Nmenth 289 Posted November 15, 2009 I just found this amusing equation: x = …999 Then, 10x = …990 So, 10x = x − 9 x = −1 Therefore, if 0.999… = 1 and …999 = -1, …999.999… = 0 And here's a joke for you all: How many mathematicians does it take to screw in a light bulb? A: 0.999…. Share this post Link to post

Guest Stevie_K Posted November 15, 2009 I was never good at maths... but I don't understand the concept that different numbers must have a number in between??? Can I have the george-bush-for-president simple answer? I don't get this either. I just found this amusing equation: x = …999 Then, 10x = …990 So, 10x = x − 9 x = −1 Therefore, if 0.999… = 1 and …999 = -1, …999.999… = 0 And here's a joke for you all: How many mathematicians does it take to screw in a light bulb? A: 0.999…. LoL Share this post Link to post

Inferno 22 Posted November 15, 2009 (edited) I was never good at maths... but I don't understand the concept that different numbers must have a number in between??? Indeed. 0.999 is simply the closest thing possible to 1, but that still doesn't make it 1. Also 1/3 isnt 0.3333. It's simply the closest thing to 1/3. We can't write 1/3 precisely in numbers. Edited November 15, 2009 by Inferno Share this post Link to post

kastro 0 Posted November 15, 2009 (edited) You misunderstand our purpose here. We are not trying to prove that 1 ≠ 0.999…, but rather are trying to disprove that 1 = 0.999…. I know, it sounds like a contradiction of rephrased words, but it is nevertheless true. These are equivalent. I.e., I say it is law that infinity cannot be multiplied by anything other than 1 and 0, which your side refuses to acknowledge because you claim that infinity is not comparable to infinite repetition. And it is not. An infinite sum may well converge to a finite number, and hence is not at all comparable to 'infinity' in the unbounded sence. In any case, infinity is not a real number, and operations on it are not defined so multiplying it by anything does not make sense. Your side says that it is law that if there is no possible other number between two existent numbers, they must be equal, which our side refuses to acknowledge because this is an arbitrary rule that only exists to prove your point (I realize I make your argument sound stupid, but I am not writing unbiased here, you know). Again, this is precisely part of the definition of the real numbers. You cannot 'refuse' to acknowledge it. So what laws are true? Well, some mathematicians may be so arrogant as to say that the universe conforms to the laws of mathematics. This, however, is false. It is mathematics that conforms to the laws of the universe. The universe is what it is, and math cannot change it. Math is a concept and the universe can change it. Therefore, the laws of mathematics can only be shaped by observations of reality and has no truth in of itself, but is subjected to the universal truths. Because humans are not omniscient and are quite fallible and not one of us has seen infinity, it is hard for us to form the mathematical rules governing it. Don't really know where you're going with this. Mathematics isn't physics, and doesn't attempt to explain real-world phenomena. I was never good at maths... but I don't understand the concept that different numbers must have a number in between??? Can I have the george-bush-for-president simple answer? Given any two distinct numbers, the average of the two exists, and lies strictly between them. I just found this amusing equation: x = …999 Then, 10x = …990 So, 10x = x − 9 x = −1 Therefore, if 0.999… = 1 and …999 = -1, …999.999… = 0 So basically x was defined to be infinity. It should be obvious why this argument is false. Indeed. 0.999 is simply the closest thing possible to 1, but that still doesn't make it 1. Also 1/3 isnt 0.3333. It's simply the closest thing to 1/3. We can't write 1/3 precisely in numbers. No such 'closest number' exists--there's always a closer one. Edited November 15, 2009 by kastro Share this post Link to post

sith_wampa 6 Posted November 16, 2009 (edited) If you have not already read them, I believe these articles (0.999... and Real Numbers and Decimal Representation) provide valuable information. It's important to understand the idea that a term 0.333... can be understood as the fraction 1/3. Considering that the threes in 0.333... do in fact continue infinitely (although they are not written), it is mathematically equivalent to 1/3. That is fact. If you cannot understand that, I am sorry. Now keeping in mind that 0.333...'s digits continue infinitely, it is possible to multiply 0.333... by 3. Considering 3x3=9, multiplying 0.333... by 3 will change our infinite number of threes into an infinite number of nines, now 0.999.... When we multiply 1/3 by 3, we get 1. This allows us to say that when we multiply 0.333... by 3, we get 1. 0.333... multiplied by three also gives us 0.999..., which is simply another notation for the value of one. I know this is a walkthrough of one of the proofs presented before. Hopefully this will appeal to your "logical" reasoning, as you don't seem to have any mathematical reasoning for your claim. The mathematical evidence for 0.999... equaling 1 has been provided. You can either provide your own mathematical evidence to counter that, or accept it. Or you could just keep denying it. You are entitled to be wrong. Edited November 16, 2009 by sith_wampa Share this post Link to post

