Download MEYL624 TUTOR NOTES Module 2

MEYL624
Module 2
The Open University,
Centre for Mathematics
Education
and the
Esmée Fairbairn Foundation
Mathematics/Education
MEYL624 Interfaculty Course, Level 3
Developing Mathematical
Thinking
Module 2: Targeting thinking
TUTOR NOTES
MEYL624 TUTOR NOTES Module 2
Tutoring tips—Task 2a
(Time allocation = 45 minutes)
Note: This session is likely to be interrupted for a 15-minute briefing from the host
institution.
The aims of this introductory session are:
 to start the course immediately with mathematics.
 to stimulate creative thinking within the context of mathematics by providing for
different ways of thinking
 to continue this creative thinking and communication
i.
Whether the folding is done lengthwise or widthwise, the area is same (1/3 of
page), but the length and height have changed.
Volume changes, the larger being about 1½ times the other.
iii.
Tutors are expected to feel more at ease than the students and should
therefore be prepared to ‘expose’ more of themselves.
Tutoring tips—Task 2b
(Time allocation = 45 minutes)
Facts and skills are usually concerned with factual content and standard procedures in
arithmetic, algebra, geometry, measurement and statistics.
Mathematical thinking is concerned with communicating, problem solving, logical reasoning,
information processing, being creative, making connections, and selecting tools.
There is no right answer to this task. The division of cards into the two groups will be
subjective as it will depend on what you expect pupils to do automatically, and what you
believe they will really have to think about. The debate that follows is expected to raise more
issues than solutions.
Tutoring tips—Task 2c
(Time allocation = 40 minutes)
i.
72 km/h. Assume the towns are, say, 90 km apart. The outward trip takes 1
hour, the return trip takes 1.5 hours, so the total journey takes 2.5 hours.
Average speed = total distance ÷ total time which is 180 km ÷ 2½ which
equals 72 km/h).
ii.
(At the time of writing a first class stamp costs 28p.) No more than 17 stamps
(£5 ÷ 28p = 17.85…, but you can’t buy 0.85 of a stamp, so round down to
17).
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iii
5 buses (217 ÷ 47 = 4.6…, but you can’t hire fractions of a bus, so it is
rounded up to 5).
iv.
72 (trial and error, only 8 pairs to try).
v.
Somewhere between 0 and 6, and unlikely to be 0 or 6 or even 1 or 5. Using a
tree diagram should reveal 64 options—note: there are many more options
that lead to 3 heads and three tails than, say, 6 heads.
vi.
There are several standard geometric proofs. A useful demonstration of the
result (but not a proof!) is to cut out a triangle that has been drawn on paper,
tear off its corners and then fit the corners to form a straight line.
vii.
7:3:2 is the same as 3.5:1.5:1, so 3.5 litres of loam and 1 litre of sand.
viii.
Too many variables to be considered here.
ix.
With numbers, 6 + 8  2 (although on a clock face, 8 hours later than 6
o’clock is 2 o’clock).
x.
Always, because 2(x + 3) multiplied out equals 2x + 6 for all values of x.
xi.
Many of these problems included stages 1, 2, 3, or 4 from the “problem
solving cycle” shown below:
Real world problem  1  Mathematical problem


4
2


Real world solution  3  Mathematical solution
where 1 = modelling, 2 = solving, 3 = interpreting, and 4 = evaluating.
1, 3, & 4 involve ‘mathematical thinking, and
2 is usually use of a ‘technical’ procedure.
(Some people put another stage between the real world and the mathematical world, this is
the ‘mental image world’, and similarly going back through solutions this intermediate step
can be added.)
If the examples were statistical, a modification of this model might be
Real world problem  1  Statistical problem


