# 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 $$1$$. When analyzed rigorously, we find that $0.999\ldots = 1.$ Both numbers are exactly equal and their difference is exactly $$0$$.

## Algebric Demonstration

Before we get into a proper proof, 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 get $0.999\ldots = 1.$ That was quite straightforward but not rigorous. We will discuss more about it in a moment.

## Another Algebric Demonstration

Let us see another similar demonstration. We let $$x = 0.999\ldots$$ and obtain $10x = 9.999\ldots = 9 + 0.999\ldots = 9 + x.$ Subtracting $$x$$ from both sides, we get $9x = 9.$ Dividing both sides by $$9$$, we get $x = 1.$ Thus $$0.999\ldots = 1.$$

Now it is worth noting here that these algebraic demonstrations are not rigorous proofs because we have not shown that elementary rules for addition and multiplication extend to repeating decimal numbers too. They do but we have not demonstrated that here. However, this is a good start 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.

## Infinite Geometric Series

Let us denote $$0.999\ldots$$ as $0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \ldots$ Note that the RHS is an infinite geometric series. From the study of real analysis, we know that $a + ar + ar^2 + \dots = \frac{a}{1 - r}$ where $$a$$ and $$r$$ are real numbers and $$\lvert r \rvert < 1.$$ Now let $$a = 9/10$$ and $$r = 1/10$$ to get $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \dots = \frac{\frac{9}{10}}{1 - \frac{1}{10}}.$ The LHS is $$0.999\ldots$$ and the RHS is $$1$$, so we get $0.999\ldots = 1.$ Now one might argue that here too we did not prove the formula for geometric series. That is true but the point of this article is not to establish the basic theorems of real analysis but instead to show that we can use known theorems to convince ourselves beyond any doubt that $$0.999\ldots$$ is indeed exactly equal to $$1.$$ Relying on the formula for geometric series provides us a convenient starting point without compromising on rigour. In particular, this analytic proof depends on the convergence of the geometric series when $$\lvert r \rvert < 1$$. The proof of convergence can be found in any introductory book on real analysis.

## Avoiding Infinity

In this section, we will take a close look at the LHS above. The LHS seems to have a sum of an infinite number of terms but we cannot literally add infinite terms together, can we? What does the LHS mean then?

Adding infinite number of terms is not a valid operation. Infinity is not a number. It is a notion, a notion that we can better understand, in this context, using limits. When we write $\frac{9}{10} + \frac{9}{100} + \dots = 1$ what we really mean can be expressed formally as follows: $\lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1.$ By the definition of limit of sequence, what the above equation means is that for every real $$\epsilon > 0,$$ there exists a positive integer $$N$$ such that for all $$n \ge N,$$ we have $\left\lvert 1 - \sum_{k=1}^n \frac{9}{10^k} \right\rvert < \epsilon.$ In simpler words, we can say that no matter how small a positive real number we choose for $$\epsilon$$, we can find large enough threshold for $$n$$ such that for all values of $$n$$ equal to or above that threshold, the difference between $$1$$ and $$\sum_{k=1}^n \frac{9}{10^k}$$ becomes smaller than $$\epsilon$$. To summarize, we have shown using the definition of limit of a sequence that $0.999\ldots = \lim_{n \to \infty} \sum_{k=1}^n \frac{9}{10^k} = 1.$ This is one way of several ways to show that $$0.999\ldots = 1.$$ There are several other proofs too but this is one of the very approachable ones.