where else do such heated arguments of the insignificant need 750 comments with people acting like not anyone not agreeing is a total moron? This type of vitriol = religion in the same way that .999... = 1.

      @ simpleid It's my dislike of religion that makes me compare this argument to it, not the other way around.
  • bchoi, on 10/12/2007, -5/+7You know, as geeks we should probably understand this better than most, cuz of us working in multiple based digits and what not. For example, some numbers in decimal cannot be represented in binary neatly.
  • Hobbeswidget, on 10/12/2007, -1/+17And thats why mathematics don't like endless decimals, but use fractions, its not presice enough. Thats also why we use letters and signs for non-repeating decimals like pi, i, ... otherwise you decide where it stops, but it doesn't stop.
    • Friend, on 10/12/2007, -2/+8sorry to be pedantic, but i's not irrational, it's complex
      e, perhaps?
    • benhocking, on 10/12/2007, -3/+6@Friend - OK, if you're going to be pedantic, I will be, too. First of all, I should be capitalized. Secondly, it's I am, not I is. ;)
    • superal1394, on 10/12/2007, -0/+1i=squareroot(-1)
    • benhocking, on 10/12/2007, -1/+3@superal1394: There was a reason I had the emoticon ;) at the end of my post...
  • extreme101, on 10/12/2007, -18/+9so does 3.3333333333....... = 3.4 or 3.3334 or 3.3333333333334... where does it end?
    this guy is forgetting basic principles of irrational numbers.
    recurring numbers are infinite..... do you know what that means?
    • Hobbeswidget, on 10/12/2007, -5/+113.333333... does not end, and it's NOT 3.3334 if you round it off, if you would round off it would be 3.3333, only if you have 5 or higher you round of above.
      But rounding off endless decimals is not good, because you make your number less precise and your answer less accurate.
    • benhocking, on 10/12/2007, -3/+6@extreme101: See, the 9 here is a very special digit. 3.3333999999... = 3.3334, but 3.3333... just equals 1/3. Similarly, 3.99999... = 4. However, if you were using base 2 math, 0.1111... = 1, etc.
    • wunch, on 10/12/2007, -7/+1@Hobbeswidget: Rounding off your numbers actually is good in some situations. If you take a measurement with a low precision, then do some sort of calculation with that measurement and end up with a higher precision answer, then that doesn't make sense; any number based off of a less precise number cannot gain precision simply via mathematical operations. You're limited by the precision of your measuring instrument. In science, this is the idea of significant figures; computer scientists might call it GIGO :)

      Also, rounding your answer does not necessarily make it less accurate. Precision and accuracy are two different things.

      Finally, I agree that it's mathematical trickery. What many people seem to be missing is that the question is fairly moot for most obvious applications. If I say that an object weighs 0.9999 repeating kg, then would you really argue that it is not the same as 1 kg? The distinction depends on the application.
    • CosmicJustice, on 10/12/2007, -5/+2.3333...... does not equal 1/3. It approximates 1/3, but it does not equal it.
    • Azurensis, on 10/12/2007, -2/+5@CosmicJustice

      You are completely and utterly wrong. .333... is exacty equal to 1/3. There is nothing approximate about it. Now, .3333 is approximate, and .333333333 is approximate, but as soon as you put the bar over the last 3 or add a '...', it is an exact representation of 1/3.

      Why do people feel the need to argue about subjects they obviously know nothing of?
  • Jakelshark, on 10/12/2007, -22/+20This guy also has no concept of a limit, when you let .9999 go on and on it DOES approach 1. Each 9 you add on to the end does bring the number closer to 1.

    Also 1/3 plus 2/3 = 1 not .9999...
    .3333... = estimation of a non-terminating fraction
    1/3 = exact number
    same for 2/3

    this guy is a moron...unless he is trying to make people think hard about something and prove it to themselves without taking it as truth, then hes a genious as far as teachers go
    • Jakelshark, on 10/12/2007, -11/+7I decided to put my proof to the problem in general terms

      if and only if 1/x is a terminating sequence of decimal places can its decimal value be used in a mathmatical equation. if 1/x is a non terminating sequence of decimal places, the estimated value of the sequence may not be used interchangably with 1/x

      I have fundamentally proven the whole problem incorrect due to his representation of numbers causing different data than the original source.
    • Hobbeswidget, on 10/12/2007, -10/+11It brings it closer but never you never get there.
      Like y=1/x never reach 0 it gets closer and closer but never even touch it.
      "Also 1/3 plus 2/3 = 1 not .9999..." Thats because you think with fractions and not decimals.

      If you even consider thinking abouth this, you would realise this guy has a good point.
    • Jakelshark, on 10/12/2007, -13/+7Fractions are the ultimate precision when it comes to decimals, thats why I use them. If you lose precision, you get 1
    • Jakelshark, on 10/12/2007, -13/+4I was looking at his algebra equation and found that its fatal flaw is in the assumption that all the 9's cancel out. It seems like you could, but when multiplying by the 10 you remove 1 less 9 from the infinite number of 9s...if that makes sense. Thus you have 9.000000....000009 = 9x which divded by 9 = .999...
    • interiot, on 10/12/2007, -2/+10@Jakelshark: You've just posited that fractions are bunk, you haven't proved it.

