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This page was the subject of an MfD discussion closed on 12 May 2007, with a keep result. Xoloz 18:36, 12 May 2007 (UTC)
OK... I learned about 0.(9) being the same as 1 in high school too...
But now I have this question.
1 = 0.99999... right?
2 = 1.99999... every ok so far.
does this mean that
0.99999... = 0.999999999...888... ?
or, to put it into words, is 0.(9) the same as 0. followed by infinite 9s AND infinite 8s? is that even possible?
or, even without knowing what number it would be, one could argue the following:
Since a number, like 1, is equal to the infinitely slightly "previous" number, there are only 3 numbers: zero, an infinite positive number and an infinite negative number.
Think about it... every number has an infinitely small 'equal'. At the same time, this 'equal' should have its own infinitely small equal, and so on...
Comments? —Preceding unsigned comment added by 201.246.25.69 ( talk) 05:00, 18 January 2010 (UTC)
You have to be sure though when claiming that it is one, to point out that it is really (1 - .0000...1)
While the difference between .999... and 1 is infinitely nothing, it cannot be dismissed because it is everywhere.
.000...1 is the first thing there is greater than zero, and it's between every change between every two numbers all the way up to one. It's everywhere, but it's nothing. I believe that it is the
graviton number, but I cannot prove it. --
Neptunerover (
talk)
16:31, 21 January 2010 (UTC)
This is from the infinitesimal article: "In the 20th century, it was found that infinitesimals could also be treated rigorously. Neither formulation is wrong, and both give the same results if used correctly." My suggestion is that if both formulations work, maybe one formulation might work better for certain calculations. Using a number set that contains all infinities is better than one with no upper limit to contain anything, at least when you have to deal with infinities. In figuring out gravity, they might be better off using a set with limits, although still infinite. -- Neptunerover ( talk) 06:05, 22 January 2010 (UTC)
How does (.999...9998) differ from either (.999...) or (.999...9997)? -- Neptunerover ( talk) 09:22, 23 January 2010 (UTC)
There's the old story of "How long would it take to get to X if each day you traveled 1/2 of the remaining distance between you and X?" According to the question itself though, each day when you are through traveling, there will still be a remaining distance to the 'goal' equal to that which you just traversed that day. I don't see how this could be any different if you went 9/10 of the distance each day rather than only half way, other than getting closer faster, because the entire distance to the goal is still never spanned. I suppose that's very different than going 9.999.../10 of the distance each day. In that case the second day would be a very puzzling day. -- Neptunerover ( talk) 12:19, 26 January 2010 (UTC)
Neptunerover - there is no difference really. But I'm not sure what conclusion you're trying to draw? The halfway example presents a geometric series: 1/2 + 1/4 + 1/8th + 1/16th etc...and is equal to 1. Do you contest that? Because the same is true for 9/10: 9/10 + 9/100 + 9/1000.... = 1. Your specific application with walking I think is related to http://en.wikipedia.org/wiki/Zeno%27s_paradoxes but not really related to any equality or inequality with the given series. 76.103.47.66 ( talk) 09:11, 24 February 2010 (UTC)
Why is the following not valid? .9 does not equal 1; .99 does not equal 1; .999 does not equal 1; etc.
I think there is a paradox here. Proofs to the contrary do not make the proof above invalid. Why is this intuition not acceptable? Tristan Tondino ( talk) 02:01, 5 February 2010 (UTC)
a =√4 a=2 a=-2 2a=-2a a=-a This may be contextual of course, or just wrong. Tristan Tondino ( talk) 02:35, 6 February 2010 (UTC)
I'm sorry if this has already been discussed, but can't we just say that 0.999... does not equal one because of this function? I'm not arguing anything, just asking a question.
Floor m.n where m and n are strings of digits, is always m. So doesn't that mean floor (0.999...) = 0? Floor (1) = 1. —Preceding unsigned comment added by Goldkingtut5 ( talk • contribs) 06:49, 11 February 2010 (UTC)
Here's another way. Since 1/0 equals infinity, then 1 = 0*infinity, and 1/infinity = 0. 0.999... plus 1/infinity = 1, then 0.999... plus 0 equals 1, and 0.999... = 1. 24.1.201.172 ( talk) 01:56, 21 May 2010 (UTC)
Firstly, 1/0 is not infinity but is undefined. Secondly 0*infinity is indeterminante. 1/x as x->inf is 0. So 0.999...+ 1/x as x->inf = 0.999... = 1. This is just saying x + 0 = x because 0 is the additive identity. 86.172.185.252 ( talk) 16:15, 26 August 2010 (UTC)
I have a TI-34 II calculator at home and it has the ipart function and the fpart function. The ipart stands for integral part, and if 0.999... was inputed into the calculator with the ipart function thingamajig (like this: ipart(0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...)), it will still be 0. If you do 1, it will come up as 1. Therefore, they aren't the same. 24.1.201.172 ( talk) 02:07, 21 May 2010 (UTC)
I give reason against my brother that 0.999... is equal to 1. The reason why is because:
0.999... = the infinite series 9/10 + 9/100 + 9/1000 + 9/10000 + ... , which converges to 1. Sixeightyseventyone ( talk) 01:52, 22 May 2010 (UTC)
I see that many people are struggling to understand why 0.999... is equal to 1
I see many arguments that 0.9 is NOT 1, neither is 0.99 or 0.999 or 0.99999999999 ect... so why is 0.999... (repeating 9's) the same as 1?
It is because many people have a hard time truly understanding what infinity is. It is not a REAL number, there is no way in a string of digits like 489238902 to express infinity. What is happening with the above question is in your mind your are assigning and end to the string of 9's at some REALLY far away point. You are picturing (whether you know it or not) infinite 9's as being really really long, but subconsciously you are picking a distant point and stopping the string to say well if 0.99999 is not 1, why is 0.9999999(billions and billions of digits later)999 equal to 1? But it is not - the moment the infinite string of 9's ceases, it is NOT equal to 1.
A very important aspect of Decimal Notation (which is the number system we use) that you should understand is what numbers really mean. For example, the number 257 - we dont have a digit for that number, instead we use the single digit numbers 0-9 to express all other numbers in the system, and the way we used those digits to represent numbers was by creating "place values" - the one's place, the ten's place, the hundreds place and so on. Also, each place value is measured in magnitudes of the number 10. The ones place is 1, the tens place is 10, the hundreds place is 100 and so on
So what does 257 really mean? 2*100 + 5*10 + 7*1 which is 200 + 50 + 7 = 257 What about a number with non-0 digits after the decimal point? For example the number 624.7 - 6*100 + 2*10 + 4*1 + 7/10. For any non single digit value, you are REALLY expressing a sum.
The decimal place values are the tenths, hundredths, thousandths, and so on. They are expressed like this - 0.256 is 2/10 + 5/100 + 6/1000. So 0.999... is expressed - 9/10 + 9/100 + 9/1000 + 9/10000...
Ok? Now here is the part where you ignore your intuition. If you are absolutely stuck that all the mathematicians are wrong - then there is no point in trying to understand. All I am trying to do is explain it to people who want a way to understand it without using insane proofs that people with a limited math skill can't understand. I am going on the principle that people reading this at least a decent arithmatic skill and understand math to a very limited degree.
Ok - so the next thing about our number system you want to notice is exactly how we express decimals as fractions. Most people understand that 0.3 is the same thing as 3/10. How about 0.47? It is 47/100 right? You can go as far as you like - 0.12398 is 12398/100000 , 0.119 is 119/1000 and so on.
You can also express the value of a REPEATING fraction in a similar way. Before we used the place value to find the denominator (as we saw 0.47 was the fraction 47/100 because the "7" is the end of the fraction and it sits in the hundredths place). In dealing with repeating decimals, we use the maximum value of those place-values. An easier way to say that, is we use the place value minus 1. For example - we all know what 1/3 is. We also know it is the same as 3/9 - right? well 1/3 is expressed in decimals as 0.333... So you see the repeating digit there is "3" and it is only one digit. so it is not 3/10, that is wrong. It is 3/9 (which can be simplified to 1/3).
Take time, make sure you understand this part, convince yourself it is true with some examples. how would we write the following as a fraction? 0.454545.... Well we have two repeating digits this time - "45" is repeating, and that takes up 2 place values, but its not 45/100, that is wrong - The correct fraction is 45/99. (just a quick note, that is why before I said the max value of the place holder, because in the hundredths place, the max value is 0.99... it is also 100-1 if that is easier for you to remember). How about 0.111... Well we know its only one digit repeating, and so we use the one's place to determine our denominator because there are no other repeating digits after the first "1". So the proper fraction is 1/9. Go ahead and type 1/9 (1 divided by 9) in a caclutor and see what you get. It will be 0.111... One more because this is essential - what about 0.888... how do you think we write that as a fraction? Well you probably should know by now otherwise you really wont understand the rest. The answer of course is 8/9. Go ahead and check this out for yourself.
So what does all this mess have to do with 0.999... = 1? Well I want to make sure you understand what these digits REALLY mean.
Because you will see that what I have told you is true in ALL cases. Period.
So... what is 0.333... It is 3/9 right, we learned that. that simplifies to 1/3 what about 0.666... It is 6/9 right - and it simplifies to 2/3 finally how about 0.999... It can only be... 9/9 - right? and that simplifies to 1.
Does it make sense at first? No. It is super easy to understand? Of course not.
Sometimes you have to simply put your "common sense" notions aside. Because common sense only applies to what YOU are used to. It does not obey the laws of nature. Lots of things do not SEEM like the make sense, but when a proof is put down, it must be considered true - unless someone finds a way to completely disprove it. But really, once you get used to it, this idea is not that hard to see.
Try to understand that Mathematics was not invented in one quick swoop. No one just threw all this into existence overnight. We needed a number system, just a way to number things, count stuff, you know, basic easy ways to help keep track. Then we thought, well what if we want to use these numbers to figure something out? for example - multiplication didnt just exist out of no where. Numbering was invented, and people figured out rules for adding those numbers, because adding is helpful to our daily life. Then someone noticed when you have to add the same number over and over and over again it was tedious. 2 + 2 + 2 + 2 + 2 Who feels like doing that over and over and over again. So a shorthand was invented called multiplication.
Over time, we began to explore these numbers more and more. Finally we ran into strange numbers like pi and e. We ran into things that we couldn't make sense of, like dividing by 0. And we found unusual things that challange our common sense notions, like 0.999... = 1.
You see? It doesn't have to make sense. It's just a product of our invention of mathematics. Perhaps if some other system was invented we would not encounter things like this. It only exists because of OUR construction of the decimal notation system. It's not some official rule of the universe that would be here even if humans weren't - its just some funny little outcome that happened to exist due to our methods and design of the decimal notation numbering system.
I hope at least SOME people can understand better. I tried my best to explain it very slowly and simple - If you couldn't understand than I apologize, I am very advanced in Mathematics and sometimes going back to the basics of the basics is harder than the ridiculous stuff I am researching now. Alex DeLarge 10:03, 25 May 2010 (UTC) —Preceding unsigned comment added by A.DeLarge23 ( talk • contribs)
What are you talking about? That wasnt any self-promotion, or opinion piece, it was an explanation of why 0.999...= 1 Without complicated proofs so that people who are not math experts can understand. I can offer any number of proofs for you, but unless you have a background in Mathematics you will not have a clue what I am talking about. I noticed a lot of people were confused and bothered by this fact, so I took time out of my day to help all those people who never went far beyond arithmetic or basic algebra to at least have some sense of understanding. How are the facts that 0.3 is 3/10 and 0.333... is 3/9 ect.. opinions or self promotion? that is ridiculus and uncalled for to accuse me of this. I simply stated FACTS. And then I followed up with a brief explanation of why it seems like it violates common sense and why these things pop up to begin with, by explaining that the number system was our invention that was originally needed for counting, and we kept taking more and more steps.
Those are not opinions. This is an argument page anyways, so isn't this a place to make your arguements based on 0.999... being equal to 1? Well that is what I did. And Tkuvho didnt have any point, how could I miss it?? He said "can you summarize" and I said no, not without adding proofs that people won't understand. However Trovatore if you think I am violating Wikipedias terms than I am removing this entry - now no one will get an easy explaination, they can go take 5 years of math and understand the proofs. And from now, I won't add anything that is 100% factual based, which is .... everything I would write. 173.62.181.145 ( talk) 04:32, 1 June 2010 (UTC)
But this IS the arguement page - and I wasnt debating any mathematical facts. everything I said IS a fact, there was no debate. It was just an argument of why this is true, using simple terms. How is what I said ANY different than the people who pose the other proofs? I just did it in a more simple way because I saw many people here asking WHY because they dotn get it, because they cant read the proofs. I was offering my help to those people by stating the FACTS of mathematics, but in simple terms. NOTHING I said is not a fact when it came to the math.
0.3 IS 3/10, 0.333... is 3/9 and 0.999... is 9/9 or 3/3 or 1. These are not my interpratation of the facts. These are FACTS. These are the basic facts of our number system. It is all true, and it is a simple way for people to see it, as opposed to the algebra proofs and calculus proofs. Not everyone has this knowledge, so there are NO Facts I am debating, that is ridiculous. I have been studying Mathematics for my whole life. I Can offer you all kinds of complex proofs that are also facts - how is that different than offering proofs in a simple form, which is already on the Argument Page, this is NOT The talk page, that is why I posted here - it is another argument of why 0.999... is 1 - using the most basic of facts dealing with our number system. Not everyone took calculus or algebra to any real extent. What about those people? Are the screwed? Thats bullshit. I believe EVERYONE has the right to understand this. Thats what math should be about, it should be for everyone, not just someone who has a extended background. Go look in a 3rd grade text book and you will see these facts, that 241 is REALLY a statement of place value. Its how our number system works. And look up how we express decimals and fractions - .333 is 3/9 and .45454545454545.... is 45/99. It was just a way to help people. If anyone is going to argue that I am debating facts, or that I am biased, well thats ridiculous and wrong, this is an arguement page so thats what I did, i proposed another argument to the truth of 0.999...=1 using simple terms. If Im going to be accused of non-sense like trying to put up mathematical facts for debate then I want my entry deleted, and I Will keep deleting it over and over until it stays deleted. I took time out of my day because I wanted to help those who truly are bothered by this notion and want it to make sense. If wikipedia is not the place for people to find and understand information, then I will take my time elsewhere. 173.62.181.145 ( talk) 15:50, 3 June 2010 (UTC)
Note that if you use binary, 0.1111... equals 1, and for octal: 0.777... = 1, ternary: 0.222... = 1, hexadecimal: 0.FFF... = 1, etc. 24.1.201.172 ( talk) —Preceding undated comment added 18:50, 6 June 2010 (UTC).
That section just proved that infinity equals -1. 24.1.201.172 ( talk) 19:05, 6 June 2010 (UTC)
Huon's argument [ above] that ends with ".999 does not equal 0.999...;" attempts to prove that the argument form is invalid. Instead it seems to say to me that .999... actually has no meaning at all. That seems sensible to me: How can you use infinity to define a real number when real numbers do not include infinity? Algr ( talk) 09:49, 9 August 2010 (UTC)
It's simply a falacy. 1 = 0.9999... is not true. If you assume that false axiom as true then anything can be demonstrated. In fact, 1 is aproximately equal to 0.999... If you substitutes "=" with "~=" then axiom is true and maths match again. Even clearer:
1 > 0.999...
And then every affirmation is true and everything recovers sense again. —Preceding unsigned comment added by 88.0.131.223 ( talk) 10:38, 15 October 2010 (UTC)
I don't think it's needed. It's not really a proof of anything, more a challenge to an incorrect assertion. It could be used as a basis for longer and more formal a proof by contradiction, but such are usually used only needed if a more straightforward proof is unavailable, and there are plenty of more straightforward proofs in the article. And it's OR, or at least I've not seen it written down anywhere.-- JohnBlackburne words deeds 14:37, 15 October 2010 (UTC)
On the talk page, 210.4.96.72 asked if someone could prove the following statement:
There's one caveat: I'm not quite sure what he means by "summation of the sequence". Let's first introduce some notation: Let a0=0.1, for k>0 and for k>=0. Then the sequence (a0, a1, a2, a3, ...) is the sequence (0.1, 0.01, -0.1, 0.001, -0.01, ...). I'll prove that where I use the standard definitions for "sum of a series", which may not be what 210.4.96.72 had in mind. Namely, by definition, where finite sums can be formed simply by adding the terms. Despite the similar notation, infinite series actually aren't formed by adding the terms, since addition of infinitely many terms isn't defined. Furthermore, by definition of the limit, we say that for a sequence (sn), we have if for every ε>0 there exists an n0 (depending on ε) such that for all n>n0, we have |sn-s|<ε.
So with these definitions and the notation from above, what I'll have to prove is that for any ε>0 there exists an n0 such that for all n>n0, we have
Next, note that a0+a2 = 0.1-0.1 = 0, while for any i>1 we have a2i-1+a2(i+1) = 10-(i+1)-10-(i+1)=0. Thus, quite a few terms of the finite sums I'm about to look at will cancel out:
Now we can combine the parts: Let n>n0 be any (natural) number.
