The following eight unit fractions are given: \[\dfrac12\quad\dfrac13\quad\dfrac14\quad\dfrac15\quad\dfrac16\quad\dfrac17\quad\dfrac18\quad\dfrac19\]

- What is the probability that a pair of distinct fractions, chosen at random from these, have a difference that is also a unit fraction?

If I consider every possible pair of cards that I might select and record their differences I obtain the following results:

\(\frac{1}{2}\) | \(\frac{1}{3}\) | \(\frac{1}{4}\) | \(\frac{1}{5}\) | \(\frac{1}{6}\) | \(\frac{1}{7}\) | \(\frac{1}{8}\) | \(\frac{1}{9}\) | |
---|---|---|---|---|---|---|---|---|

\(\frac{1}{2}\) | X | \(\frac{1}{6}\) | \(\frac{1}{4}\) | \(\frac{3}{10}\) | \(\frac{1}{3}\) | \(\frac{5}{14}\) | \(\frac{3}{8}\) | \(\frac{7}{18}\) |

\(\frac{1}{3}\) | \(\frac{1}{6}\) | X | \(\frac{1}{12}\) | \(\frac{2}{15}\) | \(\frac{1}{6}\) | \(\frac{4}{21}\) | \(\frac{5}{24}\) | \(\frac{2}{9}\) |

\(\frac{1}{4}\) | \(\frac{1}{4}\) | \(\frac{1}{12}\) | X | \(\frac{1}{20}\) | \(\frac{1}{12}\) | \(\frac{3}{28}\) | \(\frac{1}{8}\) | \(\frac{5}{36}\) |

\(\frac{1}{5}\) | \(\frac{3}{10}\) | \(\frac{2}{15}\) | \(\frac{1}{20}\) | X | \(\frac{1}{30}\) | \(\frac{2}{35}\) | \(\frac{3}{40}\) | \(\frac{4}{45}\) |

\(\frac{1}{6}\) | \(\frac{1}{3}\) | \(\frac{1}{6}\) | \(\frac{1}{12}\) | \(\frac{1}{30}\) | X | \(\frac{1}{42}\) | \(\frac{1}{24}\) | \(\frac{1}{18}\) |

\(\frac{1}{7}\) | \(\frac{5}{14}\) | \(\frac{4}{21}\) | \(\frac{3}{28}\) | \(\frac{2}{35}\) | \(\frac{1}{42}\) | X | \(\frac{1}{56}\) | \(\frac{2}{63}\) |

\(\frac{1}{8}\) | \(\frac{3}{8}\) | \(\frac{5}{24}\) | \(\frac{1}{8}\) | \(\frac{3}{40}\) | \(\frac{1}{24}\) | \(\frac{1}{56}\) | X | \(\frac{1}{72}\) |

\(\frac{1}{9}\) | \(\frac{7}{18}\) | \(\frac{2}{9}\) | \(\frac{5}{36}\) | \(\frac{4}{45}\) | \(\frac{1}{18}\) | \(\frac{2}{63}\) | \(\frac{1}{72}\) | X |

From this I can see that there are \(56\) pairs of fractions I could pick (where I treat picking \(\frac{1}{2}\) first and then \(\frac{1}{3}\) as different from picking \(\frac{1}{3}\) first and then \(\frac{1}{2}\)), \(28\) of which have differences which are unit fractions. Therefore the probability that a pair of distinct fractions, chosen at random from this list, have a difference that is a unit fraction is \(\frac{28}{56}=\frac{1}{2}\).

What can you say about the pairs of fractions that have a difference that is a unit fraction?

Can you generalise your findings?

When I was completing the table of differences I noticed that there are two ‘types’ of fraction pairs that have a difference that is a unit fraction. First, there are those that ‘immediately’ result in a unit fraction, for example \[\frac{1}{2}-\frac{1}{3} = \frac{3}{6}-\frac{2}{6}=\frac{1}{6}.\] Then there are those that result in a unit fraction once simplified, for example \[\frac{1}{2}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{2}{6},\] which then simplifies to the unit fraction \(\frac{1}{3}\).

