Chapter xx
Negative Numbers
Certain mathematical topics taught in schools are a persistent source of frustration for most students,
because they appear to be totally divorced from anything meaningful in their world. This paper identifies
negative numbers as one major problematic area, and outlines the difficulties that historically even the best
mathematicians had in understanding them. Then the use of Piaget’s concrete and formal operational
thinking in teaching mathematics is outlined and applied to a new model for learning negative numbers that
will avoid the problems history shows all primary and secondary school students inevitably have in
understanding and using them.
Negative Numbers: Some History
The use of positive numbers in algebra has long been well accepted. For example the Babylonians had
algorithms to solve quadratic equations like, in modern notation, x(x + 3) = 28 that would reliably give the
positive number solution; in this case x = 4. However, they had no idea or, at least, rejected any talk about
the negative number solution x= 7 that we would now expect secondary school students to accept. With
the occasional exception - an early user of negative numbers were the Indians in the seventh century €
negative solutions historically have been regarded as nonsensical. This pattern of puzzlement about the
nature of negative numbers and their rejection, certainly in the European tradition, continued until very
€
recent times. For example, the famous mathematician Blaise Pascal (1623 - 1662) regarded the subtraction
of 4 from 0 giving a negative number as nonsense:
−
I know people who cannot understand that when you subtract four from zero what
is left zero. (Quoted in Hefendehl-Hebeker (1991))
Only recently has the idea that literal equations can allow coefficients to be real numbers i.e. they allow
the coefficients to be negative has emerged. This meant that there were multiple forms where there is now
one form. For example consider these two equations that historically were regarded as different:
x(x − b) = c
x(x + b) = c
€
€
Here b and c had to be
positive; this effectively
avoided the need for
negative numbers. Now
€
literal coefficients allow for
negative numbers i.e.
negative b and c are allowed.
So the pair of equations can
now be written as one€
equation:
b ± b2 − 4ac
x=
2a
x(x − b) = c
The use of negative coefficients is normally glossed over in secondary
school. For example consider this equation that is be solved using the
standard quadratic equation formula:
2x 2 − 3x − 17 = 0
In fact students need to be understand that, to use the formula, this
equation has to be seen as:
2x 2 +− 3x+− 17 = 0
Then the coefficients 2, -3 and -17 can be inserted in the formula:
−
€
The late invention of literal equations allowing negative coefficients should alert teachers to the deep
problems students will have in understanding algebra where negative numbers are involved.
€
Hefendehl-Hebeker (1991) provides voluminous examples from history of problems with understanding
negative numbers. She quotes Stendahl (1783-1843) on the confusion felt by most learned people at any
time in history about why negative times negative is positive; this is the bellwether problem in the history
of algebra that makes otherwise intelligent adults believe mathematics is arbitrary and foolish:
I thought that mathematics ruled out all hypocrisy, and, in my youthful
ingenuousness, I believed that the same must be true of all sciences which, I was
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told, used it. Imagine how I felt when I realized that no one could explain to me
why minus times minus yields plus … That this difficulty was not explained to me
was bad enough… What was worse was that it had been explained to me by means
of reasons that were obviously unclear to those who employed them.
The problem that most teachers do not really understanding why negative times negative is positive seems
to have changed little over the centuries. This is despite many maths teachers claiming they teach it well.
This issue is returned to later when the model for teaching negative numbers is detailed.
Concrete and Formal Operational Thinking
Piaget’s concept of concrete operational thinking is extremely helpful for primary school teachers
because most primary school students think about the world in this way. What characterises their thinking
is logical reasoning that is confirmed by the student’s experience of the world. For example, suppose
students told that all ducks are birds, which they know is true from their understanding of what birds are,
and are not, and further told that all the birds on an island are hungry because there is not enough food
they can logically conclude that all the ducks on the island are hungry. But it would be a mistake to think
students can reason this independently of context. However formal operational thinking is fundamentally
more abstract because it uses logic that does not rely on using the meaning of statements in any argument.
