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

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

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.

- 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.
- 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.

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.

I appreciate that this makes the important point that the proof that 0.9˙=1 is actually non-trivial, and that the commonly given arguments substitute mathematical trickery for addressing what's actually hard about the proof.

ReplyDeleteBut I think one should go further. The primary issue isn't so much the complexity of the proof as the fact that it digs deeper into the formal definition of the real numbers than many people are used to. It's very clear that most of the people who struggle with this are actually feeling their way towards an alternate axiomitization of the real numbers. With the benefit of a few centuries of work, we know to be skeptical of the naive theory of infinitesimals that people seem to find intuitive - there's no way to capture exactly the structure people seem to find intuitive.

The accepted formal definition of the reals is a formalization imposed by mathematicians on an informal notion, and we should stop assuming that when non-experts talk about the reals, they automatically mean the same thing mathematicians do. In fact, they usually mean the naive notion, ill-defined as it is. (And some people do genuinely seem to mean something closer to what we call the hyperreals than to what we call the reals.)

Yes. I agree with a lot of this, and the observation that it delves deeply into the definition of the reals is certainly part of my argument. The difficulty lies in the gap between the naive notion of the reals (which generally isn't precise enough to admit a precise argument) and the formal standard mathematics one (which is precise enough, but takes a lot of effort to come to terms with).

ReplyDeleteThere are certainly contexts in which non standard analysis seems closer to intuition than standard, but the relationship (transfer principle) there is also extremely subtle.

Thanks for your comments!

For the skeptical students I often ask them "If zero point nine repeating is less than one, what number is between them?"

ReplyDeleteWhich should get into an interesting discussion about the structure of the reals...

ReplyDeleteI think I was persuaded by a variant of your second argument when I was in school, some 40 years ago. However the argument was constructed using only rationals. First we learned that rationals give rise to recurring or terminating decimal expansions. Then we learned that the algorithm in the second algorithm would recover the original rational from any recurring or terminating decimal expansion (the 10 in the example should be 1 followed by one 0 for each digit in the recurring sequence). Finally we were introduced to reals by observing that there were digit sequences that neither recur nor terminate.

ReplyDeleteHas anyone tried this approach to convince skeptical students?

ReplyDelete1

= 0.9+0.1

= 0.9+0.09+0.01

= 0.9+0.09+0.009+0.001

etc.

I don't know: of course, there are lots of non-rigorous arguments, and different ones will convince different people. If you try it out, I'd certainly be interested in how successful you find it.

ReplyDeleteIf 0.9999etc was a price and i paid with a pound the till would calc my change for ever(boring and I'd died of old age) skip to eternity no change thou. If the price was a pound I'd be out the shop, can be more different

ReplyDeleteIf 0.9999etc was a price and i paid with a pound the till would calc my change for ever(boring and I'd died of old age) skip to eternity no change thou. If the price was a pound I'd be out the shop, can be more different

ReplyDeleteI can see that the explanation with 1/3 has a big assumption at the beginning...but I am very curious as to what is wrong with the 9S version.

ReplyDeleteAside from seeming sound it also is v similar to the (now alleged!) proof I use for the sum of a geometric sequence...

Is my (limited) knowledge all built on sand and I am just regurgitating maths comforters to the masses?

It's mostly the (easily fixed) problem of multiplying an infinite sum by a constant giving the same result as multiplying each term by a constant. The distributive law only tells you that for finite sums, so there's a proof of convergence needed. (And if you're very picky, subtracting infinite decimal expansions; the algorithm never terminates, so again you have to assume that something 'obvious' in this particular case, but a little technical, is true.) Not built on sand, just lacking a little bit of detail.

ReplyDeleteThe usual gaps in the argument for the infinite geometric series are this 'infinite distributive law' and showing that a^n really does give 0 in the limit.

So whatever you do, it all ultimately rests on some properties of limits and convergence. With the 0.999... example, the properties are so 'obvious' it's hard to remember that you are using them.