If I consider those pairs that ‘immediately’ result in a unit fraction, then I can look for any patterns: \[\begin{align*} \frac{1}{2} \quad &\text{and} \quad \frac{1}{3},& \frac{1}{2} \quad &\text{and} \quad \frac{1}{4},\\ \frac{1}{3} \quad &\text{and} \quad \frac{1}{4},& \frac{1}{3} \quad &\text{and} \quad \frac{1}{6},\\ \frac{1}{4} \quad &\text{and} \quad \frac{1}{5},& \frac{1}{4} \quad &\text{and} \quad \frac{1}{6},\\ \frac{1}{4} \quad &\text{and} \quad \frac{1}{8},& \frac{1}{5} \quad &\text{and} \quad \frac{1}{6},\\ \frac{1}{6} \quad &\text{and} \quad \frac{1}{7},& \frac{1}{6} \quad &\text{and} \quad \frac{1}{8},\\ \frac{1}{6} \quad &\text{and} \quad \frac{1}{9},& \frac{1}{7} \quad &\text{and} \quad \frac{1}{8},\\ \frac{1}{8} \quad &\text{and} \quad \frac{1}{9}. \end{align*}\]Among these, I notice that there are lots of pairs of fractions whose denominators are consecutive. In fact, every possible such pair appears in this list, so it looks as if whenever two unit fractions have denominators that are consecutive then their difference will be a unit fraction. In order to decide whether this is always true I need to prove it.

Two unit fractions with consecutive denominators can be represented as \[\frac{1}{n} \quad \text{and} \quad \frac{1}{n+1}.\] In order to find the difference between these two fractions I need to find a common denominator. In this case I might chose to multiply the denominators together giving \(n(n+1)\) as the common denominator.

Writing each fraction with a denominator of \(n(n+1)\) gives \[\frac{1}{n}=\frac{n+1}{n(n+1)}\quad\text{and}\quad \frac{1}{n+1}=\frac{n}{n(n+1)}.\]

I can now find the difference between the two fractions \[\frac{n+1}{n(n+1)} - \frac{n}{n(n+1)} = \frac{1}{n(n+1)},\] which is a unit fraction!

I’m not yet sure what to do with the other pairs of fractions, though. There are some patterns with some of them, but not all. I think I’ll pick some of those where I can spot a pattern, and leave the rest for the time being.

Now I’ll consider the other pairs of fractions that have a difference that is a unit fraction. I notice that for some of these pairs, the denominator of one is half (or double) the other. For example, \(\frac{1}{2}\) and \(\frac{1}{4}\); \(\frac{1}{3}\) and \(\frac{1}{6}\); \(\frac{1}{4}\) and \(\frac{1}{8}\) all have a difference that is a unit fraction.

I could represent a pair of unit fractions of this form as \[\frac{1}{n} \quad \text{and} \quad \frac{1}{2n}.\] Then finding the difference I have \[\frac{1}{n}-\frac{1}{2n}=\frac{2}{2n}-\frac{1}{2n}=\frac{1}{2n}\] which is a unit fraction. So this type of pair always works.

I still have some pairs of fractions to consider, some which have a difference that is a unit fraction once simplified, others of which give a unit fraction straight away. As it’s not clear to me what to do with the remaining examples in the above list, I’ll shift my attention to the one example we have which gives a unit fraction once simplified.

There is one pair of fractions: \[\frac{1}{2} \quad \text{and} \quad \frac{1}{6},\] which gives a unit fraction after simplification. How does this happen?

We have \[\frac{1}{2}-\frac{1}{6} = \frac{3}{6}-\frac{1}{6}=\frac{2}{6}= \frac{1}{3}.\] After pondering for a bit, I’ve noticed that the factor of \(2\) which cancelled at the end was a factor of the original denominators, and in fact was their highest common factor (HCF). That’s making me look back at the original list of fraction pairs to see if there are any other examples where the denominators have a highest common factor of \(2\).