For example, suppose students are told that all men wear dresses, and also Julia is a man, and are asked “is
Julia a man?” The formal operational thinkers will set aside the practical problems that all men don’t wear
dresses - at least in most countries - and Julia is recognisably a woman’s name, then reason that Julia wears
a dress by the logic of the problem and not the content. However, the concrete operational thinker will be
unable to reason their way through the problem because they are confused by the two premises that do
not correspond to their knowledge of the world.
Any wise primary teacher will exploit students use of concrete operational thinking in learning to
understand mathematics relationally (Skemp, 19xx). For example, suppose young children are learning
4 + 3 = 7. They need repeated exposure to a wide variety of materials that show, in all cases, four objects
plus three more objects of the same kind will always give seven objects of the same kind. This may appear
to be straight-forward to adults but it is not for young children. Consider, for example, children being
presented with four large multi-coloured counters to which three small purple counters are added. The
problem from the child’s point of view is to see past the differences in the materials - shape, size, colour
etc. to generalisation that four plus three always equals seven.
The intent of most mathematics learning in the primary school should be to make the generalisations
that allow students to solve problems without reference to materials that they may have used in the
learning process. For example a year six student might well understand that working out 1432 - 999 can be
achieved by adding 1 to both numbers. This originally might have been established by comparing a pile of,
say, $1432 in play money, and another pile of $999. Adding $1 to both piles does not alter the difference.
The teacher needs to observe when the students can ignore any material and reason the answer using this
generalisation:
a − b = (a + c) − (b + c) ∀a,b,c
€
(This of course does not mean students use this explicit algebraic formulation with letters but their correct
use of this principle in solving problems does mean that understand generalisations, which are necessarily
algebraic.)
So in preparing a model for learning negative numbers we should try to teach initially to the extend that
concrete operational thinking will work; some things will and some things won’t. To determine which
things will work we will adopt a sensible principal from Kline (1971):
Mathematics is best taught by helping students to solve problems drawn from
their own experience.
For the things that don’t work using this principle of linking problems to experience inevitably formal
operational thinking will be required. This in turn will mean that the students (and adults) who do not
reach this level of thinking will never understand the most important mathematics.
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Before proceeding to the model for teaching negative numbers it is useful to consider the very real
pedagogical difficulties involved in reasoning outside the realm of experience that relies on logical reasoning
using definitions that are inevitably arbitrary. Deciding upon which definitions to use will be determined by
their usefulness not whether they seem correct from experience. A famous Lewis Carroll character
understood this well:
“When I use a word,” Humpty Dumpty said, in a rather scornful tone, “it means just
what I choose it to mean, neither more nor less.” “The question is,” said Alice,
“whether you can make words mean so many different things.” “The question is,”
said Humpty Dumpty, “which is to be master - that's all.” (Carroll, 1872).
Pickering (1995) gives a brilliant example of the usefulness of students ranging over many arbitrary
definitions and determining the consequences, and hopefully finding a useful one; doing this requires formal
operational thinking. He shows how William Rowan Hamilton tried over many years to extend complex
numbers, which Hamilton understood as two-dimensional vectors, and multiplication by the “imaginary i”
representing rotating these vectors by π / 2 clockwise. He wondered whether he could three dimensional
complex numbers were possible where these “hypercomplex” numbers would also have a geometrical
rotational interpretation. He introduced j where j 2 =− 1. So his three dimensional objects looked like
3 + 4i − 6 j
€
€
Immediately Hamilton ran into a major problem. Formally multiplying two hypercomplex numbers together
€ replaced both i2 and j 2 by −1 whenever they occur - left the
- he used the laws for real numbers and
mysterious objects ij and ji for which he had no meaning. It took many years of struggle for Hamilton to
find a viable definition. Eventually he hit on a really wild solution; he arbitrarily creating a fourth dimension
and added these definitions:
€
€
−
−
k 2 =− 1, ij€= k , ji=
€ k , jk = i , and ki= j .
€
These objects are quaternions, and they look like 2 - 6i + 3j - 2k.
€
There are a number of lessons in this tale. Firstly, quaternions broke the long held view that multiplication
- many more non-commutative algebraic systems would soon follow. Secondly,
€ be commutative
€
€ must
€
teachers need to understand that, if a system can only be understood at a formal operational level, letting
students engage in seeking sensible but arbitrary definitions is a vital part of their mathematics education.
