Chapter 7
Decimal Number Place-Value
Naming Decimal Fractions and Decimal Numbers
Just as it is desirable to call of a whole number and a fraction like 4 31 a mixed number rather than a
mixed fraction, a whole number and decimal fraction like 45· 05 will be called a decimal number rather than
decimal fraction. However, a number like 0·76 may be called a decimal fraction as it has no whole number
part.
€
All students will arrive at their first experience with decimal numbers having been exposed to the
“point” in a wide variety of contexts that have nothing to do with them. Here are some of them:
The Full Stop
This is the most common use of the point. This is so obvious that it is easy for teachers to forget that
students have been endlessly exposed to it before the introduction decimal numbers.
Money
The most common number experience with the point is in money contexts where the point actually
separates two whole numbers. Children will have been exposed from a young age to numbers like $3· 45
at the supermarket; here the point separates a whole number of dollars from a whole number of cents
rather than a whole number of dollars plus a decimal fraction of a dollar.
In money contexts there is the added complication in that there is a base hundred operating in addition
to the normal base ten whole numbers. For example, adding $4· 55 to $2· 95 requires adding 55 cents and
95 cents to get 155 cents, which is 100 cents plus 55 cents i.e. $1· 55. This is another reason not to use
money in introducing decimal numbers. (It is quite common for teachers to begin teaching decimal numbers
through money, but it will cause serious problems for students.)
Times
Another common use of the point is in TV guides. While 30 minutes past 1 o’clock should be written
using a colon, that is to say 1:30, newspapers, often in the same edition, also use the point. So 30 minutes
past 1 o’clock is written as 1.30. Notice, however, times given in hours, minutes and seconds always the
colon rather than the point is normally used. For example, a trip taking 6 hours, 45 minutes and 12 seconds
will be written as 06: 45:12 rather than 06.45.12.
Dates
In New Zealand there are the main two ways to indicate a date like 23 October 2011:
• 23/10/11 - this would be 10/23/11 in the United States
• 23.10.11
This is yet another meaning of the point is introduced that has nothing to do with decimal numbers.
Classifying Books
The Dewey decimal system for labelling books uses the point. For example 658.31124 BEH identifies a
book about management. Here the point has nothing to do with decimal numbers or, indeed, any kind of
number.
Sections, Sub-Sections, and Sub-Sub-Sections
Multiple points are often used to reference sections within sections within sections. For example, in a
certain printer manual 4.2.1. Clearing the jammed paper meant:
• Go to section 4 in the manual, then
• Locate the subsection 4.2 called Paper Jams, then
• Locate the sub-subsection 4.2.1 called Clearing the Jammed Paper
Again this is nothing to do with decimal numbers.
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Multiplication
While not common in New Zealand teachers should be aware that students from other countries may
well use the point to indicate multiplication. For example, 5· 3· 7 would mean 5 x 3 x 7. Its use is not
recommended.
Repeated Points
In mathematics repeated points indicate there are terms in a sequence that are not written down. For
example, 1, 3, 5, 7, … 107, 109 is a way of writing down all the odd numbers from 1 to 109. This is also
used in essays to indicate in essays that some part of an original quotation has been left out.
Cricket
An over in cricket consists of six deliveries. When the TV coverage of a game shows there are 6· 4
overs to go in an innings this means there are six overs and four deliveries yet to be bowled. And, if the
point in this case were to be regarded as indicating the presence of fractions, the digit after the point would
mean sixths not tenths.
Decimal Numbers and the Point
When students come across the point indicating they are working with decimal numbers for the first
time, it is strongly recommended that teachers use this definition:
The digit to the immediate right of the decimal point is in the tenths column
Then teachers and students should read numbers like 34· 7 as “thirty-four and seven tenths” not “thirtyfour point seven” for a considerable time. This emphasises the fact that “decimals” are special find of
fraction
Extending this definition to two decimal place numbers we have:
The digit two columns to the right of the decimal point is in the hundredths column
Then, for some time, teachers and students should read numbers like 34· 76 as “thirty-four, seven tenths,
and six hundredths” not “thirty-four point seven six”, and certainly never as “thirty-four point seventy-six”.
Eventually this needs to be extended to thousandths, ten-thousandths and so on.
Inevitably teachers and students will drop reading the decimal fraction words in a decimal number. So
23· 87 will move from being read as twenty-three plus eight tenths plus seven hundredths to being read as
“twenty-three point eight seven”. However, it is strongly recommended that every time a new concept in
decimal numbers is introduced that teachers revert to using the decimal fraction words for a while.
Typesetting the Decimal Point
Although not nearly as common now as it used to be, the presence of decimal numbers may usefully be
indicated by vertically centring the decimal point rather that writing it at the bottom of a line:
• Write “·” rather than “.”
This symbol is called an interpunct; it is available in Word for a Mac through Option-Shift 9 and on a PC
through the Insert Menu then Symbol. The interpunct is particularly desirable for sentences ending with a
decimal number and a full stop. For example “The cost of the cake is $4· 90.” is clearer than “The cost of
the cake is $4.90.” There is good case for constantly using the interpunct; it emphasises that, whatever
other meaning for the point students may have, students are now dealing with decimal numbers.
Point and Comma
While the point is used for decimal numbers in New Zealand it should be noted that much of the world
uses a decimal comma. For example, the decimal point is used in English-speaking countries as well as most of
Asia, while the decimal comma is used in South Africa and most of continental Europe.
Systeme Intèrnationale
Because the comma is used in decimal numbers in many parts of the world the standard agreed Systeme
Intèrnational (SI) notation for numbers leaves out commas entirely to avoid any possible confusion. Where
New Zealanders would have put commas in large numbers these are replaced by a half-space. So 3 762 056
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330· 98 is SI compliant, but 3,762,056,330· 98 is not. The details of which units are used in SI are left to the
chapter on Geometry and Measurement.
Introduction of Decimal Numbers
Teachers at all levels are prone to teach decimal numbers without relating them to fractions. Indeed
two things are very common:
• Decimal numbers are taught before fractions in a given year
• Decimal numbers are introduced through money.
Neither is logically possible, and both are pedagogically unsound. Firstly “decimal” is an adjective and
therefore requires a noun to be associated with it; this noun is “fraction”. Teachers are strongly advised to
introduce “decimal fractions” rather than just “decimal”- “decimal number” can be introduced later. This
emphasises to teachers and students that decimal fractions are a kind of fraction, and therefore cannot be
learned before fractions are well understood. So “decimals” should not precede “fractions” in any school
scheme.