      Anyway, so you're saying that 3/10 plus 3/100 plus 3/1000 plus 3/10000 ... (expressed as a proper sigma that repeats to infinity) != 1/3?
    • Jakelshark, on 10/12/2007, -13/+31/3 can only be represented in one way and that is 1/3, you can make a whole bunch of theoretical equations like yours to make it equal a decimal equvilance but it is still 1/3

      1/3 is an absolute number, .333... is not

      3(1/3) = 1
      3(.333...) = .999...

      its a postulate of math if I remember my math courses right
    • lbjazz, on 10/12/2007, -17/+8EXACTLY - My brain was bleeding through the whole article because this guy has no concept of the diff. between a rational and irrational and the concept of limits. Hell, all I've got is hs calculus and I know enough to call BS on this one. .9999 . . . is the lim as x approaches 1. It cannot be expressed in fractions.
    • Virak, on 10/12/2007, -5/+16"Each 9 you add on to the end does bring the number closer to 1."

      How, exactly, would you add something on to the end of something that HAS NO ***** END? IT'S INFINITE!
    • drwtsn32, on 10/12/2007, -4/+13"but when multiplying by the 10 you remove 1 less 9 from the infinite number of 9s"

      Um, how do you have one less than infinite? If one less than infinite is not infinite, then what number is it exactly?

      I thought this guy saying 0.999999...=1 must obviously be wrong when I first saw this on digg, but after reading his reasons they certainly do make sense.
    • Jakelshark, on 10/12/2007, -10/+1its called a limit, the closer a number gets to a value (say by adding 9 infinite times to the tail end) it approaches a number and never passes it or equals it
    • pornel, on 10/12/2007, -2/+8No, 0.(3) is an exact representation of 1/3.
    • Jakelshark, on 10/12/2007, -12/+0how can a number be exact if it cannot be written explicitly

      1/3 is exact, I can put 3 on the end of .3333... and never have the exact same value
    • Jakelshark, on 10/12/2007, -10/+1nevermind i made a stupid mistake on this comment
    • spyrochaete, on 10/12/2007, -8/+4"Um, how do you have one less than infinite? If one less than infinite is not infinite, then what number is it exactly?"

      Not all infinites are created equal. Sure, infinite is infinite, but compare these two series for example:
      x = 1, 2, 3, 4, 5.....
      2x = 2, 4, 6, 8, 10.....

      Both go on to infinite, but the second series increases at double the rate. Therefore, the second infinite is twice as large as the first, assuming they increase at the same rate.
    • yournamehere, on 10/12/2007, -11/+1doesn't matter. if you let the 9's go on it does get closer to 1 but it still will never be 1
    • Jakelshark, on 10/12/2007, -9/+0Someone pointed out a good point,

      Conceptually there is a difference between .999... and 1, but when .999... becomes infinitly long the difference between the two becomes infinitly small and thus it comes to the point of no distinguishible difference. Alot of you argue that the difference is minute enough to ignore, I argue that you cannot ignore a difference no matter how small because it is still a difference between the numbers. I finally understand completly, I had an idea of what you were trying to say, with the limit approaching 1 but then again it is a limit approaching 1 and will never equal 1. That is what a limit does.

      The limit of a/(1 - r) with a = 9/10, and r = 1/10. is 1. and by the definition of a limit, it will never equal 1. It just gets so damn close you can interchange it with 1, even though it still isnt 1
    • benhocking, on 10/12/2007, -2/+2@spyrochaete: No, there are exactly the same number of even integers as there are integers. However, there are different kinds of infinities. For example, the number of integers is the infinity known as Aleph_0 (pronounced Alef-null), and the number of irrationals is Aleph_1.
    • LucasOman, on 10/12/2007, -2/+14LISTEN PEOPLE. We're not talking about limits here. This number doesn't APPROACH 1. It's a number, not a series. It IS 0.999... followed by infinite 9s. It doesn't "grow". It doesn't "get longer". It IS long. Infinitely long.

      If .999... is not equal to 1, then there HAS to be a number between it and 1.0. Find it. You can't.
    • mentholmoose, on 10/12/2007, -1/+5@yournamehere

      There are already an infinite number of zeros in .9 repeating. It repeats infinitely. It's not as if nines are added to something like .999999900000 after a certain time. The nines already exist throughout every part of the number.
    • Jakelshark, on 10/12/2007, -4/+1.9999999...=1
      1=1

      1-.888.......=0.11111111.......2
      .9999....-.88888......=0.11111111......1

      how can one have a 2 at the end and one have a 1 at the end? or is .1111...1 the same as .1111....2
    • Jakelshark, on 10/12/2007, -1/+1nevermind, the difference between the two is .00...001 which is 0

      I admit defeat and will proclaim the word, seeing as I could make people suffer with dealing with this knowledge
    • brilliantshadow, on 10/12/2007, -5/+0I agree. I think this is where the whole thing really does fall apart.. or you get into philosophy, whichever you prefer.
      I would phrase it like this.

      if 1/3 = .333333... then the proof holds water for me. But I don't think 1/3 = .3333333.. as has been pointed out the infinitely repeating 3's are a decimal approximation (or a limit, take your pick) of 1/3. So the proof does not work, because 1/3 != .333333, strictly mathematically speaking.
    • viking, on 10/23/2007, -5/+3"LISTEN PEOPLE. We're not talking about limits here. This number doesn't APPROACH 1. It's a number, not a series. It IS 0.999... followed by infinite 9s. It doesn't "grow". It doesn't "get longer". It IS long. Infinitely long.

      If .999... is not equal to 1, then there HAS to be a number between it and 1.0. Find it. You can't."