Both cases need fact 1 for the very last equation (though case 1 could do with a weaker version), and each case also needs one of the other two facts for the second equation. QED. Huon ( talk) 20:32, 20 October 2010 (UTC)
I would nominate this article for "worst article on wikipedia". This article shows how the ignorant rabble can outvote the truth. No matter hown many people say the earth is flat, it's a sphere. No matter if Zeno says Achilles can't beat the tortoise, Achilles will win. No matter how many people continue to assert that 0.999 ... = 1, they are not equal equal. The *** limit *** of 0.999 ... is 1. That's it. Unfortunately, most people don't understand limits. It's not easy for the non-mathematician to understand. Most are so numerically challenged, they cannot even understand simple math concepts. Limits are far beyond them.
The article makes silly, pseudo-intellectual statements such as "the question of how two different decimals can be said to be equal at all". The article supposedly addresses "alternative number systems", but fails to mention other base systems. The reason many "initially question or reject it" is because it is not true. The business about "math educators" trying to figure out why students see through the fallacy is hilarious. The "math educators" are just showing their math deficiencies. "Has long been accepted by mathematicians" has no citation. The "proofs" shown on the page are ludicrous, and prime examples of logical fallacies, on the order of Zeno's paradox. The "proofs" remind me of the proofs of God's existence. Only one proof should be needed. Instead, many proofs are offered, because each one is flawed.
This article is a classic, and would be extremely humorous for any mathematician to view if it were satire. Since it's presented as fact, it's very disheartening to view. For those who would say "then edit it!", I tried that long ago. It's a total waste of my time. wikipedia is not peer-reviewed. It's rabble-reviewed. 174.31.157.82 ( talk) 17:24, 8 December 2010 (UTC)
Well, 174.31.157.82, since you think we are numerically challenged, allow me to pose a numeric challenge to you. Here's a table.
Numerator | Denominator | Decimal representation |
---|---|---|
1 | 9 | 0.111... |
2 | 9 | 0.222... |
3 | 9 | 0.333... |
4 | 9 | 0.444... |
5 | 9 | 0.555... |
6 | 9 | 0.666... |
7 | 9 | 0.777... |
8 | 9 | 0.888... |
x | 9 | 0.999... |
Solve for x. Presumably you think x is not 9, so I would be curious to know what you think it is. 28bytes ( talk) 20:55, 8 December 2010 (UTC)
A simple question: |
---|
y>1 |
y+x≥1 |
Solve for x |
This question does not involve any infinities or "strange objects" or even any repeating decimals, and yet the real set falls apart over the idea. Algr ( talk) 04:27, 4 January 2011 (UTC)
In reply to the "simple question": The set of solutions is the set of pairs (x,y) such that x≥1-y and y>1. That's a well-defined set of real numbers. Obviously there is not a single set of x that will be the full set of solutions independent of y, though any non-negative x will solve the equation for every admissible y. So what?
Concerning limits and numbers: By definition the number a is the limit of the sequence (an) if for every real number ε>0 there exists a natural number nε such that for every n>nε we have |a-an|<ε. Thus by definition limits are numbers.
Concerning the "wrong" equations: Algr, what do you think is the decimal representation of 1/3? Do you think that 1/3 doesn't have any decimal representation? Wouldn't a system of representations that does not represent all your numbers be quite useless?
Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation, but doing so will hardly involve 0.999... - rather, one can use long division to divide one by three. The technical details may be hidden, but there is no circular reasoning.
Huon (
talk)
11:56, 4 January 2011 (UTC)
"Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation -Huon" But that is the very definition of circular. When the meaning of infinitely recurring decimals is questioned, it is insulting to simply point to another infinitely recurring decimal and assume that that one shouldn't be questioned as well. Algr ( talk) 08:54, 31 January 2011 (UTC)
As I said three years ago:
Removed. There is no reason for Wikipedia to be providing a platform for Anthony; it's time for him to go elsewhere. I'll be removing all his nonsense as it appears. -- jpgordon ∇∆∇∆ 15:40, 23 October 2007 (UTC) [1]
-- jpgordon ::==( o ) 17:52, 30 January 2011 (UTC)
So explain this,
We take it as given that .9999... has to continue for infinity in order to equal 1? Why is that? Isn't it because no ammount of 9s after the decimal point could represent 1 exactly? Therefore it must continue for infinity in order to be considered an exact representation. But isn't inability to physically stop and the need to continue appending 9s the very definition of an approximation? Not unlike we consider 3.14159... to be an approximation of pi. It doesn't matter that .999... follows a pattern when .314159 does not. Both could be theoretically calculated to an infinite precision. This suggests that there is a difference between the representation .999... and actual endless .999999999999999999999999999999999999999. But both .999... and the longform both remain with the exact same problem: while .999... conveniently adapts non-practical idea of infinity even at infinity still remeains but an approximation because there is no ends for infinity. (This, by the way suggests that infinity itself isn't really a number but a process but I won't go into that.).
Also:
1/3 = .333333333... therefore .3333333333...*3 =.999999999... = 1 is an example of an something that doesn't really explain anything. At least not to everyone. Why should a person who doesn't take .999999999... to be a number that represents 1 feel any differently about .33333333333... accurately representing a third of 1? In other words, this proof starts out with an assamption that is just as questionable as what it is trying to prove.
Why should 1/0 = x therefore x*0 = 1 be considered to be inaccurate whereas the example above is taken to hold true? I know the result of 1/0 is undefined but it is a precedent for laws of division and mathematical operators being not 100% reliant. From the point of view of mathematical operations we have a very similar set-up.
Another thing, to people insisting that .999... is a number and not a process. I have no problem grasping the concept of infinity and infinite repetition but answer me this then, if the fact that the full form version of that number (i.e. not the shortcut .999... form) with endless nines cannot actually be be finished being written down (or said, or, what that, matter USED) than how can you say it is a number in this when you cannot even finish it in this form?
Again, the obvious conclusion is that something that cannot be finished is an approximation by definition (again, I'm not talking about .999... form here but the full meaning behind the form). —Preceding unsigned comment added by 198.30.78.254 ( talk) 16:03, 9 February 2011 (UTC)
The point is this. Such important feature as invariability of significant number length for changes of this number in result of some mathematical operations is not taking into account in these calculations. Multiplication of decimal fractions on 10 is execute as well as for integers. This implyies what besides transposition of the decimal point on one numeral to the end of number the zero is added. Distinction is that non-significant zero in decimal fraction can be cutting. Interestingly that the infinite circulating decimal can be written in the form when direct mathematical operations with low-order digit are possible. The example of manipulation with numbers must be written as follows:
x=0,999... x=0,(9)9 //Recording of infinite circulating decimal in the fit form for mathematical operations with the low-order digit// 10x=9,(9) //Number shift in the register, the low-order digit accepts zero value and cutting// 10x-x=9,(9)-0,(9)9 9x=9-0,(0)9 x=1-0,(0)1 0,999...=0,999...
Consequently "a" was equal 0,9999... and has remained equal 0,9999... because such important value as infinitesimal decimal unit was missing in the previous calculations. It was infinitesimal blunder. — Preceding unsigned comment added by Leonid 2 ( talk • contribs) 09:00, 4 March 2011 (UTC)
I have read about my explanations - "I find it quite impossible to fathom what this is supposed to mean". "I belive this is wrong". Probable cause is my bad English or the explanation was too short in spite of clearness. There are Hemming’s words - "The purpose of computing is insight, not numbers".
The program of information processing called Digit manipulation contain very gross bug connected with infringement of the order of performance of mathematical operations. Mathematical operations of addition, subtraction and multiplication of decimal fractions execute in the certain order and it is indifferent when these operations execute by a digit machine or on a sheet of paper. It is the order of processing from younger meaning bit to the senior. If addition of infinite periodic decimal fraction with an integer still can be written down in the form of 9,(0)+0,(9) having added to the integer an infinite chain of zeroes then in case of multiplication the problem gets fatal sense.
It seems that multiplication of decimal fraction on 10 is simple mathematical operation connected with carry of a decimal point on one digit. Actually for calculation it is necessary to execute some the consecutive strictly certain instructions. It is necessary to write down number, necessarily to add 0 by the end of record (initialization of operation of multiplication on 10, but at calculations on a paper usually it is not considered) and to add a sign on executed operation (for example, ^).
But number 0,999... hasn't the end. Correct operation of multiplication is prohibitive in this case.
If number was written 0,999...=0,(9) or such as of dimensionless linear array 0,(9)=0,(999...999) correct operation of multiplication is prohibitive too.
10*0,(9)=>0,(9)0^=>0,(9) //empty operation because shift on step =0 is impossible act//
For reception of an opportunity of shift of all array you must create control element (buffer entry). You must receive rights for changes most lower digit. Number 0,999... is infinitely, but if it had the end, it would come to the end on 9. Therefore number 0,999... write such as 0,(9)9 or execute the operation of global unconditional assignment.
x=0,(9)9
Now indifferent if quantity of nines in the array (9) is endless, this array will be in computing such as unit and addition even only one 9 activate of the array overflow.
10*0,(9)9=>0,(9)90^=>9,(9)0=>9,(9) //execution of instruction//
But after operation of global unconditional assignment it is impossible to get rid from buffer digit for operations of addition and subtraction.
10х-х=9,(9)-0,(9)9=9,(9)0-0,(9)9
After the further calculations we receive very important volume - infinitesimal decimal unit - 0,(0)1.
x=1-0,(0)1=1,(0)0-0,(0)1
It is interesting element with original features. It is the mathematical tool which allows to formulate precisely many the existing mathematical proofs and it will be useful in exact numerical calculations.
Really 9*0,(9)=8,(9)1. The operation 9x=9 in programm called Digit manipulation is the operation of local un/conditional assignment x=1.
I found the bug in programm called Digit manipulation. Corrected result has the dead level 0,999...999==0,999...999 and verification can be. Probably I made a mistake. People said in Ancient Rome - "Errare humanum est". I ask argue away computing without words "I believe" and "I trust what 0,(9)=1".
My English isn't Shakespeare's language. Probably some considerations is no correct in English. I left text of this message on Russian on my talk page in English Wiki.
Leonid 2 (
talk)
11:10, 7 March 2011 (UTC)
"You seem to do mathematics from the perspective of computer science." Yes, of course. Computer science is the part of mathematics. Properties of part can't differ from properties of whole.
9*0,(9)=? //Question. How many will be, if multiply 0,(9) by 9?// 9*0,9=8,1 9*0,99=8,91 9*0,999=8,991 9*0,9999=8,9991 9*0,99999=8,99991 ............... 9*0,(9)=8,(9)1 9*0,(9) no equivalent 9 //Verification//
Do you have more questions? Leonid 2 ( talk) 12:25, 11 March 2011 (UTC)
I'm not want the discuss about the full article 0.999... . I do not care 0,(9) is equivalent to 1 or not. But the section of the article entitled Digital manipulation contains an obvious error which must be corrected. This error can be detected by direct calculations on a sheet of paper. In the multiplication 0.999 ... by 9 you will never get 9, no matter how many nines after the decimal point will be, and this should be displayed on the main page of the article. Leonid 2 ( talk) 08:08, 19 March 2011 (UTC)
I asked to refute my computations. I got the misty arguments about the properties of infinity about the properties of a real numbers and so on, instead of other computations. But I'm not want the discuss about the full article 0.999... and catch the wind in a net. Therefore once more.
9*0,999...=? //Question. How many will be, if multiply 0,999... by 9?// 9*0,9=8,1 //Start of the computing// 9*0,99=8,91 //Continuation of the computing// 9*0,999=8,991 9*0,9999=8,9991 9*0,99999=8,99991 ................... //Continuation of the computing// ..................... ....................... //Stop when you will be tired or are finished these computations// Шутка 9*0,999...=8,999...9991 //Answer. Any can enter the infinite number of nines instead of the dots by oneself//
Such decision is possible even for a scholar from an elementary school. In the multiplication 0.999 ... by 9, you will never get 9, no matter how many nines after the decimal point will be. The section of the article entitled Digital manipulation contains an obvious error. 9x=9, if x=0.999... it is the blunder. I wrote about this in detail above how this blunder arises.
Don't you agree with me? It may be. Which means that the listing must be of the your step-by-step computations 9*0,999...=9.
1/0.(0)1=1(0).0
Leonid 2 ( talk) 09:11, 22 March 2011 (UTC)
Thank you. You have talked so much about the properties of rational numbers for me. Finally, I understood everything. The number 0.999... is not rational. It is the recurrent decimal fraction. It can not be represented as simple fraction. In converting simple fractions to decimals you may lose some information irreversibly. The numbers of multiples of 10 can not be divided by 9 or 3 evenly. 10/9=1+1, 100/9=11+1, 1000/9=111+1, 10000/9=1111+1, ... (10/3=3+1, 100/3=33+1, 1000/3=333+1, 10000/3=3333+1, ...).
The infinitesimal decimal unit 0,(0)1 proved to be useful once again.
1/9=0,(111...111)1+0,(000...000)1 1/3=0,(333...333)3+0,(000...000)1 0,111...=1/9-0,(0)1 0,333...=1/3-0,(0)1 0,999...=1/1-0,(0)1
The numbers 0.111... and 0.333... is not rational too. Addition and subtraction with decimal infinitesimals for the such numbers is admissible mathematical operation. Special form of recording is compulsory regulation in this case. Some features of the infinitesimal decimal unit 0,(0)1 on my talk page -
Russian and English text.
But I do not care 0.999... is equivalent to 1 or not as before.
Leonid 2 (
talk)
10:08, 30 March 2011 (UTC)
I calculated quantity of solutions for distinction 0,999... and 1 on my talk page
"Quantity of solutions for distinction 0,999... and 1" for two cases (if quantity of nines after decimal separator is infinity and if quantity of nines after decimal separator is the more of infinity). General quantity of solutions for distinction 0,999... and 1 is equal ∞+2 solutions including solution in Lightstone's extended decimal notation. Probably 0,999... is not equal 1. Confidence is more 100%. I ask argue away these calculations only by other calculations.
Leonid 2 (
talk)
07:10, 10 April 2011 (UTC)
ok, and apologies if this has been covered before. An informal 'handwaving' argument but one that might help overcome some intuition blocks:
It should be obvious to all that 1.1 > 1. Similarly, 1.01, 1.001, 1.0...01 etc are all >1 for all such numbers with a (finite) string of '0's.
So what happens with an infinite string of 0's? It should be obvious that 1.0...=1 (there being no 'last place' for the digit 1)
Note that the difference of 1&1.1 is the same as the difference of 1&0.9
Similarly for 1.01 & 0.99; 1.001 & 0.999 and so on.
It (informally) follows that the difference of 1 & 1.000...(equality) is the same as the difference of 1 & 0.999... Yossarian68 ( talk) 23:10, 23 April 2011 (UTC)
http://uncyclopedia.wikia.com/wiki/0.999...
In mathematical theory .9 is a specific number. It is not half or a fraction of a number. It is .9 and it remains .9 in the equation. Unless an equation calls for rounding of the numbers, then .9 remains solely .9 in the equation. Therefore 0.999_ is a infinitely repeating number.
In mathematical practice there are no exact numbers. As reality is incapable of producing an exactly equal number in relation to something. So in mathematical practice 0.999_ cannot equal 1 regardless.
In mathematical theory where rounding occurs then 0.999_ equals 1.
If rounding is not established or .9 is specifically established as being an exact number, then no amount of crying and bitching changes it to 1.
You've been Uncyclopediaed, yo. 58.7.214.181 ( talk) 03:50, 19 May 2011 (UTC) Harlequin
Perhaps you should go back and read where I specifically, in detail, referred to equations where it wasn't being "rounded". But if all you can come up with is "no, no...if they are not infinitely repeating, then it totally equals 1", it's just further proving my point. 58.7.214.181 ( talk) 04:03, 19 May 2011 (UTC) Harlequin
I'm using Uncyclopedia as a reference? Haha, well done at further proving my point and again failing to read. Strange, seems I explained exactly how 0.999_ does not equal 1 without anything relating to the Uncyclopedia article on the subject. I merely pointed out another site that makes fun of the bitching and crying those like yourself are doing.
I gave the equation separate from any other site. Which you should have known had you, you know, actually read anything. Hahaha.
0.999_ repeating in mathematical theory can only be 1 if it is not infinitely repeating. Your claim here is that somehow an infinitely repeated number will magically have an end and that it will magically round up once it has reached that end, even when the equation it is in specifically states it doesn't. Like ive stated, no amount of bitching and crying changes that fact that an infinitely repeating number does not have an end.
It's pretty hilarious really. I even explain that where it isn't infinitely repeating and is in an equation that "rounds" the numbers that it would equal 1 (again, had you actually read anything) and yet you seem to think that an equation where it can't even reach an end to "round" would somehow still do so even when it's specifically mentioned that none can occur?