Secondly money is not a decimal fraction – if it were then $4· 75 would be read as four dollars plus
seven tenths of a dollar plus five hundredths of a dollar, whereas it is actually read as four dollars and
seventy-five cents. So the “·” in money is a separator of two whole numbers – the number before the point
refers to dollars, and the number after the point refers to cents. Students who are initially taught decimal
numbers using money will have such little understanding of decimal numbers that it will often prevent them
being numerate by the time they leave secondary school.
Linking Fractions to the Place-Value System
Two independent sets of ideas are prerequisites for students’ understanding of decimal numbers. Firstly,
the key ideas of place-value that are relevant for whole numbers apply exactly to decimal numbers.
Secondly, students need to understand the meaning of fractions.
Many teachers create a place-value problem for students
by initially introducing, decimal fractions like 0· 68 as meaning
Ones
Tenths
Hundredths
68 hundredths; it does not. If it did mean 68 hundredths then
0
0
68
68 would be in the hundredths column – see Figure 1. This in
turn would mean 68 hundredths, which would be written as
0· 068. This is all very confusing because 0· 68 is equivalent to
68 hundredths, but it does not mean 68 hundredths; 0· 68
Figure 1
means 6 tenths plus 8 hundreds just in the same way that 68
ones means 6 tens and 8 ones. However, 6 tenths is equivalent to 60 hundredths – this is where
understanding equivalent fractions is vital – and 60 hundredths plus 8 hundredths equals 68 hundredths.
This subtle distinction between the place-value meaning of a number and an equivalent form is exactly
the same as the distinction that is absolutely vital for calculations with whole numbers. For example, in the
canonical form, the iron rule that the maximum number in any column is 9, so 345 means 3 hundreds plus
4 tens plus 5 ones. Now in order to divide 345 by five, 345 must be seen in an alternative non-canonical
form namely 34 tens and 5 ones. Then 34 tens can be divided by five, and the calculation proceeds. This
logic exactly applies to division of whole numbers that produce decimal fraction answers.
These general statements applies to all calculations whether with whole numbers or decimal numbers:
.
• Numbers always come in the canonical form, that is to say the maximum number in any column is 9
• To proceed with a calculation temporary non-canonical forms are usually necessary
• The final answer must be turned back into its canonical from
For example, consider working out 0· 35 ÷
• the canonical form for 0· 35 i.e. 3 tenths and 5 hundredths needs to be transformed into an
equivalent non-canonical form namely 35 hundredths.
• 35 things divided by 7 always equals 5 things from the multiplication tables
• 35 hundredths ÷ 7 equals 5 hundredths.
Therefore we can write 0· 35 ÷ 7 equals 0· 05.
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In essence, whatever works for operations with whole numbers will also work for decimal numbers,
which are composed of a whole number part and a decimal fraction part. This is the source of the power of
the place-value system; it enables us to do any calculation involving whole numbers and decimal numbers.
Beginning to Learn Decimal Numbers
Fractions arise from division. This needs to be well understood before attempting problems involving
decimal numbers. It is hard to underestimate the problems that many teachers and students are have at this
point because they have an inadequate understanding of this important idea:
• Everyone needs to understand why 5 wholes ÷ 6 = 5 sixths of one whole before moving on to decimal
numbers.
The difficulty that this presents to most students should not be underestimated, yet mastering the idea will
lead rapidly to students understanding decimal numbers.
Teachers should begin teaching decimal numbers with division problems that have one decimal place
answers. It is advisable to proceed very, very, slowly through this work as many students find connecting
the place-value system for whole numbers with fractions a considerable challenge. Indeed, if students
appear to struggle with one decimal place numbers it is very likely that they either do not understand
place-value, or they have an inadequate view of fractions, or both. If the teacher discovers either of these
problems then comprehensive reteaching is needed.
At the start of teaching decimal numbers, careful selection of the numbers involved in division problems
is important. Consider these two problems:
• 2 apples ÷ 5 equals 2 fifths of an apple,
• 2 apples ÷ 3 equals 2 thirds of an apple.
They are of equal difficulty as fraction problems, yet only one of them is suitable for introductory decimal
numbers:
•
As a decimal number 2 ÷ 5 = 0· 4 – a simple one decimal place number,
•
2 ÷ 3 = 0· 666666666 – an infinitely recurring, difficult decimal fraction.
So 2 ÷ 5 is suitable for introducing decimal fractions but 2 ÷ 3 is not. Teachers should check that all the
introductory division problems they use have exactly one decimal place answers.
Why Not Use Equivalent Fractions?
One way to introduce decimal numbers is to use equivalent fractions. So, for example, 2 ÷ 5 equals 2
fifths, and using equivalent fractions, 2 fifths = 4 tenths, which we write as 0· 4. However, there is an
important reason not to use this approach: the conversion method does not generalise to all problems. For
example, while 2 ÷ 3 equals 2 thirds it is inconceivable that students just learning about decimal numbers
will convert 2 thirds by equivalent fractions to 0· 666666666… in any meaningful way.
Teaching One Decimal Place Numbers
It is suggested that teachers begin teaching decimal numbers with material that has the characteristic of
all place-value material used for introducing whole numbers, namely a unit can easily be broken into ten
pieces. It is useful to point out again that real money is not suitable for any place-value work because, by
this criterion, it does not follow the canon of place value, namely ten parts is
swapped for one whole. For example, five twenty cents makes a dollar, two
Figure 2
fifty dollar notes makes a hundred dollars, and so on. A very useful material
to introduce decimal numbers are bars made up of ten Multilink or Unifix cubes that are wrapped in thin
cardboard (Figure 2). The purpose of wrapping the cubes is to focus students’ initial attention on the whole
“packets”, and when they are unwrapped to refocus their attention on the ten parts•
that are pieces of
the original whole. Teachers may start with this problem:
• Bars of chocolate have ten pieces in a bar. Six bars are shared out
among five people. What is each person’s share?
The solution is that everyone receives one whole bar leaving one bar to
share out. This bar now has to be unwrapped to produce ten pieces of
chocolate. Since 10 ÷ 5 = 2 everyone gets 2 pieces. So each person has
one whole bar and two pieces of chocolate (Figure 3).