      I agree with you until that in between part...there doesn't "have" to be a number between contiguous numbers...by your account *simplifying* 1int = 2int because there is no number between 1 and 2...taken to an *absurd* extreme two people standing close together *all left of one touching all right of the other* would in fact be the same person since there can't possibly be anyone between the two pressed bodies...

      this isn't an actual mathematical problem but an example of just how little we understand the concept of infinity...we make assumptions based on limited understanding and/or shortcutting *to make things easier ;)*...for instance a basic principal is that 1+1=2 which is inherently ambiguous because there is only 1 "1" so every time you calculate 2 in this manner you would have to define which "1"s you were using...*same to be said for all numbers*

      in fact this whole debate is a way math found itself to be ironic...
      an infinite number of arguments...to argue infinity
    • iiftmlis, on 10/12/2007, -0/+4When you are talking about real numbers then you can't talk about their contiguousness. No two real numbers are contiguous.
    • LucasOman, on 10/12/2007, -0/+5@viking

      Friends don't let philosophers do math. Please, for your sake, hand over the mechanical pencil.
    • Azurensis, on 10/12/2007, -0/+4@spyrochaete

      Aieeeee!!!

      Those two infinities are *exactly* the same size. They have the same number of entries. This can be proved by matching up the each element on one list with each element on the next list, as so:

      1 2 3 4 5 ...
      2 4 6 8 10 ...

      You can clearly see that regardless the size of the entries on the second list, there are exactly the same number of entries on each. This infinity is called Aleph-0. There are larger infinities (like the real numbers), but the magnitude of the items doesn't matter at all.
    • rderveloy, on 10/23/2007, -3/+1"If .999... is not equal to 1, then there HAS to be a number between it and 1.0. Find it. You can't."

      1-(1^(-∞)) = .99999... therefore .99999... + 1^(-∞) = 1

      1^(-∞) = 0.00000....1 where there are an infinite number of zeroes between the decimal point and one.


    • thebellmaster1x, on 10/12/2007, -0/+4@rderveloy

      0.000...1 means absolutely nothing. It's not possible to have an infinite number of 0's and then just a 1. That just means that you have a lot of 0's...and then a 1. Having an infinite number of something and then ending it would completely debase the concept of infinity; or, at least, it would IF IT WOULD STILL BE INFINITY.

      0.000...1 is a finite number. I'd suggest going back to fourth grade and asking what the "..." means.
      (Protip: No more numbers come after an ellipsis.)
    • thebellmaster1x, on 10/12/2007, -0/+2@viking

      Actually, there is number between 1 and 2.
      It's called "one and a half."
      Alternately, you could pick one from the infinite set of numbers that occur between 1 and 2.

      Ass.

      @rderveloy

      Oh, by the way...
      One to the infinite power is indeterminate.
      http://functions.wolfram.com/Constants/Indeterminate/27/
  • prKsh, on 10/12/2007, -19/+7As per google calc
    0.9999999999 + 0 =1

    verify
    http://www.google.com/search?hl=en&lr=&rls=GGGL%2CGGGL%3A2005-09%2CGGGL%3Aen&q=.9999999999+%2B0&btnG=Search

    ten or more 9's after decimal is equal to 1 as per google
    check it out!!!!!!!!
    • Hobbeswidget, on 10/12/2007, -3/+9Thats bacause calculators don't use endless numbers. They round it off. And if you round 0.9 off, you get 1, doens't matter how much 9 you put after it.
    • griz, on 10/12/2007, -2/+9Holy CRAP, then it must be true; as is everything that Google tells me.
    • d3m3, on 10/12/2007, -2/+11 - .9999999999 Where does google get an 8 from?
    • superal1394, on 10/12/2007, -0/+4or you could solve....

      let x=.999999999999999 repeating
      10X=9.9999999999999 repeating
      10X-X=9
      9X=9
      x=1

      I am getting tired of typing this
    • rderveloy, on 10/12/2007, -6/+1"let x=.999999999999999 repeating
      10X=9.9999999999999 repeating
      10X-X=9
      9X=9
      x=1

      I am getting tired of typing this"

      Um... NO!!!!
      10X - X does not equal 9, 10X - X = 9X
      if x=.999999999999999 repeating then, 9X = 9(.999999999999999...) which is APPROXIMATELY equal to 8.9999999999....

      You keep using APPROXIMATION, after APPROXIMATION, after APPROXIMATION!

      Did you completely flunk algebra or calc?
    • Thuktun, on 10/12/2007, -0/+3For people unable to comprehend simple progressions of equations, let me annotate.

      "let x=.999999999999999 repeating"

      Pretty simple, I assume everyone can understand this.

      "10X=9.9999999999999 repeating"

      This is a transformation of the previous line, multiplying each side by 10.

      "10X-X=9"

      This is a result of subtracting the above two equations, the difference of the left-hand sides equated to the difference of the right-hand sides, NOT a goofy pre-algebra mistake.

      "9X=9
      x=1"

      These are stepwise simplifications of the above steps to ensure you don't get lost, though apparently some people muyst get lost in their own bathrooms.
  • grampajoe, on 10/12/2007, -13/+16.9 repeating = 1. Take a class, learn about limits, kthxbye.
    • Jakelshark, on 10/12/2007, -11/+2he said he wasnt using limits if you read it because .999... and 1 have two different yet equal places on the numberline
    • benhocking, on 10/12/2007, -3/+9"Different yet equal places on the numberline"? What does that even mean? Do 2/2 and 3/3 also have different yet equal places on the number line? What does this phrase mean?
    • Jakelshark, on 10/12/2007, -7/+0the number 3/3 is simplfied into 1, it shares a place with 1

      .9999... isnt 1, its on a different spot on the line

      and I mean equal and seperate in that every possible number has a spot on the numberline and each number on the numberline has the same signfigance as any other spot (or 1 does not cover more on the numberline than say 1.5202052525323561)
    • benhocking, on 10/12/2007, -1/+3@Jakelshark - so, does 1.0000... also have a "different yet equal place on the numberline"?
    • Jakelshark, on 10/12/2007, -3/+0its still 1, you are just adding 0's which do nothing to the value
    • Otto, on 10/12/2007, -1/+4Jake: I'm sorry, but you're wrong. You're just wrong.