To further illustrate this at a school level we return the centuries old bugbear: why is negative times
negative positive?
Initially students need to understand why a positive times a negative is negative. For example why is
3× 5=− 15? A general pedagogical principle is that students should always use concrete operational thinking
if this is possible, and this is even true for those students who are sophisticated formal operational thinkers.
In this case it is possible; by returning to the meaning of multiplication that the students already possess, it
is simple to see a suitable problem would be:
−
€
Three students each owe the bank five dollars. Altogether they owe fifteen dollars.
It works. So students easily make this generalisation:
a×− b=− ab , where a and b are regarded as positive.
Now is there a sensible story that shows why a negative times a positive equals a negative? Trying to
relate it to a real situation fails this time:
€
Negative three students each have five dollars in the bank. Altogether they owe the bank fifteen dollars.
This is absurd. Since no concrete operational explanation can work students must seek a definition in the
arbitrary world of formal operations. And it is never obvious how to find a suitable definition as Hamilton
eventually found with quaternions.
The solution comes from the need for the commutative law to apply to negative numbers as well as
positive numbers. It will be very inconvenient, though logically possible, in algebra if the commutative law
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does not also apply to a negative number times a positive number. So we define − 5 × 3 to equal 3×− 5 . So
we use this general definition, which preserves the commutative law:
−
€
a × b=− ab , a and b are regarded as positive.
€ is both arbitrary and sensible.
It is important that students understand this is a free choice that
Unfortunately many teachers we have worked with over the years think the above reasoning is a proof that
negative times positive is negative; they simply do not understand the notion of free choice.
At last we come to why negative times negative is positive. In a forlorn hope that there is a concrete
operational meaning of multiplication we attempt to find a story that shows why − 3×− 5 =15 :
Negative three students each owe five dollars in the bank. Altogether they have fifteen dollars in the bank.
This is even crazier than the negative times positive example. Again the formal operational solution is not
obvious. In fact the need for the distributive law to work for all real numbers
is the powerful free choice.
€
Assuming the distributive law applies to real numbers we can play with the symbols. This will lead to a
definition of negative times negative:
−
a(b+− b)=− a × b+− a×− b
The left hand side equals 0, and the right hand side becomes − ab+− a×− b by the definition of negative times
positive. So:
€
0=− ab+− a×− b
Adding ab to both sides leads to ab=− a×− b
€
€
There we have it. Making the free choice that negative times negative is positive leads to the distributive
law being true for all real numbers and not just the positive real numbers. This is what we need for algebra.
In our model for teaching
€ negative numbers we ask the question why anyone would want to know this. The
answer will require finding real problems that require algebra to solve them. Such problems are typically
only solvable fully three years after students are introduced to skills with negative numbers without
context. This creates a massive problem for teachers who face teaching this to many highly disengaged
students.
Integers and Notation
Before introducing the model for teaching negative numbers it is desirable to acknowledge talking about
more than integers despite the fact that many textbooks implicitly or explicitly use them rather than real
numbers. For example a textbook might show this in integers:
−
3+− 4=− 7
But equally we should expect students to generalise the ideas with integers to rational and irrational
numbers:
€
−
5 −
6
2 11
+ 8 111 =− 10 11
−
4 2+− 8 2=− 12 2
So this paper refers to negative numbers rather than just negative integers.
€
€
A non-convention notation has been be used earlier: unlike the usual algebraic convention a letter like b
will follow the historical norm of meaning a positive number; then -b will indicate a negative number. This
will aid in stating generalisations involving negative numbers. For example, in the problems above, the
computational principle can be expressed in this way:
−
a+− b=− (a + b) for all a,b
€
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Teaching Negative Numbers: the Beginning
We begin with a set of realistic problems that many primary school students can
solve. Most secondary school teachers of mathematics are unlikely to know that even
quite young primary school students can cope with some aspects of negative numbers,
although admittedly the range is not great. For example, the occasional eight year old can
do subtractions as indicated in Figure 1. They reason three minus seven is minus four,
then thirty minus four is twenty-six. We now spell out what contextually based negative
number problems older primary school students can effectively do. And we suggest
secondary school students, who are unfamiliar with negative numbers, should begin at the
same place.
