0.999... = 1  Prove Me Wrong
Voting Style:  Open  Point System:  7 Point  
Started:  6/18/2018  Category:  Education  
Updated:  3 years ago  Status:  Post Voting Period  
Viewed:  864 times  Debate No:  115666 
0.999... = 1
 x = 0.999... 10x = 9.999... 10x  x = 9.999...  0.999... 9x = 9 x = 1  1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999...  What fraction/decimal number comes in between 0.999... and 1. Any two numbers that are different must have a difference of something. That something will be somewhere in between the two numbers Ex. 1.5 is in between 1 and 2  Put in 0. with 20 (it must be 20 or it may not work, 9's after it in a calculator and press =. It will say it equals 1. Each of these 4 arguments masquerading as proofs reflect a fundamental misunderstanding of numbers, algebra, and mathematics in general. I will speak to them one at a time: 1. a) x = 0.999... b) 10x = 9.999... c) 10x  x = 9.999...  0.999... d) 9x = 9 e) x = 1 This proof is a knockoff of that given by William Byers (also cited on Wikipedia) in 'How Mathematicians Think: Using Ambiguity, Contradiction, and Paradox to Create Mathematics (2007)'. What Pro perhaps fails to realize is that Byers did not intend this as a bona fide proof. Rather, he intended it as a trick to illustrate ambiguity and contradiction. "Students who did not accept the first argument sometimes accept the second argument, but, in Byers' opinion, still have not resolved the ambiguity, and therefore do not understand the representation for infinite decimals." It is in this category that I place Pro. There is a gross fallacy being committed here in the movement from step c to step d. One cannot simply subtract an irrational number from another and expect to get a rational number. In fact, mathematical laws tell us that given an irrational number x and another irrational number y, if x + y = a and x  y = b, it is not possible for both a and b to be rational. You can find the proof here, if you'd like: https://goo.gl... Suppose for a second that steps a through d are all correct and that 10x  x = 9x = 9. Here, to use the previous illustration, we will say that x = 10x and y = x. As has been proved by mathematicians far smarter than I, it is not possible for both x + y (10x + x) and x  y (10x  x) to be rational. Simplified, that means it is not possible for both 11x and 9x to be rational. Since Pro claims that 9x = 9 and thus x = 1, it must be 9x that is rational. Therefore, the only choice, based on a combination of mathematical proof and Pro's own allegations, is for 11x to be irrational. However, Pro used this "proof" to show that x = 1. If x = 1, then 11x = 11, which is rational. Since this is simply not mathematically possible, we have a proof by contradiction that x does not equal 1 (at least by this logic). 2. 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... This is, again, faulty logic. One cannot say that "1/3 = 0.333..." and then multiply each side by 2 to get "2/3 = 0.666". This is totally disregarding the nature of irrational, repeating decimals. In order to use this actual logic, most likely by induction, Pro would have to lay out steps like the following, placing ends to the unending decimals and showing that the same result follows. 2a. 1/3 = 0.3 2/3 = 0.7 3/3 = 1.0 Since 0.3 * 2 != 0.7, and 0.3 * 3 != 1.0, this does not work. 2b. 1/3 = 0.33 2/3 = 0.67 3/3 = 1.00 Since 0.33 * 2 != 0.67 and 0.33 * 3 != 1.00, this does not work. 2c. 1/3 = 0.333 2/3 = 0.667 3/3 = 1.000 Since 0.333 * 2 != 0.667 and 0.333 * 3 != 1.000, this does not work. I think you get the point... 3. What fraction/decimal number comes in between 0.999... and 1. Any two numbers that are different must have a difference of something. That something will be somewhere in between the two numbers Ex. 1.5 is in between 1 and 2 With all due respect, this is just silly. As an example  After 0.999999 there could be a 8. Then, clearly, there would be a number between it and 1 (0.9999998 < 0.9999999 < 1). To this, you, as an astute mathematician may say, "wait a second, you can't just remove the '...' at some point and replace it with a value!" To that, I would say that this point only holds if you can show that there is a fraction that leads to the number 0.999.... Surely, 1/3 = 0.333... but I would ask, what fraction equals 0.999... Of course, you cannot say that 1/1 = 0.999... because that would be using the conclusion to prove the conclusion, otherwise known as the logical fallacy, "begging the question." 4. Put in 0. with 20 (it must be 20 or it may not work, 9's after it in a calculator and press =. It will say it equals 1. The response to this should be evident to anyone who has ever taken basic math and used a calculator. This is simply a result of the limitations of whatever computation device/calculator you're referencing. Apparently, with this calculator, 20 digits beyond the decimal place is the point at which it chooses to round. On an old calculator, that might have been at the 3rd decimal place, or the 5th, or the 19th. Using a sufficiently old calculator, typing in 0.6 = would result in a 1. I don't think Pro wishes to argue that 0.6 = 1. I believe I have sufficiently taken each of Pro's 4 arguments/"proofs" and refuted each one using sound, established, proven mathematical and computational theory. I welcome any rebuttal or comments from the voting public. Cheers! 