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Figure 3
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Students can now copy the following equation and fill in the spaces:
•
6 bars ÷ 5 people =  whole bars +  tenths of a bar
Students need to repeat this kind of problem until the point where they can predict the answer and explain
their reasoning without using material.
Using Words For Decimal Fractions
One important feature with this teaching, and it is a recurring theme in this book, is that the initial
learning of a new idea involves talking about it then recording answers in words rather than the compact
symbols. So far in the learning of decimal fractions the decimal point had been deliberately omitted to
enable students to concentrate on the ideas of one decimal place numbers, and not be distracted by other
meanings for the “point” that they are certain to bring to this endeavour. However, when they have a solid
understanding of how tenths can arise in division problems, the simple definition, mentioned earlier, is
worth introducing:
When you write a decimal point the digit to the
right of the point is the number of tenths column.
Students will readily see the value of this – it saves space to write 3 wholes plus 7 tenths as 3· 7. Teachers
are advised to continue to say 3· 7 as “three and seven tenths” to help imbed the meaning of the symbol.
As time passes a mix of saying “three point seven” and “three and seven tenths” can be introduced, and
eventually just “three point seven”
Reversing the Process
A further powerful way to help cement the meaning of one decimal place numbers is to get students to
reverse the process of creating decimal numbers from division and get students to solve simple one
decimal place problems in addition, subtraction, multiplication and division problems on materials that are
given in the decimal point form. This forces them to connect what the decimal point means by modelling
the number on materials, and it also links decimal numbers to the rules that apply to the whole number
place-value system that they are familiar with. For example, suppose students work out 4· 5 – 1· 8:
• Students would have to get out 4 whole bars and 5 pieces from a bar
• Students must realise that one bar must be swapped for ten smaller pieces i.e. tenths,
• After subtraction the answer is two whole bars and seven tenths of a bar
This shows why 4· 5 – 1· 8 = 2· 7.
Teaching Two Decimal Place Numbers
The next development of decimal numbers is to introduce whole number division problems that have
two decimal place answers. Using the same wrapped material used for
one decimal place answers it will quickly be seen that this material is no
longer suitable. For example, consider working out 5 ÷ 4:
• With the wrapped material each person gets one whole bar
• This leaves one whole bar to be shared among four people
Figure 4
• After unwrapping this bar there are ten pieces to be shared
• Each person gets two pieces i.e. two tenths
• This leaves two pieces to be shared.
Now commonly students will imagine cutting each piece into two because 2 ÷ 4 = one half (Figure 4).
Students then commonly write 5 ÷ 4 = 1⋅ 2 12 . This is all perfectly logical when dealing with fractions, but we
are dealing with a very special class of fractions here, namely those that arise from breaking a unit in some
column of a place-value table into ten parts. To reiterate the central point for the place-value system: a unit
must be broken into ten pieces – no other breakage is permitted. This canon of whole number place-value
€ decimal place numbers. It cannot be overemphasised how important this idea is.
must now be extended to two
It will allow all the ideas learned for calculation with whole numbers to be extended to apply to decimal
numbers, and in so doing enable students to do any estimation and numerical calculation that the world can
throw at them.
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Returning to 5 ÷ 4, and solving it properly, we need to focus on how to solve two tenths divided by
four:
• The first action on material must be to imagine the two pieces each being cut into ten smaller pieces
– there is no option to this.
• And this action is not possible with the Unifix bars – this is why two decimal place number problems
should not use the wrapped cube material.
• In imagination the canon of place-value requires the two pieces to be cut into twenty smaller pieces.
Now comes a critical connection between place-value and fractions in the minds of the students; they
simply have to make these connections:
• Each piece has been cut into ten smaller pieces
• Each smaller piece is a tenth of one tenth of the original whole
• The fraction name of the smaller piece is one hundredth of the original whole
Connecting all this we have 5 ÷ 4 = 1 whole plus 2 tenths + 5 hundredths
The “Oneths” Column
A common student error is to regard the column to the right of a decimal point as the “oneths”
column; this indicates that students are viewing the decimal point as the centre of a number in which pairs
like the tens column and the tenths column are not symmetrically located. So they pair the ones column
and the oneths column symmetrically on either side of the decimal point. For example, students wrongly
suggest that the digit in the “oneths” column for 45· 62 is 6.
A useful correction to this problem is to inform students that the centre of a decimal number is in fact
the ones column not the decimal point. Then the oneths column is identical to the ones column, with the
tens column and the tenths column symmetrically located on either side of the ones column. The term
“oneths” can now be discarded entirely.
Thousandths and the Division Algorithm
When extending decimal numbers to three decimal places and
beyond the whole numbers division algorithm can be very usefully
adapted when fraction words are added that help the
understanding of decimal numbers For 21 ÷ 8 the place-value
thinking is shown in Figure 5. This is a very useful idea as it forces
students to understand what decimals numbers are. In particular it
establishes the fact that as we move to the right in a decimal
number the value of the digits is reducing sharply. This idea is
really important as many students are totally lost when asked to
explain why, for example, 45· 4 is more than 45· 39999999. Such
students are reduced to procedures when rounding decimal
numbers
Connecting Money and Decimal Numbers
Figure 5
As mentioned before a very common introduction to “decimals” uses money. Yet money has nothing to
do with decimal fractions – it is two whole numbers. For example, $3· 45 means three whole dollars and
forty-five whole cents. This strongly suggests that teachers do not introduce decimal numbers through
money. Nevertheless, at some point, students need to connect decimal number answers, typically obtained
from a calculator, to the original money problems. For example, we work out
$4 ÷ 5:
•
The decimal fraction answer is $0· 8 not $0· 80
• 1 tenth of a dollar is ten cents
• So eight tenths of a dollar is eighty cents or $0· 80
Here it is tempting to add a zero so $0· 8 = $0· 80 = 80 cents. But this is conceptually quite wrong, and
it leads to serious errors: consider returning to a whole number problem like 10 x 83. Students think
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adding a zero gives the answer, which it does, but it is a procedure that will lead to problems with decimal
numbers. Consider the common wrong answers the “add a zero” students get for 10 x 83· 2:
•
Adding a zero gives 83· 20 which is clearly nonsense – how could a numerate person suggest
10 x 83· 2 = 83· 20?
