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0.999...

Alternative decimal expansion of 1

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Through Real number, Decimal representation and Natural number

Despite common misconceptions, 0.999... is not "almost exactly 1" or "very, very nearly but not quite 1"; rather, "0.999..." and "1" represent exactly the same number.

There are many ways of showing this equality, from intuitive arguments to mathematically rigorous proofs. The intuitive arguments are generally based on properties of finite decimals that are extended without proof to infinite decimals. An elementary but rigorous proof is given below that involves only elementary arithmetic and the Archimedean property: for each real number, there is a natural number that is greater (for example, by rounding up). Other proofs generally involve basic properties of real numbers and methods of calculus, such as series and limits. Why some people reject this equality is a question studied in mathematics education.

The equality 0.999... = 1 is a special case of the fact that every non-zero terminating decimal has two equal representations (for example, 8.32000... and 8.31999...).

Everything that precedes refers to the decimal number system, that is to base 10. For other bases, everything remains true with 9 replaced with the largest digit. For example, in the binary number system, one has 0.111... = 1. In some other number systems, 0.999... can have the same meaning, a different definition, or be undefined.

Elementary proof

It is possible to prove the equation 0.999... = 1 using just the mathematical tools of comparison and addition of (finite) decimal numbers, without any reference to more advanced topics. The proof given below is a direct formalization of the intuitive fact that, if one draws 0.9, 0.99, 0.999, etc. on the number line, there is no room left for placing a number between them and 1. The meaning of the notation 0.999... is the least point on the number line lying to the right of all of the numbers 0.9, 0.99, 0.999, etc. Because there is ultimately no room between 1 and these numbers, the point 1 must be this least point, and so 0.999... = 1.

Intuitive explanation

If one places 0.9, 0.99, 0.999, etc. on the number line, one sees immediately that all these points are to the left of 1, and that they get closer and closer to 1. For any number x that is less than 1, the sequence 0.9, 0.99, 0.999, and so on will eventually reach a number larger than ⁠x⁠. So, it does not make sense to identify 0.999... with any number smaller than 1.

Meanwhile, every number larger than 1 will be larger than any decimal of the form 0.999...9 for any finite number of nines. Therefore, 0.999... cannot be identified with any number larger than 1, either.

Because 0.999... cannot be bigger than 1 or smaller than 1, it must equal 1 if it is to be any real number at all.

Rigorous proof

Denote by 0.(9)n the number 0.999...9, with n nines after the decimal point. Thus 0.(9)1 = 0.9, 0.(9)2 = 0.99, 0.(9)3 = 0.999, and so on. One has 1 − 0.(9)1 = 0.1 = ⁠110⁠, 1 − 0.(9)2 = 0.01 = ⁠1102⁠, and so on; that is, 1 − 0.(9)n = 110n for every natural number ⁠n⁠.

Let x be a number not greater than 1 and greater than 0.9, 0.99, 0.999, etc.; that is, 0.(9)n < x ≤ 1, for every ⁠n⁠. By subtracting these inequalities from 1, one gets 0 ≤ 1 − x < ⁠110n⁠.

The end of the proof requires that there be no positive number that is less than 110n for all ⁠n⁠. This follows from the Archimedean property, which can be expressed as, "for every real number, there is a natural number that is greater". By computing the reciprocal, this implies that for every positive real number, there are natural numbers whose reciprocals are smaller. Therefore, for any positive real number, there must be some ⁠n⁠ such that 110n is smaller. This property implies that if 1 − x < ⁠110n⁠ for all ⁠n⁠, then 1 − x can only be equal to 0. So, x = 1 and 1 is the smallest number that is greater than all 0.9, 0.99, 0.999, etc. That is, 1 = 0.999..., as claimed.

This proof relies on the Archimedean property of rational and real numbers. Real numbers may be enlarged into number systems, such as hyperreal numbers, with infinitely small numbers (infinitesimals) and infinitely large numbers (infinite numbers). When using such systems, the notation 0.999... is generally not used, as there is no smallest number among the numbers larger than all 0.(9)n.

Least upper bounds and completeness

Part of what this argument shows is that there is a least upper bound of the sequence 0.9, 0.99, 0.999, etc.: the smallest number that is greater than all of the terms of the sequence. One of the axioms of the real number system is the completeness axiom, which states that every bounded sequence has a least upper bound. This least upper bound is one way to define infinite decimal expansions: the real number represented by an infinite decimal is the least upper bound of its finite truncations. The argument here does not need to assume completeness to be valid, because it shows that this particular sequence of rational numbers has a least upper bound and that this least upper bound is equal to one.

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