kastro 0 Posted November 16, 2009 Now keeping in mind that 0.333...'s digits continue infinitely, it is possible to multiply 0.333... by 3. Considering 3x3=9, multiplying 0.333... by 3 will change our infinite number of threes into an infinite number of nines, now 0.999.... Nmenth's objection to this is that infinite sums cannot be multiplied by anything. This is in general true, and in order to make sense of multiplication on infinite sums we require an additional requirement, that it be convergent. In our case, 0.999... certainly converges. Share this post Link to post

sith_wampa 6 Posted November 16, 2009 Nmenth's objection to this is that infinite sums cannot be multiplied by anything. This is in general true, and in order to make sense of multiplication on infinite sums we require an additional requirement, that it be convergent. In our case, 0.999... certainly converges. You're right. But knowing that 0.333... and 1/3 are the same value, we can determine that 0.333... multiplied by 3 is 1. Share this post Link to post

Nmenth 289 Posted November 16, 2009 These are equivalent. They really aren't, I explained why. Apparently you adhere to the 'since all mammals are animals, then all animals must be mammals' logic. So basically x was defined to be infinity. It should be obvious why this argument is false. I didn't invent it, it is called a p-adic number and is used by mathematicians in number theory. If you accept that 0.999... is a valid number to be used in equations, then so is ...999. It is a known proof that ...999 = -1 in the same why that it is a known proof that 0.999... = 1. Since ...999.999... = 0 doesn't make sense, it is basically ignored when working with both p-adic and real number equations. In other words, even the mathematicians admit that their proof doesn't make sense, but for some reason, that isn't good enough to counter-prove that their equations are false. Of course, I don't believe 0.999... = 1, so naturally, I don't believe ...999 = -1 or ...999.999... = 0 either. Share this post Link to post

kastro 0 Posted November 16, 2009 They really aren't, I explained why. Apparently you adhere to the 'since all mammals are animals, then all animals must be mammals' logic. I didn't invent it, it is called a p-adic number and is used by mathematicians in number theory. If you accept that 0.999... is a valid number to be used in equations, then so is ...999. It is a known proof that ...999 = -1 in the same why that it is a known proof that 0.999... = 1. Since ...999.999... = 0 doesn't make sense, it is basically ignored when working with both p-adic and real number equations. In other words, even the mathematicians admit that their proof doesn't make sense, but for some reason, that isn't good enough to counter-prove that their equations are false. Of course, I don't believe 0.999... = 1, so naturally, I don't believe ...999 = -1 or ...999.999... = 0 either. Not quite. I'll explain why, but to be honest if you dispute that 0.999... = 1 (and I promise you there isn't a single PhD-holding mathematician in the world that disagrees with this) then you don't really have any right to even know what p-adics are. In any case, 0.999... isn't even 10-adic, since p-adics MUST have a finite set of non-zero integers to the right of the p-adic 'decimal'. But even if this result were valid, it would hold true only for the field of 10-adics; attempting to extend it to the reals would be rediculous. Share this post Link to post

Nmenth 289 Posted November 16, 2009 then you don't really have any right to even know what p-adics are. Oh, I didn't realize I wasn't allowed to know things. I guess I'll have to scrub it from my brain. What's the penalty for knowing things I'm not supposed to? Am I looking at prison time time here? Does this fall under mathematical espionage or something? Share this post Link to post

kastro 0 Posted November 16, 2009 (edited) Sorry, not meant to offend--but yes, it might be close to mathematical espionage. To be honest I know next to nothing about p-adics; any extensive study of them is probably reserved for an advanced topics grad-level course. Edited November 16, 2009 by kastro Share this post Link to post