I don't understand how proof 1 is faulty at all. Use common sense you can subtract an irrational number from an irrational number to get a rational number.
Ex. 9 +0.999... = 9.999...  0.999... = 9.999... Your website makes little sense. It depends on what those irrational numbers are. Are they all the same number such as 0.999..., do they have no patterns like pi, or do they, etc. It really depends. Here is a proof 9.999...  0.999... = 9 9 + (0.999...  0.999...). No you're wrong actually I didn't say 10x  x = 9x = 9 I said 10x  x = 9x x You're making this a lot more complicated than it has to be. What idiot says 0.333... doesn't equal 1/3. I didn't multiply I added. 0.333... + 0.333... = 0.666... It's very simple. Even if you did multiply 0.333 x 2 = 0.666... 2a. Makes absolutely no sense. How does 0.3 + 0.3 or multiplied by 2 = 0.7? 2a2. 1/3 = 0.3 2/3 = 0.6 3/3 = 0.9 Again, how does 0.3 x 2 = 0.7? 2b2. 1/3 = 0.33 2/3 = 0.66 3/3 = 0.99 2c2. 1/3 = 0.333 2/3 = 0.666 3/3 = 0.999 No, I don't get the point. I really think you're trying to misinterpret my proofs now. I'll make it toddlerly simple: 0.999... < x < 1 What is x? If you admit there is no answer then you admit 0.999... = 1. There is no fraction to represent 0.999... Just like there is no fraction to perfectly represent pi. Now this question is silly. For the sake of the debate let's use the same calculator aka Google's calculator. You can find it by typing "calculator" in Google or Chrome. I said may not work as I assumed 20 would be enough for any calculator to show one. I don't think it's a limitation rather an understatement. At a certain number of 9's the calculator understands that you don't mean 0.9, you mean 0.999... to infinity. It's simply the threshold between thinking the person means 0.9 or 0.999... to infinity. Google's calculator does not show 0.6666 as 1. I just tapped the 6 key for two minutes and it equaled 2/3, further proving 2/3 = 0.666666... I believe I have sufficiently taken each of Con's multiple arguments/"disproofs" and refuted each one using sound, established, proven mathematical and computational theory. I welcome any rebuttal or comments from the voting public. Cheers! P.S. My Email is aden.m.kamathatgmaildotcom I usually make it a habit to not engage in arguments with people who don't seem open to reason or alternate opinions, which clearly is the case here. However, as I'm fairly new to Debate.org, I'll press on. I ask that you kindly be sure to respond to all of the claims made herein. Please bear in mind that as the Affirmative, the burden of proof lays with you. As such, it is incumbent upon you to prove, beyond a reasonable doubt, that each and every one of my contentions is faulty for one reason or another, ideally using logical arguments.
"Use common sense you can subtract an irrational number from an irrational number to get a rational number" Yes, you can. But you can also get an irrational number. The law I referenced has to do with the relationship between the simultaneous addition and subtraction of the same two irrational numbers (x and y). To rephrase: Take an irrational number x and another irrational number y...if (x  y) is rational, then (x + y) must be irrational; conversely, if (x  y) is rational, (x + y) must be irrational. "No you're wrong actually I didn't say 10x  x = 9x = 9 I said 10x  x = 9x x" Two glaring problems here. 1) You did say 10x  x = 9x = 9. That is literally the definition of your steps c and d, as copied below: c) 10x  x = 9.999...  0.999... d) 9x = 9 2) Let's suppose you didn't, and that you actually said the mathematical statement you just posed in round 2: "I said 10x  x = 9x x" Let's do some basic algebra: 10x  x = 9x  x Simplify both sides 9x = 8x Divide by common factor (x) 9 = 8 I'm quite certain 9 != 8 and I don't think you or anyone on this forum believes that it does. So, not sure where this was going. "What idiot says 0.333... doesn't equal 1/3. " Despite the hilarious irony if you calling me an idiot, I also need to take exception to this point from the point of view of a person reading an English sentence. My contention was: "One cannot say that "1/3 = 0.333..." and then multiply each side by 2 to get "2/3 = 0.666"." In your attempt to discredit me as an "idiot", you conveniently left off the entire second half of my statement. This would be tantamount to saying that John Lennon once said, "You may say I'm a dreamer, but I'm not!" "I didn't multiply I added. 