• Much better many students who almost understand will say 10 x 83· 2 = 830· 2, but it is still wrong
These connections need careful teaching once students have thoroughly mastered decimal numbers.
Procedurally based students will never understand this.
Estimation
The key part of number sense that numerate adults need has always been estimation; the numerate
adult needs quickly to mentally check any calculations made whether it is done by pencil-and-paper, a
calculator, or an abacus, in order to detect errors. For example, suppose this problem was inputted into
calculator: 378· 89 + 34· 45 + 2590· 80. The display shows answer 6414·15. People with good number
sense would immediately know this must be wrong. There are a range of possible mental strategies here:
• One way could be to notice 378· 89 + 34· 45 + 2590· 80 is about 400 + 30 + 2600
• 400 + 30 + 2600 is about 3000, which is obviously nowhere near 6414· 15
Number-Lines
A major feature in using decimal numbers is the need to be able to construct scales or number lines,
and be able to use them for rounding answers. To learn to do this a much more challenging and significant
exercise than many teachers realise, and needs a great deal of careful attention.
To begin with a very powerful introduction to decimal number lines is to take on a strip of paper, which
is a little more than 2 metres in length;
• Use a metre ruler to mark out two metres at 10 cm intervals.
• Label the number line at 0·1 metres, 0· 2 metres, up to 2· 0 metres
• Draw lines across the strip (Figure 6).
0·0
0·1
0·2
0·3
0·4
0·5
0·6
0·7
0·8
0·9
1·0
1·1
1·2
1·3
1·4
1·5
1.6
1·7
1·8
1·9
2·0
Figure 6
Consider what is involved in marking, say, 1· 923 metres on the number line:
• Students need to understand how to construct the number line from 0 to 2 metres
• Students need to realise that 1· 923 metres consists of 1 metre plus nine tenths of a metre, plus two
hundredths of a metre plus three thousandths of a metre
• Students now need to divide the gap between 1· 9 metres and 2· 0 metres into ten equal pieces i.e.
each small piece is a hundredth
• Understanding that each of these pieces is equal to one hundredth of a metre, students mark a point
two hundredths of a metre beyond 1· 9 metres.
• Now the three thousandths needs to be added.
• The gap between 1· 92 and 1· 93 is divided into ten equal pieces and, so each of these pieces is equal to
one thousandth of a metre
• Students mark a point three thousandths of a metre beyond 1· 92 metres.
Teachers and students alike find this very hard. Such activities deserve as much time as a teacher can give
them as understanding will emerge only slowly.
Rounding
When a workable view of decimal number lines has been developed rounding becomes so much more
reliable. Consider a problem that is now easily solved with number line thinking: Round 12· 487 to the
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nearest whole number. The choices are either 12 or 13. On the number line 12· 487 is closer to 12, so the
rounded answer is 12 is the answer. Notice that this was answered without reference to the rule “five and
over round up”. While this rule is very common, students should never use it, as it is conceptually wrong;
and it sometimes leads to wrong answers.
Consider applying the “round up” rule in the example above:
• Round 12· 487 to 12· 49
• Round 12· 49 to 12· 5
• Round 12· 5 to 13
And the answer 13 is wrong! – it is actually 12. This is an excellent example of the pernicious effect of
procedural thinking compared with real understanding, and why learning procedures should not be
encouraged.
Multiplication and Division with Decimal Numbers
How multiplication and division of decimal numbers work is one of the most difficult ideas in all of
number. For this reason it is very easy for teachers to reduce calculation to procedures. Unfortunately
these procedures are very prone to failure. For example we can solve this problem procedurally. Find
3· 56 x 45· 4:
• Ignore the decimal points and multiply 356 x 454 – this gives 161 624
• Total the number of places following the decimal point – it is 2 + 1 = 3 here
• Count back 3 places in 161 624 – this gives 161· 624
While the answer is correct there are a number of problems with this method. No one seriously suggests
that three digits times three digits should be calculated by pencil and paper. And realistically problems often
have too many significant figures to be solved this way. What is needed is estimation ability to check
calculator answers. For example suppose, by calculator, suppose 3· 866 x 13· 89 = 53· 69874. The key
thing here is to quickly estimate the answer as a check. Here 3· 866 x 13· 89 ≈ 4 x 14 = 56 so the answer
is OK.
Division with decimal numbers is somewhat more problematic than multiplication. Of particular concern
is how to estimate when the divisor is less than 1. Consider this problem: 0· 097 kg of almonds costs
$4· 02. What is the cost per kilogram? The steps are:
• Calculate 4· 02 ÷ 0· 097
• Round the calculator to the nearest cent
• Check the answer by estimation
The problem now becomes how to estimate 4· 02 ÷ 0· 097 as a check on the calculator. And now most
students do not understand why the answer must be more than 4· 02 because the divisor is less than 1.
Students need to understand the following:
• 4· 02 ÷ 0· 097 ≈ 4 ÷ 0· 1
• Work out how many packets of almonds each weighing one tenth of a kilogram can be made from 4
kilograms of almonds
• The answer is 40
This 40 can now be checked against the calculator answer
Percentages: Two Different Ideas
Just as young children find the notion of fraction like ¾ as “3 out of 4 ” beguiling easier than three
quarters percentages like 34% are initially most easily understood as 34 out of a hundred. In both fractions
and percentages this enables students to think in whole numbers, which is much easier than fractions.
Thinking about a problem like “34% of the voters support the Blue Party” it is easy to imagine what 34 out
of every hundred voters means. It is far easier to understand than saying “0· 34 of voters support the Blue
Party”. But, just as fractions as “out of” is not helpful for fraction operations, so too operations involving
percentages becomes a mess. Many students who struggle with the meaning of fractions will have to adopt
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calculation rules that are meaningless procedures that are hard to remember. For example, finding the
percentage is typically achieved by a rule:
•
In a box of 454 oranges 128 are not ripe. What percentage of the apples is not yet ripe?
128 100
. But many students who do not understand this – how can they –
×
454 1
128 454
search their memories and apply another similar but wrong rule. They calculate
.
×
100
1
The procedural solution is
Similarly procedural thinkers equally struggle to decide on the correct method to find a percentage of a
€ this problem. 87% of the apples are of export quality. The orchardist picks 4580 apples.
number. Consider
How many are of export quality? They cannot decide between using two methods without lots of boring
repetitive practice:
€
•
•
87 4580
×
100
1
87 100
×
4580 1
So if students only view of percentages is only “out of” difficulties in making calculations is normal.