Try again. This time without trying to take credit for other sites material. Especially when the best "argument" (or lack of) you can come up with is that "some guys" created the other article (despite it's refutation of "their" supposed claim that 0.999_ ALWAYS equals 1) and because "they did" it is therefore magically wrong. 58.7.214.181 ( talk) 04:39, 19 May 2011 (UTC) Harlequin
I notice you don't know what infinite means. There is no "end", so there can't be a "last 9". Try again. Especially since I didn't state anything about Uncyclopedia being correct and specifically provided a detailed explanation of why the claim "hurr 0.999_ always equals 1 no matter what" is a false statement. 58.7.214.181 ( talk) 04:39, 19 May 2011 (UTC) Harlequin
This is such a fascinating argument to me. As someone who's knowledge of mathematics goes about as far as long division I'm perfectly willing to accept the fact that 0.9 recurring equals 1, simply because I defer to the wisdom of people much smarter than me in this subject. If mathematics professors say it's so, who am I to judge? But the moment someone get's a little bit of knowledge about this subject they think they have what it takes to prove the experts wrong. Why not just accept it and move on? -- 86.182.1.71 ( talk) 18:39, 15 June 2011 (UTC)
.9 repeating is just another way of writing 1 but with all its lesser place values being presented. Usually one is simply given the maximum place value: 1/1, but obviously the number can contain any combination of lesser place values. 1/10ths, 1/100ths etc. For example 100/100 is 1 too but for the 1/100ths. 900/1000 + 10/100 = 1 representing place values 1/100ths and 1/1000ths. Of course there is obviously an infinite amount of lesser place value delimitations intrinsic to the number 1 and these will add up to 1 just like any smaller finite combination. And when we list out these additions we get 9/10 + 9/100 + 9/1000... etc. Unless you think it is mathematically impossible to represent every possible lesser decimal place value you basically have to admit .9 repeating is exactly 1. 76.103.47.66 ( talk) 00:30, 6 August 2011 (UTC)
1 is not an "infinitely repeated number" and "Harlequin" obviously did not know anything about mathematics. — Preceding unsigned comment added by 91.185.1.174 ( talk) 12:56, 6 September 2012 (UTC)
Hello. I am not terribly knowledgeable in mathematics. Some years ago I posted here a comment, which used non-conventional notation, trying to reason that 0.(9) couldn't be equal to one, because then all numbers would be equal to all other numbers (if you nullify the difference, even infinitesimal, between two numbers, then what's to stop you from generalizing that to the whole real axis?). I received then two comments. One simply said that I was using non-standard notation and so my argument didn't follow. Another one said that it was possible I used it, in so far as I was actually able to use it and convey my meaning, and that there might be some reason to my assertion. He then directed me to some higher mathematics I couldn't understand (p-adics).
But enough with introductions. I have revised my thoughts in various ways, and I am also now able to present my simple argument using conventional notation. So here it is:
Consider the sequence 0.9, 0.99, 0.999, 0.9999, 0.99999, etc. It is obvious to anyone that the limit of this succession is one. But the definition of a limit tells us that this limit can never be actually reached. Therefore there is never actually an equality between 0.(9) and 1, except in the limit itself.
The same holds true to the sequence 0.1, 0.01, 0.001, etc. It's limit is zero. Therefore the difference between 1 and 0.(9) is only zero in the limit.
I think that a parallel could be drawn to the point in the exact center of a wheel. I think it can both be said that it rolls or that it doesn't roll, depending on the exact context. There may be a confusion regarding two different "levels" of reality, that is, a confusion between an assertion and a meta-assertion.
Joao.g.madureira ( talk) 16:48, 17 October 2011 (UTC)
Here's a different reply to João, which I hope will help convince him. João, there's an incorrect assumption in what you say, namely that a "limit can never be actually reached". That is 'often' true of limits, but not always. As a counterexample, consider the sequence a(n) = (1+(-1)^n)/n = 0, 1, 0, 2/4, 0, 2/6... Its limit is clearly 0, and a(n) does take on this value, infinitely often in fact. I suspect that this misconception about limits is part of the trouble many people have with the equality discussed in this article.
Another point, already made by Huon, is that by "0.(9)" we really mean the limit of the sequence (0.9, 0.99, 0.999, 0.9999, ...), rather than the sequence itself. So in a sense you're right that there are two distinct concepts at play here, the sequence on one hand and its limit on the other. What you (and many others) appear to misunderstand is that "0.(9)" by definition denotes the latter, not the former. Let me repeat that: by definition, 0.(9) = lim(0.9, 0.99, 0.999, 0.9999, ...) -- not 0.(9) = (0.9, 0.99, 0.999, 0.9999, ...)! Though this is a technical point, and so perhaps not as persuasive as my previous one. Regards. FilipeS ( talk) 14:49, 6 February 2012 (UTC)
I'm sorry, I do not normally log in to Wikipedia because I do not edit information, mostly because I only come here to learn about things I do not know about (and how can I add to a topic I don't know about?) but also because I do not know how all of this actually works. So if I am doing something completely and horribly wrong, then I apologize, I just needed to say this. I also do not really have more than a high school education in mathematics, so I am not any sort of mathematician expert, nor even proficiently learnéd in math. Although, I have tended to have an unusually extraordinary understanding of mathematics, especially such basic mathematics as converting fractions to decimals.
However, I cannot accept the fact that 0.99999... = 1 and after reading the proofs, I understand why. The proofs all make sense, but I have the reason/problem which causes this to happen, and it is because infinite decimals are not perfectly accurate representations of fractions, which is why they are infinite. They go on forever because those fractions cannot be created into decimals. 1/3 become 0.33333.. and so on forever, because this is the -closest- representation of it that we can create with a decimal, but it isn't perfect. Therefor, 0.9999... is the closest representation of a fractional 1 that we can produce with a decimal, however it does not truly equal 1 because it is more like an estimation or rounded number.. it is like rounding 4.9 and 4.9 both to 5 and adding them together to make 10, and saying that 9.8 = 10 because 4.9 + 4.9 actually equals 9.8. That is the same as "rounding" 1/3 into 0.333... and so on forever. Therefor the actual number 0.999... itself is NOT truly equal to 1, it only seems that way if you multiply 1/3 * 3 as 0.333... * 3 equaling 0.999... because 0.333.. is "rounded" and therefor like the 4.9's I rounded to 5. I don't know if that makes any sense to anyone but it makes perfect sense to me, and I can try to explain it further if no one is understanding me. I am also unsure if there is any way to make a proof for this because there isn't much of a way that I can show that infinite decimals are rounded or estimations, other than by using the proofs already here.
Obviously if 0.333... * 3 =/= 1, then 0.333... is not a perfect representation of 1/3 in decimal form. It is just the closest. — Preceding unsigned comment added by Sweetnaivety ( talk • contribs) 06:43, 27 October 2011 (UTC)
Short answer first: 0.999... - 0.09 = 0.90999... You are correct that I assumed the equality 0.999...=1 in order to calculate 1 - 0.111..., but I need not do so. Consider the following:
There is no missing "9" at the end because there is no end to the eights. For a more formal proof, you could employ limits and use that 0.111... is the limit of the sequence (0.1, 0.11, 0.111, ...) while 0.888... is the limit of the sequence (0.9, 0.89, 0.889, ...).
Next, you claim simultaneously that 0.333... x 3 = 1 and that "1/3 is not representable in decimal form". That obviously cannot both be true. If 0.333... is the number which when multiplied by 3 equals 1, it is by definition the decimal representation of 1/3. I assume you meant that 0.333... x 3 equals 0.999..., but not 1.
The remainder of your argument says that an infinite decimal cannot properly represent anything. We could argue whether the number represented by 0.333... is indeed 1/3 or whether it is a little less than 1/3 (and it's the former, by the Archimedean property line of reasoning I gave above). But now you seem to say that 0.333... simply does not represent any number. Why not? Why should we disallow an infinite decimal to represent something? After all, the entire point of having decimals in the first place is to have them represent certain real numbers - and the claim that just by being infinite they can no longer do so seems more of a philosophical than a mathematical objection. And it leaves you with lots of representations which do not represent anything, and conversely with lots of numbers without representations. That strikes me as rather unhelpful. Huon ( talk) 17:17, 10 November 2011 (UTC)
0.999... is equivalent to 1 in real number equivalence. They are NOT truly equivalence in mathmatical sense. We MUST use the curly equal sign to denote their equivalence, not the normal equal sign. If we mix up the two, then we can come up with a whole lot of bullshit.
Ex: If 0.999... = 1. Then 0.999... + 0.0...1 = 1 + 0.0...1 -> 1 = 1.0...1.
Since we already know 0.999... = 1, replace 1 in the above equaltion and you get 0.999... = 1.0...1.
Repeat the step infinitely, then you prove that 1 = infinity. — Preceding unsigned comment added by Ssh83 ( talk • contribs) 21:10, 1 November 2011 (UTC)
As a matter of fact, Ssh83 has it backwards. According to standard mathematical terminology, 0.999... and 1, just like 1/2 and 0.5, or 3^2+4^2 and 5^2, are, in each case, both equal and equivalent. FilipeS ( talk) 15:03, 6 February 2012 (UTC)
0.9999... is not equal to 1. Remember when you were in primary school? Well, in my primary school, they taught that a number can only be represented in one and only one way as a decimal. Anything they teach in schools is true. Sure 1 can be represented in many different ways, but as a decimal, it can only be written only one way. I am aware that 1, 01, 1.0000..., and 1.0 are the same number, but that is still the same way of representing a number in decimal form, becuase (if you remember in primary school again) if you have a number with any amount of digits in it, then there will be an infinite amount of zeroes both behind it, and after the decimal point. — Preceding unsigned comment added by Nominalthesecond ( talk • contribs) 19:16, 25 November 2011 (UTC)
Add 0.1 to 0.999...
0.1 + 0.999... = 1.099...
Now
1.099... mod 1.1 = 1.099...
Try the same with 1
Add 0.1 to 1
1 + 0.1 = 1.1
Now
1.1 mod 1.1 = 0
What now.... :)
Identity:9C7C71F43DD9DC1E604F3F21BB760A402630D9B8404B33587193D16CE90439F25C1757C2FE6C10C612BC0BF0A3CBD1FBEB5FF20B2C4F2350C85C0CBBA523464C — Preceding
unsigned comment added by
129.97.236.182 (
talk)
05:39, 2 February 2012 (UTC)
Sure, you assumed 1.1 == 1.099... and then got the result of 0, don't assume that and actually do the division using any division algorithm, You'll get 1.099...
Its true that 1.099... will act in much the same way as 1.1 in any conceivable operation other than modulus(only one I can think of at the moment).
Saying that 0.999... = 1 has with it much the same hazards as assuming 1/infinity = 0,
You're assuming that a real number can be created by dividing by something that isn't a number, and that an infinitesimally small value, regardless of its composition, is the same as the lack of a value.
By that logic x = 0.00...1 = 0.00...2 = 0.00...3 = 0.00...9 = 0.0...10 and so on, all of which equal 0
Or equivalently
That
x = 1/infinity = k/infinity,
Take the limit as K approaches infinity and you end up with indeterminate, but we've already said that 1/infinity = 0
Thus
x = 1/infinity = k/infinity = k * 1/infinity = k*0 (Which applies because you assumed that 1/infinity = 0)
Take the limit as K approaches infinity and you end up with 0, not indeterminate,
This problem arises because
0 != indeterminate != infinity
0 is a concept itself, granted, but you cannot simply equate 1/infinity to 0, and likewise 0.999... to 1 It works for all practical purposes I can think of other than modulus but it's not theoretically correct.
It is however correct to say that: The limit as n -> infinity 1/n = 0 or The limit of n/1 as n approaches 1 is 1 — Preceding unsigned comment added by 129.97.236.182 ( talk) 10:05, 2 February 2012 (UTC)
What do you mean by "1.099... mod 1.1"? Usually both arguments of the modulo operation are integers. FilipeS ( talk) 16:24, 24 February 2012 (UTC)
Thanks. Overall I agree with you, but I'm a little confused. You wrote that 1.099... mod 1.1 = 0. I can accept that, if we assume that the remainder must be an integer. But if we're using a fractional base then we should probably allow fractional remainders... FilipeS ( talk) 17:07, 7 April 2012 (UTC)
This argument will go on for a long time...all because we use base 10. Anyway...
.9999999999....=1-.0000000...1 and .9999999999...+.0000000...1=1 — Preceding unsigned comment added by Phillies9513 ( talk • contribs) 00:53, 24 February 2012 (UTC)
Am I right? You cannot deny that this isn't correct.
Plus, the common proof that since 1/9=.11111... and so on until 9/9=.99999999=1 is false because the only reason that 1/9=.11111... is because that is the closest number you can get to a number multiplied by 9 that will equal 1, in base 10. So i'm saying that 9/9 does equal 1 (duh) but 9/9 does not equal .999999.... and so therefore, 1 does not equal .999999999... — Preceding unsigned comment added by Phillies9513 ( talk • contribs) 00:52, 24 February 2012 (UTC)
In kindergarten I was told that 2 comes just after 1.Now I know that's not true,because there is 1.5,but 1 does come just after 0.999999999999............ There's just no other number between them!Think of 1/6. It equals 0.1666............6666666667 — Preceding unsigned comment added by 86.104.51.117 ( talk) 15:59, 21 May 2012 (UTC)
Every number minus itself =0 1-0.999...=0.000.....000....01 so 1 doesn't equal 0.999... — Preceding unsigned comment added by 188.25.224.87 ( talk) 08:40, 23 May 2012 (UTC)
So you mean that 0.000.....01=0 You say that 1=1.000....01.If 0.999...=1 then 0.999...=1=1.000...01 so 0.999...=1.000...01,but this is false because there is a number between them.So 0.000...01 NOT = 1,then 1-0.999... NOT = 0 and so 0.999... NOT = 1 — Preceding unsigned comment added by 188.25.220.81 ( talk) 18:59, 23 May 2012 (UTC)
Take x=0.999... and y=1. 2x=1.999..98 (that is 1.999... - 1/infinity) and 2y=2 Because there is a number between them those two are not equal so 2x≠2y. Therefore x≠y and 0.999...≠1 — Preceding unsigned comment added by 188.25.224.223 ( talk) 18:22, 12 June 2012 (UTC)
YOU DO NOT UNDERSTAND DIVISION.
A REPEATING DECIMAL IS THE QUOTIENT PORTION OF AN INCOMPLETE DIVISION - THE REPEATING DECIMAL DOES NOT INCLUDE THE REMAINDER - EVEN AS IT BECOMES INFINITELY SMALL - THAT KEEPS THE QUOTIENT FROM FULLY EXPRESSING THE VALUE OF THE FRACTION.
1/3 IS NOT EQUAL TO .333... BECAUSE .333... IS ONLY THE AMOUNT THAT IS REPRESENTED IN THE QUOTIENT.
PLEASE DO NOT DISMISS THESE FACTS BECAUSE YOU ARE NOT YET ABLE TO UNDERSTAND THEM. — Preceding unsigned comment added by Mjs1138 ( talk • contribs) 23:37, August 24, 2012 (UTC)
I know you guy are sick to death of people saying this but I do believe I'm speaking sense on this one.
The problem here is that people are visualising 0.999... as a finite concept and use finite equations to link it to a finite number (1). Although it is a finite number in essence, the representation uses a infinite concept (this is because each digit represents as value and it is the infinte chain of these values that give the number it's identity). Now this is were the equation 0.999...=1 provokes Galileo's paradox, because that we're essentially trying to do is force a infinite concept into a finite value, because taking infinity as a fixed concept always leads to contradictions somewhere along the line that just don't make snese.
Although 0.999... does appear to be exactly where 1 is what we don't take into accound is the concept that 0.999... is not the same type of number, it is construced using infinity, so as I've already said the two, following this logic, never be mixed. There are other paths of logic to follow of course that do show 0.999... to be equal to 1 but as to many things the desired 'answer' can be different depending on which logic you follow (see Exponentiation#Zero to the power of zero), which is the point I'm trying to get across. Wikipedia should never adapt one angle of view and use it as their dependance; doing as is no better than bias. All views should be taken into acount and the focus should be on the conflicts in arguments. Robo37 ( talk) 22:52, 4 October 2012 (UTC)
I am a high school student and In class we had to change 0.9999........ in to a fraction. When I done this I got 9/9 which is 1 and I got confused and thought I might research about it. I simply can't grasp the fact that 2 numbers equal the same number, I'm no mathematician but It simply makes no sense. I also think that it can't possibly be an infinite decimal otherwise it could never be represented by a fraction but reoccuring decimals can never be accuratly written (just an idea). Danjel101 ( talk) 12:02, 10 October 2012 (UTC)
Please DO NOT delete this section. It belongs here. I know you do not like it because it refutes all your silly arguments. But this is an arguments page,... https://www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c — Preceding unsigned comment added by 60.1.42.128 ( talk) 03:25, 23 January 2013 (UTC)
Furthermore, I am not proposing ANY changes to the article. In case you had not noticed, this is the ARGUMENTS page. I do not for one moment believe that ignoramuses would change their wrong opinions. Let me ask you this - why don't you delete all the other sections that are no arguments at all? On a power trip here my little man? 60.1.42.128 ( talk) 08:12, 23 January 2013 (UTC)
60.1.42.128 ( talk) 08:12, 23 January 2013 (UTC)
Your analogy of bags is so hopelessly flawed that one can't even know where to begin explaining. If a set is empty and it can contain another empty set as an element, then it is no longer an empty set. It is an ill-defined object that contains a subset and an element that are identical.
10/10 is equal to 1, so you would get 1. But 0.999.. is not equal to 1. What part of this are you unable to comprehend? k/k is ALWAYS a well-defined number, provided k is measurable.