53
- 27
- 4
30
26
Figure 1
The most understandable models for negative numbers are bank overdrafts and temperatures below
zero. The more desirable one is money based, and it is the only one that most adults will need. In the first
instance it is not difficult for primary school students to solve problems like this:
Jill has $6 in the bank. She withdraws $8. What does she owe the bank?
It is straightforward for students to see that Jill now overdrawn by $2. An important teaching point need to
be made:
Students should solve problems like 6 - 8 through stories not symbols.
Teachers are often surprised that students don’t really understand what the symbols in equations really
refer to. For example, when primary teachers to ask students to write a story for, say, 4 x 3 = 12 and many
students would say “I had 4 apples and I bought 3 more, Now I have 12 apples.” And secondary school
maths teachers should expert a nasty shock if they ask similar story-construction problems about negative
numbers.
Another important idea is this: students should move on from using small numbers that can be modelled on
material to larger numbers like 300 and 800 where the computation is easy. For example:
300 – 800 = -500
The implicit generalisation here is:
a − b=− (b − a), where a < b
€
This method of overdrafts being generated by withdrawing more money than a person has in the bank
has the large advantage that in no sense is this there “negative” money involved - someone, somewhere has
the money that created the overdraft. All the negative sign indicates is how much is owed by the borrower
to the bank. And there is no reason at this stage why student should not find overdrafts with spreadsheets
using the negative sign, red numbers or with brackets around the number. So an overdraft of $5 can
variously be shown as -$5, $5, and ($5).
The next step is to solve problems that operate on overdrafts by either depositing some money or
withdrawing some more. The various problem types are presented here with an example and the
associated generalisation:
I owe the bank $6 and I borrow another $3. I owe $9, that is to say -6 – 3 = -9.
Generalisation: − a − b=− (a + b)
I owe the bank $6 and I deposit $4. I still owe $2, that is to say -6 + 4 = -2.
Generalisation: − a + b=− (a − b), where a > b
€
I owe the bank $6 and I deposit $10. I now have $4 in the bank, that is to say -6 + 10 = 4.
Generalisation: a + b = b − a, where a < b
−
€ may seem somewhat complicated but, in fact, older primary school children can cope with this
All this
mixture of negative and positive number provided special care is taken in the teaching not to go too fast –
students need time to make the generalisations or rules for themselves.
€
In summary overdraft ideas will allow this class of real number problem to be solved:
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A negative number plus or minus a positive number i.e. ± a ± b .
And this is all almost anyone will ever need to know about negative numbers. For example an
accountant will only need to know this in preparing spreadsheets.
Another very similar model for elementary
€ negative numbers is using a vertical thermometer with
special emphasis on temperatures on both sides of zero degrees. Teachers may prefer to use problems like
this:
The temperature in Antarctica was 4º C and it fell overnight by 6º. What is the temperature now?
The generalisations that can be made are the same as the overdraft context. Teachers may choose to
reinforce the negative number arithmetic using this model of negative numbers but need to remember that
such problems are of little practical use when compared with money problems.
Teaching Negative Numbers: the Middle
The next step in teaching negative number operations is quite unsuitable for primary school students as
the contexts will be beyond them; the problems are real numbers plus or minus real numbers.
Unfortunately the methods generally taught do not link to students’ reality.
One such common model for adding or subtracting negative numbers is a “hills and dales” or “bumps
and holes” method. Figure 2 illustrates why 3+ 4= 1 in this model.
−
+
−
€
Figure 2
Students find it easy to see why the three “hills” fill three “dales” leaving one “dale” empty. So
3+ 4=− 1.
−
€
There are at least two other models for addition of negative numbers that are similar to the cancellation
of holes. Firstly a physics model in which positive and negative positive charges
cancel is easily understood. So, for example, for three positive charges plus four
negative charges the three positives cancel out three negatives leaving one
negative charge. Again 3+ 4= 1.
−
−
Secondly an open abacus can be used (Figure 3); one post represents negative
numbers with red rings, and the other post represents positive numbers with
black rings. The
€ dotted line shows negative three and positive three “cancelling”
leaving negative one above the dotted line. Again this shows why 3+ 4= 1.