€ However, there is a better 100% reliable way to calculate percentages.
Let percentages simply be defined as hundredths, that is to say as fractions with denominator 100. So
€ 34% and 34 hundredths are the same thing by definition. This needs to be connected to decimal fractions in
order to do calculations on the calculator.
synonymous for purposes of calculation:
All they need is regard all the forms of number below as
34% = 34 hundredths =
34
= 0· 34
100
Inevitably one of these forms is supplied in a problem and another form is needed in the calculation. This
requires that students having an excellent understanding of decimal numbers which, in turn, requires a deep
€ even in year eleven students in secondary schools.
knowledge of fractions, which is often sorely lacking
It is common for teachers, knowing many of their students struggle to understand decimal numbers,
bypass teaching them properly and go straight to rule-based percentage rules mentioned earlier. But the
long-term consequence is that, as adults, students taught this way will be functionally enumerate.
However, students who can flexibly change between these four forms of percentages can solve any
percentage problem simply. For example here are the two most common types of problem:
• 34% of $234·55 is found by working out 0· 34 x 234·55
• The percentage profit on $20 451 profit for sales of $234 890 is 8· 7% because
20451
≈ 0· 087 = 8· 7%
234890
No rules like is “this over this” or “times 100” are needed because percentages are merely a special kind of
decimal fraction with denominator 100.
€
Percentages Over A Hundred
An unintended consequence for students who only regard a percentage as a number out of a hundred is
that a percentage more than 100 mean nothing. (This is the same situation is encountered with fractions.)
However, students who understand that percentages over 100 are possible are able to work out efficiently
the balance on a bank account with $210. In it a year with interest rate 5· 9%: the new balance is calculated
this way:
• 105· 9% = 1· 059
• The new balance is 1· 059 x 210 ≈ $222· 39
Students who only consider percentages as “out of a hundred” will not be able to do this.
Common Unit Fractions and Percentages
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Estimating answers is a vital to help check percentage calculations that are made on a calculator;
students need to memorise these unit fractions’ percentage equivalents:
• 10% = 1 tenth
20% = 1 fifth
25% = 1 quarter
50% = 1 half
Other percentages can then be estimated using a combination of these unit fractions. For example, we can
check whether 15·3% of $23998· 95 = $3671· 84 is correct by estimating the answer:
•
15· 3% of $23 998· 95 ≈ 15% of 24 000
•
10% of 24 000 = 1 tenth of 24 000 = 24 000 ÷10 = 2400
•
5% of 24 000 = half of 10% of 24 000 = 1200
•
2400 + 1200 = 3600
•
$3671· 84 ≈ 3600
So the calculator answer is reasonable, and so should be accepted as correct. Other common percentages
can be worked out from these. For example, 40% of a quantity is double whatever 20% of the quantity is.
Percentages of What?
It is typically poorly understood that percentages may computed on either the initial amount or the final
amount. And which one is sensible to use varies with context. For example consider this problem:
• A town has projected 50% increase over five years from today’s population of 10 000. What will it new
population be?
Clearly this gives a new population of 15 000. It is obvious that the percentage calculation is made on
today’s population rather than the final population as the percentage is given on the current situation.
But consider the percentage profit made in retailing:
• A shop makes a gross profit of 20% on everything it sells. At the end of the day there is $550 in sales.
What is the gross profit?
Firstly gross profit means the amount left for the shopkeeper after paying for the items just sold. 20% of
550 = 1 fifth of 550 = $110. So the gross profit is $110. Unlike the population problem above, which starts
with initial amount and adds on a percentage, the gross profit question begins with the final amount and
takes a percentage off. Which way we do a problem needs careful thinking.
Consider this problem a problem where a retailer pays $100 for a coat then marks it up by 25%. What
is the percentage gross profit:
• $25 is the gross profit
• The fraction of gross profit is calculated on the selling price
•
€
25 1
= is the fraction of gross profit , which equals 20%
125 5
So depending on one’s viewpoint the percentage profit might be considered to be 25% of cost price or
20% of selling price and the two numbers are different. In this case business people consider percentage
gross profit to be more sensible.
A splendid problem to sort out the depth of understanding that students have is this:
• Mary buys a blouse that is discounted in sale by 20%. When she gets home she notices that the sale
docket reminds her that she paid the shop $100. She wonders what the original price before the
discount would have been.
Here the common mistake is to add 20% on to $100 i.e. the normal price was $120. But this is wrong. .
And it is far from obvious that the calculation required here is 100 ÷ 0· 8, which gives $125.
An even better example of whether percentages refer to numbers the start or end of a calculation is a
problem ascribed to a famous IT company in the States to test would-be employees on their ability to
think:
• In a tank containing 200 fish 99% of the fish are goldfish. Some goldfish are removed, and now 98% of
the fish are goldfish. How many goldfish were removed?
Most students will say 2 goldfish is the answer. But the answer is 100 goldfish. This seems to defy all
logic but it is correct.
Rounding with Percentages
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It is possible for calculated percentages not to add up to exactly 100%. For example, three percentages
might add up to 100·1%. This is not because an error has been made but in this case rounding percentages
to the nearest one decimal place has lead to an answer that very close to 100% but not exactly equal to it.
Compounding Percentages
Understanding how compound interest and compounding inflation work is for the numerate adult.
Unfortunately it requires a good understanding of place-value, fractions, the meaning of percentages and an
ability to think in a multiplicative mode. So teachers should expect many students to find compounding
ideas very difficult.
Albert Einstein once claimed “the most powerful force in the universe is compound interest”. A
calculation shows what he means:
• Janice earns show 10% annual interest. She began with $10 000. How much does she have in her term
deposit account after 10 Years?
The answer is $10000 ×1⋅17 = $19487 ⋅17 .
This illustrates the rule off 70. This states:
•
To find the number of years for money to double work out 70 ÷ the percentage interest rate
In this case 70 ÷ 10 = 7 years. This rule can also be applied to inflation. So for example if the inflation is 5%
a year
€ then prices double in 70 ÷ 5 = 14 years.