0.333... + 0.333... = 0.666... It's very simple." I think you must realize that adding a number to itself is quite literally the definition of multiplication. Regarding your whole attempt to refute point 2, I think you (again) missed or ignored a key part of the argument. Please note the part that says, "placing ends to the unending decimals", which may also be known as rounding. It seems this entire instance of logical induction was lost on you. I never said that 0.3 + 0.3 equaled 0.7. My entire case here is that it does not. If it did, then perhaps your argument would hold water. However, the whole point is that 0.3 + 0.3 != 0.7. More precisely, the point is that ROUND(1/3,1) + ROUND(1/3,1) != ROUND(2/3,1). This holds for ROUND(1/3,2) + ROUND(1/3,2) != ROUND(1/3,2) and, most importantly, ROUND(1/3,N) + ROUND(1/3,N) != ROUND(2/3,N). Let me know if you'd like an Excel file proving this. "I really think you're trying to misinterpret my proofs now. I'll make it toddlerly simple: 0.999... < x < 1 What is x? If you admit there is no answer then you admit 0.999... = 1." Despite the fact that "toddlerly" is not a word, I do appreciate you simplifying the argument so that someone of my lowly intelligence can understand it. I will attempt to respond with a marginally less condescending but, hopefully, more effective line of reasoning (or two). Marginally less condescending line of reasoning #1: Let's assume for just for a second that 0.999... is actually less than 1. If we can use deduction to start with that and show that there actually is a number between it and 1, your point will have been successfully (and irrefutably) defeated. Let's also use the variable h to represent an infinitely small number...the absolute smallest number you can think of. Perhaps this is (1e((1e100 * 1e100)*(1e100 * 1e100))). Actually, no, let's say it's one onequadrillionth of that. No, wait...one onequadrillionth of THAT. We could keep going ad infinitum here, but you get the point. 1) 0.999… < 1 [This is starting with the assumption that 0.999... != 1] 2) 0.999… + h = 1 [h is that tiny, tiny, little, infinitely small number that makes up the gap] 3) 0.999… = 1 – h [This says that we can take 1, subtract an infinitely small number from it, and get 0.999...] Marginally less condescending line of reasoning #2: This type of thinking borrows from discrete math and number theory, the same exact theory used to decide if sets are countably infinite or not (e.g. rational numbers vs. irrational numbers). Since searching for the smallest number possible is the same as searching for a number composed of an integer that follows an infinite number of zeros, it is perfectly suitable that we use the same type of thinking to show that there is a value between 0.999... and 1. I would ask that you give me a particular decimal string starting with 0.9 and continuing to the right with as many 9's as you'd like. Go ahead, I'll wait. For any number of 9's you give me, I will be able to give you a number that is between your number and 1. If you had all the time that existed in the history of the universe, you still would not be able to complete such a task. As the burden of proof lies with the Pro, I would ask how it can be that you allege this is true beyond a reasonable doubt if you cannot pass a simple test such as this. "For the sake of the debate let's use the same calculator aka Google's calculator. You can find it by typing "calculator" in Google or Chrome..." First of all, you don't need to type "calculator" into Google or Chrome. You can just type the math equation and it will calculate. To the rest of this point, I am trying my best to explain that reaching a computational limit is not the same as reaching a mathematical fact. I think we can show this very clearly with a basic example: If you were to type, into Google, 0.88^9=, you would get 0.31647838182. However, using just an eversoslightly more powerful calculator, you would find that it actually equals 0.316478381828866. I think this should be a very simple proof of the fact that Google calculator should not be taken as the ultimate truth provider. In case you have doubts or think this is merely anecdotal, I would provide a few more examples:
Why are these examples particularly relevant? The first two examples reflect cases where not only are Google's answers truncated, they don't even round properly. If the answer to the first problem is going to be cut at the 11th decimal place, it should be 0.37439308349, not 0.37439308348. If the answer to the second problem is going to be cut at the 11th decimal place, it should be 0.49125890426, not 0.49125890425.
The third example reflects Google's limited ability to deal with irrational numbers. I don't think anyone would deny that 3 * pi is a number that goes on infinitely. However, Google is not able to in any way indicate or reflect this. Rather, the answer is again truncated at the 11th decimal place. "I believe I have sufficiently taken each of Con's multiple arguments/"disproofs" and refuted each one using sound, established, proven mathematical and computational theory. I welcome any rebuttal or comments from the voting public." This comment is clearly on par with the general condescending nature of the rest of the argument. Though it would surely be inappropriate to speak this way in any case, it seems particularly imprudent to do so after using repeated arguments devoid of understanding of the underlying topic...especially when it's one chosen by the writer himself. 

