Recurring decimals are cool!

I was thinking about maths teaching the other day (I’m sure you’ll not be surprised to know this is something I often do) and in particular about the different approaches teachers bring to the subject. Teaching maths is a challenge because of the need to balance the development of basic numeracy skills through practice (something that unavoidably leads to a degree of tedium) with the creativity, fun and excitement that is inherent in the subject — but often lying just below the surface where it is easily missed by teachers and students alike.

Time pressures, syllabus constraints, focus on rote assessment, and prior negative experience and expectations all act to push this balance too far away from “fun” maths, and that makes it very difficult for the love of mathematics to grow in many students.

So I further thought about how to quickly assess a teacher’s readiness to redress this imbalance, and came up with the following simple question:

Tell me three reasons why recurring decimals are cool.

My own answers came as I formed the question, but the point is not what specific things someone believes are interesting about recurring decimals, but rather that they should be able to quickly and comfortably describe cool things about pretty much any aspect of mathematics.

Anyway, I suppose you’d like to know how I’d answer — well here are my three things.

Answer 1: Recurring decimals touch infinity

A recurring decimal’s decimal representation repeats forever. Forever!. Have you ever stopped and just thought about that? It’s amazing! It it never, ever stops. How many things in life go on forever? As I’ve said elsewhere in this blog, only in mathematics do we get to play with infinity. That’s very cool (and a bit freaky).

Consider the following sequence of equalities that shows us how we can understand that an infinitely long expression has a finite value by seeing how each time we are adding something smaller and smaller.

    \begin{eqnarray*} 0.3 & = & \frac{3}{10} \\ 0.33 & = & \frac{3}{10} + \frac{3}{100} \\ 0.333 & = & \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} \\ \vdots & & \vdots \\ 0.3333\ldots & = & \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \frac{3}{10000} + \ldots \\ & = & \frac{1}{3} \end{eqnarray*}

Each step we get closer to one-third, but only at the end, when we have added in an infinite number of terms do we actually get there. Thats right — you need an infinity of terms. If you stop earlier, no matter how far along, you will always be a little short of one-third.

And this is the case for all recurring decimals. The pattern in their decimal expansion goes on forever, but because it repeats, they will always end up with a nice fractional form.

That’s definitely one interesting thing. What else can we find?

Answer 2: One-seventh is recurring

The decimal expansion is \frac{1}{7} = 0.\dot{1}4285\dot{7}. This is interesting for a few reasons. Firstly it is a little unexpected — we tend to focus on denominators of 3, 6, 9 and other multiples of 3 when thinking about recurring decimals. This is a reminder that there are many others.

Secondly, it has a cycle of size 6 — one less than the denominator 7. That is the longest cycle possible for dividing by 7. In general, if dividing by n results in a recurring decimal, it can never result in a cycle longer than n-1.

But the most interesting thing of all is apparent in the following table:

    \begin{eqnarray*} \tfrac{1}{7}~~= & 1 \times 0.142857\ldots~~= & 0.142857 \ldots \\ \tfrac{3}{7}~~= & 3 \times 0.142857\ldots~~= & 0.428571 \ldots \\ \tfrac{2}{7}~~= & 2 \times 0.142857\ldots~~= & 0.285714 \ldots \\ \tfrac{6}{7}~~= & 6 \times 0.142857\ldots~~= & 0.857142 \ldots \\ \tfrac{4}{7}~~= & 4 \times 0.142857\ldots~~= & 0.571428 \ldots \\ \tfrac{5}{7}~~= & 5 \times 0.142857\ldots~~= & 0.714285 \ldots \end{eqnarray*}

Did you notice the pattern? Look at how the decimal values are cycling.