So if someone invests $10000 today at an interest rate of 10% the investment doubles every 7 years. After
42 years the value of the investment start at $10 000 and goes doubles every 7 years:
•
$20 000, $40 000, $80 000, $160 000, $320 000, $640 000
So $10000 becomes $640 000. This is Einstein’s miracle.
(Unfortunately this is not as good as it seems because inflation eats into the value of the investment:
Suppose inflation is running at 5% over the 42 years. By the rule of 70 prices double every 14 years.
After 42 years goods costing $10 000 double every 14 years:
•
$10 000, $20 000, $40 000, $80 000
So $10 000 invested becomes $640 000 over 42 years and $10 000 worth of goods and services then cost
$80 000. So the investor in reality has 640 000 ÷ 80 000 = 8 times as much money as he or she started
with- still a very good return.)
The Continuity of the Number Line
Most students of any age find it very difficult to understand that between any two numbers, no matter
how closer they are together, there is an infinite number of numbers. For example some 12 year old
students were asked this question:
• How many different numbers could you write which lie between 0· 41 and 0· 42?
7% realised there where is an infinite number, 22% thought there were 8, 9, or 10, and 17% thought there
was only 1 (Hart, 1981).
A powerful way for students to appreciate the infinity of numbers involved is for them to imagine an
object moves continuously along a number line, so it moves through and infinite number of points
Terminating and Recurring Decimal Numbers
Relatively few divisions have terminating decimal number answers, so most calculator divisions are
actually have recurring decimal number answers even if students don’t realise this. For example the answer
to 78 divided by 119 worked out on an on-line calculator is 0· 6554621848739496. It does not appear to
recur but it must. There is interesting and very challenging mathematics involved in determining whether a
decimal number that arises from a division will be terminating or recurring.
It turns out that termination occurs if the only prime factors of the divisor are 2 or 5. For example is
34867
is a terminating decimal because the only prime factors of 80 000 are 2 and 5. The following shows
80000
how to find its decimal representation:
•
€
€
34867 34867 34867 53 34867 ×125 4358375
=
=
× =
=
= 0 ⋅ 4358375
80000 2 7 × 5 4 2 7 × 5 4 53
10000000
10 7
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A codicil is needed to this rule about which divisions have a terminating decimal number answer.
According to the direct application of the rule 3 ÷ 15 should not be a terminating decimal number because
the divisor has a factor 3; in fact it equals 0· 2. Here the problem is that 3 and 15 have a common faction 3.
When this is removed, by noting 3 ÷ 15 = (3 ÷ 3) ÷(15 ÷ 3) = 1 ÷ 5, the terminating decimal number rule
now does work.
Decimal Number Multiplication
Consider working out say 1· 6 x 1· 7. As always we should seek
to connect this problem to existing knowledge. As we have done
previously for multiplication of fractions we may connect this
problem to area:
1·7 metres
• A builder cuts out a hardboard rectangle that is 1· 6 metres by
1· 7 metres. What is the area in square metres?
1·6 metres
When doing this with students it is desirable to get them to
construct this actually sized rectangle as shown in Figure 7. Notice
each small square is actually 10 centimetres by 10 centimetres, and
the heavier lines indicate a metre in both directions.
It is evident that there are 17 x 16 squares so the area will be the
same as the area of 272 – this is where the rule “multiply by the
whole numbers ignoring the decimal point comes from. Now the
question is where to put the decimal point. Well the area is more
Figure 7
than 1 metre by 1 metre i.e. 1 square metre, and less than 2 metres
by 2 metres i.e. 4 square metres. So the decimal point can only put in to 272 as 2· 72.
So decimal number multiplication can be attempted as whole number multiplication with the decimal
point inserted after a quick estimate of the answer.
Division with Decimal Numbers
A deep understanding of decimal number division is difficult, so it suggested that teachers adopt the
sensible rules of finding the whole number division answer then use a story to estimate the answer and so
to insert the decimal point in the right place. Consider this example:
• Find 24 ÷ 0· 4
Firstly 24 ÷ 4 = 6.
Now a story will reveal the answer. Here a story like “ 24 apples are shared among 0· 4 children” is
nonsense. So the alternative meaning of division is needed:
•
24 kg of flour are packed into 0· 4 kg packets for sale the supermarket. How many bags can be
made?
Obviously 1 kg of flour produces more that 2 packets and less than 3 packets. So 24 produces between 48
and 72 packets. 24 ÷ 0· 4 only has a 6 in it so that the answer is 60
Ratio Calculations
Techniques that work for whole number ratio problems are of restricted use. Here is an example of a
whole such number thinking:
• If 16 kg of flour needed to be mixed with 12 kg of sugar how much flour would be needed to be added
to 9 kg of sugar?
Here whole number reasoning could like this work:
•
3 kg of sugar needs 4 kg of flour
•
Tripling the amounts we have 9 kg of flour would need 12 kg of flour.
Yet here is a much more realistic ratio problem:
• A chemist makes up a solution by adding 135 ml of water to 125 ml of ethanol. If she wanted to make
up a solution of the same strength how much water would she need to add to 235 ml of ethanol?
The unit method is useful here:
• 135 ml of water is added to 125 ml of ethanol then 1 ml of ethanol needs 135 ÷ 125 ml of water
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• 235 ml of ethanol needs 235 x (135 ÷ 125) ml of water ≈ 254 ml.
Comparisons: Rates
In a number of real situations comparisons are made through rates. A judicious choice of what to
compare is needed. For example, suppose we wanted to compare the value of a 550g packet of cornflakes
costing $3· 78 and another packet of cornflakes weighing 880 grams that costs $ 5· 34. Suppose it was
decided to compare the cost of 1 kilogram for both sizes:
• For the first packet the cost of 1 kg is 1 ÷ 0· 55 x 3· 78 ≈ $6· 87
• For the second packet the cost of 1 kg is 1 ÷ 0· 88 x 5· 34 ≈ $6· 07
So the second packet is better value. However supermarkets find it more meaningful for their customers
not to quote the cost of a kilogram of food but prefer to give the cost of 100 grams. Then the calculations
above are altered:
• For the first packet the cost of 100 grams is 3· 78 ÷ 5· 5 ≈ $0· 69
• For the second packet the cost of 100 gram is 5· 34 ÷ 8.8 ≈ $0· 61
Inverse Rates
In marathons runners plan to run “splits” where they run at different speeds over parts of the race. For
example here is a possible plan:
• Run the first five kilometres at a rate of 4 minutes 30 seconds per kilometre
• Run the first five kilometres at a rate of 4 minutes 20 seconds per kilometre
• And so on
This kind of rate is the inverse of speed. Some care is required in finding the inverse speed from the speed.