−
−
-
+
Figure 3
There is no doubt these three highly visual representation can be very helpful
in learning negative number skills. So why not use these methods to complete
the learning of real number addition and subtraction?€Because, invoking the axiom previously stated, this
“cancellation” approach does not link the learning skills in manipulating negative numbers with any
conceivable real application. So we suggest don’t use any of them. So where are the realistic problems that
addition and subtraction of real numbers solve?
A remarkably simple answer lies in adding and subtracting two dimensional vectors. But first students
must learn that in all vectors the direction is arbitrarily assigned a positive and a negative direction in each
dimension. This needs some explanation. Consider a town at O (Figure 4) with one straight road through
it. A visitor wants to visit a friend at a place marked A and asks for directions in the town. An instruction
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to go 4 kilometres will send her off to the right. Then − 4 + 3 can readily be interpreted as moving 4
kilometres to the right followed by 3 kilometres to the left i.e. a one-dimensional vector; it is simple for
students to understand that the red arrows show − 4 + 3=− 1
−
€
€
€
4
3
2
1
A
-1
O
-2
-3
-4
-5
Figure 4
The use of the reverse of the conventional directions for the x –axis is only used here to show that the
positive and negative directions are arbitrarily assigned. This means that negative is not, in any sense, an
absence of quantity i.e. a dale in a hills and dales model of integers. This is particularly important when
physicists create equations of motion to problems: for example, in kinematics, when throwing a ball
vertically into the air they could define “up” as positive but then must stick to this; in this case the initial
speed of the ball must as be positive and acceleration due to gravity negative. Of course if “down” is
defined as positive then the initial speed of the ball is negative and acceleration due to gravity positive.
In physics vectors are used for displacements, velocities, forces, accelerations,
momentum, angular momentum, magnetic fields, electric fields, and so on. Their
importance is not in doubt but, as an initial application of negative numbers, most
except displacements, suffer from the fact that they act at a point (Figure 5). And for
vectors acting at a point it is not obvious in this diagram why vector addition works:
Figure 5
⎛a ⎞ ⎛ c ⎞ ⎛ a + c ⎞
⎜ ⎟ + ⎜ ⎟ = ⎜
⎟ where a,b,c,d are all real numbers
⎝b ⎠ ⎝d ⎠ ⎝b + d ⎠
But displacements, i.e. vector distances, help establish this vector
addition law. Figure 6 shows that a trip from A to B, followed by a
trip from B to C, is the same as a single journey directly from A to
C. This amount to adding the x-co-ordinates and adding the y-coordinates separately; this is contextual and easy for students to
understand.
€
B
A
C
Figure 6
Displacement problems methods will allow student to extend to
extend their computational generalisations from displacement
problems and then, in physics, more generally to other vector quantities. Then these generalisations for
each component of a vector are used:
a+− b=− (a + b)
•
−
•
a+− b=− (b − a),b > a
a+− b = a − b,a > b
€
•
€
•
a−− b = a + b
•
−
a−− b=− (a − b),a > b
•
−
a−− b=− (b − a),b > a
€
€
€
€
Teaching Negative Numbers: Endgame
All that remains in teaching negative numbers is their multiplication and division. Previously we argued
that understanding these operations was essentially a matter of thinking in a formal operational way i.e.
students cannot understand these ideas using materials or contextual examples but must understand that
that the choice of definitions is arbitrary but informed by choosing ones that allow the algebraic principles,
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such as commutativity and distributivity to apply to negative as well as positive numbers. Yet the notion of
applicability still applies. In Physics is often important to model a situation in equations so we require
algebra to work in order to solve the realistic problem: Suppose a ball is thrown vertically up above a well
at 20m/s and eventually the ball drops into a 25 metre well (Figure 7). A reasonable question would be how
long is it before the ball hits the bottom of the well.
To solve this problem this standard kinematic equation is needed:
d = ut + 12 at 2 where
d is the distance travelled in metres
u is the initial speed of the ball in metres/second
t is the time of travel in seconds
a is the acceleration of the ball in metres/second2
The important vector idea linked to negative quantities is to arbitrarily assign positive and negative
directions to these quantities. One way would be say “up” is positive then u = 20 m/s. Then there is now
no option about assigning directions to the other quantities:
d = -25 metres and
€
a = -10 metres/second2
So d = ut + 12 at 2 leads to 25 = 20t − 5t
−
2
Both these quadratic equations are the same, as must obviously be the case - we can’t have different
equations just because we assign positive and negative directions in different directions.