4
2


Real world solution  3  Statistical solution
where 1 = posing the question, 2 = collecting and analyzing the data, 3 = interpreting,
and 4 = evaluating.
xii.
theoretical problem
modelling/applied problem
design problem
iv, vi, ix (numbers), x
i, ii, iii, v, vii, ix (time)
viii
3
xiii.
no answer
only one answer
small number of answers
numerous answers
always true
ix (if numbers)
i, iii, iv, vii, ix (if time)
ii, v
viii
vi, x
Tutoring tips—Task 2d
(Time allocation = 40 minutes)
Some possible ways are: using equation(s), using logic, acting, drawing a picture, and
guessing and checking.
Equation
If c is number of chickens then 26 – c is number of sheep.
2c + 4(26 – c) = 74 (number of legs).
Solve for c, and find 26 – c.
Equations
Let c = number of chickens, and s = number of sheep.
Then, c + s = 26, 2c + 4s = 74.
Solve for c and s, algebraically or graphically.
Logic
Each of the 26 heads has at least 2 legs. That uses 52 legs, which means that
74 – 52 = 22 are left over. So, 11 have 4 legs; i.e. there are 11 sheep. The
remaining 15 (26 – 11) are therefore chickens.
Acting
26 people stand up on 2 feet, that uses 52 legs.
Now one at a time as I count in twos, one person is to move to the sheep ‘pen’
and raise their arms. 54, 56, 58, … , 74. Stop! Now count the sheep and the
chickens.
Picture
Draw 26 heads O O O O O O O O O O O O O
O O O O O O O O O O O O O
Now add 2 legs to each, then add more in twos,
counting as you go from 52 until you reach 74.
Guess
Guess: ½ each, 13 chickens (26 legs) & 13 sheep (52 legs).
Total legs = 78, too many, try one less sheep and one more chicken, 14
chicken, 12 sheep, (2x14 + 4x12)=76 legs, still too many, try 15 chicken, 11
sheep, …
General strategies for problem solving include:
guess and check;
make a list;
organise systematically;
draw a picture, table, or graph;
use a number sentence;
act out a problem;
make a model;
find a pattern, relationship, or rule;
try some simpler cases;
eliminate possibilities;
work backwards;
try extreme cases;
restate the problem;
check for hidden assumptions;
explain why it works;
change the point of view;
check regularly;
recognise when procedures are
appropriate (and justify them).
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Tutoring tips—Task 2e
(Time allocation = 40 minutes)
In discussion, it should emerge that many solutions are possible! The groups can share
examples. Add the various solutions to the wall poster (see Module 2 page 19) which should
remain for some time so that further solutions can be added.
It is possible to categorise solutions to ensure they are really different. Is 2x32, 3x22, 6x12
(Fig 1) the same as 2x32, 6x12, 3x22 (Fig 2) and 3x42, 2x62, 6x22 (Fig 3)?
Fig 1
Fig 2
Fig 3
Has anyone used only two sizes of squares, three sizes, any with four, five, … ?
Is it true for numbers other than 11? (Answer; yes, it is true for all even numbers except 2,
and for all odd numbers except 3, and 5.)
Can you prove this?
If a ‘proof’ does not emerge, then suggest the following:
- The solution is trivial for 1 and possibly for 4, 9, 16, …
- For even numbers greater than 2, use a square surrounded by an L-shape of smaller squares.
The L must contain 3, 5, 7, 9, … squares, and the total number of squares will be 4, 6, 8, …
(fig 4).
- For odd numbers, start with a 2x2 square, put an L shape with 3 squares around it,
then another with 5, 7, 9, … squares, and totals are 7, 9, 11, …(fig 5).
Fig 4
Fig 5
The important notions are trying ideas (specialising), detecting patterns (generalising),
looking for rules (conjecturing), and convincing yourself and others (proving).
Note: A proof has not been given here that it is impossible for 2, 3, and 5.
Tutoring tips—Task 2f ???
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Tutoring tips—Task 2g
(Time allocation = 30 minutes)
Some questions to ask
i.
The left-hand figure is composed of two squares and four congruent rightangled triangles. The right-hand figure is composed of one (oblique) square
and four identical right-angled triangles. Removing the right-angled triangles,
it follows that the sum of the two squares on the left hand side is equal to the
right hand (oblique) square.
ii.
What is it that is true (or untrue)?
How can I be sure it is not true from the diagram?
(Is the diagonal in the second figure a straight line?
Is the gradient of the long side of trapezium A the same as the gradient of the
hypotenuse in triangle C?)
Tutoring tips—Task 2h
(Time allocation, 30 minutes)
i.
A prime number has no factors, or, ‘has no factors apart from 1 and itself’.
Put another way, a prime number is a counting number that has exactly two
factors.
ii.
This is not always true, so it is false.
This is because when n = 1 then 2n = 2 and 2 is prime.
This is a proof by counterexample that the statement is false.
iii.
False; when n = 9, 6n + 1 is 55 which is neither prime nor square.
iv.
Starts looking good as one puts in numbers 1 to 10, and even for many other
numbers, but it is false for 11 and 12 (also 22, 23 then 33, 34 and so on).