Here is an example:
• A runner travelling at 15 kph travels
15 1
= kilometre in one minute
60 4
• The runner takes 4 minutes to run 1 kilometre
So 15 kph is equivalent to taking 4 minutes to run 1 kilometre.
€ per kilometre tells us how long we take to travel one kilometre. These
• whereas 4 minutes 30 seconds
are usefully related. For example:
Rates and Averages
Students often incorrectly solve rate problems by averaging. For example, suppose a race car averages
90 kph on the first lap of a two lap race then averages is 60 kph on the second lap. The average speed
would appear to be this:
• (90 + 60)/2 = 75 kph:
Actually the average is 72 kph:
• Imagine a track that is 90 x 60 = 5400 kilometres long – this length is unrealistic but mathematically
desirable!
• The first lap takes 5400 ÷ 90 = 60 hours
• The second lap takes 5400 ÷ 60 = 90 hours
• So 2 x 5400 = 10 800 km takes a total of 60 + 90 =150 hours
• 10 800 ÷ 150 = 72 km/h.
The message here for students is to be very wary of averaging rates or averages.
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Decimal Numbers Domain
Stage 5: Early Part-Whole Addition and Subtraction
Essential:
• Introduce decimal numbers that arise from division of whole numbers with one d.p. answers – answers
should be presented in words not with the decimal point
• Introduce the decimal point as a shorthand for tenths
• Mix up randomly reading the decimal point as “point” and “tenths”
• Solve 1 d.p. addition and subtraction problems that are presented in the decimal point form
• Order sets of one decimal place numbers
• Solve one digit number times a one d.p. number
• Solve one d.p. divided by a one digit numbers with a one d.p. answer
• Solve whole number money division problems with one d.p. answers converted to dollars and cents
Stage 6: Advanced Part-Whole Addition and Subtraction
Essential:
• Introduce two decimal place numbers arising from division of whole numbers with two d.p. answers
where answers presented in words not with the decimal point
• Solve two d.p. addition and subtraction problems that are presented in the decimal point form
• Solve multiplication problems with two d.p.s times a whole number using the vertical form
• Solve division problems with two d.p. answers using the vertical form
• Round one d.p. and two d.p. numbers to the nearest whole number
• Check answers to one and two d.p. addition and subtraction problems by whole number estimation
• Solve whole number money division problems with two d.p. answers converted to dollars and cents
• Which zeroes must be suppressed in numbers like 00300· 00800
• Understand why, if a calculator has, say, 4· 000 entered in it, pushing any operation button converts the
display to 4
• Order sets of two decimal place numbers
Hard: Examples with stories
Money stories are not desirable since money is not really a decimal number - it is two whole numbers
for dollars and cents separated by a point. The main user of decimal numbers are found in metric system
problems. For two decimal place numbers lengths measured in metres and centimetres are desirable.
Students should create word problems for the equations that involve lengths, and write down their
question before solving problems like these:
• 4· 63 +  = 12·56
•
 - 6· 09 = 11·4
• 9· 67 -  = 4·4
•
 + 4· 98 = 34·64
Percentages
• Define percentages as “out of a hundred”
• Compare simple numbers by percentages – e.g. John score 7 out of ten baskets and Kevin score 36 out
of 50. Converting both these numbers to percentages we see John succeeds 70% of the time while
Kevin succeeds 72% of the time
Optional:
• Order sets of three decimal place numbers
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• Introduce with material three decimal place numbers arising from division of whole numbers with three
d.p. answers – answer presented in words not with the decimal point
• Solve three d.p. addition and subtraction problems
• Solve multiplication problems with a whole number times up to three d.p. numbers
• Solve division problems with up to three d.p. numbers divided by a whole number, where the answers
are not recurring
• Solve problems with whole number division problems that have 3 d.p. answers using the vertical setting
out with attached fraction words
• Round up to three d.p. numbers to a given number of decimal places
• Understand which zeroes must be suppressed in numbers like 00300· 00800
• Understand the effect of multiplication and division by 10, 100, and 1000 where the digits move not the
decimal point. In particular avoid the rule of “adding zeroes” leading to mistakes with decimal numbers
e.g. 10 x 56 = 560 so 10 x 56· 3 = 56·30 or 560· 3
Stage 7: Advanced Part-Whole Multiplication and Division
Essential
• Introduce with material three decimal place numbers arising from division of whole numbers with three
d.p. answers – answer presented in words not with the decimal point
• Introduce with material three decimal place numbers arising from division of whole numbers with three
d.p. answers – answer presented in words not with the decimal point
• Solve three d.p. addition and subtraction problems
• Solve multiplication problems with a whole number times up to three d.p. numbers
• Solve division problems with up to three d.p. numbers divided by a whole number, where the answers
are not recurring
• Solve problems with whole number division problems that have three d.p. answers using the vertical
setting out with attached fraction words
• Construct number lines with up to three d.p.s
• Round up to three d.p. numbers to a given number of decimal places
• Understand the effect of multiplication and division by 10, 100, and 1000 where the digits move not the
decimal point. In particular avoid the rule of “adding zeroes” leading to mistakes with decimal numbers
e.g. 10 x 56 = 560 so 10 x 56· 3 = 56· 30 or 560· 3
• Solve three d.p. addition and subtraction problems
• Solve multiplication problems with a whole number times up to three d.p. numbers
• Solve division problems with up to three d.p. numbers divided by a whole number, where the answers
are not recurring
• Solve problems with whole number division problems that have three d.p. answers using the vertical
setting out with attached fraction words
• Check multiplication and division answers by estimation
• Round up to three d.p. numbers to a given number of decimal places
• Understand the effect of division by a decimal number less than 1 increases rather than decreases the
answer
• Understand the effect of multiplication by a decimal number less than 1 decreases rather than increases
the number
• Order sets of three and more decimal place numbers
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• Learn these facts: 10% =
1
10
, 20% =
1
5,
25% =
1
4
, and 50% =
1
2
.
