# 0.999… is 1

Here’s one funny fact in Math and numbers which puzzled me so much the first time I learned it. A very simple equality that, for first timers, simply looks wrong:

$0.\overline{9} = 1$

In even simpler terms:

$0.999... = 1$

You might be thinking: “Has he gone crazy?”, the answer is no I haven’t, though I completely understand your astonishment as I experienced the same exact feeling the first time I saw that equation. So let’s prove it and see how entertaining this can be.

### The simple way

Did you remember when, in Elementary school, you studied how to convert a decimal number into a fraction? Yes those esotheric rules involving adding zeros and nines at the denominator of the fraction. Well, this proof can be achieved by a 10 year old student and it involves the most basic of those rules:

A simply repeating decimal number is equivalent to a fraction having as numerator the recurring part, and as denominator as many 9s as the number of digits of the recurring part.

If we apply that:

$0.\overline{9} = \frac{9}{9} = 1$

Ta da!

### Still simple, but slightly more elaborate proof

You still might not believe this, after all who remembers those rules? What if they are fake? Well, we can still go raw by using very simple equation handling techniques in a proof which can still be performed by that same 10 year old kid.

1. We start with a simple imposition: $x = 0.\overline{9}$.
2. Now let’s multiply both sides by $10$: $10x = 9.\overline{9}$.
3. Both equations are equivalent, we can get another equivalent equation by subtracting the first from the second, getting: $9x = 9$ which gives us: $x=1$!

So we have: $x = 0.\overline{9} \wedge x = 1 \implies 0.\overline{9} = 1$ which proves our point.

## Let’s use cannons to shoot mosquitos

For those of you who must have it the hard way, here we go with some proofs employing Calculus. They are a bit more sophistifacted, so I would not ask the same 10 year old kid to run this Math.

### Using series

We can employ Discrete Calculus and consider the following:

$0.\overline{9} = 0.999\dots = 0.9 + 0.09 + 0.009 + \cdots = 9 \cdot 10^{-1} + 9 \cdot 10^{-2} + 9 \cdot 10^{-3} + \cdots = \sum_{n=1}^{+\infty} 9 \cdot 10^{-n}$

We get a Geometric Series, which we can develop further:

$0.\overline{9} = \sum_{n=1}^{+\infty} 9 \cdot 10^{-n} = 9 \cdot \sum_{n=1}^{+\infty} 10^{-n} = 9 \cdot \sum_{n=1}^{+\infty} \frac{1}{10^{n}} = 9 \cdot \sum_{n=1}^{+\infty} \left(\frac{1}{10}\right)^{n}$

We know that the Geometric Series converges to $\sum_{n=0}^{+\infty} r^n = \frac{1}{1-r}$ only when $|r| < 1$ (convergence region). However our series starts from $1$ not $0$, thus we need to manipulate this a little bit:

$\sum_{n=0}^{+\infty} r^n = \left. r^n \right|_{n=0} + \sum_{n=1}^{+\infty} r^n = 1 + \sum_{n=1}^{+\infty} r^n = \frac{1}{1-r} \implies \sum_{n=1}^{+\infty} = \frac{1}{1-r} - 1 = \frac{r}{1-r}$

So, back from where we left:

$0.\overline{9} = 9 \cdot \sum_{n=1}^{+\infty} \left(\frac{1}{10}\right)^{n} = 9 \cdot \frac{\frac{1}{10}}{1-\frac{1}{10}} = 9 \cdot \frac{1}{10} \cdot \frac{10}{9} = \frac{9}{9} = 1$

And there we go.

### Getting full overboard with Non-Standard Calculus

And finally, just for the fun of it, we complete the specturm of proofs in this short post by also mentioning a solution involving an advanced technique: Hyperreal Numbers in a field of Mathematics known as Non-Standard Analysis. Needless to say, things now are going to get pretty abstract here as we are going to prove this simple mathematical fact with a theory developed during the same time as Ordinary Calculus, but that we don’t really study at school (and in some rare courses at university).

Without getting too technical in the Hyperreal model, we introduce the concept of hyperreal number $H \in {}^\ast\mathbb{R}$ (where ${}^\ast\mathbb{R}$ denotes the field of hyperreal numbers) as an extension to real number $x \in \mathbb{R}$.

An hyperreal number is a number greater than any number that can be expressed in the form $1 + 1 + \dots + 1$ (for any finite number of terms). Such a number is an infinite number, its reciprocal is defined to be an infinitesimal number.

An hyperreal number is either infinite or finite, this discrimination is better formalized by introducing an important application:

The Standard Part Function $\mathrm{st}: {}^\ast\mathbb{R} \mapsto \mathbb{R}$ is a function associating to an hyperreal number $H \in {}^\ast\mathbb{R}$ its closest unique real number $x \in \mathbb{R}$.

Convince yourself that $\mathrm{st}(\cdot)$ is nothing more than a rounding function. We can then express $0.\overline{9}$ as an hyperreal number:

$0.\overline{9} = 1 - (0.0\dots01) = 1 - \left(0.00\dots;0_{H-1}0_H0_{H+1}\dots1\right)$

The semicolon inside the expression of an hyperreal number is used to separate digits of finite ranks from digits of infinite ranks. Every digit $d_H$ is an hypernatural number. We can express this hyperreal number in a more compact way:

$0.00\dots;0_{H-1}0_H0_{H+1}\dots1 = 10^{-H} = \frac{1}{10^H}$

Since $10^H$ is an infinite number, its reciprocal $10^{-H}$ is an infinitesimal number. Now, since we want to express real number $0.\overline{9}$ into another real representation, we need to use $\mathrm{st}(\cdot)$ to prove our thesis:

$0.\overline{9} = \mathrm{st}\left(1 - 10^{-H}\right) = 1$

Leibniz’s Law of Continuity basically states that: “Whatever works for $\mathbb{R}$, also works for ${}^\ast\mathbb{R}$”. This allows us to extend many concepts from $\mathbb{R}$ to ${}^\ast\mathbb{R}$, like, for example, the fact that if $a,b \in \mathbb{R}$ then $a + b \in \mathbb{R}$, which is a basic consequence of the fact that $\mathbb{R}$ is a field. And since $\mathbb{R}$ is a field, by the Law of Continuity, also ${}^\ast\mathbb{R}$ is, hence implying that: $H_1,H_2 \in {}^\ast\mathbb{R} \implies H_1+H_2 \in {}^\ast\mathbb{R}$. This proves why $1 - 10^{-H}$ is an hyperreal number given as the sum of two hyperreal numbers (every real number can be considered an hyperreal number whose standard part is the same as the number itself). The fact that $\mathrm{st}\left(1 - 10^{-H}\right) = 1$ comes from the way $\mathrm{st}(\cdot)$ works in the special case of the sum between a real and an infinitesimal number:

$\mathrm{st}\left( x + H \right) = x, \forall x \in \mathbb{R}, \forall H \in {}^\ast\mathbb{R}$

The round off will exclude the infinitesimal component in the sum, leaving only the finite factor in the sum.

## Some phylosophical facts

It might still come as a surprise that $0.\overline{9} = 1$ and it is perfectly normal (especially if you just finished reading the last paragraph). But this is not anything strange at all. Mathematics is made of models, each of which targets a specific problem. In Number Theory, we have Natural Numbers and Decimal Numbers; these two models describe two different kinds of numbers, but sometimes the same number can be described by more models (and this is good as it proves that the different models we have built align with each other).

So, basically number $1$ can be described as both a natural number and a decimal number. And this is not valid for $1$ only: every $n \in \mathbb{N}$ (with $n \neq 0$) can have such dual description. The following equations are all true: $1.\overline{9} = 2$, $2.\overline{9} = 3$, and so on. The proof is simple as every $n > 1$ can be represented as:

$n = 1 + (n - 1) = 0.\overline{9} + (n - 1) = [n-1].\overline{9}$

And that’s one of the reasons why I love Math!