Sunday, 25 March 2018


This week's set of tasks use a mapping diagram to represent liner functions. This model may not be familiar to some teachers and students but this can be advantage - it allows us to take a fresh look at the characteristics of linear functions and (though we don't do it here) can help us look at the Cartesian graph of linear functions with fresh eyes.
MONDAY: Here we are asked to read a mapping diagram and to determine the function that the given information represents. 
TUESDAY: Here we take a more detailed look at Monday's mapping diagram (which was for the function y = 3x – 2) and consider how its features can be used to identify the linear function.
WEDNESDAY: I was amazed by this phenomenon when I first came across it (thanks to JH). Of course, as with many mathematical phenomena, the more you think about it the more obvious it becomes. But this one also becomes richer. All in all, an exciting and lovely result!
We can visualise parts i and ii by translating point P and the arrows one unit to the right (part i) or one unit down (part ii). Alternatively, move both axes one unit to the left (part i) or one unit up (part ii). Which approach do you prefer?
PS: You will probably have noticed that P can be thought of as the centre of enlargement for the linear function. Image points (ie points on the y axis) are 3 times the distance from P as object points (ie the corresponding points on the x axis). So the scale factor of the 'enlargement' is ×3. Similarly, the distance between a pair of points on the y axis is 3 times the distance of corresponding points on the x axis.
In part i, the same distance-relationships hold, ie the scale factor is still ×3. However, in part ii the distance-relationships change, and therefore so does the scale factor. Has it got smaller or larger?
THURSDAY 1: Here we extend the notion that the arrows on mapping diagrams of linear functions emanate from a 'centre of enlargement', by considering centres that lie between the two axes. What does that tell us about the 'scale factor', ie the value of m in y = mx + c ?
THURSDAY 2 (instead of Good Friday): Here we consider the inverse of our original function.
The inverse takes us back to where we started. One way of representing this there-and-back journey is by adding a second mapping diagram that is a reflection of the original, as here:
Note: the inverse function itself is represented by just the second mapping diagram, but with the axes re-labeled (x axis at the top, y axis below), and with the arrows pointing downwards. As can be seen, the 'common point' (or 'centre of enlargement') is vertically below P, two units below the y axis. The scale factor here is 1/3, and if we look carefully, we can see that 0 is mapped onto 2/3, so the inverse function is y = 1/3 x + 2/3, or y = 1/3 (x + 2). We can verify this algebraically, by transforming the original function y = 3x – 2 into x = .... (and then transposing x and y).