      ".9999... isnt 1, its on a different spot on the line"

      Very well. Considering a line can be divided infinitely, please name a point on the line between 1 and .9 repeating. When you can identify such a point, then I will acknowledge your correctness.
  • vileS, on 10/12/2007, -2/+22THERE ARE FOUR LIGHTS!!
    • umrgregg, on 10/12/2007, -2/+3LOL. Good one, Jean-Luc.
    • MMNManiac, on 10/12/2007, -8/+3i wonder how many people will get your picard-TNG reference :)
    • kevin_qnn, on 10/12/2007, -9/+42+2=5
    • ArmandoM, on 10/12/2007, -4/+2"i wonder how many people will get your picard-TNG reference :)"

      On a tech site like digg? I'd bet most of them.
    • master_of_fm, on 10/12/2007, -1/+4"four lights" is actually a reference to 1984
  • interiot, on 10/12/2007, -2/+11Proof... that beating a dead horse sometimes IS fun. :) Critics pwned.
  • disco_stuburt, on 10/12/2007, -13/+4This would contradict so many things in mathematics if indeed .9999999=1. Pure logic negates this. Assymtotes wouldnt matter, positive and negative numbers would be interbreeding!!!! All hell would break loose!!
    • outhouseinput, on 10/12/2007, -2/+6How? It makes perfect sense to me. Asymptotes still make sense, because they have nothing to do with .999999..., they just approach a certain number.
    • sonofagunn, on 10/12/2007, -2/+4It would totally invalidate algebra, and hence just about every mathematical proof in existence, if it wasn't true that 1=.999999...
    • pfunked, on 10/12/2007, -1/+5Asymptotes and limits involve some variable; .(9) is a constant and doesn't approach anything. It is simply equal to 1.
    • pfister_, on 10/12/2007, -5/+5The headline doesn't say ".999999..."

      It just says .9999999. No repeating. dirtyfratboy is claiming that seven 9s after the decimal point is enough to equal 1.
  • antoniojvr, on 10/12/2007, -13/+6PLEASE stop treating infinity as a number! It can't added or subtracted. It is an approximation... an idea.... NOT a number.
    • exipolar, on 10/12/2007, -1/+3but if it can be expressed in way that it is practical to the problem and it's solution, it really doesn't matter either way
    • aluhain, on 10/12/2007, -3/+1except in the extended real number system
    • Hobbeswidget, on 10/12/2007, -2/+4But yout treat pi and i and all those infinite numbers represented by simbols as numbers why this one not?
    • lmushl, on 10/12/2007, -9/+3Pi is approximately 3.14159. i is the square root of negative one. You start writing out infinity and let us know when you have something we could use to make it easier to represent.

      The problem is that infinity cannot be grasped. It's unfathomably large. Pi at thevery least is somewhere between three and four. i is on another plane, but we can represent it using a function. Infinity just cannot be touched.
  • zentraedi, on 10/12/2007, -4/+14His proofs are 100% right. I had no idea math educuation has fallen so far.

    I knew that before I even went to university for math. I remember losing a percent on my grade 10 biology mark because my teacher didn't understand it.
    • clickwir, on 10/12/2007, -9/+1No, he's wrong because he makes a basic assumption. 1/3 does not equal a real number. It equals a theoretical number. Try writing out 1/3 in decimal form. You cannot because it goes on forever, or at least beyond our comprehension. No, .999 repeating does NOT equal 1, because it is not 1. It is close, yes, and the articles writer assumes or rounds at some point to reach the conclusion. His "examples" have the same truthfulness as this 43x21+42-1xPi(some miracle)-1=2.415243

      It just simply does not work out that way.1=1 .999 = .999
      • Dantizzle, on 10/10/2007, -0/+1ummm...lets check the definition of a real number. one that can be written in the form a/b where both are integers and b does not equal 0. last time i checked...1 and 3 were integers buddy. im with otto on this one
    • Otto, on 10/12/2007, -1/+5What?!?

      1/3 is absolutely a real number and your teachers should be ashamed of themselves for letting you think otherwise.
  • SirBriggs, on 10/12/2007, -5/+3My math teacher told us when i was in middle school. Forgot the rationale for it, though.
    • spyrochaete, on 10/12/2007, -9/+2replied to the wrong comment. mod down please.
  • exipolar, on 10/12/2007, -2/+15What I really find funny is the fact that, since our numbering system is a base 10 exponential system, it can allow for such definitions of different numbers.

    what people don't realize is that .99999 does not equal 1, duh! but when we say .99999 repeating, that DOES equal 1. it's the fact that we express it as an infinite term that allows it to be equal to 1. that said, if we had a base 4 numbering system, we would argue whether or not .3 repeating was equal to one, which in a base 4 system it would be.