You are the one who needs to avoid logical fallacies. Your lack of reasoning is absolutely pathetic. 173.11.135.234 ( talk) 09:21, 25 January 2013 (UTC)
I'll try to focus on the 0.999...≠1 proof because I believe the entire set theory discussion is rather irrelevant to it - no matter how we define them, we probably all agree on what the natural numbers should be.
On page 7 you argue (and I agree) that for all natural numbers k we have 1-1/10k<1. So far, so good. You then state (and again I agree) that "academics" will point out that we need to consider the limit. It's rather obvious that there's no natural number k such that 1-1/10k=0.999..., so we indeed need to do something more.
At the top of page 8 you write: . I still agree with the implication: If we had , then we would arrive at the obviously wrong conclusion that 1<1. So the premise must be wrong, and we arrive at and thus 0.999...≥1.
I had assumed (maybe wrongly) that you intended to argue that since 1-1/10k<1 for all k, we must also have . This implication is wrong. If for some sequence (Sk) we have Sk<x for all k, all we can say about the limit is that . I gave an explicit counterexample to the wrong implication that's a direct analogy to what I believed you intended to argue: For all k we have , but the fact that all elements of our sequence are less than does not imply that the same is true for the limit of the sequence. So there's no reason to believe that the fact that all elements of our sequence are less than 1 should imply something about the limit being less than 1. Huon ( talk) 17:35, 23 January 2013 (UTC)
"...but the fact that all elements of our sequence are less than does not imply that the same is true for the limit of the sequence."
In the same way, the lim (9/10+9/100+...) does not imply that the sum 9/10+9/100+... is the same as the limit of the sequence.
Look, I know that you find it absolutely intriguing that I chat with you on these pages. But really, I do not want to nurture a discussion with anyone. My suggestion is that you do not remove the link to the article, so that others can learn a view that is different to yours.
I do not care for your opinion or any of the Wiki sysop or admin opinions. So, please hold your two cents. I know your flawed arguments well. I also know that I am correct and have no need to change your opinion which is of no significance to me. Your opinion is all over the article. So take a deep breath, relax and learn that NO means NO. I do not want to have a discussion with any of you. 173.11.135.234 ( talk) 09:29, 25 January 2013 (UTC)
For anyone who may benefit from this, here is a page that was built after a lengthy discussion over on the Talk page: Visual proof of 0.999... = 1
Sorry Tdadamemd, but you have just done the same thing visually that all these other 'proofs' do - you have your conclusion hidden as an assumption within your proof. In the circle example, every cut you make has an 'inside' and an 'outside', and only when these are added together does the circle equal one. Every cut is defined as excluding part of the circle. But then you propose that continued to infinity, there would be a cut that contradicts what you just defined and would exclude nothing. But you give no reason why this last cut would be any different from the others - you just declare the answer that you are supposedly proving. So the proof fails like all the others. Algr ( talk) 07:05, 11 February 2013 (UTC)
See the following article but read especially pages 29 and 30. thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf You may NOT quote anything from this PDF because it has not been published in paper form. 173.228.7.80 ( talk) 15:20, 7 February 2013 (UTC)
I agree that .999... is not equal to one, but unfortunately that kind of writing is no way to go about proving anything. Algr ( talk) 12:53, 9 February 2013 (UTC)
There is no valid proof that 0.999... equals 1.
www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-171.html
www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c
I think you ought to delete this article as it contains false information. 197.79.0.4 ( talk) 16:19, 4 April 2013 (UTC)
I absolutely loved the part where Euler was wrong and where it is said: "mathematicians don't really understand limits that well. In fact all of them don't know what is a number." I'm glad we have you to enlighten us. 93.102.161.97 ( talk) 08:06, 7 April 2013 (UTC)
197.79.0.4, you really should read Gödel, Escher, Bach, the chapter on formal systems and the infinite will be enlightening. It also backs you up, sort of. You'll feel vindicated as well as corrected. Diego ( talk) 09:07, 7 April 2013 (UTC)
(1)*(x)= (x)
therefore we can deduce
1*(0.99...)= (0.99...)
If it is true that 1=0.99... that would mean that
2*(0.99...)= (1.99...)= 2
1*(0.99...)=(0.99....)=1
AND
(0.99...)*(0.99...)= (0.99...)
If we applied this logic consistently, then we would also have to say that (0.99...)*(0.99...)=(0.99...) which obviously poses a problem since
(0.99...)*(0.99...)=(0.99...)=1
[(0.99...)*(0.99...)]*(0.99...) ........... = 1
And so on.
If (0.99...) truly equals 1 then one could theoretically infinitally reduce a given number by multiplying it an infinite number of times by (0.99...), and that number would also equal 1 by this logic, since it would automatically have to equal (0.99...) which again equals 1. As an extension of this, 20 could equal 3, 5232 could equal 91, and so on. -- 88.73.3.251 ( talk) —Preceding undated comment added 20:53, 27 October 2013 (UTC)
(I hope I got the decimal notation for infinitely small but not zero correct... a guess after not being able to find it. 0.000...1 (infinitely zeroes before the 1)
a\ true or false? 0.999... = 1
b\ true or false? 0.000...1 = 0
c\ true or false? 1 - 0.000...1 = 0.999...
d\ true or false? 0.999... - 0.000...1 = 0.99...8
e\ true or false? 0.999... = 0.99...8 = 0.99...7 = 0.99...6
f\ true or false? the difference between every neighbouring real number would be 0.000...1?
g\ what's the total sum of all differences between every neighbouring real number? infinity or zero? - ZhuLien
— Preceding unsigned comment added by 202.168.6.200 ( talk • contribs) 04:49, 17 January 2014 (UTC)
"Neighbouring numbers" such as "the largest number less than 1" don't exist in the commonly used number systems that contain at least the rational numbers. You technically could construct a custom-made number system with such a property, but I don't think it would be very useful - for example, you will run into problems with division. There are several different number systems that have something resembling decimal representations with infinitely long strings of digits followed by something else, where numbers might arguably be called 0.000...1 with infinitely many zeros before the "1", among them the hyperreal numbers pointed out above. However, the standard real numbers, the numbers most commonly used, don't have such infinitesimal numbers. Among the reals 0.999... indeed equals 1. An asymptote of a curve, in this context, is a line that gets "arbitrarily close" to the curve, that is, for every positive real number ε there exist points c on the curve and l on the line so that their distance d(c, l) is less than ε. Take for exampe, for positive x, the curve y=1-1/x. The line y=1 is asymptotic to that curve because for every positive ε there exists an xε>1/ε, and the points c=(xε, 1-1/xε) and l=(xε, 1) satisfy the property above. (Actually I'd also have to require that curve and line don't just get close anywhere, but "near infinity"; I could modify the definition accordingly, but that would complicate the calculations without helping to clarify the main issue.)
Regarding "the never ending summation of the smaller numbers", how exactly would that be defined? That's a crucial question here. In the context of the real numbers, the answer is a limit: If 0.9+0.09+0.009+... is to be a real number, it's defined as the limit of the sequence (0.9, 0.9+0.09, 0.9+0.09+0.009, ...), and that limit can be shown rather easily to be equal to 1. In the context of the hyperreals, it's possible to define such infinite sums so that 0.9+0.09+0.009+... no longer equals 1 - but at least the way I can think of would lead it to also differ from 0.99+0.0099+0.000099+..., and from 0.9+0.099+0.000999+0.0000009999+..., and so on - even from 0+0.9+0.09+0.009+0.0009+...! Which of those numbers is supposed to be 0.999...?
Some final food for thought: Why should 0.9+0.09+0.009+... be less than 1? If the answer is, "because all the finite sums 0.9, 0.9+0.09, 0.9+0.09+0.009 and so on are less than 1", then I'd like to point out that, however it's defined, 0.999... also is larger than all those finite sums. Huon ( talk) 01:59, 21 January 2014 (UTC)
ZhuLien's theory of infinitesimal usefulness.
I am a number. My value is x. I might be positive or negative.
I have many neighbours depending on where they are positioned in my number space.
x+0.0...1 and x-0.0...1 are my nearest valued neighbours yet I cannot see them as they are infinitely far away but I do have an infinite number of neighbours closer than they are.
What I can do is using my value and my nearest neighbours or perhaps nearer neighbours, is work out how crowded we are.
I can compare my crowded space with other number's crowded spaces to see statistics in different ways.
I can zoom in and out indefinitely to my nearest valued neighbours to allow my crowd and other crowds to be visualised at different resolutions.
There may be mathematical operations on both individuals, neighbours and crowds which can be useful in rounding us up, or grouping us together in sets.
These sets can be useful in highlighting patterns or boundaries or just separating us by an arbitrary amount of space. ZhuLien 116.240.194.132 ( talk) 05:42, 6 February 2014 (UTC)
Trovatore says that "0.999...;...000..." is forbidden. I find that bizarre, as it comes across as a bigger failing then my Base Infinity idea. Also, how many times have people been told that .000...1 is nonsense when all that is needed is a semicolon? So anyway, without "0.999...;...000...", then what is the answer to this?
About a decade ago, the two sides of the 0.999... debate clashed regularly, and the main article was edited back and forth. Mathematicians and hobbyists on both sides. But one side was particularly more aggressive (the side that thought the two numbers are equal). And now after a decade, it's turned into full-blown suppression. You can't even post a dissenting view in the talk section? That is utterly ridiculous. God forbid you edit the main article. Even after all these years, I simply am shocked that people still buy into these false proofs. The first proof, alone, is simple to see right through. Multiply both sides by 10. It's a little magic trick. It's like something you'll see in a chain email that tricks the uneducated. But this tricks the educated, too. The more accurate version of the "Magic Trick Proof" is thus:
x = 0.9999... 10x = 9.9999...-9/∞ 10x - x = 9.9999...-9/∞ - 0.9999... 9x = 9-9/∞ x = (9-9/∞)/9 x = 1-1/∞ The point of this equation is that you can't multiply 0.999... by 10, it's not a valid mathematical equation. I'm not suggesting that we start multiplying and subtracting using infinity, I'm saying we *shouldn't.*
Follow the rules of Wikipedia and allow people to talk about this topic in the talk section. -- JohnLattier ( talk) 16:51, 18 April 2014 (UTC)
Infinity is NOT undefined; statements: "There is an infinite amount of natural numbers" and "Infinity itself is not a natural number; every natural number is finite" are by no means contradictory. (In order to be able to discuss infinite sets, we'd have to go to set theory; this is a topic I am going to avoid for now.)
You need to understand that a repeating decimal can be expressed without expressly invoking infinity. 0.999... is actually a shorthand for 0.9 + 0.09 + 0.009 + ...; which is itself a shorthand for a limit of a sequence:
See? At no point did I need an infinity-th element, or anything similar; in fact, I have (just for clarity) specifically excluded infinities and infinitesimals from the definition. - Mike Rosoft ( talk) 23:39, 12 June 2014 (UTC)
To some extent communication requires a shared agreement on what is meant by certain things. We usually agree that if x is a "number" then also 10x is a number and that from a+x=b+x we can conclude a=b. What about x=0.999... If you want to say that x is not 1 then you need to breaking the properties of numbers in such a way that x is fairly useless. You suggest that we can multiply x by 10 but we shouldn't. I'd say that EITHER we can multiply and then 10x=9+x OR ELSE x is a thing which can not be multiplied by 10 at all. If you take the second position (or even your stated position) then you have to abandon saying 1/3=0.3333... which means rejecting the possibility of having any decimal equal to 1/3. This leads to a situation where you have 5/8=0.675 but most fractions have no decimal expansion. IF infinite decimals are anything then it would have to be something where we can't add them together or multiply by 10. In your proposal that 10x = 9.9999...-9/∞ what is the purpose of the thing at the end subtracted? All I can see is that we come up with something new mainly because it gives us room to say that 10x is something other than 9+x. But can we then divide both sides by 10 to come back to x=0.999...- 0.9/∞? and then we have x+0=x+0.9/∞. Can we subtract x from both sides? Doesn't look so good. So in sum, saying that x=1 makes things work very smoothly and does not break any of what we do with numbers. Saying x is not 1 means that it is a thing where multiplying by 10 creates trouble and is of no use. Gentlemath ( talk) 02:30, 16 June 2014 (UTC)
Think of "1" as representing a basket of all the different functions that converge at 1. If "0.999..." is interpreted as a series expansion, then it's one of those functions. But under that interpretation, the statement "0.999... = 1" is like saying "apple = basket of all possible apples". Thus if someone writes "0.999... = 1", you can reasonably assume they're comparing baskets (limits) and not apples-to-baskets. It doesn't make any sense otherwise. But that said, you can part ways with that interpretation temporarily. E.g., to justify shifting the decimal point as a shortcut for multiplying by 10, you first reinterpret .999... as (.9 + .09 + ...), distribute the 10, use the associative rule to extract the 9, and convert back to decimal notation. This breaks the convention of immediately "taking the limit" of 0.999..., or its decimal expansion would be (1 + 0 + 0 + ...). So that's the real answer: if someone compares a convergent series to a number, it's convention to assume good faith that they're secretly referring to the limit rather than incorrectly comparing apples to baskets. But you can still temporarily delay "taking the limit" when it's useful (like in, oh, all of differential calculus). You can't "prove" this convention, and all of these supposed proofs are invalid by circular reasoning, temporarily breaking the convention that is supposedly being proven, etc. Maghnus ( talk) 11:40, 27 July 2014 (UTC)
The system of real numbers includes two kinds of zero that I will demonstrate are different. Imagine a machine that generates random rational numbers between 0 and 1. It is 'fair' in the sense that any rational number has the same probability of appearing as any other rational number. What is the chance of this machine generating a 2? Well, it is zero, because we defined it as only going from 0 to 1.
So, what is the chance of this machine generating a .5?
Well normally you would figure (specified outcome) / (All possible outcomes). But in this case the denominator is infinite. One might say that 1/∞ is not a number, but taking the limit of larger and larger denominators gives you an apparent result of zero. Prob(.5) and Prob(2) both seem to equal zero, but they do not equal each other, because one is possible and the other is not! Thus, two different zeros.
This closely relates to the .999... problem because you have an infinite number of elements, (digits for .999..., numbers for Prob(.5)) evaluated with a finite result. The simple solution would be that 1/∞ = the infinitesimal unit, ®1. If you insist that 1/∞ is not defined, then remember that 1/2 is not defined in the set of Natural numbers for the same reason. 2 can't be generated by my machine, but it still exists.
BTW: 1/3 in base infinity is (1/3)® or .1base 3®. Algr ( talk) 21:56, 14 August 2014 (UTC)
As I showed above, here with more details and explanations:
If that's wrong, at what step? Or is 1 smaller than itself? Is equality not transitive? Huon ( talk) 13:13, 16 August 2014 (UTC) Sorry, I'm to sick to do math this week. Seen two doctors already. Algr ( talk)
Here I will try to show that the following well known proof is not correct (or actually is correct concerning only the real numbers, although still with some flaws). I chose it because in it the mistakes are harder to be noticed.
x = 0.999...
10*x = 9.999...
10*x-x = 9.999... + 0.999...
9*x = 9
x= 1 = 0.999...
Let's take 1/3 and try to represent it as a decimal fraction.
1/3 = 0.3 + 1/30 = 0.33 + 1/300 = 0.333 + 1/3000 = ...
or
1/3 = 0.3 + 1/(3*10) = 0.33 + 1/(3*10^2) + 0.333 + 1/(3*10^3) = ... =
= 0.333...3<n times> + 1/(3*10^n) = ... = 0.333... + 1/(3*10^∞)
Now let's substitute the last expression in above mentioned proof.
x = 3*(0.333... + 1/(3*10^∞))
10*x = 30*(0.333... + 1/(3*10^∞))
10*x-x = 27*(0.333... + 1/(3*10^∞))
x = 3*(0.333... + 1/(3*10^∞))
It is the same. Even if we not use the infinitesimal part, which is not a real number (this does not mean it does not exist, but only that it is not a real number), still nothing changes.
What will happen if we get rid of the brackets? Let's try.
10*x = 30*(0.333... + 1/(3*10^∞))
10*x = 9.999... + 30/(3*10^∞)
10*x-x = 9.999... + 30/(3*10^∞) - 0.999... - 3/(3*10^∞)
9*x = 9 - 27/(3*10^∞)
x = 1 - 3/(3*10^∞)
x = 1 - 1/10^∞
1/3 = 0.333...3<n times> + 1/(3*10^n)
So it is all about using or not this infinitesimal part and keep in mind not to break the correlation between the number of the repeating digits and power n in this infinitesimal part. Because if we break this correlation we will have some other fraction and not 1/3 as in the example. The well known proof is a speculation but it is actually not needed, because there is no room for infinite and infinitesimal numbers in the real numbers line (which will make the difference between 0.999... and 1). This is the way mathematicians defined it, so it does not matter I disagree.
Now about some statements. I read some arguments that 9*0.999...is 8.999...91 but they are also wrong. It can be easy understood considering for exemple: 2*0.666... = (2*2)/3 = 4/3 = 1.333... not 1.333...32. Even if we include the infinitesimal part we'll have:
2/3 = 0.666... + 2/(3*10^∞)
4/3 = 1 + 1/3 = 1.333... + 1/(3*10^∞)
which is more than both 1.333... and 1.333...32
S — Preceding
unsigned comment added by
95.42.185.235 (
talk)
12:40, 11 October 2014 (UTC)
-- 95.42.185.235 ( talk) 14:50, 15 October 2014 (UTC)
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This page was the subject of an MfD discussion closed on 12 May 2007, with a keep result. Xoloz 18:36, 12 May 2007 (UTC)
OK... I learned about 0.(9) being the same as 1 in high school too...
But now I have this question.
1 = 0.99999... right?
2 = 1.99999... every ok so far.
does this mean that
0.99999... = 0.999999999...888... ?
or, to put it into words, is 0.(9) the same as 0. followed by infinite 9s AND infinite 8s? is that even possible?
or, even without knowing what number it would be, one could argue the following:
Since a number, like 1, is equal to the infinitely slightly "previous" number, there are only 3 numbers: zero, an infinite positive number and an infinite negative number.
Think about it... every number has an infinitely small 'equal'. At the same time, this 'equal' should have its own infinitely small equal, and so on...
Comments? —Preceding unsigned comment added by 201.246.25.69 ( talk) 05:00, 18 January 2010 (UTC)
You have to be sure though when claiming that it is one, to point out that it is really (1 - .0000...1)
While the difference between .999... and 1 is infinitely nothing, it cannot be dismissed because it is everywhere.
.000...1 is the first thing there is greater than zero, and it's between every change between every two numbers all the way up to one. It's everywhere, but it's nothing. I believe that it is the
graviton number, but I cannot prove it. --
Neptunerover (
talk)
16:31, 21 January 2010 (UTC)
This is from the infinitesimal article: "In the 20th century, it was found that infinitesimals could also be treated rigorously. Neither formulation is wrong, and both give the same results if used correctly." My suggestion is that if both formulations work, maybe one formulation might work better for certain calculations. Using a number set that contains all infinities is better than one with no upper limit to contain anything, at least when you have to deal with infinities. In figuring out gravity, they might be better off using a set with limits, although still infinite. -- Neptunerover ( talk) 06:05, 22 January 2010 (UTC)
How does (.999...9998) differ from either (.999...) or (.999...9997)? -- Neptunerover ( talk) 09:22, 23 January 2010 (UTC)
There's the old story of "How long would it take to get to X if each day you traveled 1/2 of the remaining distance between you and X?" According to the question itself though, each day when you are through traveling, there will still be a remaining distance to the 'goal' equal to that which you just traversed that day. I don't see how this could be any different if you went 9/10 of the distance each day rather than only half way, other than getting closer faster, because the entire distance to the goal is still never spanned. I suppose that's very different than going 9.999.../10 of the distance each day. In that case the second day would be a very puzzling day. -- Neptunerover ( talk) 12:19, 26 January 2010 (UTC)
Neptunerover - there is no difference really. But I'm not sure what conclusion you're trying to draw? The halfway example presents a geometric series: 1/2 + 1/4 + 1/8th + 1/16th etc...and is equal to 1. Do you contest that? Because the same is true for 9/10: 9/10 + 9/100 + 9/1000.... = 1. Your specific application with walking I think is related to http://en.wikipedia.org/wiki/Zeno%27s_paradoxes but not really related to any equality or inequality with the given series. 76.103.47.66 ( talk) 09:11, 24 February 2010 (UTC)
Why is the following not valid? .9 does not equal 1; .99 does not equal 1; .999 does not equal 1; etc.
I think there is a paradox here. Proofs to the contrary do not make the proof above invalid. Why is this intuition not acceptable? Tristan Tondino ( talk) 02:01, 5 February 2010 (UTC)
a =√4 a=2 a=-2 2a=-2a a=-a This may be contextual of course, or just wrong. Tristan Tondino ( talk) 02:35, 6 February 2010 (UTC)
I'm sorry if this has already been discussed, but can't we just say that 0.999... does not equal one because of this function? I'm not arguing anything, just asking a question.
Floor m.n where m and n are strings of digits, is always m. So doesn't that mean floor (0.999...) = 0? Floor (1) = 1. —Preceding unsigned comment added by Goldkingtut5 ( talk • contribs) 06:49, 11 February 2010 (UTC)
Here's another way. Since 1/0 equals infinity, then 1 = 0*infinity, and 1/infinity = 0. 0.999... plus 1/infinity = 1, then 0.999... plus 0 equals 1, and 0.999... = 1. 24.1.201.172 ( talk) 01:56, 21 May 2010 (UTC)
Firstly, 1/0 is not infinity but is undefined. Secondly 0*infinity is indeterminante. 1/x as x->inf is 0. So 0.999...+ 1/x as x->inf = 0.999... = 1. This is just saying x + 0 = x because 0 is the additive identity. 86.172.185.252 ( talk) 16:15, 26 August 2010 (UTC)
I have a TI-34 II calculator at home and it has the ipart function and the fpart function. The ipart stands for integral part, and if 0.999... was inputed into the calculator with the ipart function thingamajig (like this: ipart(0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...)), it will still be 0. If you do 1, it will come up as 1. Therefore, they aren't the same. 24.1.201.172 ( talk) 02:07, 21 May 2010 (UTC)
I give reason against my brother that 0.999... is equal to 1. The reason why is because:
0.999... = the infinite series 9/10 + 9/100 + 9/1000 + 9/10000 + ... , which converges to 1. Sixeightyseventyone ( talk) 01:52, 22 May 2010 (UTC)
I see that many people are struggling to understand why 0.999... is equal to 1
I see many arguments that 0.9 is NOT 1, neither is 0.99 or 0.999 or 0.99999999999 ect... so why is 0.999... (repeating 9's) the same as 1?
It is because many people have a hard time truly understanding what infinity is. It is not a REAL number, there is no way in a string of digits like 489238902 to express infinity. What is happening with the above question is in your mind your are assigning and end to the string of 9's at some REALLY far away point. You are picturing (whether you know it or not) infinite 9's as being really really long, but subconsciously you are picking a distant point and stopping the string to say well if 0.99999 is not 1, why is 0.9999999(billions and billions of digits later)999 equal to 1? But it is not - the moment the infinite string of 9's ceases, it is NOT equal to 1.
A very important aspect of Decimal Notation (which is the number system we use) that you should understand is what numbers really mean. For example, the number 257 - we dont have a digit for that number, instead we use the single digit numbers 0-9 to express all other numbers in the system, and the way we used those digits to represent numbers was by creating "place values" - the one's place, the ten's place, the hundreds place and so on. Also, each place value is measured in magnitudes of the number 10. The ones place is 1, the tens place is 10, the hundreds place is 100 and so on
So what does 257 really mean? 2*100 + 5*10 + 7*1 which is 200 + 50 + 7 = 257 What about a number with non-0 digits after the decimal point? For example the number 624.7 - 6*100 + 2*10 + 4*1 + 7/10. For any non single digit value, you are REALLY expressing a sum.
The decimal place values are the tenths, hundredths, thousandths, and so on. They are expressed like this - 0.256 is 2/10 + 5/100 + 6/1000. So 0.999... is expressed - 9/10 + 9/100 + 9/1000 + 9/10000...
Ok? Now here is the part where you ignore your intuition. If you are absolutely stuck that all the mathematicians are wrong - then there is no point in trying to understand. All I am trying to do is explain it to people who want a way to understand it without using insane proofs that people with a limited math skill can't understand. I am going on the principle that people reading this at least a decent arithmatic skill and understand math to a very limited degree.
Ok - so the next thing about our number system you want to notice is exactly how we express decimals as fractions. Most people understand that 0.3 is the same thing as 3/10. How about 0.47? It is 47/100 right? You can go as far as you like - 0.12398 is 12398/100000 , 0.119 is 119/1000 and so on.
You can also express the value of a REPEATING fraction in a similar way. Before we used the place value to find the denominator (as we saw 0.47 was the fraction 47/100 because the "7" is the end of the fraction and it sits in the hundredths place). In dealing with repeating decimals, we use the maximum value of those place-values. An easier way to say that, is we use the place value minus 1. For example - we all know what 1/3 is. We also know it is the same as 3/9 - right? well 1/3 is expressed in decimals as 0.333... So you see the repeating digit there is "3" and it is only one digit. so it is not 3/10, that is wrong. It is 3/9 (which can be simplified to 1/3).
Take time, make sure you understand this part, convince yourself it is true with some examples. how would we write the following as a fraction? 0.454545.... Well we have two repeating digits this time - "45" is repeating, and that takes up 2 place values, but its not 45/100, that is wrong - The correct fraction is 45/99. (just a quick note, that is why before I said the max value of the place holder, because in the hundredths place, the max value is 0.99... it is also 100-1 if that is easier for you to remember). How about 0.111... Well we know its only one digit repeating, and so we use the one's place to determine our denominator because there are no other repeating digits after the first "1". So the proper fraction is 1/9. Go ahead and type 1/9 (1 divided by 9) in a caclutor and see what you get. It will be 0.111... One more because this is essential - what about 0.888... how do you think we write that as a fraction? Well you probably should know by now otherwise you really wont understand the rest. The answer of course is 8/9. Go ahead and check this out for yourself.
So what does all this mess have to do with 0.999... = 1? Well I want to make sure you understand what these digits REALLY mean.
Because you will see that what I have told you is true in ALL cases. Period.
So... what is 0.333... It is 3/9 right, we learned that. that simplifies to 1/3 what about 0.666... It is 6/9 right - and it simplifies to 2/3 finally how about 0.999... It can only be... 9/9 - right? and that simplifies to 1.
Does it make sense at first? No. It is super easy to understand? Of course not.
Sometimes you have to simply put your "common sense" notions aside. Because common sense only applies to what YOU are used to. It does not obey the laws of nature. Lots of things do not SEEM like the make sense, but when a proof is put down, it must be considered true - unless someone finds a way to completely disprove it. But really, once you get used to it, this idea is not that hard to see.
Try to understand that Mathematics was not invented in one quick swoop. No one just threw all this into existence overnight. We needed a number system, just a way to number things, count stuff, you know, basic easy ways to help keep track. Then we thought, well what if we want to use these numbers to figure something out? for example - multiplication didnt just exist out of no where. Numbering was invented, and people figured out rules for adding those numbers, because adding is helpful to our daily life. Then someone noticed when you have to add the same number over and over and over again it was tedious. 2 + 2 + 2 + 2 + 2 Who feels like doing that over and over and over again. So a shorthand was invented called multiplication.
Over time, we began to explore these numbers more and more. Finally we ran into strange numbers like pi and e. We ran into things that we couldn't make sense of, like dividing by 0. And we found unusual things that challange our common sense notions, like 0.999... = 1.
You see? It doesn't have to make sense. It's just a product of our invention of mathematics. Perhaps if some other system was invented we would not encounter things like this. It only exists because of OUR construction of the decimal notation system. It's not some official rule of the universe that would be here even if humans weren't - its just some funny little outcome that happened to exist due to our methods and design of the decimal notation numbering system.
I hope at least SOME people can understand better. I tried my best to explain it very slowly and simple - If you couldn't understand than I apologize, I am very advanced in Mathematics and sometimes going back to the basics of the basics is harder than the ridiculous stuff I am researching now. Alex DeLarge 10:03, 25 May 2010 (UTC) —Preceding unsigned comment added by A.DeLarge23 ( talk • contribs)
What are you talking about? That wasnt any self-promotion, or opinion piece, it was an explanation of why 0.999...= 1 Without complicated proofs so that people who are not math experts can understand. I can offer any number of proofs for you, but unless you have a background in Mathematics you will not have a clue what I am talking about. I noticed a lot of people were confused and bothered by this fact, so I took time out of my day to help all those people who never went far beyond arithmetic or basic algebra to at least have some sense of understanding. How are the facts that 0.3 is 3/10 and 0.333... is 3/9 ect.. opinions or self promotion? that is ridiculus and uncalled for to accuse me of this. I simply stated FACTS. And then I followed up with a brief explanation of why it seems like it violates common sense and why these things pop up to begin with, by explaining that the number system was our invention that was originally needed for counting, and we kept taking more and more steps.
Those are not opinions. This is an argument page anyways, so isn't this a place to make your arguements based on 0.999... being equal to 1? Well that is what I did. And Tkuvho didnt have any point, how could I miss it?? He said "can you summarize" and I said no, not without adding proofs that people won't understand. However Trovatore if you think I am violating Wikipedias terms than I am removing this entry - now no one will get an easy explaination, they can go take 5 years of math and understand the proofs. And from now, I won't add anything that is 100% factual based, which is .... everything I would write. 173.62.181.145 ( talk) 04:32, 1 June 2010 (UTC)
But this IS the arguement page - and I wasnt debating any mathematical facts. everything I said IS a fact, there was no debate. It was just an argument of why this is true, using simple terms. How is what I said ANY different than the people who pose the other proofs? I just did it in a more simple way because I saw many people here asking WHY because they dotn get it, because they cant read the proofs. I was offering my help to those people by stating the FACTS of mathematics, but in simple terms. NOTHING I said is not a fact when it came to the math.
0.3 IS 3/10, 0.333... is 3/9 and 0.999... is 9/9 or 3/3 or 1. These are not my interpratation of the facts. These are FACTS. These are the basic facts of our number system. It is all true, and it is a simple way for people to see it, as opposed to the algebra proofs and calculus proofs. Not everyone has this knowledge, so there are NO Facts I am debating, that is ridiculous. I have been studying Mathematics for my whole life. I Can offer you all kinds of complex proofs that are also facts - how is that different than offering proofs in a simple form, which is already on the Argument Page, this is NOT The talk page, that is why I posted here - it is another argument of why 0.999... is 1 - using the most basic of facts dealing with our number system. Not everyone took calculus or algebra to any real extent. What about those people? Are the screwed? Thats bullshit. I believe EVERYONE has the right to understand this. Thats what math should be about, it should be for everyone, not just someone who has a extended background. Go look in a 3rd grade text book and you will see these facts, that 241 is REALLY a statement of place value. Its how our number system works. And look up how we express decimals and fractions - .333 is 3/9 and .45454545454545.... is 45/99. It was just a way to help people. If anyone is going to argue that I am debating facts, or that I am biased, well thats ridiculous and wrong, this is an arguement page so thats what I did, i proposed another argument to the truth of 0.999...=1 using simple terms. If Im going to be accused of non-sense like trying to put up mathematical facts for debate then I want my entry deleted, and I Will keep deleting it over and over until it stays deleted. I took time out of my day because I wanted to help those who truly are bothered by this notion and want it to make sense. If wikipedia is not the place for people to find and understand information, then I will take my time elsewhere. 173.62.181.145 ( talk) 15:50, 3 June 2010 (UTC)
Note that if you use binary, 0.1111... equals 1, and for octal: 0.777... = 1, ternary: 0.222... = 1, hexadecimal: 0.FFF... = 1, etc. 24.1.201.172 ( talk) —Preceding undated comment added 18:50, 6 June 2010 (UTC).
That section just proved that infinity equals -1. 24.1.201.172 ( talk) 19:05, 6 June 2010 (UTC)
Huon's argument [ above] that ends with ".999 does not equal 0.999...;" attempts to prove that the argument form is invalid. Instead it seems to say to me that .999... actually has no meaning at all. That seems sensible to me: How can you use infinity to define a real number when real numbers do not include infinity? Algr ( talk) 09:49, 9 August 2010 (UTC)
It's simply a falacy. 1 = 0.9999... is not true. If you assume that false axiom as true then anything can be demonstrated. In fact, 1 is aproximately equal to 0.999... If you substitutes "=" with "~=" then axiom is true and maths match again. Even clearer:
1 > 0.999...
And then every affirmation is true and everything recovers sense again. —Preceding unsigned comment added by 88.0.131.223 ( talk) 10:38, 15 October 2010 (UTC)
I don't think it's needed. It's not really a proof of anything, more a challenge to an incorrect assertion. It could be used as a basis for longer and more formal a proof by contradiction, but such are usually used only needed if a more straightforward proof is unavailable, and there are plenty of more straightforward proofs in the article. And it's OR, or at least I've not seen it written down anywhere.-- JohnBlackburne words deeds 14:37, 15 October 2010 (UTC)
On the talk page, 210.4.96.72 asked if someone could prove the following statement:
There's one caveat: I'm not quite sure what he means by "summation of the sequence". Let's first introduce some notation: Let a0=0.1, for k>0 and for k>=0. Then the sequence (a0, a1, a2, a3, ...) is the sequence (0.1, 0.01, -0.1, 0.001, -0.01, ...). I'll prove that where I use the standard definitions for "sum of a series", which may not be what 210.4.96.72 had in mind. Namely, by definition, where finite sums can be formed simply by adding the terms. Despite the similar notation, infinite series actually aren't formed by adding the terms, since addition of infinitely many terms isn't defined. Furthermore, by definition of the limit, we say that for a sequence (sn), we have if for every ε>0 there exists an n0 (depending on ε) such that for all n>n0, we have |sn-s|<ε.
So with these definitions and the notation from above, what I'll have to prove is that for any ε>0 there exists an n0 such that for all n>n0, we have
Next, note that a0+a2 = 0.1-0.1 = 0, while for any i>1 we have a2i-1+a2(i+1) = 10-(i+1)-10-(i+1)=0. Thus, quite a few terms of the finite sums I'm about to look at will cancel out:
Now we can combine the parts: Let n>n0 be any (natural) number.