• Find mentally 5%, 10%, 20%, 25%, 40%, 50%, 60%, 75%, and 80% of easy numbers by using these facts:
10% =
1
10
1
1
1
, 20% = 5 , 25% = 4 , and 50% = 2 .
€
€
€
€
• Calculate the percentage of a number using a calculator and round the answer sensibly. For example,
12· 6% of $45 674· 67 = 0·126 x $45 674·67 = $5755· 00842 ≈ $5755· 01 to the nearest cent.
€
€
€
€
• Calculate percentages by using a calculator. For example, the fractional profit for $4520 profit on sales
of $13 908 is 4520 ÷ 13908 = 0· 324992809 ≈ 0· 324 = 32·4%.
• Estimate percentages of large numbers. For example, Renee says 15· 08% of $319 890· 67 is
$33 159·51 by use of her calculator. Explain why she must be wrong by estimating the correct answer.
• Select and apply percentages to initial or final amounts as appropriate. For example, Jack buys a suit for
his clothes shop. It costs him $100 and adds on 50% as his mark up. The suit does not sell so he marks
the selling price down to his cost price i.e. $100. Explain why he has discounted the selling price by
33 31 % not 50%.
• Division leading to recurring decimal numbers e.g. 11 ÷ 7 = 1· 571428571428571428571428…
• Construct scales to show numbers like 4· 5492 as exactly as possible
€
• Round numbers to any given number of significant figures or decimal places
• Make metric system conversion between units e.g. 12 000 mg = 12 grams
• Create word problems from given equations, and write down their question before solving problems
like these:
4· 6 ÷  = 0·23
• 9 x  = 5·67
•
 ÷ 28 = 13
•
 x 4 = 34·64
• Select the correct operation from a story then proceed to use a calculator to solve the problem. Finally
round the answer sensibly. For example: 0· 289 kg of cheese costs $7· 09. We want to find the cost of
1 kilogram of cheese. Which calculation is appropriate for this problem?
0· 289 x 7·09
0· 289 ÷ 7·09
0· 289 ÷ 7· 09
0· 289 + 7· 09
Then solve the problem with a calculator rounding sensibly
• Understand that there is an infinite number of decimal number, and indeed fractions, between any two
numbers on a number line
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Stage 8: Advanced Equivalent Fractions and Ratios
Essential
• Understanding how to work out a decimal number times a decimal number for any number of decimal
places
• Understanding how to work out a decimal number divided by a decimal number for any number of
decimal places
• Solve proportion and ratio problems involving decimal numbers that cannot be solved by simple factors.
• Solve compounding situations e.g. population growth, money
• Understand how to work out a decimal number divided by a decimal number for any number of decimal
places
• Understand the need for standard form of numbers e.g. why write 4 673 000 000 000 as 4 ⋅ 673 ×1012
• Use standard form to do calculations
€
Rates
Again the key idea here is to understand why rates cannot be averaged.
• Murray drives half way from Auckland to Hamilton at an average speed of 90 kph, then he drives the
second half of the trip at an average speed of 60 kph. Explain why the average speed for the whole trip
is 72 kph.
• John and Jackie decide to have a two-lap car race. They have to work out before the race what speeds
they will do for each lap so that the numbers add up to 200. So, for example, lap one at 120 kph and lap
two at 80 kph is OK, but lap one at 130 kph and lap two at 60 kph is not OK. John decides to travel the
first lap at 198 kph and the second lap at 2 kph. Jackie decides to travel the first lap at 100 kph, and also
travel the second lap at 100 kph. Who will win the race. It is not a tie!
• An ace goal-shooter averages 24 goals in the first twelve games of the season and 18 goals in the last six.
Show her season average is 22 goals per game.
Inverse Ratios and Proportions
Inverse ratio problems are perversely difficult for many students. They often fail to realise the problems
cannot be solved by addition and subtraction; such problems need students need to have mastered a high
level of multiplicative thinking.
• Working day and night 8 machines will take 6 days to fill an order for widgets. The customer urgently
needs the widgets to be made in 3 days. How many machines will the factory manager have to allocate
to the production to get the widgets ready on time?
Here the concept of “machine-days” is useful. It will take 48 machine-days of work to produce the
widgets, so it needs 48 ÷ 3 = 16 machines to produce the widgets in 3 days.
• 12 workers take 15 days to paint 18 houses. An urgent job requires 6 houses to be painted in 6 days.
How many workers are needed?
The concept of “worker-days” is useful. It will take 12 x 15 = 180 worker-days to paint 18 houses so
one house needs 180 ÷ 18 = 10 worker-days to be completed. So 6 houses need 6 x 10 = 60 workerdays to be completed. This must take 6 days so 60 ÷ 10 = 10 workers are needed
Problems
Most ratio and proportion problems in science and technology do not have numbers that have nice
whole number solutions spotting factor. Typically they require a “unit” solution method that differs from
the kind immediately above.
• A chemical formula calls for mixing 455 millilitres of chemical A with 345 millilitres of chemical B What
quantity of chemical B is needed when using 675 millilitres of chemical A?
Every millilitre of chemical A needs 345 ÷ 455 millilitre of chemical B. So 675 millilitres of chemical A
needs 675 x 345 ÷ 455 ≈ 512 millilitres of chemical B
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Constructing Sensible Measures of Comparison
After the invention of the printing press the nuns at the convent of San Jacobi charged 3 florins to print
1025 copies of Plato’s Dialogues. At this time a monk-scribe would charge 1 florin to hand write a single
copy. How would we compare prices? It would hardly be sensible to work out that a printed copy cost
0· 0029 of a florin. Better would be to realise that handing writing was about 340 times as expensive as
printing. To reason in this way students need to be thoroughly multiplicative in their thinking
A common supermarket comparison is where the cost per hundred grams is shown on the product
card. This enables shoppers to compare value.
Indices
Indices like the Consumer Price Index (CPI) are constructed to help economists and others to track
inflation. In the case of the CPI the prices of a set “basket” of goods and services are tracked typically four
times a year and an inflation rate announced. Originally any index is set at a nice number like 1000 and the
increase/decrease in this index is tracked over time. For example suppose a new CPI was started:
• A basket of goods and services costs on average $434· 56 per week – call this a CPI of 1000
• A year later the basket of goods and services costs on average $450· 58 per week
• The new CPI is 450· 58 ÷ 434· 56 x 1000 ≈ 1037, so inflation is 3· 7%
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