    whoever, the real fact of the matter is that we have to live with the fact that there are more than one way of expressing the same thing. numbers the way we represent them are simply compressed base 10 poly nomials
    like 4829= 4*10^3+8*10^2+2*10^1+9*10^0
    when you realize that, it's just obvious that .9 repeating is just another way of saying
    sum from n=1 to infinity 9/10^n
    which is what the article goes over

    when you think about it, it's just an odd way of saying one
    just like e^(i*pi/2+2pi*n) is a very odd way of saying 1
    • cavemanf16, on 10/12/2007, -10/+10I completely agree. The blogger erred from the very beginning with his equation and how he set it up to "look". On his blog, he has it in the normal 3rd grader algebra style with the numbers being subtracted below the numbers that are being subtracted. Also, his equation is predicated on two assumptions (as you have already stated) that .999999... = 1. THAT is a bad assumption, because he's automatically saying that a discrete, integer number is equal (not simply equivalent) to a "real" number. Secondly, when he performs his equation he is completely neglecting the precision issue, stating in essence that .9 repeating can be subtracted from itself. In essence, that's saying that you can subtract 0 from 0, or infinity from infinity (because, after all, .9 repeating goes on for infinity), and we all know that those are mathematical fallacies, or at least "not defined." In programming parlance, this is the NULL case - undefined.

      I call ***** on this crackpot. No digg!
    • spyrochaete, on 10/12/2007, -6/+1Thanks for making the point I had in mind. Does this mean that a hexadecimal acimptote is closer to 1 than a decimal equivalent?
      0.FFFFFFFF... > 0.9999999...?

      What if we state it in trinary (is that a word?) instead?
      0.99999... > 0.22222...?

      Binary?
      0.999999... > 0?
      0.99999... = 0?
      0.99999... = 1?
      1 = 0???
    • jewster, on 10/12/2007, -0/+1e^(i*pi/2 2pi*n)=1
      Where did you get this from? Does this work for all values of n. Is it derived from Euler's formula?
      e^(x*i)=cos(x) i*sin(x).
      I'm not busting your balls or anything. Just curious.
  • JayRod, on 10/12/2007, -8/+5I remember a while back on the show AOTS, the Wonder Years girl came on. She's a math wiz and she proved that .9999999 = 1. Haven't watched that show in a while, is it still on?
    • umrgregg, on 10/12/2007, -2/+4Your comment made my brain bleed.
    • wunch, on 10/12/2007, -2/+4He's referring to the G4TV show "Attack of the Show". Danica McKeller (Winnie Cooper from The Wonder Years) was a guest one episode, and she also happens to be a mathematician. On her visit, she used a whiteboard and wrote down one of the same proofs that this blog article uses in its argument.
    • spyrochaete, on 10/12/2007, -3/+2I saw that episode! She was awesome! The sexiest female organ is the brain and she had a DD! :>

      Yes, the show is still on. No, it's not worth watching. The last time I flipped to it they were interviewing some eating contest champion. On the set. Sarah Lane left and now they have some British bimbo doing The Feed which is now about celebrity dating. RIP TSSAOTS.
    • clickwir, on 10/12/2007, -8/+1nope, cute or smart or not... it's not 1. It's .999. It's very simple and these "examples" really don't prove anything other than the authors assumption.
  • griz, on 10/12/2007, -8/+7In the article, he states....
    >>"Really???? Infinitely many zeros and then after the infinite list that never ends, there's a 1????"

    The problem here is that he uses the phrase "After the infinite". How can you have "after infinite"? If you have an "after", then it is finite.
    • lmushl, on 10/12/2007, -2/+9It's supposed to be a sarcastic 'after'
    • drwtsn32, on 10/12/2007, -2/+10That's exactly his point! He is showing that there can't possibly be something "after" an infinite number of 9's.
    • aluhain, on 10/12/2007, -3/+1look up the concept of the first uncountable ordinal sometime if you think that blows your mind
  • dclarktandem, on 10/12/2007, -16/+4This argument is 99 percent snake oil and .99 percent rounding error. It is also likely to be eaten alive in the arena of 0 and 1 that it just walked into. Our computers can't even truly represent .999.... with infinite 9s. That is how non '1' it is.
    • LucasOman, on 10/12/2007, -1/+8You don't need a computer. Technological limitations have absolutely NOTHING to do with the fundamentals of mathematics.

      *sigh* This is the age we live in. Everyone now thinks that if it can't be punched into a calculator, then it can't be proven.
    • toxicredm, on 10/12/2007, -0/+2That's only 99.99%, what is the other .01%? Or did you mean .99...% which would be 1%, giving you a total of 100%.
  • dBLiSS, on 10/12/2007, -13/+30.9 (repeating) is basically a limit as n approachs 1. So, it is essentially one, and not one at the same time depending on how you are using it. So, both parties are partly right.
    • LucasOman, on 10/12/2007, -0/+11It's NOT a limit. It's NOT a series. It's a NUMBER. It doesn't "grow". It doesn't "get longer". It is what it is. It is a number that has nothing on the numberline between itself and 1, meaning that it is equal to 1.
    • clickwir, on 10/12/2007, -7/+1LucasOmen: nothing inbetween it doesn't mean that it's equal. Just because there is nothing between .8888 and .9999 does not mean .8888 equals .9999. .9999 repeating is a theoretical number, it is infinate, it cannot be expressed accuratly and get a result in an equaiton. .999 repeating does not equal 1. It's very very close, but not it.
    • benhocking, on 10/12/2007, -1/+6@clickwir - 0.9 is between 0.8888... and 0.9999...
    • LucasOman, on 10/12/2007, -0/+7You are incorrect. There IS something between .888... and .999... In fact, there is an infinite set of numbers between those two. That's why they're NOT equal. If, in some alternate number system, there happened to be nothing between them on the numberline, then they would in fact be equal. Read a pre-algebra or algebra text book.
  • viclopez, on 10/12/2007, -18/+0saying .9999...= 1 is incorrect. Altought the number with every 9 added does get closer and closer to 1, it will never reach 1. Its simple logic.
    • drwtsn32, on 10/12/2007, -2/+5Did you even read the article?
    • Norweed, on 10/12/2007, -4/+15LOLOLOL....

