missjsmathematicalmadness

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Coordinates and Linear Relationships

How does a grid represent the values of two variables?

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Algebraic Reasoning

Algebra? Isn’t that secondary school mathematics?

Nah - Most mathematics educators, by contrast, would argue that genuine algebraic thinking starts in the exploration of patterns in both shape and number in the early years of primary school. For example, the Australian mathematics curriculum for the foundation stage (ages 4–5 years) includes, under the heading of ‘patterns and algebra’, the expectation that children should ‘sort and classify familiar objects and explain the basis for these classifications; copy, continue and create patterns with objects and drawings’ (ACARA, 2016). Below are some examples of things that children up to the age of 7 years might do that are the beginnings of thinking algebraically:

●    make a repeated pattern using coloured beads on a string;

●    explore, discuss and continue patterns in odd and even numbers;

●    understand a statement like ‘all numbers ending in 5 or 0 can be grouped into fives’ and check whether this is true in particular cases, using counters;

●    on a grid 3 squares wide, colour in the numbers 1, 4, 7, 10, 13, 16 ��� and describe the pattern that emerges (see Figure 19.1);

●    recognize the relationship between doubling and halving;

●    put the numbers from 1 to 10 in a ‘double and add 1’ rule and discuss the pattern of numbers that emerges.

Examples such as these, none of which uses letters for numbers, are about making generalizations.  A generalization is an assertion that something is always the case.

So what do the letters used in algebra actually mean?

Algebra - A branch of mathematics in which letters are used to represent variables in order to express generalizations.

When you introduce older primary children to the use of letters to express general statements, emphasize the idea that a letter in algebra stands for ‘any number’ or ‘whatever number is chosen’, that is, a variable.

Avoid the fruit-salad approach to explaining algebraic statements; for example, referring to 3a as ‘3 apples’ and 5b as ‘5 bananas’, or anything that reinforces the idea that the letters stand for objects or specific numbers.

So what is an equation?

Introduce primary school children to equations by using ‘what is my number?’ puzzles, expressed in words and then in algebraic symbols. For example, ‘I am thinking of a number. Subtract my number from 30 and multiply my number by 2 and you get the same answer. What is my number?’ In symbols, this becomes the equation: 30 – n = 2n. Children solve this by trying various values of the variable n.

Primary school children meet the basic idea of an equation when they are faced with finding the number that should go in the box in statements like 3 +  = 5 (which is equivalent to the equation 3 + x = 5). They can be later introduced to equations by using ‘what is my number?’ puzzles, expressed first in words and then in algebraic form. For example, ‘I am thinking of a number. If you add 7 to my number and subtract my number from 11 you get the same answer. What is my number?’ This is, of course, the equation discussed above put into words: x + 7 = 11 – x. Primary school children would solve this simply by trying various values of the variable, as below:

Try x = 0. Left-hand side = 0 + 7 = 7; right-hand side = 11 – 0 = 11.

Try x = 1. Left-hand side = 1 + 7 = 8; right-hand side = 11 – 1 = 10.

Try x = 2. Left-hand side = 2 + 7 = 9; right-hand side = 11 – 2 = 9.

So the solution is x = 2.

How are equations solved by trial and improvement?

Trial and improvement - A procedure for finding the solution to a mathematical problem by means of successive approximations (trials) which gradually close in on the required solution; it can be used to find solutions to simple equations.

We would not usually introduce formal algebraic processes for solving equations in primary school – especially since the techniques involved can so easily reinforce the idea that the letters stand for ‘things’, or even specific numbers, rather than variables. What is appropriate, however, is to give children the opportunity to develop their algebraic reasoning by solving problems through a trial and improvement approach. These can be purely numerical problems that cannot be solved by a simple arithmetic procedure, or they can be equations that arise from ‘What is my number?’ puzzles, or they can be equations that arise from practical problems. Here is an example:

If you subtract 19 from my number and multiply by my number, you get 666. What is my number?

Expressed algebraically, this problem asks us to solve the equation (n – 19) × n = 666. Figure 19.3 shows how this might be solved by trial and improvement, using a calculator to do any tricky multiplications, thus allowing the child to focus on the process and mathematical reasoning involved. The first column shows the value of the variable n being tried; the second shows the corresponding value of n – 19; the third shows the product of these. We are aiming to get 666 in the third column! In this case, the first value tried is n = 50. The result (1550) is far too large. So, next the value n = 30 is tried, giving a result that is too small (330). So, we try something in between: n = 40, which gives 840. That’s too large. And so on, with n = 35, then 36 and finally 37, which gives the answer required. So, we have solved the equation: n = 37.

What about equations with two variables?

