Sunday, 24 June 2018

ALG 20

Algebraderci! This post might be the last ALG post (at least for a while). Fittingly, perhaps, we devote this week's task to rules of single and double false position, which are methods that go back about 2000 years and that surfaced in many parts of the world - China, Egypt, the Arab world, India and, eventually, Europe. It is not clear whether the methods spread from one region to another, or were discovered/invented spontaneously in several regions. They were generally used to solve practical problems and as such were known not just to mathematicians but to people such as merchants who would use them as algorithms, ie as rules that worked rather than as rules that needed to be justified and proved. The rules apply to situations that we would now represent with linear or affine equations, ie equations of the form y = ax and y = ax + b respectively. However the rules were devised many centuries before symbolic algebra was invented, and continued to be used for some centuries thereafter.
MONDAY: We start with a problem taken from the Eygyptan Rhind Mathematical Papyrus (RMP), which dates from about 1550 BC and was written by a 'copyist', Ahmose, who claims to have been copying work from one or several centuries earlier.
This particular problem is quite straightforward, and certainly doesn't need formal algebra, though that provides one way of solving it. We present it here to illustrate the rule of single false position, which is the way it is solved in the RMP.
The above problem is Problem 26 in the RMP. Problems 24, 25 and 27 are similar, though interestingly they would seem more complex to us than Problem 26, since their solutions are all fractional. In fact, the first three problems appear to get progressively easier! Something to think about, for the small-stepping devotees of 'intelligent practice'.
Here are the problems, expressed in modern form:
24: x + x/7 = 18
25: x + x/2 = 16
26: x + x/4 = 15
27: x + x/5 = 21.
TUESDAY: We show how a problem can be solved using the rule of double false position - or rather, we show the first two steps, and leave it to the reader to complete the method. We provide some help with the final step in Wednesday's version of the task.
Note: We came across this Apples task in a very useful conference paper by Schwartz, presented in 2004. Schwartz gives the source of the task as the Liber Augmenti et Diminutionis: "This book, probably from the 12th Century, is the translation of some now-lost earlier work attributed to one 'Abraham' and generally believed to have been written in Arabic or Hebrew" (ibid). The task can also be solved algebraically, though it wouldn't have been in its day. Interestingly, an algebraic approach turns out to be very cumbersome here, compared to the use of double false position.
WEDNESDAY: We revisit Tuesday's task but use a diagram to help make sense of the double false position rule.
In this example, the double false position rule can be written as (100×21 – 204×8)÷(21 – 8), which reduces to 36. An alternative version, which can perhaps be derived more directly from the diagram, is this:
100 – 8/(21–8) of (204–100)
= 100 – 8/13 of 104
= 100 – 64
= 36.
 The general form of the rule is (xe₂ – xe₁)÷(e₂ – e₁) which, presented rhetorically, would in ancient times probably have been treated simply as a practical algorithm to be learnt and applied.
THURSDAY: We apply the double false position rule to parts of a fish.... This problem was probably written by de la Grange. It appears in the 1822 translation into English of Euler's Elements of Algebra. The intention would have been to solve the task algebraically, using simultaneous equations. However, it also lends itself very nicely to the double false position approach.
Here, the weighted average of 11 and 31 can be written as (2×11 + 8×31)÷(8 + 2) = (22+248)÷10 = 27, so the head weights 27 lb. [In turn, the fish weighs 72 lb, as stated in the text.]
Here's a nice bar-model approach (thanks to Bernie Westacott). One can see how it corresponds to an algebraic approach using simultaneous equations.

FRIDAY: Friday night is bath night and we take a (classic) bath.... This type of problem goes back centuries, though this one is only from 1936 (it is quoted in a paper co-authored by Arcavi). The problem is remarkably easy to solve using single false position. A greater challenge is to find an informal method - to start with, it might be helpful to simplify the task slightly, by ignoring the waste-pipe, ie by putting the plug in the bath.
Valediction: 20 sets of task, that's 20 weeks or almost 5 months of algebra tasks, 100 in all. Time to say Algebraderci (at least for a while).