      This thread is soooo funny. I can see all these people that think they know math, but in fact are just guessing and spouting off stupid crap. Take a few college level math classes and it'll become PAINFULLY obvious that infact 3/3 = .99999... = 1

      I don't really know how to dumb it down any more than this, but see if you understand this. The fact that there are an infinite number of 9's makes it equal one. If there were ANY finite number of 9's then it would NOT equal 1, however once this list is declared as going on forever it equals one.

      Hell, this is pre-calc limits stuff. Look up any number of series that approach a number. They will never equal that number untill the limit is taken to infinity. When this happens it actually GETS to the number, not just really really close. Hence why 3/3 = .9999999..... = 1

      If you don't get it then at least stop making yourself look stupid.
  • IQ70, on 10/12/2007, -8/+4Limit x->1, where x is always < 1.
    Stating that 0.9bar=1 is approximating it for calculations.
    In theory it can approach 1 but never be equal to 1.

    If his theory was correct that by regression one could prove 0=1 since any infinitesimally small difference approaching infinity needs to be ignored. And we all know that is not correct. :-)
    • IQ70, on 10/12/2007, -5/+1Remember, this is talking about limits. Limits in terms of integration?
      Integrate all the errors of infintely small proportions for infinite number of observations.
      Then come back tell me if its 0 or 1.
    • Otto, on 10/12/2007, -0/+6>>>"In theory it can approach 1 but never be equal to 1."

      You clearly don't understand the concept of "limits". Limits have actual, defined answers. Just because they approach something as they extend towards infinitely does not mean they never reach it. They do reach it, at infinity. That's the whole point of limits.
  • mystagogue, on 10/12/2007, -4/+19i'm surprised that so many on digg have a problem accepting this. maybe just because i have a math degree, but i remember encountering this before college. between every two rational numbers is another rational number. if .9999... didn't equal one then tell me what number is between it and one?
    • IQ70, on 10/12/2007, -20/+3I will tell you what is between 0.9999.... and 1.
      It is 0.9999....5.

      Happy?
    • mystagogue, on 10/12/2007, -3/+12no. where does the five go? the nines are repeating forever.
    • postapoc, on 10/12/2007, -6/+1.1^oo
    • IQ70, on 10/12/2007, -11/+1The 5 goes right after the 9 you just thought off.
      There is absolutely nothing wrong in saying that there exists a number greater than one with an infinitely large decimal trail.
    • benhocking, on 10/12/2007, -2/+9@IQ70: Yes there is. You see, you can never get to the end of "an infinitely large number trail".
    • Jaq524, on 10/12/2007, -0/+8Don't go bashing on IQ70, he's only living up to his name :P
    • IQ70, on 10/12/2007, -6/+1Wow, I can hardly argue with people who not only will call me dumb but will also go bury my posts.
      How cool is that?

      Why cant you get to the end of the number trail? Did you mean mathematically?

      Ofcourse, there is no end to the 9s, but I want to add a 5 to the trail, you can keep expanding the 9s in the middle.
      Why do you think that the trail only exists at the end? The expansion of 9s can begin at the decimal end.

      So, 0.95 can grow to 0.995, 0.9995, 0.99999999.....95.
      At any given time, 0.999...9995 is larger than 0.9999...999
      I know it looks whacky when seen on paper, but sit down and think abt it.
      :-)
    • solarpowered, on 10/12/2007, -5/+1mystagogue: Find a number between 0.9999999... and 1. Find it!

      me: OK, let's start here: There is a number between 0.9 and 1, I choose 0.99

      mystagogue: How about the number between 0.99 and 1? I DEMAND YOU FIND IT!

      me: This is going to take some time, I can see. There is a number between 0.99 and 1, I choose 0.999

      (You can see where this is headed... I always have a reply... and this does take forever)


    • benhocking, on 10/12/2007, -0/+4@solarpowered: so you're claiming the number between 0.9999... and 1 is 0.9999...? Or is it 0.99999... or 0.9999...9 (which doesn't make sense) or something that just can't be represented? You're so close to that AHA! moment...

      @IQ70. An interesting way of representing it. By that representation, however, 0.999...9995 (where the ... represents an infinite number of 9's) still equals 1. That's because in the limit there's no difference from 1, and we're in the limit.
    • simpleid, on 10/12/2007, -0/+4IQ70, .9- is already infinite. There is no place for a breaking point, stopping point, there is NO +1 to infinity, as that just re-encompasses itself. It's, Infinite, Already.
  • Desolite, on 10/12/2007, -17/+1i don't think it equals 1, i think its "next to" 1 in the number line but its close enough that it doesn't actually matter anyway. unfortunately, whoever wrote this is too stubborn to change his stance on this anyway after commiting so much energy to convince people of the notion that its different.
    • pfunked, on 10/12/2007, -3/+14The trick here is that it's not "next to" 1 at all. It actually is "1".