€
So, solving the equation
with the help of a little algebra we have:
€
25=− 20t + 5t 2 ⇒ 5(t − 5)(t +1) = 0
€
The solution of this equation is t = 5 or -1. As an important aside physicists need to make sense of the
negative time solution that would have been regarded as impossible until quite recently. The solutions can
be interpreted this way:
t = 5 means it the time of the flight of the ball was 5 seconds
t = -1 means that the ball could have been projected from the bottom of the well 1 second before time 0
seconds
Hidden away in the algebra in the problem above is the reason why negative times negative is positive is
−
2
the only sensible yet still arbitrary choice that mathematicians must make. Solving 25= 20t + 5t needs a
number of steps that require the use of the distributive law on a number of occasions.
Therefore the arbitrary definition, and there can be no mistake that it must always arbitrary but sensible, is
negative times negative must be positive to be consistent with the distributive law. This is not saying it is
€
true, merely convenient.
References
Carroll, L. (1872). Through the looking-glass, and what Alice found there. London: Macmillan and Co.
Hefendehl-Hebeker, L. (1991). Negative numbers: obstacles in their evolution from intuitive to
intellectual constructs. For the learning of mathematics 11(1). British Columbia: FLM Publishing Association.
Pickering, A. (1995). The mangle of practice. Chicago: University of Chicago Press.
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The Negative Numbers Domain
Stage 4: Advanced Counting
Objective 1
Calculate a simple whole number minus a simple number using counters where the answer is a negative
number.
Diagnosis
• For 3 - 7 = 4 make up a word story, with a question and its answers, about withdrawing money from a
bank . Explain how to get the answer
• Work out 3 - 5
Learning Experiences
Consider examples like this: 4 - 5.
In pairs or groups one student acts as the banker. He or she has a pile of counters representing dollar
coins. Another student has 4 counters that are included in the bank’s pile of counters. The customer
withdraws five coins. The banker records the customer balance with a negative sign i.e. 1.
Generalisation:
Repeat with similar problems until the students can work out, say, 6 - 9 without needing to use counters.
Stage 5: Early Part-Whole Addition and Subtraction
Objective 1
Calculate a simple large whole number minus a simple large number where the answer is a negative
number.
Diagnosis
• Work out 300 - 700, 7000 - 8000
Learning Experiences
Do problems like 40 - 50, 400 -700, and 7000 - 9000 using ten-dollar, hundred-dollar, and thousand-dollar
play money.
Generalisation:
Repeat with similar problems until the students can work out, say, 6000 - 9000 without needing to use play
money.
Objective 2
Calculate a simple negative number plus a simple positive number where the answer is a positive number.
Diagnosis
• For 4 + 6 = 2 make up a word story, with a question and its answers, about depositing money in a bank
• Model 3 + 5 on counters and explain how to get the answer
• Work out 30 + 80, 600 + 700
Learning Experiences
In pairs or groups one student acts as the banker. Another student owes the bank a small sum of money
and he or she deposits an amount that wipes out the debt. The banker records the customer’s positive
balance.
Generalisation:
Use numbers that are too large to be done on materials but simple enough to work out using basic
subtraction facts. For example:
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Suitable: 30 + 50, 400 + 700, 4000 + 9000
Unsuitable: 49 + 79, 409 + 798, 4870 + 7834
Objective 3
Calculate a simple negative number plus a simple positive number where the answer is a negative number.
Diagnosis
• For 4 + 1 = 3 make up a word story, with a question and its answers, about depositing money in a
bank
• Model 6 + 5 on counters and explain how to get the answer
• Work out 80 + 30, 600 + 400, 9000 + 2000
Learning Experiences
In pairs or groups one student acts as the banker. Another student owes the bank a small sum of money
and he or she deposits an amount that wipes out the debt. The banker records the customer’s positive
balance.