And indeed there are! The pairs \[\frac{1}{4}\quad\text{and}\quad\frac{1}{6};\qquad \frac{1}{6}\quad\text{and}\quad\frac{1}{8}\] also have this property.

So we can rewrite the three pairs of fractions from above as follows: \[\begin{align*} \frac{1}{2 \times 1} \quad &\text{and} \quad \frac{1}{2 \times 3},\\ \frac{1}{2 \times 2} \quad &\text{and} \quad \frac{1}{2 \times 3},\\ \frac{1}{2 \times 3} \quad &\text{and} \quad \frac{1}{2 \times 4}, \end{align*}\]We could even add \(\frac{1}{2}\) and \(\frac{1}{4}\) to this list, even though it is in the form \(\frac{1}{n}\) and \(\frac{1}{2n}\), which we have already dealt with.

These pairs of fractions are in the form \(\dfrac{1}{2a}\) and \(\dfrac{1}{2b}\), where \(a\) and \(b\) are coprime.

Calculating the difference between this pair of general fractions gives \[\frac{1}{2a} - \frac{1}{2b}=\frac{b}{2ab}-\frac{a}{2ab}=\frac{b-a}{2ab}.\] This will be a unit fraction if \(b-a=1\) or if \(b-a\) is a factor of \(2ab\).

This explains all of the examples here: for the first pair, \(a=1\) and \(b=3\) differ by \(2\); for the other pairs (\(a=2\) and \(b=3\); \(a=3\) and \(b=4\)), the difference is \(b-a=1\).

What possible values could \(b-a\) take if this does simplify to a unit fraction?

One pair of fractions remains unaccounted for: \(\frac{1}{6}\) and \(\frac{1}{9}\). These denominators have an HCF of \(3\), so it might be interesting to think in general about a pair of unit fractions that have denominators with an HCF other than \(1\) or \(2\).

A pair of fractions that have an HCF of \(c\) can be written in the form \[\frac{1}{ac} \qquad \text{and} \qquad \frac{1}{bc}\] where \(a\) and \(b\) are coprime. Calculating their difference I find that \[\frac{1}{ac}-\frac{1}{bc}=\frac{b-a}{abc}.\] I see that, as before, this will be a unit fraction if and only if \(b-a=1\) or if \(b-a\) is a factor of \(abc\). In fact, as \(1\) is a factor of \(c\), we can just say that this will be a unit fraction if and only if \(b-a\) is a factor of \(c\).

As before, what possible values could \(b-a\) take if this does simplify to a unit fraction?

Now I understand what I first observed. If I write my pair of unit fractions as \[\frac{1}{ac}\quad\text{and}\quad\frac{1}{bc}\] where \(c\) is the HCF of the denominators, so that \(a\) and \(b\) are coprime, then this will simplify to a unit fraction if \(b-a\) is a factor of \(abc\) (in fact, if it is a factor of \(c\)). It will give a unit fraction immediately if \(b-a=1\), that is, if the denominators differ by their highest common factor, or are consecutive multiples of their HCF. They will give one after simplification if \(b-a>1\), but is still a factor of their HCF.

The fractions in this problem are all unit fractions. Can you find a pair of fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\) where \(a\neq 1\) and \(c\neq 1\), which have a difference that is a unit fraction?

Can you expand on your answer to the previous question – under what conditions will a pair of fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\) where \(a\neq 1\) and \(c\neq 1\), have a difference that is a unit fraction?

If I consider the more general fractions \(\frac{a}{b}\) and \(\frac{c}{d}\) then I can identify the conditions under which they will have a difference that is a unit fraction.

Calculating the difference between the two general fractions gives \[\frac{a}{b} - \frac{c}{d}=\frac{ad-bc}{bd}.\] For this to be a unit fraction I require that \(ad-bc=1\) or that \(ad-bc\) is a factor of \(bd\), so that the difference will simplify to a unit fraction.