      Imagine conversely trying to represent an "infinitely small" number. 0.000... (zeroes going on forever). Is this equal to 0 or "next to" 0? We know we can just as easily write 0.0 as 0.000..., and there's no value at the end of those infinite zeroes, so 0.000... *is* zero. The same concept with 0.9999...., just a bit harder to swallow.
    • Desolite, on 10/12/2007, -12/+1if your saying the next infinitely close number to 1, then its 'next to' 1. if your saying the infinitely small number next to 0 with a 1 at the "end", its "next to" zero.

      maybe you're having trouble swallowing the concept of infinity.
  • banglogic, on 10/12/2007, -15/+6Of course it's not correct. They are two different numbers. It is pretty funny though.
    • halik, on 10/12/2007, -8/+1LOL
      I wish digg had slashdots marking system. Put me down for +5 funny on this one
  • lmushl, on 10/12/2007, -4/+11There have been a lot of intelligent replies that cite facts to support claims, but a lot of these facts are either close-but-not-quite-right or just wrong. When the author of the article cites information on 'infinite series' he's not trying to claim that the number IS a series - only that it can be REPRESENTED as a series. Just as it can be represented by .9 repeating. The number itself is just a number - not moving, not APPROACHING anything. .9 repeating is not ".9 ... then .99 ... then .999 ...." and so on - it is just .9 repeating. No matter how many 9's one CHOOSES to 'put on the end' to represent the number, the fact is that there IS no limit to how many there are. Additionally, .3333 is NOT an estimation of a fraction, it is meant to be interpreted as .3 repeated, which, similarly, has an infinite number of 3's 'on the end' of the decimal. Likewise with .6666. The author is representing the repeating decimal with a short selection of the repeating decimal in order to have it more easily visualized, and is assuming that the audience will be knowledgeable enough to understand that you could write it with one decimal point with a line over it, or a thousand digits stretching across the page. The ellipses denote the repetition. It's not basic-level math. It requires calculus. And God forbid a little bit of open-mindedness.
    • eggo, on 10/12/2007, -0/+3It seems to me this will be one of those things our descendants will look back on and laugh. That is, if their machine bodies are capable of laughter, which they aren't.
  • GODSUN, on 10/12/2007, -17/+20.9999999 + 0.0000001 = 1
    So by the logic of the blog, 0.0000001 = 0 ????

    The day you will proove me 0.00000001 = 0 maybe ill beleive it, until then, this is a huge scam/haox.
    • exipolar, on 10/12/2007, -7/+6.000000001 does not equal zero
      but .0000....infinitatly many 0's....1 does equal zero
    • drwtsn32, on 10/12/2007, -2/+11Wrong. 0.99999...... + 0.0000000.... = 1. You can't have a number like 0.00000000....1. How can an infinitely repeating zeros suddenly change and end with a 1?
    • exipolar, on 10/12/2007, -6/+4precisly my point! .00000....1 would technically be 1/x as x approaches infinity, which would make it 0 anyways.
    • GODSUN, on 10/12/2007, -7/+1Ridiculous!
    • WinterSolstice, on 10/12/2007, -5/+2All I can say is that I hope if I'm ever in a space craft that has to make a multi-billion km trip that the computer doesn't have a rounding error :)

      Of course, that's what course correction is for, but still.

      -WS
    • LucasOman, on 10/12/2007, -1/+8'scam/hoax'? (typos fixed)

      You're talking about this like it's an alien abduction headline on the National Enquirer. READ a pre-algebra text book. I took that class in 6th grade. This isn't difficult stuff.

      If you don't understand it, that's ok. But just like you don't look over the shoulder of surgeons and question what artery they're clamping, don't argue with people who actually know, study, and teach mathematics.
    • dorkafork, on 10/12/2007, -1/+30.1 = 1/10, 0.01 = 1/100. 0.000...1 is 1/(10^n) where n=infinity. The limit of 1/(10^n) as n goes to infinity = 0.

      1 - [ lim (n->inf) 1/(x^n) ] = 0.999...
      1 - 0 = 0.999...
      1 = 0.999...
  • grum, on 10/12/2007, -12/+3"I'm not a magician, I'm a mathemagician!"

    seriously though, my maths teacher would throw these maths tricks at us in class back in the day for the pure reason of opening up students to a healthy debate. he would stubbornly keep up the false notion that a certain formula or concept was correct and would let the students battle it out - much like the comments in this story, until months later revealing it was all just an elaborate mind bender.

    Is it just me or is the same thing happening here?
  • PowerCow, on 10/12/2007, -4/+15http://mathforum.org/dr.math/faq/faq.0.9999.html
    http://www.cut-the-knot.org/arithmetic/999999.shtml
    it is not like he invented this or pulled it out his butt.
    This is famous, well proven and well accepted
    find me one proof against the arguement... one not made up by a commenter.
    Do you know what it means to be infinately close to one? means there is no finite difference between the two. There is a meta phyisical one but not a finite one.
    • IQ70, on 10/12/2007, -13/+3To accept mathematics as it exists without challenge is to rot in the 5th BC.
    • mystagogue, on 10/12/2007, -0/+4@IQ70, the beauty of mathematics is that it's not a matter of opinion. it's fact. undebatable once it's proved or disproved.
  • manumitx, on 10/12/2007, -11/+2The value of .9999999 is worth a whole (1) only in situations that only deal with wholes. Otherwise the value is still that of less than a whole. Its a subjective thing based on the perception of what is needed for value in the system that the number is in.
  • spyrochaete, on 10/12/2007, -7/+12"Pi is exactly 3!" - John Frink
  • MrWashy, on 10/12/2007, -10/+1One question: how many significant digits should there be? If there are infinite sig digs, then .9999999... is not equal to 1. If you have less than infinite sig digs, then it rounds to one. In either case, with real quantities, .999999.... is not equal to one.

    Granted the difference is infinitesimal, but there's still a difference if you're dealing with measured quantities. Hence the problem with math and why many kids can connect with it - it deals with pure numbers, not measurements, and with pure numbers you get such things as .999... = 1.