As I've said so many times to so many people, 0.999 and 0.999... are different numbers. GO figure out yourself, I don't have time to explain it again.
1. As I've said before it depends what x and y are that determine if you can subtract or add them 2. I'm sorry I didn't know you simplified the problem to 10x  x = 9x = 9. But yes if you plug in x you will see this equation is true. 3. Sorry I mistyped that I meant 10x  x = 9.999...  0.999... 4. Once again you can't take away the ... it simply means the number goes on with 9's forever. 0.3 + 0.3 = ).6 how could it equal 0.7, please show me a proof. Give me the Excel please. By the way then lets not round. 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... There it goes on forever. The way people get is they ROUND. 1/3 + 1/3 = 0.333... + 0.333. = 0.666... Since you round up the 6 becomes a 7 giving you 0.666...7. 5. This proof of yours makes no sense. Exactly you admit you couldn't give me a number as you gave me an "h". What is "h"? I don't think you understand the concept of infinity as there is no "tiny, tiny, little, infinitely small number" to make up the difference as ... means it goes on forever. Forever doesn't have a gap. 5a. Your number that you are waiting for is 0.9999... which is equal to 0.999... the ... means it goes on infinitely. The number of nines in ... is so much it's not even a number it's a concept. 6. Okay fine don't use Google use any calculator you want it says 0.999999999... equals 1. The calculator really doesn't matter. I would like to kindly ask that Pro remove his/her condescending tone and aggressiveness from this debate. It has no place in logical discourse and is neither productive nor appropriate. But I digress...let's go step by step here. Note, I think Pro mislabeled their counterarguments here in terms of numbering, so I will try my best to match them up with mine.
"1. As I've said before it depends what x and y are that determine if you can subtract or add them" I am not talking about whether or not x can be subtracted from y, I am calling attention to an established theorem that regards the results of subtracting one irrational number from another. This has nothing to do with determining "if you can subtract or add them." This statement reflects a blatant misunderstanding of set theory. One can always add or subtract irrational numbers. The issue at hand is whether the result(s) are rational or irrational. You alleged that I should, "use common sense (to see that I) can subtract an irrational number from an irrational number to get a rational number." Yes, you can, but that doesn't mean you will. In fact, if you add two irrational numbers together and get an irrational sum, you can never get an irrational number when taking the difference between those two numbers. "2. I'm sorry I didn't know you simplified the problem to 10x  x = 9x = 9. But yes if you plug in x you will see this equation is true." I'm not sure we can possibly have a reasonable discussion if Pro thinks that "if you plug in x (here) you will see this equation is true." Rather, if you plug in ANY x, you will see that this equation has no solution. Let's approach this algebraically then using specific examples. 10x  x = 9x  x Simplify both sides 9x = 8x Divide by common factor (x) 9 = 8 I'm quite certain 9 != 8 and I don't think you or anyone on this forum believes that it does. So, not sure where this was going. Using a few examples:
No matter how you slice it, there is no value you can substitute for x (except for 0, which would be the trivial case) that makes this statement true. If Pro feels that this method of reasoning is proof that 0.999... = 1, I fear he/she is sorely mistaken and completely detached from elementary math. As for the Excel, I don't know if I feel comfortable sending you something from my personal email address, but it should be abundantly simple for Pro to open up Excel and verify this in less than 10 seconds. Set Theory
I think it is absolutely vital that Pro obtains even a basic understanding of these concepts before condescending toward me that my proof "makes no sense" and that I don't "understand the concept of infinity". "5a. Your number that you are waiting for is 0.9999... which is equal to 0.999... the ... means it goes on infinitely. The number of nines in ... is so much it's not even a number it's a concept." Again, please see my comments on derivatives, limits, and set theory as a complete lack of understanding thereof makes it difficult to have this conversation. "6. Okay fine don't use Google use any calculator you want it says 0.999999999... equals 1. The calculator really doesn't matter." I'm glad Pro has finally conceded that one of their arguments was completely and utterly invalid. Since Pro asks me to "use any calculator (I) want", I will do so. Here are a list of just a few calculators/computers that will not show 0.999... equals 1 after at least one million9's:
I'm hoping the voting public can see I've properly deflected Pro's generallydisrespectful attitude and also demonstrated his/her lack of a fundamental understanding of the very mathematical concepts they are invoking in addition to those that I am.
Looking forward to the next round! 

