0.999...

By Susam Pal on 09 Dec 2001

What is the difference between \( 0.999\ldots \) and \( 1? \) Note that the ellipsis denotes a never-ending sequence of 9s. Is the difference a small but positive real number? If so, what is that number? Or is the difference simply \( 0? \) It turns out that the difference is indeed \( 0. \) In fact, we can write \[ 1 - 0.999\ldots = 0. \] Intuitively, it may feel like the number \( 0.999\ldots \) gets closer and closer to \( 1 \) without ever reaching \( 1. \) This intuition is common. After all, adding each new digit brings the number closer to 1. There is a precise way to define such behaviour, using the concept of limits. When we do so, we will find that \[ 0.999\ldots = 1. \] Both numbers are exactly equal and their difference is exactly \( 0. \)

Contents

Algebraic Illustration

Before we do a formal analysis, it might be worthwhile demonstrating the equality above algebraically. Let us first write \[ 0.333\ldots = \frac{1}{3}. \] Multiplying both sides by \( 3, \) we obtain \[ 0.999\ldots = 1. \] That was quite straightforward, though not rigorous. We will return to this point shortly.

Another Algebraic Illustration

Let us see another similar demonstration. This time we follow a popular method for finding a fraction that represents a repeating decimal number. Let \[ x = 0.999\ldots \] Multiplying both sides by \( 10 \) gives us \[ 10x = 9.999\ldots \] Subtracting \( x = 0.999\ldots \) from both sides, we get \[ 9x = 9. \] Dividing both sides by \( 9 \) gives us \[ x = 1. \] Thus \( 0.999\ldots = 1. \)

It is important to note that these algebraic demonstrations do not constitute rigorous proofs because we have not shown that the elementary rules for addition and multiplication extend to repeating decimal numbers too. They do, but we have not demonstrated that here. Nonetheless, they provide a useful way to challenge any faulty intuition that might have led one to doubt the above equality. While there are many ways to prove this equality, one easy way is to rely on the formula for the infinite geometric series, which has a sound basis in real analysis.

Three Equalities

We now prove that \( 0.999\ldots = 1 \) using the following sequence of equalities: \[ 0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots = \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1. \] For brevity, we will take certain standard results from real analysis as given. Now let us take a close look at each equality.

The First Equality

The first equality is in fact the definition of the number \( 0.999\ldots. \) Every real number has a decimal expansion, which can be expressed as a finite or infinite sum of fractions. The number \( 0.999\ldots, \) when expressed as a sum of fractions, can be written as \[ 0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots \] The left-hand side (LHS) and the right-hand side (RHS) represent the same real number, expressed in two different notations. The LHS uses decimal notation, while the RHS uses series notation.

The Second Equality

The RHS of the equation above appears to involve the sum of an infinite number of terms. But we cannot literally add an infinite number of terms together, can we? Infinity is not a number. What does the RHS mean then?

Let \( (a_n) \) denote a sequence of real numbers \( a_1, a_2, \dots. \) We say that the limit of the sequence \( (a_n) \) is \( L, \) and write \[ \lim_{n \to \infty} a_n = L \] if, for every real number \( \epsilon \gt 0, \) there exists a positive integer \( N \) such that for all \( n \ge N, \) we have \( \lvert a_n - L \rvert \lt \epsilon. \) In symbolic form, this definition can be expressed as: \[ \forall \varepsilon \gt 0, \; \exists N \in \mathbb{N} \; \text{such that} \; \forall n \ge N, \; \lvert a_n - L \rvert \lt \varepsilon. \] In simple terms, saying that the limit of the sequence \( (a_n) \) converges to a limit \( L \) means that no matter how small a positive real number we choose for \( \epsilon, \) we can find a large enough threshold for \( n \) such that for all values of \( n \) equal to or above that threshold, the difference between \( a_n \) and \( L \) is smaller than \( \epsilon. \) In other words, the difference can be made as small as we like simply by considering terms that are farther and farther out.

Now consider the sequence \[ \frac{9}{10}, \quad \frac{9}{10} + \frac{9}{10^2}, \quad \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3}, \quad \dots \] The \( n \)th term has the form \[ s_n = \sum_{k=1}^n \frac{9}{10^k}. \] where \( n \) is a positive integer. We say that \[ \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \dots = \lim_{n \to \infty} s_n. \] Or if we write the expansion of the RHS, then \[ \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \dots = \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k}. \] Again, this is a definition. It does not feel like we are proving anything. We are merely saying one thing is the same as another thing because we named them so. It may not feel satisfying at first, but definitions lay the groundwork for meaningful results. After all, to get the result \( 1 + 1 = 2, \) we first need to define \( 2 \) as a successor of \( 1. \) We must assign a meaning to \( 2 \) before we can get any interesting results about it.