Equations can, of course, involve two (or more) variables. For example, ‘I am thinking of two numbers that add up to 12. What might they be?’ Using x and y for the two variables here, the equation is x + y = 12. This equation is satisfied by lots of pairs of values for x and y. For example, when x = 0, y = 12; when x = 1, y = 11; when x = 2, y = 10; when x = 3, y = 9; when x = 4, y = 8; when x = 5, y = 7; when x = 6, y = 6; when x = 7, y = 5, and so on. If we include negative and rational numbers, there are limitless possible solutions for this equation. For example, when x = 3.7, y = 8.3; and when x = 32, y = –20. We shall see in the next chapter that each pair of values for the two variables in an equation like this can be interpreted as the coordinates of a point in a graphical representation of the relationship between the variables.

How is the equals sign used in algebra?

What the equals sign means strictly in mathematical terms is not the same thing necessarily as the way it is interpreted in practice. When doing arithmetic, that is manipulating numbers, most children (and especially younger children) think of the equals sign as an instruction to do something with some numbers, to perform an operation. They see ‘3 + 5 =’ and respond by doing something: adding the 3 and the 5 to get 8. So, given the question, 3 + □ = 5, many young children (incorrectly) put 8 in the box; this is because they see the equals sign as an instruction to perform an operation on the numbers in the question, and naturally respond to the ‘+’ sign by adding them up.

In algebra, however, the equals sign must be seen as representing equivalence – one of the fundamental concepts in mathematics (see Chapter 5). It means that what is written on one side ‘is the same as’ what is written on the other side. Of course, it has this meaning in arithmetic as well: 3 + 5 = 8 does mean that 3 + 5 is the same as 8. But children rarely use it to mean this; their experience reinforces the perception of the equals sign as an instruction to perform an operation with some numbers. But in algebraic statements it is the idea of equivalence that is strongest in the way the equals sign is used. For example, when we write p + q = r, this is not actually an instruction to add p and q. In fact, we may not have to do anything at all with the statement. It is simply a statement of equivalence between one variable and the sum of two others. This apparent lack of closure is a cause of consternation to some learners. If the answer to an algebra question is p + q, they will have the feeling that there is still something to be done, because they are so wedded to the idea that the addition sign is an instruction to do something to the p and the q.

What is meant by ‘precedence of operators’?

Algebraic operating system - A system used by scientific calculators and spreadsheet software that follows the algebraic conventions of precedence of operators.

This means that they adopt the convention of giving precedence to the operations of multiplication and division. So, when you enter 3, +, 5, the calculator waits to determine whether there is a multiplication or a division following the 5; if there is, this is done first. If you actually mean to do the addition first, you would have to use brackets to indicate this: for example, (3 + 5) × 2.

Precedence of operators - A convention that, unless otherwise indicated by brackets, the operations of multiplication and division should have precedence over addition and subtraction. This convention is always used in algebraic expressions.

This is an opportune moment to mention the convention of dropping the multiplication sign in algebraic expressions: so instead of writing a × b, we can write just ab. We can do this sometimes in arithmetic calculations where brackets remove any ambiguity. For example, the calculation (3 + 5) × 2, using the commutative law, can be written as 2 × (3 + 5); and then, using the convention, this could also be written as 2(3 + 5). The multiplication sign is omitted but understood. Similarly, (x + y) × z would normally be written as z(x + y) and x + y × z would normally be written as x + yz.

How can the idea of a letter being a variable be developed?

Children should have plenty of opportunities to use letters as algebraic symbols representing variables. The most effective way of doing this is through the tabulation of number patterns in columns, with the problem being to express the pattern in the numbers, first in words and later in symbols. A useful game in this context is What’s my rule?

Figure 19.4 shows some examples of this game. In each case, the children are challenged to say what is the rule that is being used to find the numbers in column B and then to use this rule to find the number in column B when the number in column A is 100. In example (a), children usually observe first that the rule is ‘adding 2’. Here they have spotted what I refer to when talking to children as the ‘up-and-down rule’. When talking to teachers, I call it the sequential generalization. This is the pattern that determines how to continue the sequence.

Sequential generalization - When the input and output sets of a mapping are tabulated, a rule for getting the next value of the dependent variable from the previous one(s); the up-and-down rule.

Global generalization - When the input and output sets of a mapping are tabulated, a rule for getting the value of the dependent variable from any value of the independent variable; the left-to-right rule.

Asking what answer you get when the number in A is 100, or some other large number, makes us realize the inadequacy of the sequential generalization. We need a ‘left-to-right rule’: a rule that tells us what to do to the numbers in A to get the numbers in B. This is what I shall refer to as the global generalization. Many children towards the top end of the primary range can usually determine that when the number in A is 100, the number in B is 201, and this helps them to recognize that the rule is ‘double and add 1’.