Again, a proof (of falsity) by counter-example.
v.
False if you use the definition that a prime number is a counting number that
has exactly two factors. 1 only has the factor 1.
vi.
True. All properties of rectangles are true for squares. A square is usually
defined as a rectangle with adjacent sides equal in length. Note that an oblong
is an un-square rectangle.
vii.
False. One counter example will do. e.g.,
viii.
Examples are evidence, (e.g. 1/3 & 2/3, 0.2 & 0.8, and +4 & -3).
A proof could be: If a + b = 1, then a = 1 – b. Substituting this in the LHS
gives 1 – b + b2. Substituting it into the RHS gives
(1 – b)2 + b, which also simplifies to 1 – b + b2.
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Tutoring tips -Task 2i
(Time allocation = 45 minutes)
Children spend a lot of time trying to making sense of the calculator so this task is worth
spending considerable time on.
i.
This can be done in many ways for most numbers.
ii.
Again there are many ways to do this.
iii.
The ! on a calculator is called ‘factorial’
4! is 4 x 3 x 2 x 1 = 24
0! is defined as 1,
x! is not normally defined for negative or fractional numbers.
0! = 1, 1! = 1, 2! = 2, For x > 2, x! > x.
iv.
It gives the positive square roots for all non-negative numbers.
It gives an error message for negative ones because you cannot have a (real)
square root for a negative number.
When x > 1 the square root is less than x,
when x = 1 the square root equals 1,
for 1 > x > 0 the answer is larger than x, and for x = 0 the answer is 0.
For negative numbers there is no (real) square root.
The inverse of x is x2.
(If time permits, graph y = x and y = x2 on the same page.)
But the inverse of x2 is ±x, not x which is the positive value.
v.
1/x (or x-1) gives the reciprocal of x
(except 0 because 1/0 is not defined).
When x > 1 the reciprocal is less than 1, when 1 it equals 1, and for numbers
between 0 and 1 the answer is greater than 1.
For negative numbers between 0 and –1 the reciprocal is less than –1, for –1
the reciprocal equals –1, and for numbers less than –1 the reciprocals are
greater than –1.
1/x is its own inverse.
vi.
Raise to the power. 00 is undefined; all others work as per the laws of
indices—for example, that xa x xb = xa+b.
vii.
The calculator provides a context, and may encourage the student to think
about the mathematics.
Calculators can promote mathematical thinking by taking away the drudgery
and allowing pupils to concentrate on the mathematical ideas, (cf. problem
solving cycle in 2d).
Extension
Explore your calculator’s constant facility. Does it work differently for other models of
calculator?
Tutoring tips—Task 2j ????
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Tutoring tips—Task 2k ????
Tutoring tips—Task 2l
(Time allocation = 20 minutes)
i.
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iii.
Yes, a pattern does emerge, but it is not clear until you get many numbers.
The pattern goes up in ones, twos, threes, and it seems that they are grouped
in threes. One way to visualise this is to have a transparent copy of the
triangle, turn it over, then slide it over the original, and find the maximum
overlap, hence the minimum number of moves.
Record the pattern on a table using the first two headings from the table below. (Possible
extension: Suggest the 3rd heading and see what else is needed for the 1st, 4th, 7th, and 10th
terms.)
Ht. of  (n)
1
2
3
4
5
6
7
8
9
10
Minimum no. of moves (m)
0
1
2
3
5
7
9
12
15
18
(n2 + n)/6
[(1 + 1)/6 = 1/3]
[(4 + 2)/6 = 1]
[(9 + 3)/6 = 2]
[(16 + 4)/6 = 3 1/3]
[(25 + 5)/6 = 5]
[(36 + 6)/6 = 7]
[(49 + 7)/6 = 9 1/3]
[(64 + 8)/6 = 12]
[(81 +9)/6 = 15]
[(100 +10 )/6 = 18 1/3]
For the 1st, 4th, 7th, and 10th terms you need to subtract 1/3.
Note: This is not a proof of the result, but it fits with the first ten terms. You might conjecture
that it will be a general result.
Tutoring tips—Task 2m
(Time allocation = 15 minutes)
i.
10 ways (unpainted, 1 face painted, 2 touching faces painted, 2 opposite faces
painted, 3 faces painted including a pair of opposites, 3 faces of a corner
painted, 4 faces painted with a pair of opposites unpainted, 4 faces painted
with 2 touching faces unpainted, 5 faces painted, all faces painted.)
ii.
For a 333 cube: (0,1), (1, 6), (2, 12), (3, 8),
for a nnn cube: (0, (n – 2)3), (1, 6(n – 2)2), (2, 12(n – 2)), (3, 8)
iii.
Only one way.
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iv.
Possible sections are squares, rectangles, trapeziums, triangles, pentagons,
and hexagons.
Tutoring tips—Task 2n
(Time allocation = 10 minutes)
It is important to realize that different people ‘see’ things differently, and that these different
perspectives are often quite reasonable. This diversity of ‘ways of seeing’ is not just
restricted to mathematical diagrams but applies to most areas of mathematics.
Tutoring tips—Reflections
The final 15 minutes of this session should be spent writing reflective notes on the day’s
tasks.
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