Sunday, 18 March 2018


This week we look at two tasks of classic form, both of which are rather contrived but which are also quite engaging if simply treated as a puzzles. The aim is to show the power of using a symbolic algebra approach, although in the case of the first task, below, we have deliberately devised a task which is also amenable to informal methods.
MONDAY: This first task is of the classic 'How many cows and chickens' form, although we have tried to make it a bit more challenging by involving three rather than just two sets of 'elements', in this case, bikes, trikes and quadracycles.
As we spell-out in Tuesday's variant, this Cycles task can be solved algebraically by creating equations such as
T = Q + 5
31 = B + T + Q
76 = 2B + 3T + 4Q.
We can then substitute Q for T. Also, if we double the second equation, giving 64 = 2B + 2T + 2Q, we can eliminate B by subtracting this equation from the third equation, leaving us with 14 = T+ 2Q.
In Tuesday's variant, we also indicate that the task can be solved informally with this sort of argument:
Imagine that the cycles are all bikes;
we would then have 64 tyres (or wheels) which is 14 fewer than the actual number;
so we would need to replace some of the bikes with trikes and/or quads;
each trike would increase the number of tyres by 1, each quad would increase the number of tyres by 2;
so ....
There is a clear link between this narrative approach and the process that gave us the equation 14 = T + 2Q. Of course, students might not see this link immediately, though Freudenthal (in his China Lectures published in 1991) suggests that students can achieve this by solving a range of similar tasks in this informal way and generalising.
There is an interesting tension here. One of the strengths of a symbolic approach is that it 'frees' us from thinking about the context, once we have expressed the given information symbolically, and assuming we are fluent in manipulating the symbols.
On the other hand, the act of keeping hold of the context, or of referring back to it periodically, can help us (or more to the point, our students) to form the symbolisations and to check that these, and the subsequent manipulations, are valid.
TUESDAY: Here we present two methods for solving yesterday's Cylces task, one informal the other involving symbolic algebra, which we ask students to complete.
How students respond to these methods will depend on their knowledge and experience. Some students might gain a better appreciation of the power and efficiency of symbolic algebra, while others might gain some understanding of the algebraic approach by linking it to the more concrete, informal approach.
WEDNESDAY: Here we present two new tasks, both set in the context of 'sharing acorns' but where the given information is structured differently. We consider how amenable the tasks are to informal and formal approaches in the light of this difference in structure.
The first task can be solved in informal and formal ways that are very similar: the collection of 60 acorns can be partitioned into 1 plus 4 plus 5 equal shares, ie into 10 equal shares altogether; algebraically, we can construct something like this: g + 4g + 5g = 60, so 10g = 60.
The second task can be solved informally but it probably requires quite a lot of insight to do this successfully. On the other hand, it can be solved in a fairly routine way using symbolic algebra, assuming the solver has adequate technical skills.
Thus, we would argue that the second task is that elusive entity: a task (albeit a mere pointless puzzle!) where the use of symbolic algebra comes into its own. Of course, in the first task, we could also strengthen the need to record and symbolise, by increasing the number of recipients and by relating their shares to George's share in more varied ways.
Note 1: You may have noticed that the methods of sharing in the two tasks turn out to be equivalent. The second task has been made more complex by using what Dettori et al (2006) call 'circular references'. I came across this work in the excellent revue by Mason & Sutherland (2002), Key Aspects of Teaching Algebra in Schools. You might like to devise similar tasks to challenge your students, by taking a straightforward sharing task and then using circular references to recast the description of some of the shares.
Note 2: The 'circular references' task used by Dettori (and reproduced in Mason & Sutherland) is this:
Interestingly, an almost identical task occurs in an English translation (1822, p204) of a French translation of Leonhard Euler's Vollständige Anleitung Zur Algebra (1771).
The English book (below) contains 'Additions of M. De La Grange' and it is possible that the problem stems from him rather than Euler, as I can't find it in the original German edition.
THURSDAY: Here we introduce Auntie's way of sharing acorns. It is not equivalent to Grandpa's method, but it turns out to produce the same result for a particular number of acorns.
We can find the particular number of acorns using symbolic algebra, by solving the equation resulting from equating Peppa's shares (ie 4g and 2g + 20) or Chloe's shares (5g and 2g + 30) under the two ways of sharing. Both give the value g = 10, so that the total number of acorns is 100.
We have obviously carefully contrived the two sharing methods for them to produce the same outcome for all three recipients for a particular number (100) of acorns. This wouldn't happen by chance. However it raises an interesting (although rather subtle) question: given that the choice of Grandpa's or Auntie's sharing method makes no difference for 100 acorns, does it make a difference, for any individual recipient, for other numbers of acorns?
FRIDAY: Having established that Grandpa's and Auntie's method happen to produce identical shares when there are 100 acorns, today we consider what happens when there are more than 100 acorns: for each individual recipient, is one sharing method more favourable than the other?
Note: This task is quite complex and is probably more suitable as a challenge for teachers than as an activity in class.
In the case of George, he always gets one tenth of the acorns when Grandpa shares them out. This is not a large proportion, but still larger than what he gets from Auntie when there are less than 100 acorns. In the extreme case, when there are only 50 acorns, George gets none! George is better off with Auntie when there are more than 100 acorns.
The situation with Peppa turns out to be rather surprising. When there are 100 acorns, George's share is 10 acorns by either sharing method and she gets 40 acorns by either sharing method (ie 4×10 or 2×10 + 20). Now, when Georges share (g) is greater than 10, 4g will be greater than 2g + 20, which suggests that Grandpa's method will be more favourable than Auntie's when there are more than 100 acorns. However, it turns out that this is not the case! Why? Because Peppa's share depends on George's (as it's a function of g), but George's share differs depending on whether Grandpa or Auntie are sharing out the acorns. It turns out that Peppa's share comes to the same amount, whether it's Grandpa or Auntie doing the sharing, regardless of the number of acorns being shared out! For exmple, for 50 acorns she gets 20 under both methods, for 110 acorns she gets 44 under both methods. We leave it to the reader to verify/prove this.
For Chloe, it looks like Grandpa's method is more favourable than Auntie's when there are more than 100 acorns, as 5g > 2g + 30 when g > 10. But can we be sure? In the case of Grandpa's method, Chloe's share is 5/10 or 1/2 of the total number of acorns. In the case of Auntie's method, Chloe's share is 2/5 of the total plus 10 (you might like to verify this). When the total is more than 100, 1/2 of the total is greater than 2/5 of the total plus 10.