Generalisation:
Use numbers that are too large to be done on materials but simple enough to work out using basic
subtraction facts. For example:
Suitable: 30 + 10, 700 + 500, 8000 + 2000
Unsuitable: 71 + 23, 479 + 198, 4850 + 1834
Objective 4
Calculate a simple negative number minus a simple positive number.
Diagnosis
• For 4 - 1 = 5 make up a word story, with a question and its answers, about withdrawing money from a
bank where the customer already owes the bank money
• Work out 40 - 30, 100 - 400, 2000 - 1000
Learning Experiences
In pairs or groups one student acts as the banker. Another student owes the bank a small sum of money
and he or she withdraws some more money. The banker records the customer’s negative balance.
Generalisation:
Use numbers that are too large to be done on materials but simple enough to work out using basic
subtraction facts. For example:
Suitable: 30 - 10, 200 - 500, 3000 - 2000
Unsuitable: 71 - 45, 409 - 198, 4350 - 1834
Objective 5
Calculate a simple negative number plus or minus a simple positive number using a spreadsheet. Use the
various notations for negative numbers.
Diagnosis
• Insert = 4 - 1 into a cell in a spreadsheet. Use Format, Cells, then Number to locate Negative Numbers.
Alter the cell format to show the answer in these forms: -5, (5), and 5.
• Insert = 4 + 1 into a cell in a spreadsheet. Use Format, Cells, then Number to locate Negative Numbers.
Alter the cell format to show the answer in these forms: -3, (3), and 3.
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• Insert = 6 + 11 into a cell in a spreadsheet. Use Format, Cells, then Number to locate Negative Numbers.
Alter the cell format to show the answer is the same in all cases: 5
Learning Experiences
Students set up their own spreadsheets and do a series of
deposits and withdrawals (Figure 7). Negative balances should
be shown in red.
Figure 7
Stage 6: Advanced Part-Whole Addition
and Subtraction
Objective 1
Calculate a negative number plus or minus a positive number using a temperature problem.
Diagnosis
• For 4 - 1 = 5 make up a word story, with a question and its answers, about temperatures
• For 2 + 7 = 5 make up a word story, with a question and its answers, about temperatures
• For 6 + 2 = 4 make up a word story, with a question and its answers, about temperatures
Learning Experiences
Diagnosis
• Draw vertical temperature scales to work out problems like 4 - 1, 2 + 7 = 5, and 6 + 2
Generalisation:
Repeat with similar problems with larger numbers. (Notice temperatures can never be lower than
273· 15º Celsius as this is absolute zero.)
Objective 2
Find the combined displacement of positive plus negative displacements along a horizontal or vertical
number line.
Diagnosis
• Draw a horizontal number line and explain why 2 + 7 = 5
• Work out 200 + 700 = 500
• Draw a vertical number line and explain why 7 + 3 = 5
• Work out 8000 + 3000
Learning Experiences
Draw horizontal or vertical number lines and work out the following:
• 2+8
10 + 6
−
−
-
-
•
•
12 5
+
19 19
2· 4 + 5· 5
8 11
+
50 50
2· 4 + 1· 4
Generalisation:
€
For real numbers:
€
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•
−
a + b=− (a − b),a > b
•
−
a + b = b − a,a < b
€
Objective 3
€
Find the combined displacement of negative plus negative displacements along a horizontal or vertical
number line.
Diagnosis
• Draw a horizontal number line and explain why 2 + 7 = 9
• Work out 1· 8 + 0· 3
−
−
95 2
• Work out
+
97 97
Learning Experiences
Draw
€ horizontal or vertical number lines and work out the following:
• 2+ 8
10 + 6
−
•
•
€
−
-
−
990
9
+
1000 1000
-
2· 4 + 5· 5
Generalisation: €
2· 4 + 1· 4
For all real numbers:
•
€
−
12 3
+
19 19
−
a+− b=− (a + b)
Objective 4
Find the combined displacement of real number minus negative displacements along a horizontal or vertical
number line.