Both cases need fact 1 for the very last equation (though case 1 could do with a weaker version), and each case also needs one of the other two facts for the second equation. QED. Huon ( talk) 20:32, 20 October 2010 (UTC)
I would nominate this article for "worst article on wikipedia". This article shows how the ignorant rabble can outvote the truth. No matter hown many people say the earth is flat, it's a sphere. No matter if Zeno says Achilles can't beat the tortoise, Achilles will win. No matter how many people continue to assert that 0.999 ... = 1, they are not equal equal. The *** limit *** of 0.999 ... is 1. That's it. Unfortunately, most people don't understand limits. It's not easy for the non-mathematician to understand. Most are so numerically challenged, they cannot even understand simple math concepts. Limits are far beyond them.
The article makes silly, pseudo-intellectual statements such as "the question of how two different decimals can be said to be equal at all". The article supposedly addresses "alternative number systems", but fails to mention other base systems. The reason many "initially question or reject it" is because it is not true. The business about "math educators" trying to figure out why students see through the fallacy is hilarious. The "math educators" are just showing their math deficiencies. "Has long been accepted by mathematicians" has no citation. The "proofs" shown on the page are ludicrous, and prime examples of logical fallacies, on the order of Zeno's paradox. The "proofs" remind me of the proofs of God's existence. Only one proof should be needed. Instead, many proofs are offered, because each one is flawed.
This article is a classic, and would be extremely humorous for any mathematician to view if it were satire. Since it's presented as fact, it's very disheartening to view. For those who would say "then edit it!", I tried that long ago. It's a total waste of my time. wikipedia is not peer-reviewed. It's rabble-reviewed. 174.31.157.82 ( talk) 17:24, 8 December 2010 (UTC)
Well, 174.31.157.82, since you think we are numerically challenged, allow me to pose a numeric challenge to you. Here's a table.
Numerator | Denominator | Decimal representation |
---|---|---|
1 | 9 | 0.111... |
2 | 9 | 0.222... |
3 | 9 | 0.333... |
4 | 9 | 0.444... |
5 | 9 | 0.555... |
6 | 9 | 0.666... |
7 | 9 | 0.777... |
8 | 9 | 0.888... |
x | 9 | 0.999... |
Solve for x. Presumably you think x is not 9, so I would be curious to know what you think it is. 28bytes ( talk) 20:55, 8 December 2010 (UTC)
A simple question: |
---|
y>1 |
y+x≥1 |
Solve for x |
This question does not involve any infinities or "strange objects" or even any repeating decimals, and yet the real set falls apart over the idea. Algr ( talk) 04:27, 4 January 2011 (UTC)
In reply to the "simple question": The set of solutions is the set of pairs (x,y) such that x≥1-y and y>1. That's a well-defined set of real numbers. Obviously there is not a single set of x that will be the full set of solutions independent of y, though any non-negative x will solve the equation for every admissible y. So what?
Concerning limits and numbers: By definition the number a is the limit of the sequence (an) if for every real number ε>0 there exists a natural number nε such that for every n>nε we have |a-an|<ε. Thus by definition limits are numbers.
Concerning the "wrong" equations: Algr, what do you think is the decimal representation of 1/3? Do you think that 1/3 doesn't have any decimal representation? Wouldn't a system of representations that does not represent all your numbers be quite useless?
Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation, but doing so will hardly involve 0.999... - rather, one can use long division to divide one by three. The technical details may be hidden, but there is no circular reasoning.
Huon (
talk)
11:56, 4 January 2011 (UTC)
"Finally, the 1/3=.333... proof isn't circular. It assumes that you have independently verified that equation -Huon" But that is the very definition of circular. When the meaning of infinitely recurring decimals is questioned, it is insulting to simply point to another infinitely recurring decimal and assume that that one shouldn't be questioned as well. Algr ( talk) 08:54, 31 January 2011 (UTC)
As I said three years ago:
Removed. There is no reason for Wikipedia to be providing a platform for Anthony; it's time for him to go elsewhere. I'll be removing all his nonsense as it appears. -- jpgordon ∇∆∇∆ 15:40, 23 October 2007 (UTC) [1]
-- jpgordon ::==( o ) 17:52, 30 January 2011 (UTC)
So explain this,
We take it as given that .9999... has to continue for infinity in order to equal 1? Why is that? Isn't it because no ammount of 9s after the decimal point could represent 1 exactly? Therefore it must continue for infinity in order to be considered an exact representation. But isn't inability to physically stop and the need to continue appending 9s the very definition of an approximation? Not unlike we consider 3.14159... to be an approximation of pi. It doesn't matter that .999... follows a pattern when .314159 does not. Both could be theoretically calculated to an infinite precision. This suggests that there is a difference between the representation .999... and actual endless .999999999999999999999999999999999999999. But both .999... and the longform both remain with the exact same problem: while .999... conveniently adapts non-practical idea of infinity even at infinity still remeains but an approximation because there is no ends for infinity. (This, by the way suggests that infinity itself isn't really a number but a process but I won't go into that.).
Also:
1/3 = .333333333... therefore .3333333333...*3 =.999999999... = 1 is an example of an something that doesn't really explain anything. At least not to everyone. Why should a person who doesn't take .999999999... to be a number that represents 1 feel any differently about .33333333333... accurately representing a third of 1? In other words, this proof starts out with an assamption that is just as questionable as what it is trying to prove.
Why should 1/0 = x therefore x*0 = 1 be considered to be inaccurate whereas the example above is taken to hold true? I know the result of 1/0 is undefined but it is a precedent for laws of division and mathematical operators being not 100% reliant. From the point of view of mathematical operations we have a very similar set-up.
Another thing, to people insisting that .999... is a number and not a process. I have no problem grasping the concept of infinity and infinite repetition but answer me this then, if the fact that the full form version of that number (i.e. not the shortcut .999... form) with endless nines cannot actually be be finished being written down (or said, or, what that, matter USED) than how can you say it is a number in this when you cannot even finish it in this form?
Again, the obvious conclusion is that something that cannot be finished is an approximation by definition (again, I'm not talking about .999... form here but the full meaning behind the form). —Preceding unsigned comment added by 198.30.78.254 ( talk) 16:03, 9 February 2011 (UTC)
The point is this. Such important feature as invariability of significant number length for changes of this number in result of some mathematical operations is not taking into account in these calculations. Multiplication of decimal fractions on 10 is execute as well as for integers. This implyies what besides transposition of the decimal point on one numeral to the end of number the zero is added. Distinction is that non-significant zero in decimal fraction can be cutting. Interestingly that the infinite circulating decimal can be written in the form when direct mathematical operations with low-order digit are possible. The example of manipulation with numbers must be written as follows:
x=0,999... x=0,(9)9 //Recording of infinite circulating decimal in the fit form for mathematical operations with the low-order digit// 10x=9,(9) //Number shift in the register, the low-order digit accepts zero value and cutting// 10x-x=9,(9)-0,(9)9 9x=9-0,(0)9 x=1-0,(0)1 0,999...=0,999...
Consequently "a" was equal 0,9999... and has remained equal 0,9999... because such important value as infinitesimal decimal unit was missing in the previous calculations. It was infinitesimal blunder. — Preceding unsigned comment added by Leonid 2 ( talk • contribs) 09:00, 4 March 2011 (UTC)
I have read about my explanations - "I find it quite impossible to fathom what this is supposed to mean". "I belive this is wrong". Probable cause is my bad English or the explanation was too short in spite of clearness. There are Hemming’s words - "The purpose of computing is insight, not numbers".
The program of information processing called Digit manipulation contain very gross bug connected with infringement of the order of performance of mathematical operations. Mathematical operations of addition, subtraction and multiplication of decimal fractions execute in the certain order and it is indifferent when these operations execute by a digit machine or on a sheet of paper. It is the order of processing from younger meaning bit to the senior. If addition of infinite periodic decimal fraction with an integer still can be written down in the form of 9,(0)+0,(9) having added to the integer an infinite chain of zeroes then in case of multiplication the problem gets fatal sense.
It seems that multiplication of decimal fraction on 10 is simple mathematical operation connected with carry of a decimal point on one digit. Actually for calculation it is necessary to execute some the consecutive strictly certain instructions. It is necessary to write down number, necessarily to add 0 by the end of record (initialization of operation of multiplication on 10, but at calculations on a paper usually it is not considered) and to add a sign on executed operation (for example, ^).
But number 0,999... hasn't the end. Correct operation of multiplication is prohibitive in this case.
If number was written 0,999...=0,(9) or such as of dimensionless linear array 0,(9)=0,(999...999) correct operation of multiplication is prohibitive too.
10*0,(9)=>0,(9)0^=>0,(9) //empty operation because shift on step =0 is impossible act//
For reception of an opportunity of shift of all array you must create control element (buffer entry). You must receive rights for changes most lower digit. Number 0,999... is infinitely, but if it had the end, it would come to the end on 9. Therefore number 0,999... write such as 0,(9)9 or execute the operation of global unconditional assignment.
x=0,(9)9
Now indifferent if quantity of nines in the array (9) is endless, this array will be in computing such as unit and addition even only one 9 activate of the array overflow.
10*0,(9)9=>0,(9)90^=>9,(9)0=>9,(9) //execution of instruction//
But after operation of global unconditional assignment it is impossible to get rid from buffer digit for operations of addition and subtraction.
10х-х=9,(9)-0,(9)9=9,(9)0-0,(9)9
After the further calculations we receive very important volume - infinitesimal decimal unit - 0,(0)1.
x=1-0,(0)1=1,(0)0-0,(0)1
It is interesting element with original features. It is the mathematical tool which allows to formulate precisely many the existing mathematical proofs and it will be useful in exact numerical calculations.
Really 9*0,(9)=8,(9)1. The operation 9x=9 in programm called Digit manipulation is the operation of local un/conditional assignment x=1.
I found the bug in programm called Digit manipulation. Corrected result has the dead level 0,999...999==0,999...999 and verification can be. Probably I made a mistake. People said in Ancient Rome - "Errare humanum est". I ask argue away computing without words "I believe" and "I trust what 0,(9)=1".
My English isn't Shakespeare's language. Probably some considerations is no correct in English. I left text of this message on Russian on my talk page in English Wiki.
Leonid 2 (
talk)
11:10, 7 March 2011 (UTC)
"You seem to do mathematics from the perspective of computer science." Yes, of course. Computer science is the part of mathematics. Properties of part can't differ from properties of whole.
9*0,(9)=? //Question. How many will be, if multiply 0,(9) by 9?// 9*0,9=8,1 9*0,99=8,91 9*0,999=8,991 9*0,9999=8,9991 9*0,99999=8,99991 ............... 9*0,(9)=8,(9)1 9*0,(9) no equivalent 9 //Verification//
Do you have more questions? Leonid 2 ( talk) 12:25, 11 March 2011 (UTC)
I'm not want the discuss about the full article 0.999... . I do not care 0,(9) is equivalent to 1 or not. But the section of the article entitled Digital manipulation contains an obvious error which must be corrected. This error can be detected by direct calculations on a sheet of paper. In the multiplication 0.999 ... by 9 you will never get 9, no matter how many nines after the decimal point will be, and this should be displayed on the main page of the article. Leonid 2 ( talk) 08:08, 19 March 2011 (UTC)
I asked to refute my computations. I got the misty arguments about the properties of infinity about the properties of a real numbers and so on, instead of other computations. But I'm not want the discuss about the full article 0.999... and catch the wind in a net. Therefore once more.
9*0,999...=? //Question. How many will be, if multiply 0,999... by 9?// 9*0,9=8,1 //Start of the computing// 9*0,99=8,91 //Continuation of the computing// 9*0,999=8,991 9*0,9999=8,9991 9*0,99999=8,99991 ................... //Continuation of the computing// ..................... ....................... //Stop when you will be tired or are finished these computations// Шутка 9*0,999...=8,999...9991 //Answer. Any can enter the infinite number of nines instead of the dots by oneself//
Such decision is possible even for a scholar from an elementary school. In the multiplication 0.999 ... by 9, you will never get 9, no matter how many nines after the decimal point will be. The section of the article entitled Digital manipulation contains an obvious error. 9x=9, if x=0.999... it is the blunder. I wrote about this in detail above how this blunder arises.
Don't you agree with me? It may be. Which means that the listing must be of the your step-by-step computations 9*0,999...=9.
1/0.(0)1=1(0).0
Leonid 2 ( talk) 09:11, 22 March 2011 (UTC)
Thank you. You have talked so much about the properties of rational numbers for me. Finally, I understood everything. The number 0.999... is not rational. It is the recurrent decimal fraction. It can not be represented as simple fraction. In converting simple fractions to decimals you may lose some information irreversibly. The numbers of multiples of 10 can not be divided by 9 or 3 evenly. 10/9=1+1, 100/9=11+1, 1000/9=111+1, 10000/9=1111+1, ... (10/3=3+1, 100/3=33+1, 1000/3=333+1, 10000/3=3333+1, ...).
The infinitesimal decimal unit 0,(0)1 proved to be useful once again.
1/9=0,(111...111)1+0,(000...000)1 1/3=0,(333...333)3+0,(000...000)1 0,111...=1/9-0,(0)1 0,333...=1/3-0,(0)1 0,999...=1/1-0,(0)1
The numbers 0.111... and 0.333... is not rational too. Addition and subtraction with decimal infinitesimals for the such numbers is admissible mathematical operation. Special form of recording is compulsory regulation in this case. Some features of the infinitesimal decimal unit 0,(0)1 on my talk page -
Russian and English text.
But I do not care 0.999... is equivalent to 1 or not as before.
Leonid 2 (
talk)
10:08, 30 March 2011 (UTC)
I calculated quantity of solutions for distinction 0,999... and 1 on my talk page
"Quantity of solutions for distinction 0,999... and 1" for two cases (if quantity of nines after decimal separator is infinity and if quantity of nines after decimal separator is the more of infinity). General quantity of solutions for distinction 0,999... and 1 is equal ∞+2 solutions including solution in Lightstone's extended decimal notation. Probably 0,999... is not equal 1. Confidence is more 100%. I ask argue away these calculations only by other calculations.
Leonid 2 (
talk)
07:10, 10 April 2011 (UTC)
ok, and apologies if this has been covered before. An informal 'handwaving' argument but one that might help overcome some intuition blocks:
It should be obvious to all that 1.1 > 1. Similarly, 1.01, 1.001, 1.0...01 etc are all >1 for all such numbers with a (finite) string of '0's.
So what happens with an infinite string of 0's? It should be obvious that 1.0...=1 (there being no 'last place' for the digit 1)
Note that the difference of 1&1.1 is the same as the difference of 1&0.9
Similarly for 1.01 & 0.99; 1.001 & 0.999 and so on.
It (informally) follows that the difference of 1 & 1.000...(equality) is the same as the difference of 1 & 0.999... Yossarian68 ( talk) 23:10, 23 April 2011 (UTC)
http://uncyclopedia.wikia.com/wiki/0.999...
In mathematical theory .9 is a specific number. It is not half or a fraction of a number. It is .9 and it remains .9 in the equation. Unless an equation calls for rounding of the numbers, then .9 remains solely .9 in the equation. Therefore 0.999_ is a infinitely repeating number.
In mathematical practice there are no exact numbers. As reality is incapable of producing an exactly equal number in relation to something. So in mathematical practice 0.999_ cannot equal 1 regardless.
In mathematical theory where rounding occurs then 0.999_ equals 1.
If rounding is not established or .9 is specifically established as being an exact number, then no amount of crying and bitching changes it to 1.
You've been Uncyclopediaed, yo. 58.7.214.181 ( talk) 03:50, 19 May 2011 (UTC) Harlequin
Perhaps you should go back and read where I specifically, in detail, referred to equations where it wasn't being "rounded". But if all you can come up with is "no, no...if they are not infinitely repeating, then it totally equals 1", it's just further proving my point. 58.7.214.181 ( talk) 04:03, 19 May 2011 (UTC) Harlequin
I'm using Uncyclopedia as a reference? Haha, well done at further proving my point and again failing to read. Strange, seems I explained exactly how 0.999_ does not equal 1 without anything relating to the Uncyclopedia article on the subject. I merely pointed out another site that makes fun of the bitching and crying those like yourself are doing.
I gave the equation separate from any other site. Which you should have known had you, you know, actually read anything. Hahaha.
0.999_ repeating in mathematical theory can only be 1 if it is not infinitely repeating. Your claim here is that somehow an infinitely repeated number will magically have an end and that it will magically round up once it has reached that end, even when the equation it is in specifically states it doesn't. Like ive stated, no amount of bitching and crying changes that fact that an infinitely repeating number does not have an end.
It's pretty hilarious really. I even explain that where it isn't infinitely repeating and is in an equation that "rounds" the numbers that it would equal 1 (again, had you actually read anything) and yet you seem to think that an equation where it can't even reach an end to "round" would somehow still do so even when it's specifically mentioned that none can occur?
