| Frequently asked questions
- Q: Are you positive that 0.999... equals 1 exactly, not approximately?
- A: In the set of real numbers, yes. This is covered in the article. If you still have doubts, you can discuss it at Talk:0.999.../Arguments. However, please note that original research should never be added to a Wikipedia article, and original arguments and research in the talk pages will not change the content of the article—only reputable secondary and tertiary sources can do so.
- Q: Can't "1 - 0.999..." be expressed as "0.000...1"?
- A: No. The string "0.000...1" is not a meaningful real decimal because, although a decimal representation of a real number has a potentially infinite number of decimal places, each of the decimal places is a finite distance from the decimal point; the meaning of digit d being k places past the decimal point is that the digit contributes d · 10-k toward the value of the number represented. It may help to ask yourself how many places past the decimal point the "1" is. It cannot be an infinite number of real decimal places, because all real places must be finite. Also ask yourself what the value of
 would be. Those proposing this argument generally believe the answer to be 0.000...1, but, basic algebra shows that, if a real number divided by 10 is itself, then that number must be 0.
- Q: The highest number in 0.999... is 0.999...9, with a last '9' after an infinite number of 9s, so isn't it smaller than 1?
- A: If you have a number like 0.999...9, it is not the last number in the sequence (0.9, 0.99, ...); you can always create 0.999...99, which is a higher number. The limit
 is not defined as the highest number in the sequence, but as the smallest number that is higher than any number in the sequence. In the reals, that smallest number is the number 1.
- Q: 0.9 < 1, 0.99 < 1, and so forth. Therefore it's obvious that 0.999... < 1.
- A: No. By this logic, 0.9 < 0.999...; 0.99 < 0.999... and so forth. Therefore 0.999... < 0.999..., which is absurd.
- Something that holds for various values need not hold for the limit of those values. For example, f (x)=x 3/x is positive (>0) for all values in its implied domain (x ≠ 0). However, the limit as x goes to 0 is 0, which is not positive. This is an important consideration in proving inequalities based on limits. Moreover, although you may have been taught that
 must be less than  for any values, this is not an axiom of decimal representation, but rather a property for terminating decimals that can be derived from the definition of decimals and the axioms of the real numbers. Systems of numbers have axioms; representations of numbers do not. To emphasize: Decimal representation, being only a representation, has no associated axioms or other special significance over any other numerical representation.
- Q: 0.999... is written differently from 1, so it can't be equal.
- A: 1 can be written many ways: 1/1, 2/2, cos 0, ln e, i 4, 2 - 1, 1e0, 12, and so forth. Another way of writing it is 0.999...; contrary to the intuition of many people, decimal notation is not a bijection from decimal representations to real numbers.
- Q: Is it possible to create a new number system other than the reals in which 0.999... < 1, the difference being an infinitesimal amount?
- A: Yes, although such systems are neither as used nor as useful as the real numbers, lacking properties such as the ability to take limits (which defines the real numbers), to divide (which defines the rational numbers, and thus applies to real numbers), or to add and subtract (which defines the integers, and thus applies to real numbers). Furthermore, we must define what we mean by "an infinitesimal amount." There is no nonzero constant infinitesimal in the real numbers; quantities generally thought of informally as "infinitesimal" include ε, which is not a fixed constant; differentials, which are not numbers at all; differential forms, which are not real numbers and have anticommutativity; 0+, which is not a number, but rather part of the expression
 , the right limit of x (which can also be expressed without the "+" as  ); and values in number systems such as dual numbers and hyperreals. In these systems, 0.999... = 1 still holds due to real numbers being a subfield. As detailed in the main article, there are systems for which 0.999... and 1 are distinct, systems that have both alternative means of notation and alternative properties, and systems for which subtraction no longer holds. These, however, are rarely used and possess little to no practical application.
- Q: Are you sure 0.999... equals 1 in hyperreals?
- A: If notation '0.999...' means anything useful in hyperreals, it still means number 1. There are several ways to define hyperreal numbers, but if we use the construction given here, the problem is that almost same sequences give different hyperreal numbers,
 , and even the '()' notation doesn't represent all hyperreals. The correct notation is (0.9; 0.99; 0,999; ...).
- Q: If it is possible to construct number systems in which 0.999... is less than 1, shouldn't we be talking about those instead of focusing so much on the real numbers? Aren't people justified in believing that 0.999... is less than one when other number systems can show this explicitly?
- A: At the expense of abandoning many familiar features of mathematics, it is possible to construct a system of notation in which the string of symbols "0.999..." is different than the number 1. This object would represent a different number than the topic of this article, and this notation has no use in applied mathematics. Moreover, it does not change the fact that 0.999... = 1 in the real number system. The fact that 0.999... = 1 is not a "problem" with the real number system and is not something that other number systems "fix". Absent a WP:POV desire to cling to intuitive misconceptions about real numbers, there is little incentive to use a different system.
- Q: The initial proofs don't seem formal and the later proofs don't seem understandable. Are you sure you proved this? I'm an intelligent person, but this doesn't seem right.
- A: Yes. The initial proofs are necessarily somewhat informal so as to be understandable by novices. The later proofs are formal, but more difficult to understand. If you haven't completed a course on real analysis, it shouldn't be surprising that you find difficulty understanding some of the proofs, and, indeed, might have some skepticism that 0.999... = 1; this isn't a sign of inferior intelligence. Hopefully the informal arguments can give you a flavor of why 0.999... = 1. If you want to formally understand 0.999..., however, you'd be best to study real analysis. If you're getting a college degree in engineering, mathematics, statistics, computer science, or a natural science, it would probably help you in the future anyway.
- Q: But I still think I'm right! Shouldn't both sides of the debate be discussed in the article?
- A: The criteria for inclusion in Wikipedia is for information to be attributable to a reliable published source, not an editor's opinion. Regardless of how confident you may be, at least one published, reliable source is needed to warrant space in the article. Until such a document is provided, including such material would violate Wikipedia policy. Arguments posted on the Talk:0.999.../Arguments page are disqualified, as their inclusion would violate Wikipedia policy on original research.
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