# Why does 0.999… = 1? (Part 5)

Here’s one more way of convincing students that $0.\overline{9} = 1$. Here’s the idea: how far apart are the two numbers?

First off, since $1 \ge 0.\overline{9}$, we know that $1 - \overline{9} \ge 0$.

Of course, we know that $1-0.9 = 0.1$. Since $0.\overline{9}$ must lie between $0.9$ and $1$, we know that $1 - 0.\overline{9}$ must be less than $0.1$.

Second, we know that $1-0.99 = 0.01$. Since $0.\overline{9}$ must lie between $0.99$ and $1$, we know that $1 - 0.\overline{9}$ must be less than $0.01$.

Third, we know that $1-0.999 = 0.001$. Since $0.\overline{9}$ must lie between $0.999$ and $1$, we know that $1 - 0.\overline{9}$ must be less than $0.001$.

By the same reasoning, we conclude that $0 \le 1 - 0.\overline{9} < \displaystyle \frac{1}{10^n}$

for every integer $n$. What’s the only number that’s greater than or equal to $0$ and less than every decimal of the form $0.00\dots001$? Clearly, the only such number is $0$. Therefore, $1 - 0.\overline{9} = 0$, or $0.\overline{9} = 1$. I like this approach because it really gets at the heart of the difference between integers $\mathbb{Z}$ and real numbers $\mathbb{R}$. For integers, there is always an integer to the immediate left and to the immediate right. In other words, if you give me any integer (say, $15$), I can tell you the largest integer that’s less than your number (in our example, $14$) and the smallest integer that’s bigger than your number ( $16$).

Real numbers, however, do not have this property. There is no real number to the immediate right of $0$. This is easy to prove by contradiction. Suppose $x > 0$ is the real number to the immediate left of $0$. That means that there are no real numbers between $0$ and $x$. However, $x/2$ is bigger than $0$ and less than $x$, providing the contradiction.

(For what it’s worth, the above proof doesn’t apply to the set of integers $\mathbb{Z}$ since $x/2$ doesn’t have to be an integer.)

By the same logic — visually, you can imagine reflecting the number line across the point $x = 0.5$ — there is no number to the immediate left of $1$. So while $0.\overline{9}$ would appear to be to the immediate left of $1$, they are in reality the same point.