Sunday, 17 June 2018

ALG 19

This week we're looking at graphs that model 'real life' situations. The focus is on qualitative aspects of the graphs, though we do take a more analytic view at times. Pioneering work on school students' understanding of graphs was undertaken in the late 1970s and early 1980s by Daphne Kerslake, by Claude Janvier and by Malcolm Swan. I doubt whether one can better some of their outstanding tasks.
MONDAY: Mathematically, this task involves a nice context, though one that is perhaps not so familiar to today's school students - you might therefore have to tease out the crucial fact that audio tape has to travel at a constant speed (as it passes over the tape-head which picks up the sound). Most students will realise that as the left hand spool unwinds, the width (diameter) of the reel of tape decreases. This means the circumference also decreases and so each successive turn of the spool will feed out less and less tape. So if the tape passes the tape-head at a constant speed, the spool will rotate more and more quickly, and the width of the reel will decrease more and more rapidly.
This task should generate rich classroom discussion about these geometric aspects of the context and about the shape of the graph: the graph clearly slopes downwards, but is it a straight line?
As the width of the reel of tape decreases more and more rapidly (over time), the slope of the graph gets progressively steeper (the gradient is negative throughout). This raises interesting questions about the slope near points A and B. Could it be horizontal at A and vertical at B? [Well, no, not quite...] And if we continue the curve, where might it cut the x-axis, and what would this mean in terms of the context?
Note: The curve through A and B is a parabola. It turns out that it passes roughly through the point (76, 0) and that the relationship between M minutes and W mm, is given by M ≂ 76 – (W²/30) [or, more precisely, W = 78.83 – (12W²/345)*, if we assume the given measurements are accurate].
*You might like to verify this formula (please correct me if I've got it wrong).
Note 2: Here is an alternative wording for the task. Would this be better?
TUESDAY: In this task we assume that the pile of sand keeps its exact shape as it gets bigger. Students might initially think that if the time for which the sand flows is doubled (from 1 minute to 2 minutes), then the height of the pile will be doubled too. (It turns out that it takes 8 minutes rather than just 2, for the height to reach 2 metres - something we look at in Wednesday's task.)
However, it shouldn't take much discussion for students to realise that as the pile gets higher it gets wider too, so ever more sand is needed to raise the height by a given amount. This means that the graph will consist of a curve with an ever flatter positive slope. In turn this means that at the 2 minute mark, the height of the graph will be substantially less than 2 (m). However, at this juncture we wouldn't expect students to have a more precise sense of what that height will be. This is something we consider in a more analytic way in Wednesday's task.
WEDNESDAY: We consider what the height of the pile of sand will be after 2 minutes, and use this information to draw a more accurate graph, which we then use to interpolate the height at 2 minutes.
If we double the height of the pile of sand, from 1 m to 2 m, we also double the width (in two dimensions), so we have 2×2×2 = 8 times as much sand as before, so it will take 8 times as long to produce.
A good approximation for the value of the height at 2 minutes is 5/4. This is close to the cube-root of 2, since (5/4)-cubed is 125/64 = 1.953125.
THURSDAY: We step onto a travelator and think about some (nice, simple, straight line) distance-time graphs.
FRIDAY: We relate our distance-time graphs to speed-time graphs. This may not be easy - they look so very different!
Before thinking about the speed-time graph for Mario, it is interesting to consider how students might arrive at the red line - ie the distance-time graph for Mario while he is walking on the travelator. One way would be to find a specific point [such as (10, 14)] and join it to (2, 2) with a straight line. Another way would be to consider the distance travelled each second on the travelator - namely 0.5 m due to the travelator, plus 1 m due to the fact that Mario is walking at 1 m per second relative to the surface of the travelator - so the red line has a slope of 1.5 (m per sec).
Marco's and Mario's speeds are constant for most of the time (except for the brief moments when their speeds change). This means their speed-time graphs will consist primarily of horizontal straight lines. As such, the graphs look very different from the distance-time graphs, where the increase in distance (from Lisa) over time is shown by the upward slope of the graphs. In contrast to this, on the speed-time graphs, the increasing distance is shown by the increase in area under the graphs as the time increases. This looks far less dramatic.