Remomeister 0 Posted November 30, 2009 "It is better to remain silent and be thought a fool, that to post a meaningless statement and remove all doubt." All numbers that are not exactly the same, such as 1 = 1.0, have an infinite amount of numbers between them. As for an average between 1.0 and .999... , consider that the average between 1.0 and .9 is .95 and between 1.0 and .99 is .995 and so between 1.0 and .999... is .999...5 nuff said. Share this post Link to post

Guest Stevie_K Posted November 30, 2009 "It is better to remain silent and be thought a fool, that to post a meaningless statement and remove all doubt." All numbers that are not exactly the same, such as 1 = 1.0, have an infinite amount of numbers between them. As for an average between 1.0 and .999... , consider that the average between 1.0 and .9 is .95 and between 1.0 and .99 is .995 and so between 1.0 and .999... is .999...5 nuff said. I think it's opposite. Say something meaningless and be thought a fool. Good point however. So you would have to put that 5 where? At the end of an infinite decimal? Theoretically yes, and therefore your'e right in my opinion. (even though it's not really possible it theoretically is). I'm glad I'm not the only one who believes 1=1. And 0.999 = 0.999. Share this post Link to post

Doctor Destiny 41 Posted December 1, 2009 "It is better to remain silent and be thought a fool, that to post a meaningless statement and remove all doubt." All numbers that are not exactly the same, such as 1 = 1.0, have an infinite amount of numbers between them. As for an average between 1.0 and .999... , consider that the average between 1.0 and .9 is .95 and between 1.0 and .99 is .995 and so between 1.0 and .999... is .999...5 nuff said. Your logic is easily refutable. 0.999... means these nines go on without an end. There is no place for that "5" to go. Hence the reasoning is easily refuted with simple mathematics. Why does everyone forget the simple premise of infinitely many places? Share this post Link to post

Nmenth 289 Posted December 1, 2009 "It is better to remain silent and be thought a fool, that to post a meaningless statement and remove all doubt." An excellent stand-alone quote... when followed by a statement in which the validity of its context is refutable... not so good. Your logic is easily refutable. 0.999... means these nines go on without an end. There is no place for that "5" to go. Indeed, it cannot be written 0.999...5 anymore than 0.000...1 can be. Everyone ought to know neither are a legitimate number. However, 0.999... r5 or 0.000... r1 seem perfectly acceptable to me. I really wish this grave had not been unearthed though... I guess it needs to be whacked with a shovel a few more times. Share this post Link to post

Quadhelix 0 Posted December 2, 2009 If 1/3 = 0.333..., multiply each side by three. Wouldn't 3/3 = 0.999...? In other words, wouldn't 1 = 0.999...? No, you cannot multiply ∞ times 3. What you seem to fail to understand is that 0.333... is not "infinity." Indeed, infinity isn't even a number - it's more of a limit. The idea of infinity is that, if I have some number N, I can always find some other number N+1. Since, for example, 5 is greater than 0.333..., it is obvious that 0.333... is not "infinity." In any case, 0.333... is the same thing as the infinite series (0.3+0.03+0.003+...) It is quite obvious that none of the entries in that series are "infinity," even using your retarded idea of infinity. As for it being an infinite series, multiplying an infinite series by a number is the same as multiplying each entry in that series that number. In any case, it is easy to see that 3*0.333... = 3*(0.3+0.03+0.003+...) = (3*0.3+3*0.03+3*0.003+...) = (0.9+0.09+0.009+...) = 0.999... 1 - 0.999… = 0.000… because the zeroes are infinitely repeating, there is no place to put the 1 in so therefore, the answer would be 0 (just like how 1 - 1 = 0) Key words here: there is no place to put the 1. So you acknowledge there is supposed to be a 1? There is no 1, because there is no end to ∞... So you acknowledge there is supposed to be a 1? You just asked that... You just evaded answering that. Whatever, so there’s an imaginary 1, you still can’t put it anywhere. The 1 is as imaginary as the ‘fact’ that it does not exist. I do not argue the one can go on to the end of infinity, obviously it cannot, but you also cannot ignore its existence. You are rounding 0.000…1 to 0.000… because 0.000…1 isn’t a real number. However, just as true as it is not a real number, as there is no end to infinity, it is also true that 0 ≠ 0.000…1 no matter how you slice it. Actually, your refutation is quite self-evidently wrong. Take, for example, 0.99. This is obviously 1-0.01. Now take 0.999, which is obviously 1-0.001. It is obvious that you can extend this as far as you want: 0.99...99=1-0.00...01. Note, however, the way that I wrote that: any string of nines is 1 minus a string of zeroes with a 1 as the last digit. Here, however, is where your argument breaks apart: if the string is infinite, there is no last digit - that's what "infinite" means. Therefore, there is no 1 at the end of the string of zeroes: 0.999...=1-0.000..., with no rounding. Share this post Link to post