    Oh before someone asks what kind of number has infinite sig digs, definitions do. 12 in = 1 foot, both the 12 and 1 are infinitely significant. Pi = 3.14159... is infinitely significant.
  • pxa270, on 10/12/2007, -3/+90.9999999 is of course not equal to 1.
    0.9999999... does equal 1 for most reasonable definitions of the dots part. Most people who think otherwise are just missing a precise notion of what dots or what infinity means.

    About anyone who passed Calculus 101 can agree that 0.99999... is loose notation for
    sum _{j=1} ^{infinity} 9^{-j}
    which in turn is defined as
    lim_{n to infinity} sum _{j=1} ^{n} 9^{-j}

    It's not very difficult to formally proof that this limit equals 1 (look up the epsilon-delta definition of limits).
    • pxa270, on 10/12/2007, -1/+6Whoops, those should of course be
      9 * 10^{-j}

      Anyway, the point is, those who think otherwise juse have no (or a very non-standard) concept of base 10 notation with repeating digits. There is no vague "approaches but never reaches 1" or "is infinitesimally smaller than 1'. The limit IS 1, no questions about it.
  • TriZz, on 10/12/2007, -11/+2On a long enough time line, the survival rate for everything becomes 0.
  • Pakman20, on 10/12/2007, -12/+2.9 repeating does not equal exactly 1 bacause there is always that last .000...01 to add, but for the purpose of any reasonable example, real-world use, or classroom, .99999... should be accepted as 1.
    • lmushl, on 10/12/2007, -2/+6You let me know when you get to the end of infinity to add 'that last .000...01'

      ...I'll prolly still be here...
    • Norweed, on 10/12/2007, -2/+7what don't you understand about infinity? There is no room to add anything to the end. THERE IS NO END. That's the point of infinity and why it equals 1.
    • pfunked, on 10/12/2007, -1/+6but that last .000...01 never appears. Instead, we have zeroes all the way. If you have 0.000.... going on forever, that is exactly the same as zero. (Thus, .999.... is exactly the same as one).
    • zentraedi, on 10/12/2007, -5/+2That sounds a little too much like "turtles all the way"
  • digitalsin, on 10/12/2007, -13/+2He jumps through a lot of hoops to try and prove his point but seems to fall short about .1
    • digitalsin, on 10/12/2007, -1/+8I stand corrected - now I've seen the light. I'm an idiot
  • endgames, on 10/12/2007, -11/+12+2=5 then
  • repins, on 10/12/2007, -11/+20 = 1 = infinity

    take the fraction 0/0
    1) zero over any number is zero
    2) any number over itself is 1
    3) any number divided by zero is undefined or infinity

    makes sense if you leave out a few details ;)
  • swilly, on 10/12/2007, -12/+2.999... could be close enough to 1 due to its infinitely small difference to be considered the same for just about any practical purpose, however viewing it from a technical perspective they are using completely different digits and one is only expressed as a decimal.

    long story short - so in practical application there isn't much of a difference, however to be mathematically correct they are not equal
    • KilgoreCarp, on 10/12/2007, -5/+1For every possible purpose using repeating .999... or 1 will make absolutely no difference. Long story short - it doesn't matter what you decide to use.
    • interiot, on 10/12/2007, -1/+6If it's mathematically correct that they're two different numbers, then name a number that lies between them. If they're different, there should be an infinite number of numbers in between, so it shouldn't be hard...
  • PissedGodzilla, on 10/12/2007, -13/+1He's Mixing up integral math with fractional math, it is flawed logic and it is crap.
    • interiot, on 10/12/2007, -1/+5So 1.0 isn't a rational number?
    • Norweed, on 10/12/2007, -3/+6How?

      There are NO numbers between .99999... and 1. Thus, they must be the same number. Tell me a number between .999999... and 1.
    • Couchy, on 10/12/2007, -3/+3you know what else is crap? your FREAKING BRAIN!
  • jgreene777, on 10/12/2007, -6/+4"Really???? Infinitely many zeros and then after the infinite list that never ends, there's a 1???? Surely that's stranger than the possibility that .9 repeating simply does equal 1. "

    Folks had a hard time with the concept of zero a long long time ago... perhaps this ".999999...1" deal is a concept we will learn to accept in the future.
  • 1c3d0g, on 10/12/2007, -14/+5Let me be the first to call ***** on this. 0.9999999 != 1, no matter how you ***** twist it.
  • Jeebugorn, on 10/12/2007, -9/+3that guy still didn't show an accurate fractional form of 0.999...he gave 53/53 as equaling 0.9... well, i admit that i am no math wiz...hell, i actually just finished an introductory algebra class in college last week. but i do know that to convert a fraction to a decimal, you divide the numerator by the denominator. that would be 53 divided by 53. any number divided by itself is 1. so 53/53 does not equal 0.9999999999.....
    • pfunked, on 10/12/2007, -2/+5If 0.999.... is another way of saying "1", then 53/53 definitely is 0.999...., as we see in the proof.
    • Jeebugorn, on 10/12/2007, -9/+5if frogs had glass butts they wouldn't hop...but they dont
    • mcdougrs, on 10/12/2007, -8/+2His argument doesn't work backwards. When you take 1/3 to mean .33333... and add that together 3 times you get 3/3 = .99999... BUT when you convert 3/3 back to "decimal form" it comes out to 1. Its kind of circular logic.
    • pfunked, on 10/12/2007, -1/+5> Its kind of circular logic.

      Transitive, not circular ;)
  • digitalsin,