The clue is in the question
Starting with one of the mini-challenges, how many of the other mini-challenges will you invent for yourself?
Age
16 to 18
Challenge level
This problem is in two parts. The first part provides some building blocks which will help you to solve the final challenge. These can be attempted in any order. This problem can also test your powers of conjecture and discovery: As you start from one of the mini-challenges, how many of the other related mini-challenges will you invent for yourself?
This challenge involves building up a set $F$ of fractions using a starting fraction and two operations which you use to generate new fractions from any member of $F$.
Rule 1: $F$ contains the fraction $\frac{1}{2}$.
Rule 2: If $\frac{p}{q}$ is in $F$ then $\frac{p}{p+q}$ is also in $F$.
Rule 3: If $\frac{p}{q}$ is in $F$ then $\frac{q}{p+q}$ is also in $F$.
Choose a mini-challenge from below to get started. There is a lot to think about in each of these mini-challenges, so as you think about them, continually ask yourself: Do I have any other thoughts? Do any other questions arise for me? Make a note of these, as they might help when you consider other parts of the problem.
| Mini-challenge A | |
| Mini-challenge B | |
| Mini-challenge C | |
| Mini-challenge D | |
| Mini-challenge E | |
| Mini-challenge F | |
| FINAL CHALLENGE |
James Bell from the MacMillan academy sent in his impressive solution to this difficult problem. Our congratulations go to James!
Steve says, the argument runs as follows:
1. N and D map proper fractions to proper fractions.
2. The inverses of N and D are unique fractions F.
3. The numerators and denominators never decrease following a transformation.
5. There are no fixed points of $N$ and $D$.
Putting these points together show that any proper fraction is in the set $F$: pick a proper fraction and you can always work backwards in a chain which leads to the proper fraction $\frac{1}{2}$.
James writes:
Applying rule 2 to $\frac{x}{y-x}$ gives $\frac{x}{y}$
Applying rule 3 to $\frac{y-x}{x}$ gives $\frac{x}{y}$
So, $\frac{x}{y}$ is a member of F if either $\frac{x}{y-x}$ or $\frac{y-x}{x}$ are members of F.
Any rational number between 0 and 1 can be written x/y where x and y are integers that share no common factors except 1. Given any such x and y either:
1) (y-x)> x in which case x/(y-x) (as x and y share no common factors except 1 neither can x and y-x) is a rational number between 0 and 1 with numerator+denominator less than x/y (y rather than x+y), whose presence in F would imply x/y's presence by rule 2
2) (y-x)< x in which case (y-x)/x (as x and y share no common factors except 1 neither can x and y-x) is a rational number between 0 and 1 with numerator+denominator less than x/y (y rather than x+y), whose presence in F would imply x/y's presence by rule 3
or 3) (y-x)=x and y=2x so x/y=x/2x=1/2 which we know to be a member of F using these rules
Given x/y, such that x+y> 3, we can generate either x/(y-x) or (y-x)/x (less than 1 greater than 0) whose presence in F implies x/y's presence and has smaller numerator over denominator. If we repeat this algorithm numerator+denominator must eventually fall to 0,1,2 or 3 (can't be negative as numerator and (denominator-numerator) must always be positive as they were to begin with and denominator is always greater than numerator, fraction is < 1)
case 0: only possibility 0/0 which can't be reached as it is not greater than 0
case 1:either 1/0 which can't be reached because it has greater numerator than denominator or 0/1 which can't be reached because it is not greater than 0.
case 2: 2/0 (numerator> denominator), 0/2 is not > 0, 1/1 is not < 1 case 3: 3/0 (numerator> denominator), 0/3 is not > 0, 2/1 is not < 1
which leaves from all cases only 1/2 (which we know is a member of F) can be reached and therefore must be the ultimate destination of all x/y defined above as we apply the algorithm above.
And so as 1/2 is in F so are all x/y (in lowest terms and between 0 and 1)
and so all rational numbers between 0 and 1 are in F QED
Why do this problem?
This problem involves a significant 'final challenge' which is
broken down into a sequence of mini-challenges. The mini-challenges
are not arranged in any particular order and the problem is
structured such that students are likely to 'discover' some of the
mini-challenges for themselves as they strive to solve other
mini-challenges.
These notes are designed for classes who are able to work in
groups.
At the outset all challenges are hidden to the learner to
maximise the chance of discovery for the learners.
The purpose of this is two-fold: first to scaffold learners to
help them solve a difficult challenge; second to show that
mathematics is a natural subject where certain questions naturally
arise through the consideration of other questions. This will
firstly help students to structure their mathematical thinking and
secondly to help them to realise that mathematics is not externally
or meaninglessly imposed.
You can print out cards of the statements here.
Possible approach
There are 6 mini-challenges and the final challenge. Leave
these all hidden to begin with.
Throughout the challenge the focus will be on constructing
clear, concise proof and on thinking of possible extension
questions.
Very able students might wish to start on the Final Challenge,
but it will be good to give them a single mini-challenge and see
where their thinking and invention takes them. Indeed, the best and
most inventive students are likely to 'discover' the final
challenge for themselves.
It is suggested that the following approach be taken
0) The context introduced so that every one understands the
rules
1) (10 minutes) Students individually given one of the
mini-challenges to think about and work on. Spread the different
mini-challenges amongst the group -- don't allow students to see
any of the other challenges or talk about this with their
neighbours. Students are to think about their challenges and
explicitly write down any other thoughts or questions that arise.
It is not expected at each stage that the challenges will be solved
-- merely that difficulties and other questions arise. Encourage
those that think they have an answer to construct the clearest
proof possible or to think about possible extensions.
2) (5 minutes) A selection of students to describe their
challenge and some of the difficulties and other questions arising.
It is likely that some of the questions arising will be the
problems other students were working on directly.
3) (10 minutes) Ask the class to organise themselves in pairs
so as to get insight into solving their mini-challenge. It might be
that students pair with people working on the same mini-challenge
or pair with someone who was thinking about a related problem
4) Repeat step 2 and group into 4s.
5) Throughout encourage the class to propose their own
extension questions.
At some point students might solve their challenges or pose
the final challenge for themselves. When appropriate move the
discussion onto the construction of a clear proof of the final
challenge. This will be ideally a group effort.
Key questions
As you think about your mini-challenge, what questions and
extensions arise?
Complete the sentence: I am finding this task difficult
because ...
Complete the sentence: I wonder if ....
Complete the sentence: I would be more able to solve my
challenge if I knew ...
Can you explain your proof clearly in words?
Possible extension
Solution of the final challenge on its own is a tough
challenge. The main extensions might be:
What happens if you start the process with a different
starting fraction?
Can you relate this process to any other mathematics you know
about?
Possible support
This task is designed for group work -- encourage groups not
to move on until all in the group understand.
Some students might be uncomforable with posing their own
questions or verbalizing their difficulties. Encourage an
atmosphere where all questions and difficulties are valid.