Similarly, we must assign a meaning to \( 0.999\ldots \) before we can find interesting results about it. We have to assign some meaning to recurring decimal numbers. That meaning becomes the definition of the number. The equation in the previous section is how we choose to define \( 0.999\ldots. \) If you choose to define this number differently, in a way that is inconsistent with the definition here, that's fine, but then you and I will be working with different mathematical systems. The onus will be on you to demonstrate that your mathematical system is useful and consistent. In this article, however, we will work with the above definitions, which are widely accepted and free of known contradictions.

In this section, we stated that \[ \frac{9}{10} + \frac{9}{10^2} + \frac{9}{10^3} + \dots = \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k}. \] In the next section, we will examine the limit on the RHS more closely.

The Third Equality

Let us evaluate \[ \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k}. \] The RHS is an infinite geometric series. From the study of real analysis, we know that \[ \lim_{n \to \infty} \sum_{k=0}^n ar^k = \frac{a}{1 - r} \] where \( a \) and \( r \) are real numbers and \( \lvert r \rvert \lt 1. \) Now let \( a = 9/10 \) and \( r = 1/10 \) to get \[ \lim_{n \to \infty} \sum_{k=0}^n \left( \frac{9}{10} \cdot \frac{1}{10^k} \right) = \frac{\frac{9}{10}}{1 - \frac{1}{10}} \] Simplifying both sides, we get \[ \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1. \] We have not proven the formula for geometric series here. The point of this article is not to establish the basic theorems of real analysis from scratch, but instead to use them to explore what \( 0.999\ldots \) is. Relying on the geometric series formula gives us a convenient starting point without compromising on rigour. The analytic proof of the formula depends on the convergence of the geometric series when \( \lvert r \rvert \lt 1. \) The proof of convergence can be found in any introductory book on real analysis.

This limit tells us that no matter how small a positive real number we choose for \( \epsilon, \) we can find a large enough threshold for \( n \) such that for all values of \( n \ge N, \) the difference between \( 1 \) and \( \sum_{k=1}^n \frac{9}{10^k} \) is smaller than \( \epsilon. \) In other words, the difference can be made as small as we like simply by adding sufficiently many terms.

Tying It Together

The three equalities in the three sections above establish the following sequence of equations: \[ 0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots = \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1. \] The first two equalities follow from definitions; the last follows from real analysis. If this still seems unconvincing, we must ask: what is the difference between \( 0.999\ldots \) and \( 1? \) It is either zero or non-zero. Suppose the difference is non-zero. Let \[ \lvert 0.999\ldots - 1 \rvert = \delta \] where \( \delta \) is a positive real number. Therefore \[ \left\lvert \left( \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} \right) - 1 \right\rvert = \delta. \] But we have already shown that \[ \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1 \] By the definition of limit, it means that no matter how small a positive real number we choose for \( \epsilon, \) there is a positive integer \( n \) such that for all \( n \ge N, \) we have \[ \left\lvert \left( \sum_{k=1}^n \frac{9}{10^k} \right) - 1 \right\rvert \lt \epsilon. \] So if we choose \( \epsilon = \delta, \) the very number we supposed was the difference between \( 0.999\ldots \) and \( 1, \) we arrive at \[ \left\lvert \left( \sum_{k=1}^n \frac{9}{10^k} \right) - 1 \right\rvert \lt \delta. \] for all \( n \ge N. \) This contradicts our assumption that the difference was \( \delta. \) Therefore the difference between \( 0.999\ldots \) and \( 1 \) cannot be non-zero. So the only possibility is that the difference is zero, meaning \( 0.999\ldots = 1. \)

Conclusion

We have examined the equality \( 0.999\ldots = 1 \) from several perspectives: informal algebraic arguments, the geometric series formula, and the rigour of limits in real analysis. While the idea may initially seem counterintuitive, each approach confirms that \( 0.999\ldots \) is exactly equal to \( 1. \) The difference between them is not a small number. It is precisely zero. This example is a useful reminder that initial mathematical intuition may need to be revised through formal reasoning. Once that happens, the insight gained often becomes new intuition!

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