Nmenth 289 Posted December 2, 2009 your retarded idea of infinity. Choose your words wisely, I created this thread and I will take it away if the debate can't remain civil. Share this post Link to post

Quadhelix 0 Posted December 3, 2009 (edited) Choose your words wisely, I created this thread and I will take it away if the debate can't remain civil.Fair enough - I got emotional and I should have put that better. In any case, my point remains. You claim that you cannot multiply 0.333... by any number other than 1 or 0 because 0.333... is "infinity." However, this is wrong and I can prove it: 0.333... can be rewritten as the infinite series [sum from n=1 to ∞] 3*10^-n. Each element 3*10^-n is obviously finite, and can thus be multiplies by whatever real (or complex) number you desire. However, the transitive property states that multiplying each element of the series by a given number is the same thing as multiplying the entire series by that number: (N*a+N*b+N*c...)=N*(a+b+c+...). Therefore, since we can multiply 3*10^-n by 3 to get 9*10^-n, we can also multiply 0.333... by 3 to get 0.999..., which rebuts your rebuttal of the 1/3 argument. When multiplying 0.333… by 3, you will keep getting repeating 9’s, however, look over to the next decimal, what is that? Why, it’s another 3! The problem can never be completed because there will always be another 3 to be multiplied (the answer will continue to be 9 forever, but is not the final answer because there will always still be one more three to be multiplied). Looks like 0.999… to me... That’s because you are stuck looking at ∞ as if it were a finite number, there is no end to the 3’s all equaling 9, which appears to equal 999…, but as long as there are still 3’s to be multiplied, the problem is not solved. You are looking at the infinite answer while ignoring that it is an infinite problem. I think that I see where you're going with this: you are implying that, if you were to multiply the "last 3" in 0.333..., that it would be as though the "last 'digit'" were a 10, cascading through up to 1.000... However, that is the same thing as saying that an infinite string of nines (0.999...) is equal to 1. Therefore, you are assuming 0.999...=1 in order to rebut a proof of 0.999...=1. Edited December 3, 2009 by Quadhelix Share this post Link to post

Remomeister 0 Posted December 7, 2009 There are always two sides to every argument. Both sides will defend their position to the death, no matter how you prove the other side wrong. If I drop a rubber ball from 12 meters, it drops, hits the ground and ascends to 6 meters, drops, hits the ground and ascends to 3 meters, etc. there will be those who are foolish enough to argue that the ball will be bouncing forever, always half of its previous height. Those with true intelligence, will not base foolish opinions on what others have attempted to convince them, but will use true logic. Hence, if that same ball were dropped from 12 meters and it took 2 seconds to travel down to the ground and back to the acme of 6 meters, then 1 second to drop to the ground and bounce back to 3 meters, etc. it is quite obvious that the ball will stop bouncing before it ever gets to 4 seconds, a far cry from bouncing forever. If you have one pie, and three hungry mates to share it, they will readily slice the pie into three thirds, and all will be satisfied. However, if someone were to say it must be divided exactly decimally, two mates would probably punch his lights out and cut it in half. In other words, fractions are real, and decimals are mathematical imaginations. Cutting a pie into thirds is practical and useful, but where would one find practicality in dividing something into three .333... pieces? You won't be stopping within one crumb, one grain of sugar, one molecule of fruit, one atom of carbon, one electron of hydrogen, or even one third of a quark. So. if a number say, .333...and one third, is multiplied by 3 to get 1.0, you are multiplying a 1/3 at the end of your number to get a whole number, but if you multiply .333... by three, it is not the same thing to start with, and will only equal .999... and not 1.0 because you dropped off the "one third part". 3 times one third will equal 1.0 because fractions are real. but 3 times .333... will never equal 1.0 because it is purely imaginary and leaves off the essential fraction. Share this post Link to post

