Inspiring and enriching lessons at school. Gerry Leversha MA Conference, Oxford April 2016

Transcription

1 Inspiring and enriching lessons at school Gerry Leversha MA Conference, Oxford April 2016

2 Inspiring lessons Until a few months ago, the buzzword was mastery The new national curriculum, having abolished levels, set mathematics teachers the aim of ensuring that their pupils achieved this quality Now it is called greater depth And we used to call it enrichment (and still do) Whatever it is called, what does it mean? How can we use it in our teaching? This is what I shall be talking about

3 Inspiring lessons My understanding is that we should base our teaching on the belief that children can succeed in doing mathematics the value given to conceptual understanding rather than algorithmic procedures the emphasis on problem-solving And we can begin to develop the idea of.

4 Inspiring lessons an emphasis on depth of understanding rather than rapid progress through topics on a superficial level embedding enrichment in the curriculum so that the process is not driven by preparation for examinations ideas are constantly interrogated, so that pupils are encouraged to question what they have learnt and to be sceptical about methods which are presented as mindless algorithms

5 Inspiring lessons For the rest of this talk I will take these general principles as given and discuss particular ideas for lessons which encapsulate mastery and greater depth I will choose a few central topics and suggest how to achieve these at various levels of the school curriculum, from primary to A level Remember that you can revisit a primary school topic in an A level class

6 Inspiring lessons And I think it is possible to identify some common themes in this endeavour Having mastered a process, investigate the converse process Look for unexpected implications Make connections across the subject Emphasis the why over the how

7 Inspiring lessons And also a maxim It is better to solve one problem in five different ways than to solve five problems using the same method So now let us examine some key topics in school mathematics

8 Place value Let us start with primary school In my view the essential concept is place value Repeatedly we ought to emphasise what base 10 representation means This applies also when decimal fractions are introduced Suppose, however, that the task for the next few lessons is addition within 1000

9 Place value So let s suppose that this process has been explained in terms of place-value and practised How do we enrich this topic?

10 Place value Inverse operations find the missing digits

11 Place value *** *** *** Is it possible for all of the digits 1 to 9 to appear exactly once in this addition?

12 Place value *** *** *** Using each digit from 1 to 9 once, what is the largest sum you can obtain?

13 Place value *** *** *** What if the digit 0 is allowed? Can we obtain a larger sum?

14 More on place-value SEND MORE Each letter stands for a different digit. Can you reconstruct the addition? MONEY Is the solution unique? Dudeney s cryptarithm

15 More on place-value * * * * * * * * * * * * * * * * * * * * * 8 * * * * * * * * Another puzzle by Dudeney

16 Ratio and Proportion In my view this is the single most important topic in middle school mathematics I ll begin with something which does not at first sight look like this topic at all What are the formulae for the circumference and area of a circle of radius r?

17 Circumference and Area formulae Here is a parody of a lesson on this topic Teacher writes on the board A = πr 2 C = 2πr Teacher does four examples using these formulae Teacher hands out worksheet with 30 questions for the pupils to practise the formulae

18 Circumference and Area formulae This is not teaching to mastery These formulae A = πr 2 C = 2πr are not the first thing which should appear on the board They are the last thing to write on the board

19 Circumference and Area formulae In my view, the lesson should begin by talking about the perimeters of some regular figures Square of side 1 perimeter is 4 Square of side 2 perimeter is 8 Square of side 100 perimeter is 400

20 Circumference and Area formulae In my view, the lesson should begin by talking about the perimeters of some regular figures So the side and perimeter are in direct proportion the ratio is 1:4

21 Circumference and Area formulae Now talk about a pentagon The side and perimeter are proportion but the ratio is 1:5 still in direct In fact, this happens for any regular polygon, though the ratio is different.

22 Circumference and Area formulae A circle is a polygon with a very large number of sides The side length and perimeter will still be in direct proportion Unfortunately, a circle does not have a side length.

23 Circumference and Area formulae So let s reconsider polygons but measure them differently If we replace the side by the greatest width, then their perimeters are proportional to this quantity.

