# Mathematical Etudes

In the September edition of the STEM Learning magazine we look at how to tackle mathematical fluency in the classroom in a way which will give at least as good a result as traditional methods but in, maybe, a more interesting way than the use of exercises that often repeat the same kind of question over and over again?

At the British Congress of Mathematics Education, I attended Dr Colin Foster’s session which explored the work of the ‘Mathematical Etudes Project’. The project “aims to find creative, imaginative and thought-provoking ways to help learners of mathematics develop their fluency in important mathematical procedures.”

You can find more about Mathematics Etudes Project here: http://www.mathematicaletudes.com/

Here is an example from Colin’s website.

Ask students to add together as many of the six fractions below to get an answer that is as near to one as possible – using each fraction only once.

^{ 1}/_{6} ^{1}/_{25} ^{ 3}/_{5} ^{3}/_{20 } ^{4}/_{15} ^{5}/_{8 }

Here, the skill of adding fractions is embedded in a larger problem and the practice of addition is needed, but it is not the end goal of the lesson.

In the replies to this post we will add some ideas of our own. Please feel free to add examples of your own.

## Comments

### Mike AndersonWed, 2018-09-26 10:09

The RISPS resource 'Brackets Out, Brackets In' could be used as an Etude:

Pick three different integers between -4 and 4 inclusive (not 0) to place in the three blanks below,with no repeats.

** (x + _ ) ( _x + _ )**

How many different orders are there?

Write down all the possible expressions, then multiply them all out and add all the results together.

Now take this sum- can you factorise it?

What do you find?

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### Mike AndersonThu, 2018-10-11 10:26

There is a great Etudes article on the NRICH website: https://nrich.maths.org/13206

In it, Colin Foster suggests that instead of asking questions like “what are the factors of 6?” we can ask students instead to “Find some numbers that have exactly 4 factors”

Colin describes the task as “wide open (there is certainly more than one number with exactly 4 factors), which means that different students can give different correct responses. It feels like a beginning rather than an end. As I accumulate numbers with 4 factors, I begin to ask, “Why do these numbers have 4 factors? What makes that happen?” If I find numbers with other numbers of factors, that’s almost as useful, really, because the 4 is clearly arbitrary. The real question behind the question is: What determines how many factors a number has?”

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### NoelstojSat, 2018-11-24 16:01

In the solutions in the magazine it shows 3/5 + 4/15 + 3/20 = 61/60 as closest to one. But 5/8 + 4/15 + 1/16 + 1/25 = 1193/1200 which is closer to one (7/1200 away, instead of 1/60 = 20/1200 away). I don't know if that was commented somewhere else already.

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### Mike AndersonTue, 2018-11-27 11:27

Hello!

It's always good to know that someone checks the answers- so thanks for your commnet. You are of course correct, 1193/1200 is closer!

It seems there was a typo in the final version of the magazine, the 1/16 was orginally meant to be 1/6 (as above). Good spot!

Thanks again,

Michael

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## Steve Lyon

Here is another fractions idea from the 'Something in Common' collection.

Choose three, different, non-zero integers that sum to zero. E.g. 3, 5, -8

Form all the possible ordered pairs you can with these numbers (you should get six)

Turn all of these into fractions using the first number as the numerator and the second as the denominator.

Sum all of these fractions.

Find the product of all of these fractions

Can you prove any findings you make?