Quadhelix 0 Posted December 8, 2009 If you have one pie, and three hungry mates to share it, they will readily slice the pie into three thirds, and all will be satisfied. However, if someone were to say it must be divided exactly decimally, two mates would probably punch his lights out and cut it in half. In other words, fractions are real, and decimals are mathematical imaginations. Cutting a pie into thirds is practical and useful, but where would one find practicality in dividing something into three .333... pieces? You won't be stopping within one crumb, one grain of sugar, one molecule of fruit, one atom of carbon, one electron of hydrogen, or even one third of a quark. So. if a number say, .333...and one third, is multiplied by 3 to get 1.0, you are multiplying a 1/3 at the end of your number to get a whole number, but if you multiply .333... by three, it is not the same thing to start with, and will only equal .999... and not 1.0 because you dropped off the "one third part". 3 times one third will equal 1.0 because fractions are real. but 3 times .333... will never equal 1.0 because it is purely imaginary and leaves off the essential fraction. This entire argument is based off of one faulty assumption: that 1/3 is something different from 0.333... That assumption is incorrect: the truth is that 0.333... is just another way of writing 1/3. I'm going to assume that you know long division, so take 1 and divide it by 3. The first digit is zero, of course, since 3 is greater than 1. To find the next digit, you divide 10 by 3, which means that the next digit is three and you have a remainder of 1. Therefore, going to the next digit, you take that remainder, "pull down" the 0, and again divide by three, again finding that this digit is 3. It soon becomes apparent that, no matter how many times you repeat this operation, the next digit is always going to be 3. This means that the end result is going to be an infinite string of threes. This is written as 0.333... Now, both you and Nmenth mentioned the possibility of including a remainder on the number. This, however, makes no sense. Remainders only exist within the context of incomplete division: if you have a remainder, you have not finished dividing. Share this post Link to post

Inferno 22 Posted December 9, 2009 This entire argument is based off of one faulty assumption: that 1/3 is something different from 0.333... That assumption is incorrect: the truth is that 0.333... is just another way of writing 1/3. It's not the same. 0.33333 is merely a round-up of 1/3. Even with infinite decimals, it's still not completely accurate. Share this post Link to post

Remomeister 0 Posted December 12, 2009 I'm going to assume that you know long division, so take 1 and divide it by 3. The first digit is zero, of course, since 3 is greater than 1. To find the next digit, you divide 10 by 3, which means that the next digit is three and you have a remainder of 1. Therefore, going to the next digit, you take that remainder, "pull down" the 0, and again divide by three, again finding that this digit is 3. It soon becomes apparent that, no matter how many times you repeat this operation, the next digit is always going to be 3. Up to this point, you are absolutely correct. Just as 100 divided by 3 equals 33 and one third. 10 divided by 3 equals 3 and one third. 1 divided by 3 equals .3 and one third. In place of “and one third” you can say, “remainder one”. When you have an infinite string such as .333… , you technically drop the “and one third” part of the problem. If you try to multiply your infinite string of .333… by 3, (if it were possible to multiply an infinite string), you would get an infinite string of nines after the decimal. Now, both you and Nmenth mentioned the possibility of including a remainder on the number. This, however, makes no sense. Remainders only exist within the context of incomplete division: if you have a remainder, you have not finished dividing. Then you contradict yourself. In order to equal 1.0, you have to start your problem by acknowledging the “and one third”. Therefore, .333…and one third, times 3, equals .999…and three thirds. Tah dah, you now have something that does indeed equal 1.0 just as Nmenth pointed out in his proofs way back on page one of this topic, and just as you yourself admitted to in the first part of your tirade. .999… does not equal 1.0, but .999…and three thirds does equal 1.0 plain and simple. (Inferno, above, is absolutly correct)! Share this post Link to post

sith_wampa 6 Posted December 12, 2009 (edited) Wtf? .999... does not equal 1, but .999... + 3/3 (or 1) equals 1? You sir are bat**** insane. Edited December 12, 2009 by sith_wampa Share this post Link to post

comander starlin 6 Posted December 12, 2009 according to the computer it is probable at this many .3x32 0.33333333333333333333333333333333 so in the first part the equation supplied there is no physical way to get 1. Share this post Link to post