Introduction

As a high school student, mathematics was the subject I struggled the most with. However, I was rather intrigued by the mysteries surrounding math and although I had no idea what most of the symbols were, I enjoyed playing around with my calculator.

I first came across the idea that 0.999… = 1 (where the 9’s are repeating infinitely many times, of course) whilst studying Geometric Series and I must say, even after seeing the proof, I was not yet fully convinced. This was due to my lack of understanding of what “infinitely many” really meant. At the time, I needed something tangible that I could grasp.

An Informal Proof

The basic intuition I had was to use the fact that 1 is a rational number and so is 0.999…, since they are equal, then I must be able to express 0.999… as a ratio of two integers which are equal to 1. This is very easy since the ratio of any integer, except 0, with itself is equal to 1. That is,

$${a \over a} = 1, a \ne 0, a \in \mathbb{Z}$$.

Now, after spending some time playing around with the numbers on my calculator, I discovered the following pattern:

$$\frac{1}{9} = 0.111 \ldots$$

$$\frac{2}{9} = \ldots$$

$$\frac{3}{9} = \frac{1}{3} = 0.333 \ldots$$

$$\frac{4}{9} = 0.444 \ldots$$

$$\frac{5}{9} = 0.555 \ldots$$

$$\frac{6}{9} = \frac{2}{3} = 0.666 \ldots$$

$$\frac{7}{9} = 0.777 \ldots$$

$$\frac{8}{9} = 0.888 \ldots$$

Here it was reasonable to assume by induction that, $$\frac{9}{9} = 1 = 0.999\ldots$$

Rigorous Proof

First, we rewrite 0.999… as the sequence of partial sums denoted by $$s_n$$ where n is the number of terms (or number of 9’s) and is given by:

$$s_n = {9 \over 10} + {9 \over 10^2} + {9 \over 10^3} + \cdots + {9 \over 10^n}$$

Multiplying both sides by ${1 \over 10}$ gives

$${1 \over 10} s_n = {9 \over 10^2} + {9 \over 10^3} + {9 \over 10^4} + \cdots + {9 \over 10^{n+1}}$$

Subtracting $(2)$ from $(1)$ then gives

$$(1 - {1 \over 10}) s_n = ({9 \over 10} + {9 \over 10^2} + {9 \over 10^3} + \cdots + {9 \over 10^n}) - ({9 \over 10^2} + {9 \over 10^3} + {9 \over 10^4} + \cdots + {9 \over 10^{n+1}})$$

so

$$(1 - {1 \over 10}) s_n = {9 \over 10} - {9 \over 10^{n+1}}$$.

Therefore,

$$s_n = \frac { \frac {9}{10} - \frac {9}{10^{n+1}}}{1 - \frac {1}{10}} = 1 - {1 \over 10^n}$$.

Hence,

$$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } {1 - {1 \over 10^n}} = 1$$.

Thus, we conclude that 1 and 0.999… are different decimal representations of the same real number.