0.999... it just keeps on going.

If you follow any kind of mathematical discussions on the internet, you'll probably have noticed that there's one particular topic that refuses to die: \[ 0.\dot{9} = 1 \] In the way of perpetual internet conversations, although the participants change, the actual positions are pretty constant.

One side of the conversation insists that \(0.\dot{9}\) is not equal to \(1\), but is a little (even infinitesimally) less than \(1\).

The other side correctly says (but somehow unpersuasively argues) that, on the contrary, \(0.\dot{9}\) is exactly equal to \(1\), and provides various reasons for that.

Here are what my extremely scientific survey suggests are two of the commonest arguments presented in support of that true statement.

The first goes something like:

We all know that \[ \frac{1}{3} = 0.\dot{3} \] so, multiplying both sides by \(3\) we obviously have \[ 1 = 3 \times \frac{1}{3} = 3 \times 0.\dot{3} = 0.\dot{9} \]

The second is some version of:

Let \[ 0.\dot{9} = S. \] We'll find an equation for \(S\) that tells us what it is.

Clearly, \[ 10S = 9.\dot{9} \] and so \[ 10S-S = 9.\dot{9} - 0.\dot{9} \] i.e. \[ 9S = 9 \] and therefore \(S=1\).

And yet somehow the intransigent "it's a bit less than \(1\)" brigade aren't quite convinced. But why not?

Let's take a closer look at those two arguments.

The multiplications both rely on operating on an infinite decimal expansion by manipulating it in a way that looks entirely plausible. It should, because in each case it is in fact correct. But in each case the truth is not actually obvious. Why should we believe that multiplying an infinite decimal expansion by \(3\) is the same as multiplying each term by \(3\), or believe that multiplying an infinite decimal expansion by \(10\) is the same as shunting the decimal point along one place to the left? The second argument also requires the subtraction of one infinite decimal expansion from another. The algorithms we have for doing such arithmetic start with the rightmost non-zero digit, and in these cases there is no starting position.

In fact none of these calculations is trivial. The distributive law obviously holds for any multiple of any finite sum, but a non-terminating decimal is not a finite sum. It might be tempting to say that the multiplication and subtraction work no matter how many terms in the expansion we use, so they also work when the number of terms is infinite. Unfortunately for that argument, it doesn't matter after how many digits of \(0.\dot{9}\) we truncate, the result is less than \(1\), so why isn't that also true when the number of digits is infinite?

We have, then, two serious issues.

1. Most people don't actually have a clear understanding of infinite decimal expansions. They're introduced early enough in the school curriculum that everybody becomes familiar with them, but that isn't the same thing. So arguments involving them don't carry the psychological weight that they might.
2. The inductive argument makes it so 'obvious' that \(0.\dot{9} < 1\) that the other approaches, persuasive as they might seem, can't displace this conviction.

I suspect that some combination of these issues (though doubtless not clearly formulated) is what makes it so hard for people to see that \(0.\dot{9}\) is actually equal to \(1\). In the end, when faced with two incompatible statements, people will prefer the one that causes less mental discomfort.

Now, we have to take seriously the fact that it really does take a considerable amount of work to give a genuine proof that \(0.\dot{9} = 1\). You have to explain what is meant by a non-terminating decimal expansion, which means that you have to explain the meaning of the limit of a sequence, and in particular the meaning an infinite series as the limit of a sequence of partial sums. This is a serious undertaking.

Once you've done that, you have various strategies available. Amongst them are proving that the multiplications above are legitimate, and so that with this additional detail the result is established. And by the time you've done all the heavy digging about infinite series, and what a limit is, and what the real numbers are, the recipient of your wisdom will find it much harder to retain their conviction that the sum is somehow "infinitesimally" less than \(1\). (I hesitate to claim that they will find it impossible.)

It's worth noting that some of those arguing that \(0.\dot{9} = 1\) may not have a much better justification for their position that those who argue that \(0.\dot{9} <1\). You can believe the right thing for the wrong reason.

So don't be dejected if you can't persuade somebody whose mathematical preparation doesn't extend to an understanding of an infinite decimal expansion as the limit of an infinite sequence of partial sums. For such people, it is (almost) inconceivable that a sequence of approximations can all be less than \(1\) and the limit be exactly \(1\). In fact, if you have persuaded such a person by means of one of the arguments up above, you probably ought to feel a little shabby about it: although the conclusion is true, the argument as presented is no (or at least not much) more rigorous than the (mistaken) intuition that each finite expansion of \(9\)'s is less than \(1\), so the infinite expansion is still less than \(1\).

Alas, you probably ought to be dejected if one of the arguments presented above was what convinced you that \(0.\dot{9}=1\). I'm afraid you were tricked, and it really is a bit more complicated than that. On the bright side, you have a fascinating journey ahead of you if you decide to fill in the gaps.