Monday, 11 June 2018

ALG 18

This week we look at the rules of arithmetic, as expressed through the use of brackets. But we start informally, with a 'real life' task, specifically a sour dough problem, which may not be to everyone's liking....
MONDAY: A lighthearted story about pizza slices.
TUESDAY: Here we complete some numerical expressions involving brackets to describe a set of dots.
No pizzas.... No pseudo contexts...? Just dots. Simple.
Note: I have curtailed the string of expressions a bit. You might, for example, want to insert the expression 3×5×(20+?), and perhaps also 3×(5×20 + 5×?), between my second and third expressions.
WEDNESDAY: Here we look at expressions that give the total number of dots in a (recursive) dice pattern.
Note 1: we have used brackets more frequently than is strictly necessary here, so that the order of operations is explicit and unambiguous, ie so that the expressions can be read without knowing the conventional order of operations. Of course, it is important to learn these conventions but it means that the current expressions should be accessible to a high proportion of students. Also, as the expressions refer to familiar, concrete elements that are easy to group and count, the work should help demystify the use of expressions involving brackets - students don't need to refer to rules like 'do the brackets first' to make sense of the notation.
Note 2: you might want to ask students to annotate the diagram, or to draw a revised diagram, to show how the dots are being structured by the various expressions.
THURSDAY: Here we use expressions to count dots in some partitioned/compound arrays.
Again, it might be helpful to annotate the sets of dots to show how they relate to the expressions.
FRIDAY: We look at some neat and some not so neat ways of performing the calculation 15 × 22.

This slide is somewhat overloaded. It's really two tasks in one. And for the first task, it would be better to introduce the parts one at a time. Then, having worked through these exemplars, it should prove fruitful to devote time to students' methods of calculating 15 × 22, and to representing these as strings of expressions involving brackets.
The second task should help students take a more conscious, formal view of the distributive law and to think about when it applies.

Sunday, 3 June 2018

ALG 17

She was just seventeen, You know what I mean, And the way she looked Was way beyond compare. How could I dance with another? Oooh! When I saw her standing there.
One of the great early Beatles tracks. With ALG 17, we're standing on the number line and seeing how our position is related to multiples, especially multiples of 2 and 3.
MONDAY: Where do multiples of 3 occur on the number line? How can we tell whether a multiple of 3 is also a multiple of 6? And of 9??
TUESDAY: This is a classic task that I was introduced to by Lulu Healy. You might want to generate some data and look for patterns, but it is also perfectly accessible via a generic (analytic) approach by thinking about the occurance of multiples in strings of consecutive numbers.
 WEDNESDAY: more multiple fun....
This task keeps catching me out! One moment I think it's OK, the next I think I've got it wrong and it doesn't work!
Part b) can be solved by simply going through the multiples of 7 and 8 in the standard times tables, which leads us to 63, 64. However, a key feature here is the relation of these numbers to 56 (= 7×8) .... Or one can make use of 'the difference of two squares': 8×8 is a multiple of 8, 8×8 – 1  = (8–1)(8+1) is a multiple of 7. [This generalises very nicely, eg to finding consecutive numbers that are multiples of 19 and 20 respectively.]
THURSDAY: A classic 'think of a number ....' task. The task has links with Tuesday's version of ALG 17 and one can get quite a long way by using an empirical, data-generating approach. However, its attraction lies in the fact that one can use some fairly routine algebra to throw light on the underlying structure. [This is all the more noteworthy, given that we often use algebra to ease our path to an answer while ignoring structure!]
Note: I came across this particular task in a 1995 paper by Alan Bell (Purpose in school algebra, JMB, 14, 41-73). Sadly, Alan died this year. (9 April 1929 - 5 April 2018)
FRIDAY: We take an explicit look at the structure of Thursday's task.
Here the geometric and/or symbolic representation may help us see that Thursday's task results in the product of the 'outer two' of three consecutive numbers. When the middle number is odd, the outer numbers will be even so both must be a multiple of 2 and one must also be a multiple of 4 (of course, they can be multiples of other numbers too). So their product is a multiple of 8.