Answer 3: 0.9999… = 1

Yes, 0.\dot{9}=1. Exactly. Not just really, really close, but exactly equal. We saw earlier that 0.\dot{3}=\frac{1}{3}, so let’s multiply each side by 3.

    \begin{eqnarray*} 0.3333\ldots & = & \frac{1}{3} \\ \color{red}{\scriptstyle \times 3} & & \color{red}{\scriptstyle \times 3} \\ 0.9999\ldots & = & 1 \end{eqnarray*}

You can also see this by setting n=0.9999\ldots. Then see that

    \begin{eqnarray*} 10n & = & 9.9999\ldots~~~-\\ n & = & 0.9999\ldots \\ \cline{1-3}  9n & = & 9.0000\ldots \end{eqnarray*}

so 9n=9, and hence n=1.

That is a good one, and quite a bamboozler. But I’ve got an even better bamboozler to follow. Working with infinity makes it quite likely you will end up bamboozled!

Bamboozler: Grandi’s Series

In the first point above, we looked at the decimal expansion of ⅓ in the form of an infinite series (i.e. an infinitely long sum of numbers). That series works out very nicely, but summing infinitely long expressions can be quite tricky and lead to some very bamboozling results.

The infinite series


is named after Guido Grandi, who studied it in detail more than 300 years ago. What do you think it evaluates to? Have a think about it before you read ahead.

Well, what did you come up with?

Here’s one possible answer. We can group the 1s in pairs to get

    \[S=(1-1)+(1-1)+(1-1)+\ldots=0+0+0+\ldots = 0.\]

That seems reasonable enough, but what if we kept the first 1 separate, then grouped the others?

    \[S=1+(-1+1)+(-1+1)+(-1+1)+\ldots=1+0+0+0+\ldots = 1.\]

Hang on there! What kind of sorcery is this? How can the same sum equal both 0 and 1?

Furthermore, we can move a few extra 1s to the front, and get

    \begin{eqnarray*} S & = & 1+1+1+1+(-1+1)+(-1+1)+(-1+1)+\ldots \\   & = & 1+1+1+1+0+0+0+\ldots \\   & = & 4. \end{eqnarray*}

That also works, because even though we’ve brought some 1s to the front, we never run out of positive and negative terms to pair up afterwards, so all the other terms still cancel. It seems that this series can be made to equal any integer simply by rearranging the terms.

Let’s try a more formal approach. We have


but adding a single 0 at the front can’t change the value of the sum, so we also have


Now let’s add these two series term by term:

    \begin{eqnarray*} S & = & 1-1+1-1+1-\ldots~~~+\\ S & = & 0+1-1+1-1+\ldots \\ \cline{1-3}  2S & = & 1+0+0+0+0+\ldots \end{eqnarray*}

So 2S=1, and therefore S=\frac{1}{2}.

Whoa! That’s the strangest one yet. We have just shown, somehow, that


When dealing with infinite series, there are three possible outcomes. The sum may diverge — that is increase without bound and thus have no answer. Alternatively the sum may converge — that is evaluate to a finite answer. But, as we’ve seen with Grandi’s series, that’s not enough to ensure a series plays nicely. To be a truly friendly infinite series, it must converge absolutely — evaluate to the same answer no matter how you rearrange the terms. Grandi’s series converges, but not absolutely.

But does that mean that our answer S=1-1+1-1+\ldots=\tfrac{1}{2} is wrong? Well not really. When you move from finite sums to infinite sums, things change, and this just reflects a different way to understand summation when there is an infinity of terms. It is not unusual in maths to push things beyond a boundary and then extend their meaning.

Think, for example, about powers. We all learned that 32 means multiply 3 by itself 2 times, but then what can 3-2 mean? To multiply something by itself a negative number of times makes no sense at all. But extending the definition of power to include negative numbers, specifically defining a^{-n}=\frac{1}{a^n}, and exploring the implications leads to important and very meaningful results and greatly broadens our mathematical understanding.

Never forget that there is much to learn by exploring the boundaries of your mathematical knowledge. Almost certainly you know more than you realise, and a bit of experimentation can result in a wonderful journey of discovery, where seeming “nonsense” becomes clear and empowering!

Mr Wessen

Sometime teacher, sometime programmer, full time mathematician.

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