maths

# Limits and decimal expansions

When I was finishing up high school some years ago, I started visiting universities to figure out where and what I would study. In one of the open lectures I was introduced to the concept of infinite decimal expansions for integers and in particular the most famous and troublesome one,

$0.9999 \ldots = 1$.

The proof was a mind-opener for sure. Now that I am myself teaching fertile young minds the depth and beauty of mathematics, I see that many students have trouble with understanding the equation. Here is my attempt to explain it.

#### The algebraic argument

Let’s start with the classic, and perhaps least satisfying proof. Let’s start with giving it a name, let’s call $E = 0.999\ldots$ the value of the number we’re trying to figure out. If we were to multiply the number $E$ by 10, we find that it’s just the same as adding 9, so

$10 E = 9.999\ldots = 9+E$.

Subtracting $E$ on both sides, we find $9 E = 9$, so that dividing by 9 gives $E=1$. By doing basic algebraic operations, we have found our answer. The key insight here is that the string of nines is infinitely long, so that shifting all of them to the left by one decimal position simply adds a nine. However, infinity is not actually a number as we know it, and adding one to infinity yields infinity. Many students will find this argument wanting. If we were to take any number, multiply it by 10, subtract it once and divide by 9, of course we’ll end up with the same number! So why doesn’t it seem to be the case with this one?

#### A limit approach

An infinite string of nines is somehow outside our common experience. If we think of a finite string of nines, this will never be equal to one, would it? Well, this is a second way of interpreting the number $E$. Let’s say we have a sequence of numbers, called $E_n$ with $n = 1,2,3,\ldots$, so that $E_n$ is a string of $n$ nines after the decimal point. For example,

$E_1 = 0.9$,
$E_2 = 0.99$,
$E_3 = 0.999$,

etc. As you can see, the numbers $E_n$ get closer to 1 as $n$ increases. Suppose we let $n$ go to infinity, then it stands to reason that the limit, which happens to be $E$, is equal to 1. To prove this, we look at the difference,

$1-E_n = 10^{-n}$.

As $n$ becomes larger and larger, the difference becomes smaller and smaller. In fact, we can make the difference arbitrarily small. In mathematics, we like to play a game when such a situation occurs. The game goes as follows: you give me a finite number strictly larger than zero, let’s call it $\varepsilon>0$. Then, I find an integer $N$ such that $1-E_N < \varepsilon$. If we then find that for all $m>N$ the inequality still holds, we continue the game. When such an index $N$ can be found for any number $\varepsilon > 0$ we say that the sequence has a limit. This, in short, is the Cauchy defition of a limit. The value we assign to the limit is the reference number, in this case 1.

#### A series approach

So we actually find that, in some sense, writing $0.999\ldots$ is just sloppy notation. It’s shorthand for the limit of the sequence $E_n$. Perhaps this idea is too blunt, perhaps the infinite string of nines does actually carry some significance. This brings us to a third way of interpreting the number $E$, as we can represent all the numbers in the sequence $E_n$ as a special sort of sum. Let’s write

$E_1 = 0.9$,
$E_2 = 0.9 + 0.09$,
$E_3 = 0.9 + 0.09 + 0.009$,

etc. These numbers can be represented by  a geometric sum,

$E_n = \frac{9}{10} \sum\limits_{k=0}^n 10^{-k}$.

Passing to the limit of $n \to \infty$ gives the number $E$, so that it is in fact an infinite sum, called a series.
A geometric series is a special type of infinite sum, typically one of the kind

$S = 1 + r + r^2 + r^3 + \ldots = \sum\limits_{k=0}^\infty r^k$

To figure out the value of $S$, we apply a similar approach as earlier and define the sequence of sums

$S_n = \sum\limits_{k=0}^n r^k$,

so that

$S_1 = 1+r$,
$S_2 = 1+ r + r^2$,

etc. In this way, the limit of $n \to \infty$ provides us with the sum $S$. The trick we now have to perform is the observation that

$(1-r) S_n = 1+r + \ldots + r^n - (r + r^2 + \ldots + r^{n+1} ) = 1-r^{n+1}$.

If $|r|<1$, then the right-most side gets closer and closer to 1 as $n$ increases. So we find that in the limit, $S$ must satisfy

$1+ r + r^2 + \ldots = \frac{1}{1-r}$.

If an infinite sum of terms, a series, has a finite value, we call the series convergent. The geometric series is only convergent for $|r|<1$, otherwise it is divergent.
To write $E$ as a geometric series, we see that we need to take $r = \frac{1}{10}$ and multiply by $\frac{9}{10}$, resulting in

$E = \frac{9}{10} \sum\limits_{k=0}^\infty \left( \frac{1}{10} \right)^k$.

We just calculated what the value of the geometric series must be, so we can just fill in the values,

$E = \frac{9}{10} \frac{1}{1-\frac{1}{10}}$.

We can simplify this expression by multiplying both numerator and denominator by 10, and so

$E = \frac{9}{10} \frac{10}{9} = 1$.

#### So what now?

After these three proofs, and with it interpretations, what have we learned? If nothing else, it seems that at least these three interpretations are consistent. They all show that something special seems to happen in some way or other as we add more and more nines to the string. They also suggest that we should interpret the infinite string of nines as a limit, rather than an actual number.
The algebraic proof relies on the fact that any number added to infinity produces infinity. The sequence proof relies on the fact that the we can always find a value arbitrarily close to 1 by adding more nines. Finally, the number $E$ may be written as a geometric series and we have a direct way to compute the result. To me, the infinite decimal expansion $0.999\ldots = 1$ mostly seems to illustrate that our intuition breaks down when mathematical infinities are encountered. In daily life and ordinary experience we never encounter infinities and so we simply don’t know how these things behave.