Sunday, 11 March 2018


In this week's set of tasks we look at figurative patterns and express their structure in generic terms by using a quasi variable (in this case, 20). We look at equivalent ways of construing/expressing the structure.
Some of the week's patterns turn out to have a linear structure and some a quadratic structure - which raises the question,  
How can we recognise this in the pattern itself, as well as in its symbolic representation?
MONDAY: Here is the week's root task:
Note: We have chosen 20 here (and in later tasks) on the basis that it is big enough to deter most students from simply counting each individual dot. Also, if students can find an efficient method for counting a 20-chain, it is likely that they can do so for any length of chain, perhaps even a chain of n Y-shapes. So, for many students, 20 is taking on the role of a variable.
Note 2: Another reason for going for such a 'far generalisation' early on, is to encourage students to look at the structure of the 20-chain as a whole. Of course, it is still possible to use an 'incremental' (or scalar or term-to-term) approach by noting that the 2-chain has 4 more dots than the 1-chain (and so the 20-chain has 19×4 more dots than the 1-chain). It would be interesting to see how the task would work without showing the 2-chain. We decided to put it in to avoid students thinking that there might be a hidden dot underneath the dot where the Ys join (in which case a 2-chain would simply have 2×5 dots and a 20-chain would simply have 20×5 dots).
Note 3: The presence of the 2-chain does open up other interesting possibilities: some students might argue that the 20-chain will have 10 times as many dots as the 2-chain; others might simply focus on the given numerical information, 1→5 and 2→9, and look for a mapping-rule that fits both, which can then be applied to 20.
TUESDAY: Here we are given two expressions for finding the number of dots in the 20-chain. The expressions are deliberately left open so as to show how the chain of Y-shapes can be structured.
It may be helpful to make an annotated sketch, as in the examples below, when describing each structure (ie when explaining Alma's and Bojan's methods).
The given expressions are of course equivalent and you might want to challenge students to show how one expression can be transformed into the other. It is possible to structure the pattern in other (equivalent) ways, resulting in other (equivalent) expressions 'in 20'. How many can you find?
WEDNESDAY: Here we consider two further chains of Y-shapes, but this time the size of the Ys increases as the number of Ys in the chain increases. What does this do to the relationship between the number of Ys and the number of dots in a chain? Is it still linear? Can you tell from the pattern? Can you tell from the generic expression for the total number of dots?
 Note: Here we deliberately consider a far generalisation again, rather than ask for the number of dots in the 5-chain, say. The hope is that this will focus students' attention on the overall structure of the pattern rather than on the change from one chain to the next. Put another way, we are trying to see whether students will consider the pattern generically. Of course, it is perfectly legitimate to look at differences, ie to consider the pattern 'incrementally', but it's not particularly efficient here.
THURSDAY: Here we look at some more dot patterns, but this time the Y-chains have morphed into single trees.
Yesterday, the relation between chain-length and the total number of dots turned out to be quadratic, in both cases. What about today? Can you tell just by looking at the pattern? Do the generic expressions for the 20-brown-dots trees confirm this?
FRIDAY: Here we consider equivalent generic expressions for the number of dots in Fynn's 20-brown-dots tree. We try to relate them to the dot pattern - the first one is relatively straightforward, the second less so. We then try to transform one expression into the other.
We can show the structure contained in the expressions with annotated sketches like these:


Sunday, 4 March 2018


In this week's set of ALG 5 tasks, we find specific unknown values by using the so-called bar model (curved bars in our case...) and by forming and equating algebraic expressions - and we look at alternative ways of forming the expressions.
MONDAY: The root ALG 5 task. A sketch might help or you might find another way to visualise or represent the situation - or you could use trial and improvement. You can view a dynamic version of the queue here.
Note: there's plenty of scope for varying the task.
Here's an easy variant:
When has Deka queued for 11 times as long as Eric?
Note 2:  The bar model (or in this case, 'bent rod model') is a powerful device for visualising quantities, but what if you decide to make a sketch and the proportions in your drawing turn out to be not that good? That's probably OK, as long as you don't expect instant answers from the model and are prepared to slow down and modify the drawing (on paper or in your head).
TUESDAY: Here we ask for a symbolic algebra approach. Constructing the expressions can be quite challenging, though we can check to see whether they make sense by making use of the values found on Monday, ie by checking to see whether we get a value of 30 when e = 10. Then, having got the expressions, we can consider how they could themselves be used to derive the value e = 10, eg by forming an equation and solving it in some way ....
WEDNESDAY: In this variant, we solve the same problem by forming expressions in d, the time taken by Deka, instead of e, the time taken by Eric.
As well as solving the problem itself (whose solution we already know, of course), this gives us the opportunity to compare the two sets of expressions and to consider how an expression for e in terms of d is related to, and can be transformed into, the corresponding expression for d in terms of e.
THURSDAY: Here we use algebra to solve a slightly more complex problem which isn't quite as amenable to a bar-model approach as the previous task. So where previously we tried to give meaning to an algebraic approach, here we try to show its utility.
FRIDAY: Our final version of the ALG 5 task involves the same context but has a different structure. We again represent/solve it using the clock-diagram (or bar model) and in symbolic algebra - which approach do you find easier here?
We can also approach the task using a mix of trial and improvement and analysis: 
At present (with d = 20 and e = 10) Deka and Eric have queued for a total of 30 minutes. If we wind forward 5 minutes, say, this will add 10 minutes to the total, so if we wind forwards a total of 15 minutes (making d = 20+15 and e = 10+15) they will have queued for a total of 60 minutes. This movie might help students see when to 'stop' the clock: Long lunch with Deka and Eric.
Another approach would be to argue like this:
30+30 = 60
31+29 = 60
35+25 = 60.
These approaches are not purely empirical - they make use of the task's structure so they can be classed as algebraic, I think. But they don't make use of symbolic algebra, even though the last method could be said to embody d+e=30 and de = 10. This highlights a real dilemma: it is not that easy to devise accessible tasks where the need for symbolic algebra is compelling. However, as we shall see in later weeks, one strategy is to make the task more complex, so that symbolising provides a way of keeping track of the information.
Notice: There's an interesting symmetry in the clock-diagram when d + e = 60: