It never ends: 0.9999... = 1!

While a mini-vacation/wine-soaked-trip around Napa, an old high-school friend was asking me about the whole 0.9999.... = 1 thing. Sometimes this is written as $0.\bar{9} = 1$. So I thought I'd dip my toe into and write about that here. This will also server as a test run of putting math equations into the blog. Also, it ties in nicely with my $\epsilon$ tee-shirt design.

I'm usually careful to not use exclamation points in math equations because they could represent a factorial and not just the normal excitement inherent in any discussion of mathematics but in this case it is ok since $1! = 1$.

First off, the most popular "proof" that 0.9999... = 1 goes like this. Suppose that
$$n = 0.9999\ldots$$
then multiple both sides by 10, which perseveres the equality:
$$10n = 9.9999\ldots.$$
Then you can subtract the first equation from the second and get:
$$9n = 9.0000\ldots$$
which gives that n = 1. The problem with this is that it's a bit of mathematical slight of hand and hides what's really going on. The problem is what do we even mean by 0.9999...?

Wherefore art thou 0.9999...?

0.9999... is  usually thought of as an infinity/never-ending sequence of nines following the decimal point. You can write this as a sum:
$$9/10 + 9/100 + 9/1000 + \cdots = \frac{0}{10^0} + \frac{9}{10^1} + \frac{9}{10^2} + \frac{9}{10^3} + \cdots.$$
This can be succinctly written as:
$$\sum_{i = 1}^{\infty} \frac{9}{10^i}.$$
So 0.9999... isn't number. It's a never ending sum. There is no point where the process stops and spits out the final number. Since it's not a number, what does it mean to ask what number does 0.9999... equal? How can an infinite process equal a number, one is a process and the other is a number? There is a mismatching of concepts.

There is a way (several in fact) where one can assign a value to the above infinite sum. First we start by looking at the partial sum, i.e. the sum where we stop after n terms:
$$S(n) = \sum_{i = 1}^{n} \frac{9}{10^i}$$
Two observations:

• for a given value of n, the sum S(n) is a finite and has a definite value.
• for larger and larger values of n, S(n) does get closer and closer to 1.

This can be formalized by saying that no matter how strict your tolerance, there is a value of n that makes the difference between S(n) and 1 within that tolerance. Suppose that your tolerance is $\epsilon$ (the greek letter epsilon is traditionally used for small positive numbers), there exists a value of n such that $\left|S(n) - 1\right| < \epsilon$. Or if you really like symbols:
$$\forall \epsilon >0 \ \ \exists n \ \ \ \text{s. t.}\ \ \  \forall m > n \left| S(m) - 1\right| < \epsilon.$$
I think that morass of symbols obscures more than it revivals. The ideas are the thing. In this case, the difference between S(n) and 1 has a simple closed form (left to the reader to derive) and it's clear that we can pick n such that difference is less than $\epsilon$ for any positive $\epsilon$. When this is the case, we say that S(n) converges to 1 as n goes to infinity. It's important to note that n never reaches infinity (whatever that might mean).

Thus $0.9999... = \sum_{i = 1}^{\infty} \frac{9}{10^i} = 1$ is really shorthand for $\sum_{i = 1}^{n} \frac{9}{10^i}$ converges to 1.

There are lots of infinite sums that converge to 1. Here's another famous one:
$$\sum_{i = 1}^{\infty} \frac{1}{2^i}.$$
Once you wrap your head around that one, what about this:
$$\sum_{i = 1}^{\infty} (-1)^n.$$
Does that converge? If so, to what?