Diagnosis
• Work out 22 - 4
• Work out 1· 8 - 0· 3
−
5 2
• Work out 3 −
11 11
Learning Experiences
Draw horizontal or vertical number lines and show why a number minus a negative number is the same as
adding
€ a positive number:
- • 2-8
10 - 6
−
€
-
−
990
9
−
1000 1000
2· 4 - 5· 5
2· 4 - 1· 4
•
•
−
12 3
−
19 19
Generalisation: €
For all real numbers:
•
±
a−− b=± a + b
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Objective 5
Add and subtract two-dimensional displacements and link this to drawing of displacements.
Diagnosis
⎛ 3 ⎞ ⎛ − 4 ⎞ ⎛ −1 ⎞
⎝ 5⎠ ⎝ 2 ⎠ ⎝ 7 ⎠
⎛3⎞ ⎛ 2 ⎞ ⎛1⎞
• Draw a diagram that shows why ⎜ ⎟ − ⎜ − ⎟ = ⎜ ⎟
⎝2⎠ ⎝ 4 ⎠ ⎝6⎠
⎛ − 3⎞ ⎛ €
2 ⎞
• Work out 3⎜⎜ ⎟⎟ − 5⎜ − ⎟
⎝ 2 ⎠ ⎝ 4 ⎠
€
Generalisation:
• Draw a diagram that shows why ⎜ − ⎟ + ⎜⎜ − ⎟⎟ = ⎜⎜ − ⎟⎟
For all real numbers adding or subtracting two-dimensional displacements amount to adding or subtracting
the x and y components separately.
€
Stage 7: Advanced Part-Whole Multiplication and Division
Objective 1
Check whether students are operating at formal operations or concrete operations.
Diagnosis
• All humans love eating broccoli. Reggie is a human being. Does Reggie eats broccoli? Explain.
• If all nulls are nurds, and all nurds are quods and Joannie is a quod does it follow she is a null? Explain.
Learning Experiences
What is being assessed here is whether the students can set aside any logical influence from their world
and reason by the logic alone. If they are not able to do this then it is pointless to proceed to teaching
negative times positive and negative times negative.
Generalisation:
If students operate at formal operations then negative times positive must be defined to be positive in
order that the commutative law is still true. Similarly negative times negative is defined to be positive so the
distributive law is still correct. In thee circumstances all the algebra required in solving equations is correct.
Objective 2
Solve physics problems using the kinematics quadratic equation s = ut + 21 at 2 where s is the distance
travelled, u is the initial speed, a is the (constant) acceleration, and t is the time of travel. Solutions of any
quadratic equation are whole numbers.
Diagnosis
€
2
• A ball is thrown straight up in the air. u = 10 m/s and a = 10 m/s are given.
Construct the formula s =10t − 5t 2 .
Graph the formula.
From the graph solve how high does the ball go before returning to the ground
By solving a quadratic equation find how long is the ball in the air
€
• A girl throws a ball down a well at 10m/s. It hits the ground after falling 40 metres.
2
How long was the ball in the air? (Acceleration due to gravity = 10 m/s .)
Interpret physically what the negative solution to the quadratic equation means.
Learning Experiences
• Graph s against t for s = ut + 21 at 2 when u and a are given. Draw interpretations from the graph.
(Numbers are selected to allow whole number solutions to problems.)
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• Given s, a and u and knowing s = ut + 21 at 2 find the time of journeys. Numbers will be selected to give
integer times. Normally there is a negative solution that requires a physical interpretation from the
graph.
€
Stage 8: Advanced Equivalent Fractions and Ratios
Objective 1
Repeat Objective 1 from the previous stage except that the solutions of quadratic equations needs this
formula:
b ± b 2 − 4ac
at + bt + c = 0 ⇔ t =
2a
−
2
Interpret problems where b 2 − 4ac = 0 and b 2 − 4ac < 0 .
2
2
(Use g = 9· 8 m/s instead of 10 m/s .)
€
If b 2 − 4ac = 0 there is only one time solution and if b 2 − 4ac < 0 there are no time solutions possible.
€
For example, a ball is thrown up vertically at 20m/s. How long will it take to reach 100 metres above the
ground? Solution:
€
100 = 20t − 4 ⋅ 9t 2
€
4 ⋅ 9t 2 − 20t +100 = 0
b 2 − 4ac =( − 20)2 − 4 × 4 ⋅ 9 ×100 < 0
So there is no solution.
€
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