Try again. This time without trying to take credit for other sites material. Especially when the best "argument" (or lack of) you can come up with is that "some guys" created the other article (despite it's refutation of "their" supposed claim that 0.999_ ALWAYS equals 1) and because "they did" it is therefore magically wrong. 58.7.214.181 ( talk) 04:39, 19 May 2011 (UTC) Harlequin
I notice you don't know what infinite means. There is no "end", so there can't be a "last 9". Try again. Especially since I didn't state anything about Uncyclopedia being correct and specifically provided a detailed explanation of why the claim "hurr 0.999_ always equals 1 no matter what" is a false statement. 58.7.214.181 ( talk) 04:39, 19 May 2011 (UTC) Harlequin
This is such a fascinating argument to me. As someone who's knowledge of mathematics goes about as far as long division I'm perfectly willing to accept the fact that 0.9 recurring equals 1, simply because I defer to the wisdom of people much smarter than me in this subject. If mathematics professors say it's so, who am I to judge? But the moment someone get's a little bit of knowledge about this subject they think they have what it takes to prove the experts wrong. Why not just accept it and move on? -- 86.182.1.71 ( talk) 18:39, 15 June 2011 (UTC)
.9 repeating is just another way of writing 1 but with all its lesser place values being presented. Usually one is simply given the maximum place value: 1/1, but obviously the number can contain any combination of lesser place values. 1/10ths, 1/100ths etc. For example 100/100 is 1 too but for the 1/100ths. 900/1000 + 10/100 = 1 representing place values 1/100ths and 1/1000ths. Of course there is obviously an infinite amount of lesser place value delimitations intrinsic to the number 1 and these will add up to 1 just like any smaller finite combination. And when we list out these additions we get 9/10 + 9/100 + 9/1000... etc. Unless you think it is mathematically impossible to represent every possible lesser decimal place value you basically have to admit .9 repeating is exactly 1. 76.103.47.66 ( talk) 00:30, 6 August 2011 (UTC)
1 is not an "infinitely repeated number" and "Harlequin" obviously did not know anything about mathematics. — Preceding unsigned comment added by 91.185.1.174 ( talk) 12:56, 6 September 2012 (UTC)
Hello. I am not terribly knowledgeable in mathematics. Some years ago I posted here a comment, which used non-conventional notation, trying to reason that 0.(9) couldn't be equal to one, because then all numbers would be equal to all other numbers (if you nullify the difference, even infinitesimal, between two numbers, then what's to stop you from generalizing that to the whole real axis?). I received then two comments. One simply said that I was using non-standard notation and so my argument didn't follow. Another one said that it was possible I used it, in so far as I was actually able to use it and convey my meaning, and that there might be some reason to my assertion. He then directed me to some higher mathematics I couldn't understand (p-adics).
But enough with introductions. I have revised my thoughts in various ways, and I am also now able to present my simple argument using conventional notation. So here it is:
Consider the sequence 0.9, 0.99, 0.999, 0.9999, 0.99999, etc. It is obvious to anyone that the limit of this succession is one. But the definition of a limit tells us that this limit can never be actually reached. Therefore there is never actually an equality between 0.(9) and 1, except in the limit itself.
The same holds true to the sequence 0.1, 0.01, 0.001, etc. It's limit is zero. Therefore the difference between 1 and 0.(9) is only zero in the limit.
I think that a parallel could be drawn to the point in the exact center of a wheel. I think it can both be said that it rolls or that it doesn't roll, depending on the exact context. There may be a confusion regarding two different "levels" of reality, that is, a confusion between an assertion and a meta-assertion.
Joao.g.madureira ( talk) 16:48, 17 October 2011 (UTC)
Here's a different reply to João, which I hope will help convince him. João, there's an incorrect assumption in what you say, namely that a "limit can never be actually reached". That is 'often' true of limits, but not always. As a counterexample, consider the sequence a(n) = (1+(-1)^n)/n = 0, 1, 0, 2/4, 0, 2/6... Its limit is clearly 0, and a(n) does take on this value, infinitely often in fact. I suspect that this misconception about limits is part of the trouble many people have with the equality discussed in this article.
Another point, already made by Huon, is that by "0.(9)" we really mean the limit of the sequence (0.9, 0.99, 0.999, 0.9999, ...), rather than the sequence itself. So in a sense you're right that there are two distinct concepts at play here, the sequence on one hand and its limit on the other. What you (and many others) appear to misunderstand is that "0.(9)" by definition denotes the latter, not the former. Let me repeat that: by definition, 0.(9) = lim(0.9, 0.99, 0.999, 0.9999, ...) -- not 0.(9) = (0.9, 0.99, 0.999, 0.9999, ...)! Though this is a technical point, and so perhaps not as persuasive as my previous one. Regards. FilipeS ( talk) 14:49, 6 February 2012 (UTC)
I'm sorry, I do not normally log in to Wikipedia because I do not edit information, mostly because I only come here to learn about things I do not know about (and how can I add to a topic I don't know about?) but also because I do not know how all of this actually works. So if I am doing something completely and horribly wrong, then I apologize, I just needed to say this. I also do not really have more than a high school education in mathematics, so I am not any sort of mathematician expert, nor even proficiently learnéd in math. Although, I have tended to have an unusually extraordinary understanding of mathematics, especially such basic mathematics as converting fractions to decimals.
However, I cannot accept the fact that 0.99999... = 1 and after reading the proofs, I understand why. The proofs all make sense, but I have the reason/problem which causes this to happen, and it is because infinite decimals are not perfectly accurate representations of fractions, which is why they are infinite. They go on forever because those fractions cannot be created into decimals. 1/3 become 0.33333.. and so on forever, because this is the -closest- representation of it that we can create with a decimal, but it isn't perfect. Therefor, 0.9999... is the closest representation of a fractional 1 that we can produce with a decimal, however it does not truly equal 1 because it is more like an estimation or rounded number.. it is like rounding 4.9 and 4.9 both to 5 and adding them together to make 10, and saying that 9.8 = 10 because 4.9 + 4.9 actually equals 9.8. That is the same as "rounding" 1/3 into 0.333... and so on forever. Therefor the actual number 0.999... itself is NOT truly equal to 1, it only seems that way if you multiply 1/3 * 3 as 0.333... * 3 equaling 0.999... because 0.333.. is "rounded" and therefor like the 4.9's I rounded to 5. I don't know if that makes any sense to anyone but it makes perfect sense to me, and I can try to explain it further if no one is understanding me. I am also unsure if there is any way to make a proof for this because there isn't much of a way that I can show that infinite decimals are rounded or estimations, other than by using the proofs already here.
Obviously if 0.333... * 3 =/= 1, then 0.333... is not a perfect representation of 1/3 in decimal form. It is just the closest. — Preceding unsigned comment added by Sweetnaivety ( talk • contribs) 06:43, 27 October 2011 (UTC)
Short answer first: 0.999... - 0.09 = 0.90999... You are correct that I assumed the equality 0.999...=1 in order to calculate 1 - 0.111..., but I need not do so. Consider the following:
There is no missing "9" at the end because there is no end to the eights. For a more formal proof, you could employ limits and use that 0.111... is the limit of the sequence (0.1, 0.11, 0.111, ...) while 0.888... is the limit of the sequence (0.9, 0.89, 0.889, ...).
Next, you claim simultaneously that 0.333... x 3 = 1 and that "1/3 is not representable in decimal form". That obviously cannot both be true. If 0.333... is the number which when multiplied by 3 equals 1, it is by definition the decimal representation of 1/3. I assume you meant that 0.333... x 3 equals 0.999..., but not 1.
The remainder of your argument says that an infinite decimal cannot properly represent anything. We could argue whether the number represented by 0.333... is indeed 1/3 or whether it is a little less than 1/3 (and it's the former, by the Archimedean property line of reasoning I gave above). But now you seem to say that 0.333... simply does not represent any number. Why not? Why should we disallow an infinite decimal to represent something? After all, the entire point of having decimals in the first place is to have them represent certain real numbers - and the claim that just by being infinite they can no longer do so seems more of a philosophical than a mathematical objection. And it leaves you with lots of representations which do not represent anything, and conversely with lots of numbers without representations. That strikes me as rather unhelpful. Huon ( talk) 17:17, 10 November 2011 (UTC)
0.999... is equivalent to 1 in real number equivalence. They are NOT truly equivalence in mathmatical sense. We MUST use the curly equal sign to denote their equivalence, not the normal equal sign. If we mix up the two, then we can come up with a whole lot of bullshit.
Ex: If 0.999... = 1. Then 0.999... + 0.0...1 = 1 + 0.0...1 -> 1 = 1.0...1.
Since we already know 0.999... = 1, replace 1 in the above equaltion and you get 0.999... = 1.0...1.
Repeat the step infinitely, then you prove that 1 = infinity. — Preceding unsigned comment added by Ssh83 ( talk • contribs) 21:10, 1 November 2011 (UTC)
As a matter of fact, Ssh83 has it backwards. According to standard mathematical terminology, 0.999... and 1, just like 1/2 and 0.5, or 3^2+4^2 and 5^2, are, in each case, both equal and equivalent. FilipeS ( talk) 15:03, 6 February 2012 (UTC)
0.9999... is not equal to 1. Remember when you were in primary school? Well, in my primary school, they taught that a number can only be represented in one and only one way as a decimal. Anything they teach in schools is true. Sure 1 can be represented in many different ways, but as a decimal, it can only be written only one way. I am aware that 1, 01, 1.0000..., and 1.0 are the same number, but that is still the same way of representing a number in decimal form, becuase (if you remember in primary school again) if you have a number with any amount of digits in it, then there will be an infinite amount of zeroes both behind it, and after the decimal point. — Preceding unsigned comment added by Nominalthesecond ( talk • contribs) 19:16, 25 November 2011 (UTC)
Add 0.1 to 0.999...
0.1 + 0.999... = 1.099...
Now
1.099... mod 1.1 = 1.099...
Try the same with 1
Add 0.1 to 1
1 + 0.1 = 1.1
Now
1.1 mod 1.1 = 0
What now.... :)
Identity:9C7C71F43DD9DC1E604F3F21BB760A402630D9B8404B33587193D16CE90439F25C1757C2FE6C10C612BC0BF0A3CBD1FBEB5FF20B2C4F2350C85C0CBBA523464C — Preceding
unsigned comment added by
129.97.236.182 (
talk)
05:39, 2 February 2012 (UTC)
Sure, you assumed 1.1 == 1.099... and then got the result of 0, don't assume that and actually do the division using any division algorithm, You'll get 1.099...
Its true that 1.099... will act in much the same way as 1.1 in any conceivable operation other than modulus(only one I can think of at the moment).
Saying that 0.999... = 1 has with it much the same hazards as assuming 1/infinity = 0,
You're assuming that a real number can be created by dividing by something that isn't a number, and that an infinitesimally small value, regardless of its composition, is the same as the lack of a value.
By that logic x = 0.00...1 = 0.00...2 = 0.00...3 = 0.00...9 = 0.0...10 and so on, all of which equal 0
Or equivalently
That
x = 1/infinity = k/infinity,
Take the limit as K approaches infinity and you end up with indeterminate, but we've already said that 1/infinity = 0
Thus
x = 1/infinity = k/infinity = k * 1/infinity = k*0 (Which applies because you assumed that 1/infinity = 0)
Take the limit as K approaches infinity and you end up with 0, not indeterminate,
This problem arises because
0 != indeterminate != infinity
0 is a concept itself, granted, but you cannot simply equate 1/infinity to 0, and likewise 0.999... to 1 It works for all practical purposes I can think of other than modulus but it's not theoretically correct.
It is however correct to say that: The limit as n -> infinity 1/n = 0 or The limit of n/1 as n approaches 1 is 1 — Preceding unsigned comment added by 129.97.236.182 ( talk) 10:05, 2 February 2012 (UTC)
What do you mean by "1.099... mod 1.1"? Usually both arguments of the modulo operation are integers. FilipeS ( talk) 16:24, 24 February 2012 (UTC)
Thanks. Overall I agree with you, but I'm a little confused. You wrote that 1.099... mod 1.1 = 0. I can accept that, if we assume that the remainder must be an integer. But if we're using a fractional base then we should probably allow fractional remainders... FilipeS ( talk) 17:07, 7 April 2012 (UTC)
This argument will go on for a long time...all because we use base 10. Anyway...
.9999999999....=1-.0000000...1 and .9999999999...+.0000000...1=1 — Preceding unsigned comment added by Phillies9513 ( talk • contribs) 00:53, 24 February 2012 (UTC)
Am I right? You cannot deny that this isn't correct.
Plus, the common proof that since 1/9=.11111... and so on until 9/9=.99999999=1 is false because the only reason that 1/9=.11111... is because that is the closest number you can get to a number multiplied by 9 that will equal 1, in base 10. So i'm saying that 9/9 does equal 1 (duh) but 9/9 does not equal .999999.... and so therefore, 1 does not equal .999999999... — Preceding unsigned comment added by Phillies9513 ( talk • contribs) 00:52, 24 February 2012 (UTC)
In kindergarten I was told that 2 comes just after 1.Now I know that's not true,because there is 1.5,but 1 does come just after 0.999999999999............ There's just no other number between them!Think of 1/6. It equals 0.1666............6666666667 — Preceding unsigned comment added by 86.104.51.117 ( talk) 15:59, 21 May 2012 (UTC)
Every number minus itself =0 1-0.999...=0.000.....000....01 so 1 doesn't equal 0.999... — Preceding unsigned comment added by 188.25.224.87 ( talk) 08:40, 23 May 2012 (UTC)
So you mean that 0.000.....01=0 You say that 1=1.000....01.If 0.999...=1 then 0.999...=1=1.000...01 so 0.999...=1.000...01,but this is false because there is a number between them.So 0.000...01 NOT = 1,then 1-0.999... NOT = 0 and so 0.999... NOT = 1 — Preceding unsigned comment added by 188.25.220.81 ( talk) 18:59, 23 May 2012 (UTC)
Take x=0.999... and y=1. 2x=1.999..98 (that is 1.999... - 1/infinity) and 2y=2 Because there is a number between them those two are not equal so 2x≠2y. Therefore x≠y and 0.999...≠1 — Preceding unsigned comment added by 188.25.224.223 ( talk) 18:22, 12 June 2012 (UTC)
YOU DO NOT UNDERSTAND DIVISION.
A REPEATING DECIMAL IS THE QUOTIENT PORTION OF AN INCOMPLETE DIVISION - THE REPEATING DECIMAL DOES NOT INCLUDE THE REMAINDER - EVEN AS IT BECOMES INFINITELY SMALL - THAT KEEPS THE QUOTIENT FROM FULLY EXPRESSING THE VALUE OF THE FRACTION.
1/3 IS NOT EQUAL TO .333... BECAUSE .333... IS ONLY THE AMOUNT THAT IS REPRESENTED IN THE QUOTIENT.
PLEASE DO NOT DISMISS THESE FACTS BECAUSE YOU ARE NOT YET ABLE TO UNDERSTAND THEM. — Preceding unsigned comment added by Mjs1138 ( talk • contribs) 23:37, August 24, 2012 (UTC)
I know you guy are sick to death of people saying this but I do believe I'm speaking sense on this one.
The problem here is that people are visualising 0.999... as a finite concept and use finite equations to link it to a finite number (1). Although it is a finite number in essence, the representation uses a infinite concept (this is because each digit represents as value and it is the infinte chain of these values that give the number it's identity). Now this is were the equation 0.999...=1 provokes Galileo's paradox, because that we're essentially trying to do is force a infinite concept into a finite value, because taking infinity as a fixed concept always leads to contradictions somewhere along the line that just don't make snese.
Although 0.999... does appear to be exactly where 1 is what we don't take into accound is the concept that 0.999... is not the same type of number, it is construced using infinity, so as I've already said the two, following this logic, never be mixed. There are other paths of logic to follow of course that do show 0.999... to be equal to 1 but as to many things the desired 'answer' can be different depending on which logic you follow (see Exponentiation#Zero to the power of zero), which is the point I'm trying to get across. Wikipedia should never adapt one angle of view and use it as their dependance; doing as is no better than bias. All views should be taken into acount and the focus should be on the conflicts in arguments. Robo37 ( talk) 22:52, 4 October 2012 (UTC)
I am a high school student and In class we had to change 0.9999........ in to a fraction. When I done this I got 9/9 which is 1 and I got confused and thought I might research about it. I simply can't grasp the fact that 2 numbers equal the same number, I'm no mathematician but It simply makes no sense. I also think that it can't possibly be an infinite decimal otherwise it could never be represented by a fraction but reoccuring decimals can never be accuratly written (just an idea). Danjel101 ( talk) 12:02, 10 October 2012 (UTC)
Please DO NOT delete this section. It belongs here. I know you do not like it because it refutes all your silly arguments. But this is an arguments page,... https://www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c — Preceding unsigned comment added by 60.1.42.128 ( talk) 03:25, 23 January 2013 (UTC)
Furthermore, I am not proposing ANY changes to the article. In case you had not noticed, this is the ARGUMENTS page. I do not for one moment believe that ignoramuses would change their wrong opinions. Let me ask you this - why don't you delete all the other sections that are no arguments at all? On a power trip here my little man? 60.1.42.128 ( talk) 08:12, 23 January 2013 (UTC)
60.1.42.128 ( talk) 08:12, 23 January 2013 (UTC)
Your analogy of bags is so hopelessly flawed that one can't even know where to begin explaining. If a set is empty and it can contain another empty set as an element, then it is no longer an empty set. It is an ill-defined object that contains a subset and an element that are identical.
10/10 is equal to 1, so you would get 1. But 0.999.. is not equal to 1. What part of this are you unable to comprehend? k/k is ALWAYS a well-defined number, provided k is measurable.