24 Circumference and Area formulae Hence the circumference of a circle is proportional to its greatest width its diameter. Now discuss the ratio : somewhere between 3 and 4, and explain that this number is π

25 Circumference and Area formulae Now we have the formula C = πd = 2πr and we have to start thinking about the area

26 Circumference and Area formulae Take the circle and chop it up like a pizza into very thin slices (sectors)

27 Circumference and Area formulae Now reassemble the slices to obtain a sort of wobbly rectangle The height of the rectangle is approximately the radius of the circle And the width is approximately half the circumference of the circle

28 Circumference and Area formulae Now reassemble the slices to obtain a sort of wobbly rectangle So its area is r πr = πr 2 You can discuss why the rectangle looks less wobbly when the number of slices increases

29 Ratio and Proportion Now we go onto more conventional problems on ratio and proportion My approach when I was still a teacher would have been largely algebraic, or would have involved scale factors However, I have recently been very impressed by the Singapore method of bar modelling This seems to combine rigorous understanding of ratio with a pleasing simplicity of technique

30 Ratio and Proportion Let us consider a problem from year 5 in primary school There are 88 children at a school sports day. 1/3 of the boys pair up with 2/5 of the girls in a three-legged race. The rest of the children watch. How many girls watch the race?

31 Ratio and Proportion 88 children 1/3 boys 2/5 girls In this pictorial approach, ratios are represented by proportions of a bar So this bar represents boys, and the red squares are the third who take part in the race Boys This bar represents a proportion of 2/5 for girls Girls

32 Ratio and Proportion 88 children 1/3 boys 2/5 girls But the shaded parts represent equal numbers So we double the number of squares in the boys bar so we can compare them Boys Girls

33 Ratio and Proportion 88 children 1/3 boys 2/5 girls But the shaded parts represent equal numbers So we double the number of squares in the boys bar so we can compare them Boys Girls

34 Ratio and Proportion 88 children 1/3 boys 2/5 girls Now there are 11 squares representing 88 children So each square represents 8 children Boys Girls

35 Ratio and Proportion 88 children 1/3 boys 2/5 girls Hence 16 boys and 16 girls participate in the race Boys Girls 32 boys watch the race 24 girls watch the race

36 Ratio and Proportion 88 children 1/3 boys 2/5 girls I doubt that this problem would be accessible to year 5 pupils without bar modelling. Boys Girls

37 Ratio and Proportion Can bar modelling help to solve this old chestnut? Tom and Dick take 2 hours to dig a hole Dick and Harry take 3 hours to dig a hole Harry and Tom take 4 hours to dig a hole How long does it take all three of them to dig a hole?

38 Ratio and Proportion Do the obvious thing and write down some algebra T + D = 2 D + H = 3 H + T = 4 T + D + H = Why is this nonsense?

39 Ratio and Proportion Do the obvious thing and write down some algebra T + D = 2 D + H = 3 H + T = 4 So why did it happen? T + D + H = 4 1 2

40 Ratio and Proportion The variables represent times T + D = 2 D + H = 3 H + T = 4 T + D + H = Does it make sense to add times when jobs are being done simultaneously?

41 Ratio and Proportion The variables represent times T + D = 2 D + H = 3 H + T = 4 So what can we add? T + D + H = 4 1 2

42 Ratio and Proportion With a bar modelling approach, at least we have to think about what the bars represent A good choice is holes, since you can clearly add these To keep things simple, let the bars represent the number of holes dug in 12 hours Notice we don t just translate words into symbols we think carefully about what the words mean

43 Ratio and Proportion In 12 hours Tom and Dick dig 6 holes Dick and Harry dig 4 holes Tom and Harry dig 3 holes Hence, in 12 hours, Tom, Dick, Harry and their identical twin brothers dig 13 holes

44 Ratio and Proportion Hence, in 12 hours, Tom, Dick, Harry and their identical twin brothers dig 13 holes Without their twins, in 12 hours, Tom, Dick and Harry dig 6½ holes Hence it takes the trio 1 11 hours to dig one hole 13 (not an answer you would guess) Notice that a problem which caused difficulty to maths undergraduates and teachers now seems quite simple

45 Irrational numbers and surds Now take a look at surds Prove that 2 is irrational Suppose 2 = a in lowest terms b Then, manipulating, 2b 2 = a 2 So a 2 is even So a is even; put a = 2c Then, manipulating, b 2 = 2c 2 So, after two more steps, b is even But then the fraction is not in its lowest terms

46 Irrational numbers and surds Now take a look at surds What next? Prove that What next? Prove that 3 is irrational 4 is irrational What goes wrong?