How can 10x  x = 9x  x? If I said that I'm sorry but that's simply not possible. 9x  x is always something less than 10x  x.
Ex. x = 2 20  2 = 18  2 18 = 16 You are right no proof shows 0.3 + 0.3 = 0.7 I'm saying that you got 0.7 by rounding 0.333... + 0.333... What proof do you have that h exists. I could say 1 < h < 1 I checked in those calculators you gave me and they all showed 0.999... = 1. PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE watch this wonderful video, it will make everything super clear. You said "10x  x = 9x x" in Round 2. Also, it is not at all true that "9x  x is always something less than 10x  x". Simply set x equal to any negative number and you will see that 10x  x is always less than 9x  x. I don't mean to sound like a broken record, but Pro's complete lack of understanding of such basic mathematical concepts as the one(s) that would make obvious the falsity of the statement regarding (9x  x) always being less than (10x  x) is making this quite a difficult conversation.
I don't know exactly how else to explain the proof by contradiction that was based on the entire fact that 0.3 + 0.3 != 7. I'd urge you to actually go through the resources I provided as it should be very valuable to you in future debates. I am not sure I at all understand what you mean when you say, "what proof do you have that h exists(?) I could say '1 < h < 1'" The whole point of using a variable like h or x is to illustrate a later contention. Stating that h exists is an assumption, in the mathematical, logical sense, used to prove a point. I take it you did not view the documentation on the proof of derivative differentiation using limits, which is the basis of the entire realm of calculus. If you checked those calculators I gave you, you either used them incorrectly or entered too many 9's. That is not to say there is a given number of 9's that make 0.999... actually equal 1, it's just a limitation of any given calculator. Precision of calculators are specifically defined by the number of digits to which they compute. Given infinitely long precision/computing power, you will never get that result. I don't know if it's fair for you to ask me to watch your "wonderful" video that will "make everything super clear" without you having pursued ANY of the resources I provided, but you did say "please" in all caps 15 times which makes me realize that you are significantly desperate to prove a point with someone else's words that you cannot with your own. Anyway, let's look at each of the "proofs" offered by this video:
I understand Pro was unable to defend his/her very own claims after offering many illogical, selfcontradictory views. When this happens, it is not uncommon for debaters to try and use other people's words in their own defense. I'm not sure that a YouTube video should qualify as irrefutable proof of anything, but this particular video and its accompanying invalid arguments should definitely not be factored into any intelligent argument.
Have a great day! 

I now think you"re trying to not accept 0.999... = 1. To be blatantly honest you"ve just beaten around the bush this whole argument.
For the algebra x = 0.999... whatever Your making it so much more complicated that it had to be. All I asked you to do was disprove that this proof couldn"t work, or is invalid. For 1/3 = 0.333... whatever you said that "2/3 = 0.667" in round 1. For what number comes in between whatever... You gave me h. Yes h. This clearly shows you can"t find an actual number that comes in between so you gave me a variable to cover up the fact that you couldn"t come up with an answer because THERE IS NONE! And you"ve completely dodged and ducked the calculator thing. Yes I did in as you put it "put too many nines". There is no such thing as too many nines as it infinetelty repeats nines. Ask any mathematician about this and they will in fact tell you that 0.999... = 1. At this point, I don't know how much more disrespectful, grammaticallyincorrect rambling I can sort through since it is clear that this is not going anywhere.
To anyone voting, I will say:  There are many things that can be focused on in this debate. Most notably, I feel, is Pro's repeated demonstration of a complete lack of understanding of even the most basic mathematical concepts...the very mathematical concepts on which he/she bases his/her argument.  I think it has become more than abundantly clear that Pro does not have a single original thought of their own and is simply lifting arguments from outside sources without giving credit.  Further, and as a result of this, Pro is unable to absorb, understand, and refute even the most basic challenges to these arguments that he/she did not come up with in the first place Pro's repeated use of aggressiveness, invocation of condescending language, and flippant disregard for logical discourse have cast a negative shadow upon the entire debate Thanks for your time!

AKMath  BertrandsTeapot  Tied  

Agreed with before the debate:      0 points  
Agreed with after the debate:      0 points  
Who had better conduct:      1 point  
Had better spelling and grammar:      1 point  
Made more convincing arguments:      3 points  
Used the most reliable sources:      2 points  
Total points awarded:  3  0 
I'd appreciate anyone who can put aside any preconceived notions and take a look at purely the quality of debate here.
Oh and I saw one of AK's other debates with Bertrand where she said something like "P.S. Good luck in the 0.999 debate." I don't really see how that type of arrogance is cool, but especially when she is embarrassing herself this much it really is stupid.
Good day.