You are the one who needs to avoid logical fallacies. Your lack of reasoning is absolutely pathetic. 173.11.135.234 ( talk) 09:21, 25 January 2013 (UTC)
I'll try to focus on the 0.999...≠1 proof because I believe the entire set theory discussion is rather irrelevant to it - no matter how we define them, we probably all agree on what the natural numbers should be.
On page 7 you argue (and I agree) that for all natural numbers k we have 1-1/10k<1. So far, so good. You then state (and again I agree) that "academics" will point out that we need to consider the limit. It's rather obvious that there's no natural number k such that 1-1/10k=0.999..., so we indeed need to do something more.
At the top of page 8 you write: . I still agree with the implication: If we had , then we would arrive at the obviously wrong conclusion that 1<1. So the premise must be wrong, and we arrive at and thus 0.999...≥1.
I had assumed (maybe wrongly) that you intended to argue that since 1-1/10k<1 for all k, we must also have . This implication is wrong. If for some sequence (Sk) we have Sk<x for all k, all we can say about the limit is that . I gave an explicit counterexample to the wrong implication that's a direct analogy to what I believed you intended to argue: For all k we have , but the fact that all elements of our sequence are less than does not imply that the same is true for the limit of the sequence. So there's no reason to believe that the fact that all elements of our sequence are less than 1 should imply something about the limit being less than 1. Huon ( talk) 17:35, 23 January 2013 (UTC)
"...but the fact that all elements of our sequence are less than does not imply that the same is true for the limit of the sequence."
In the same way, the lim (9/10+9/100+...) does not imply that the sum 9/10+9/100+... is the same as the limit of the sequence.
Look, I know that you find it absolutely intriguing that I chat with you on these pages. But really, I do not want to nurture a discussion with anyone. My suggestion is that you do not remove the link to the article, so that others can learn a view that is different to yours.
I do not care for your opinion or any of the Wiki sysop or admin opinions. So, please hold your two cents. I know your flawed arguments well. I also know that I am correct and have no need to change your opinion which is of no significance to me. Your opinion is all over the article. So take a deep breath, relax and learn that NO means NO. I do not want to have a discussion with any of you. 173.11.135.234 ( talk) 09:29, 25 January 2013 (UTC)
For anyone who may benefit from this, here is a page that was built after a lengthy discussion over on the Talk page: Visual proof of 0.999... = 1
Sorry Tdadamemd, but you have just done the same thing visually that all these other 'proofs' do - you have your conclusion hidden as an assumption within your proof. In the circle example, every cut you make has an 'inside' and an 'outside', and only when these are added together does the circle equal one. Every cut is defined as excluding part of the circle. But then you propose that continued to infinity, there would be a cut that contradicts what you just defined and would exclude nothing. But you give no reason why this last cut would be any different from the others - you just declare the answer that you are supposedly proving. So the proof fails like all the others. Algr ( talk) 07:05, 11 February 2013 (UTC)
See the following article but read especially pages 29 and 30. thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf You may NOT quote anything from this PDF because it has not been published in paper form. 173.228.7.80 ( talk) 15:20, 7 February 2013 (UTC)
I agree that .999... is not equal to one, but unfortunately that kind of writing is no way to go about proving anything. Algr ( talk) 12:53, 9 February 2013 (UTC)
There is no valid proof that 0.999... equals 1.
www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-171.html
www.filesanywhere.com/fs/v.aspx?v=8b6966895b6673aa6b6c
I think you ought to delete this article as it contains false information. 197.79.0.4 ( talk) 16:19, 4 April 2013 (UTC)
I absolutely loved the part where Euler was wrong and where it is said: "mathematicians don't really understand limits that well. In fact all of them don't know what is a number." I'm glad we have you to enlighten us. 93.102.161.97 ( talk) 08:06, 7 April 2013 (UTC)
197.79.0.4, you really should read Gödel, Escher, Bach, the chapter on formal systems and the infinite will be enlightening. It also backs you up, sort of. You'll feel vindicated as well as corrected. Diego ( talk) 09:07, 7 April 2013 (UTC)
(1)*(x)= (x)
therefore we can deduce
1*(0.99...)= (0.99...)
If it is true that 1=0.99... that would mean that
2*(0.99...)= (1.99...)= 2
1*(0.99...)=(0.99....)=1
AND
(0.99...)*(0.99...)= (0.99...)
If we applied this logic consistently, then we would also have to say that (0.99...)*(0.99...)=(0.99...) which obviously poses a problem since
(0.99...)*(0.99...)=(0.99...)=1
[(0.99...)*(0.99...)]*(0.99...) ........... = 1
And so on.
If (0.99...) truly equals 1 then one could theoretically infinitally reduce a given number by multiplying it an infinite number of times by (0.99...), and that number would also equal 1 by this logic, since it would automatically have to equal (0.99...) which again equals 1. As an extension of this, 20 could equal 3, 5232 could equal 91, and so on. -- 88.73.3.251 ( talk) —Preceding undated comment added 20:53, 27 October 2013 (UTC)
(I hope I got the decimal notation for infinitely small but not zero correct... a guess after not being able to find it. 0.000...1 (infinitely zeroes before the 1)
a\ true or false? 0.999... = 1
b\ true or false? 0.000...1 = 0
c\ true or false? 1 - 0.000...1 = 0.999...
d\ true or false? 0.999... - 0.000...1 = 0.99...8
e\ true or false? 0.999... = 0.99...8 = 0.99...7 = 0.99...6
f\ true or false? the difference between every neighbouring real number would be 0.000...1?
g\ what's the total sum of all differences between every neighbouring real number? infinity or zero? - ZhuLien
— Preceding unsigned comment added by 202.168.6.200 ( talk • contribs) 04:49, 17 January 2014 (UTC)
"Neighbouring numbers" such as "the largest number less than 1" don't exist in the commonly used number systems that contain at least the rational numbers. You technically could construct a custom-made number system with such a property, but I don't think it would be very useful - for example, you will run into problems with division. There are several different number systems that have something resembling decimal representations with infinitely long strings of digits followed by something else, where numbers might arguably be called 0.000...1 with infinitely many zeros before the "1", among them the hyperreal numbers pointed out above. However, the standard real numbers, the numbers most commonly used, don't have such infinitesimal numbers. Among the reals 0.999... indeed equals 1. An asymptote of a curve, in this context, is a line that gets "arbitrarily close" to the curve, that is, for every positive real number ε there exist points c on the curve and l on the line so that their distance d(c, l) is less than ε. Take for exampe, for positive x, the curve y=1-1/x. The line y=1 is asymptotic to that curve because for every positive ε there exists an xε>1/ε, and the points c=(xε, 1-1/xε) and l=(xε, 1) satisfy the property above. (Actually I'd also have to require that curve and line don't just get close anywhere, but "near infinity"; I could modify the definition accordingly, but that would complicate the calculations without helping to clarify the main issue.)
Regarding "the never ending summation of the smaller numbers", how exactly would that be defined? That's a crucial question here. In the context of the real numbers, the answer is a limit: If 0.9+0.09+0.009+... is to be a real number, it's defined as the limit of the sequence (0.9, 0.9+0.09, 0.9+0.09+0.009, ...), and that limit can be shown rather easily to be equal to 1. In the context of the hyperreals, it's possible to define such infinite sums so that 0.9+0.09+0.009+... no longer equals 1 - but at least the way I can think of would lead it to also differ from 0.99+0.0099+0.000099+..., and from 0.9+0.099+0.000999+0.0000009999+..., and so on - even from 0+0.9+0.09+0.009+0.0009+...! Which of those numbers is supposed to be 0.999...?
Some final food for thought: Why should 0.9+0.09+0.009+... be less than 1? If the answer is, "because all the finite sums 0.9, 0.9+0.09, 0.9+0.09+0.009 and so on are less than 1", then I'd like to point out that, however it's defined, 0.999... also is larger than all those finite sums. Huon ( talk) 01:59, 21 January 2014 (UTC)
ZhuLien's theory of infinitesimal usefulness.
I am a number. My value is x. I might be positive or negative.
I have many neighbours depending on where they are positioned in my number space.
x+0.0...1 and x-0.0...1 are my nearest valued neighbours yet I cannot see them as they are infinitely far away but I do have an infinite number of neighbours closer than they are.
What I can do is using my value and my nearest neighbours or perhaps nearer neighbours, is work out how crowded we are.
I can compare my crowded space with other number's crowded spaces to see statistics in different ways.
I can zoom in and out indefinitely to my nearest valued neighbours to allow my crowd and other crowds to be visualised at different resolutions.
There may be mathematical operations on both individuals, neighbours and crowds which can be useful in rounding us up, or grouping us together in sets.
These sets can be useful in highlighting patterns or boundaries or just separating us by an arbitrary amount of space. ZhuLien 116.240.194.132 ( talk) 05:42, 6 February 2014 (UTC)
Trovatore says that "0.999...;...000..." is forbidden. I find that bizarre, as it comes across as a bigger failing then my Base Infinity idea. Also, how many times have people been told that .000...1 is nonsense when all that is needed is a semicolon? So anyway, without "0.999...;...000...", then what is the answer to this?
About a decade ago, the two sides of the 0.999... debate clashed regularly, and the main article was edited back and forth. Mathematicians and hobbyists on both sides. But one side was particularly more aggressive (the side that thought the two numbers are equal). And now after a decade, it's turned into full-blown suppression. You can't even post a dissenting view in the talk section? That is utterly ridiculous. God forbid you edit the main article. Even after all these years, I simply am shocked that people still buy into these false proofs. The first proof, alone, is simple to see right through. Multiply both sides by 10. It's a little magic trick. It's like something you'll see in a chain email that tricks the uneducated. But this tricks the educated, too. The more accurate version of the "Magic Trick Proof" is thus:
x = 0.9999... 10x = 9.9999...-9/∞ 10x - x = 9.9999...-9/∞ - 0.9999... 9x = 9-9/∞ x = (9-9/∞)/9 x = 1-1/∞ The point of this equation is that you can't multiply 0.999... by 10, it's not a valid mathematical equation. I'm not suggesting that we start multiplying and subtracting using infinity, I'm saying we *shouldn't.*
Follow the rules of Wikipedia and allow people to talk about this topic in the talk section. -- JohnLattier ( talk) 16:51, 18 April 2014 (UTC)
Infinity is NOT undefined; statements: "There is an infinite amount of natural numbers" and "Infinity itself is not a natural number; every natural number is finite" are by no means contradictory. (In order to be able to discuss infinite sets, we'd have to go to set theory; this is a topic I am going to avoid for now.)
You need to understand that a repeating decimal can be expressed without expressly invoking infinity. 0.999... is actually a shorthand for 0.9 + 0.09 + 0.009 + ...; which is itself a shorthand for a limit of a sequence:
See? At no point did I need an infinity-th element, or anything similar; in fact, I have (just for clarity) specifically excluded infinities and infinitesimals from the definition. - Mike Rosoft ( talk) 23:39, 12 June 2014 (UTC)
To some extent communication requires a shared agreement on what is meant by certain things. We usually agree that if x is a "number" then also 10x is a number and that from a+x=b+x we can conclude a=b. What about x=0.999... If you want to say that x is not 1 then you need to breaking the properties of numbers in such a way that x is fairly useless. You suggest that we can multiply x by 10 but we shouldn't. I'd say that EITHER we can multiply and then 10x=9+x OR ELSE x is a thing which can not be multiplied by 10 at all. If you take the second position (or even your stated position) then you have to abandon saying 1/3=0.3333... which means rejecting the possibility of having any decimal equal to 1/3. This leads to a situation where you have 5/8=0.675 but most fractions have no decimal expansion. IF infinite decimals are anything then it would have to be something where we can't add them together or multiply by 10. In your proposal that 10x = 9.9999...-9/∞ what is the purpose of the thing at the end subtracted? All I can see is that we come up with something new mainly because it gives us room to say that 10x is something other than 9+x. But can we then divide both sides by 10 to come back to x=0.999...- 0.9/∞? and then we have x+0=x+0.9/∞. Can we subtract x from both sides? Doesn't look so good. So in sum, saying that x=1 makes things work very smoothly and does not break any of what we do with numbers. Saying x is not 1 means that it is a thing where multiplying by 10 creates trouble and is of no use. Gentlemath ( talk) 02:30, 16 June 2014 (UTC)
Think of "1" as representing a basket of all the different functions that converge at 1. If "0.999..." is interpreted as a series expansion, then it's one of those functions. But under that interpretation, the statement "0.999... = 1" is like saying "apple = basket of all possible apples". Thus if someone writes "0.999... = 1", you can reasonably assume they're comparing baskets (limits) and not apples-to-baskets. It doesn't make any sense otherwise. But that said, you can part ways with that interpretation temporarily. E.g., to justify shifting the decimal point as a shortcut for multiplying by 10, you first reinterpret .999... as (.9 + .09 + ...), distribute the 10, use the associative rule to extract the 9, and convert back to decimal notation. This breaks the convention of immediately "taking the limit" of 0.999..., or its decimal expansion would be (1 + 0 + 0 + ...). So that's the real answer: if someone compares a convergent series to a number, it's convention to assume good faith that they're secretly referring to the limit rather than incorrectly comparing apples to baskets. But you can still temporarily delay "taking the limit" when it's useful (like in, oh, all of differential calculus). You can't "prove" this convention, and all of these supposed proofs are invalid by circular reasoning, temporarily breaking the convention that is supposedly being proven, etc. Maghnus ( talk) 11:40, 27 July 2014 (UTC)
The system of real numbers includes two kinds of zero that I will demonstrate are different. Imagine a machine that generates random rational numbers between 0 and 1. It is 'fair' in the sense that any rational number has the same probability of appearing as any other rational number. What is the chance of this machine generating a 2? Well, it is zero, because we defined it as only going from 0 to 1.
So, what is the chance of this machine generating a .5?
Well normally you would figure (specified outcome) / (All possible outcomes). But in this case the denominator is infinite. One might say that 1/∞ is not a number, but taking the limit of larger and larger denominators gives you an apparent result of zero. Prob(.5) and Prob(2) both seem to equal zero, but they do not equal each other, because one is possible and the other is not! Thus, two different zeros.
This closely relates to the .999... problem because you have an infinite number of elements, (digits for .999..., numbers for Prob(.5)) evaluated with a finite result. The simple solution would be that 1/∞ = the infinitesimal unit, ®1. If you insist that 1/∞ is not defined, then remember that 1/2 is not defined in the set of Natural numbers for the same reason. 2 can't be generated by my machine, but it still exists.
BTW: 1/3 in base infinity is (1/3)® or .1base 3®. Algr ( talk) 21:56, 14 August 2014 (UTC)
As I showed above, here with more details and explanations:
If that's wrong, at what step? Or is 1 smaller than itself? Is equality not transitive? Huon ( talk) 13:13, 16 August 2014 (UTC) Sorry, I'm to sick to do math this week. Seen two doctors already. Algr ( talk)
Here I will try to show that the following well known proof is not correct (or actually is correct concerning only the real numbers, although still with some flaws). I chose it because in it the mistakes are harder to be noticed.
x = 0.999...
10*x = 9.999...
10*x-x = 9.999... + 0.999...
9*x = 9
x= 1 = 0.999...
Let's take 1/3 and try to represent it as a decimal fraction.
1/3 = 0.3 + 1/30 = 0.33 + 1/300 = 0.333 + 1/3000 = ...
or
1/3 = 0.3 + 1/(3*10) = 0.33 + 1/(3*10^2) + 0.333 + 1/(3*10^3) = ... =
= 0.333...3<n times> + 1/(3*10^n) = ... = 0.333... + 1/(3*10^∞)
Now let's substitute the last expression in above mentioned proof.
x = 3*(0.333... + 1/(3*10^∞))
10*x = 30*(0.333... + 1/(3*10^∞))
10*x-x = 27*(0.333... + 1/(3*10^∞))
x = 3*(0.333... + 1/(3*10^∞))
It is the same. Even if we not use the infinitesimal part, which is not a real number (this does not mean it does not exist, but only that it is not a real number), still nothing changes.
What will happen if we get rid of the brackets? Let's try.
10*x = 30*(0.333... + 1/(3*10^∞))
10*x = 9.999... + 30/(3*10^∞)
10*x-x = 9.999... + 30/(3*10^∞) - 0.999... - 3/(3*10^∞)
9*x = 9 - 27/(3*10^∞)
x = 1 - 3/(3*10^∞)
x = 1 - 1/10^∞
1/3 = 0.333...3<n times> + 1/(3*10^n)
So it is all about using or not this infinitesimal part and keep in mind not to break the correlation between the number of the repeating digits and power n in this infinitesimal part. Because if we break this correlation we will have some other fraction and not 1/3 as in the example. The well known proof is a speculation but it is actually not needed, because there is no room for infinite and infinitesimal numbers in the real numbers line (which will make the difference between 0.999... and 1). This is the way mathematicians defined it, so it does not matter I disagree.
Now about some statements. I read some arguments that 9*0.999...is 8.999...91 but they are also wrong. It can be easy understood considering for exemple: 2*0.666... = (2*2)/3 = 4/3 = 1.333... not 1.333...32. Even if we include the infinitesimal part we'll have:
2/3 = 0.666... + 2/(3*10^∞)
4/3 = 1 + 1/3 = 1.333... + 1/(3*10^∞)
which is more than both 1.333... and 1.333...32
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