47 Irrational numbers and surds Now prove that 2 is irrational another way Usual start: a 2 = 2b 2 Look at prime factorisations The power of 2 on the left is even The power of 2 on the right is odd Here is another contradiction, as required Now try this for 3 And what happens for 4? Notice the connection in the syllabus

48 Irrational numbers and surds Now prove that 2 is irrational another way Usual start: 2 = a b Choose this fraction so that b is as small as possible Now check that 2 is also 2b a a b And that 2b a and a b are positive integers And that the denominator is smaller than b And again we have a contradiction Now try this for 3 and 4

49 Irrational numbers and surds Now prove that Usual start: Then 1 > 2 = a b and so we have 1 > 2 is irrational another way 2 1 = a b b > b Now take powers of 2 1 Each power is of the form p 2 + q and so is also greater than or equal to 1 b But they form a sequence whose limit is zero

50 Irrational numbers and surds So there are four proofs and you can find more Try a geometrical diagram: a right angled triangle with integer sides construct a smaller triangle which also has integer sides Try trigonometry Try infinite series Remember: you can keep on returning to this problem all the way through the school syllabus until Year 13

51 Limiting processes Now let us move onto A levels and integration When it comes to improper integrals, there is an interesting Byzantine case 6 What is the value of 1 6 x dx?

52 Limiting processes If you use the formula a 1 b x dx = ln a b then there is of course no problem. We have the value ln 6 = ln 1 = ln1 = 0 6 Is this correct?

53 Limiting processes Note that the graph has a discontinuity at x = 0

54 Limiting processes The ceiling function is also discontinuous at x = 0 but we can integrate it (as an area) The integral between -5 and 5 is 5

55 Limiting processes However, this function tends to as x 0 from below and to as x 0 from above

56 Limiting processes However, as the function is clearly odd, it might 6 still seem sensible to claim that 1 dx = 0 6 x

57 Limiting processes What is wrong with that? Is the injunction you cannot integrate over an asymptote just a killjoy? So let us be a little more sophisticated; let us use limits to evaluate the improper integrals.

58 Limiting processes Let ε > 0 (as ever) ε Now we calculate 1 6 x dx dx ε x then let ε 0 and We abandon the lazy use of the modulus function and use the more precise indefinite integral 1 x dx = ln x + c x > 0 ln x + c x < 0

59 Limiting processes Now 6 1 x dx x 6 ln ln ln 6 and 6 1 x dx 6 ln x ln 6 ln so, adding, we obtain zero without even needing to allow ε 0 6 and again it seems reasonable that 1 6 x dx=0

60 Limiting processes The more carefully we analyse this integral, the more it appears that its value is zero So what is the reason that we are always told that we cannot integrate over an asymptotic discontinuity? However, there is more than one way for ε to tend to zero and perhaps we ought to allow the doors to close at different speeds?

61 Limiting processes Now x dx 2 ln x ln 2 ln 6 6 and x dx 6 ln x ln 6 ln 3 3 and now the sum is ln2ε ln3ε = ln 2 3 (and again we don t even need to allow ε 0) 6 since we obtain 1 6 x dx=ln2 3

62 Limiting processes As you can see, it is possible for the integral to be any value at all, if we select the speeds of convergence appropriately so this integral does not exist and we now have a much more sophisticated idea of what makes integrals (and especially improper integrals) tick without trespassing on university analysis

63 Inspiring lessons Common themes and a maxim Having mastered a process, investigate the converse process Look for unexpected implications Make connections across the subject Emphasis the why over the how It is better to solve one problem in five different ways than to solve five problems using the same method Now do it!