Later, this can be expressed algebraically. If we use x to stand for ‘any number in column A’ and y to stand for the corresponding number in column B, then the generalization is y = x × 2 + 1, or y = 2x + 1. This clearly uses the idea of letters as variables, expressing generalizations. The statement means essentially, ‘The number in column B is whatever number is in column A multiplied by 2, add 1’.

Similarly, in Figure 19.4(b), the sequential generalization, ‘add 4’, is easily spotted. More difficult is the global generalization, ‘multiply by 4 and subtract 1’, although, again, working out what is in B when 100 is in A helps to make this rule explicit. This leads to the algebraic statement, y = (x × 4) – 1, or y = 4x – 1.

In these kinds of examples, where x is chosen and a rule is used to determine y, x is called the independent variable and y is called the dependent variable.

Independent variable - In a relationship between two variables, the variable whose values may be chosen freely from the given input set, and are then put into the rule to generate the values of the dependent variable in the output set.

Dependent variable - In a relationship between two variables, the one whose values are determined by the value of the independent variable and the rule.

What is a formula?

Formula - An algebraic rule involving one or more independent variables, used to determine the value of a dependent variable.

Here are some very simple examples that can be used to introduce formulas:

If x is a number on a hundred square and y is the number immediately below it, then y = x + 10. If n is the number of bicycles in the rack and m is the number of wheels, then m = 2n. If m is the number of matchsticks available and n is the number of triangles you can make with them, then n = m/3. (The value obtained will need to be rounded down for some values of m.) The number of squares, n, in an array of p squares by q squares is given by the formula n = pq.

Where else are tabulation and algebraic generalization used?

The experience of tabulation and finding generalizations to describe the patterns that emerge occurs very often in mathematical investigations, particularly those involving a sequence of geometric shapes.

Encourage children to tabulate results from investigations, to enable them to find and articulate patterns in the sequence of numbers obtained.

Figure 19.6 provides another example: the problem is to determine how many children can sit around various numbers of tables, arranged side by side, if six children can sit around one table. The number of tables here is the independent variable and the number of children the dependent variable.

With 2 tables we can seat 8 children; with 3 tables we can seat 10. These results are already tabulated. The tabulation can then be completed for other numbers of tables, the sequential generalization can be articulated, the answer for 100 tables can be predicted, and, finally, the global generalization can be formulated. This will be first in words (‘the number y is equal to the number x multiplied by …’) and then in symbols (y = …), with x being the independent variable (the number of tables) and y the dependent variable (the corresponding number of children). This is left as an exercise for the reader, in self-assessment question 19.6.

Take children through this procedure; expect those of differing abilities to reach different stages; tabulate results in an orderly fashion; articulate the up-and-down rule; check this with a few more results; predict the result for a big number, such as 100; articulate the left-to-right rule in words; check this against some results you know; and, for the most able children, express the left-to-right rule in symbols.

So what is a mapping?

In the examples of tabulation used above, there have always been the following three components: a set of input numbers (the values of the independent variable), a rule for doing something to these numbers and a set of output numbers (the values of the dependent variable). These three components put together – input set, rule, output set – constitute what is sometimes called a mapping. It is also sometimes called a functional relationship and the dependent variable is said to be a function of the independent variable.

This idea of a mapping, illustrated in Figure 19.7, is an all-pervading idea in algebra. In fact, most of what we have to learn to do in algebra fits into this simple structure of input, rule and output. Sometimes we are given the input and the rule and we have to find the output: this is substituting into a formula. Then sometimes we are given the input and the output and our task is to find the rule: this is the process of generalizing (as in the examples of tabulation above). Then, finally, we can be given the output and the rule and be required to find the input: this is the process of solving an equation. That just about summarizes the whole of algebra!

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Proportionality and Percentages

How do you solve direct proportion problems?

Proportion - looking at the use of this word in mathematics to describe particular kinds of relationships

Consider the following five problems, all of which have exactly the same mathematical structure:

Recipe problem 1: A recipe for 6 people requires 12 eggs. Adapt it for 8 people.

Recipe problem 2: A recipe for 6 people requires 4 eggs. Adapt it for 9 people.

Recipe problem 3: A recipe for 6 people requires 120 g of flour. Adapt it for 7 people.

Recipe problem 4: A recipe for 8 people requires 500 g of flour. Adapt it for 6 people.

Recipe problem 5: A recipe for 6 people requires 140 g of flour. Adapt it for 14 people.

These have the classic structure of a problem of direct proportion. Such a problem involves four numbers, three of which are known and one of which is to be found.

Direct proportion - The relationship between two variables where the ratio of one to the other is constant. For example, the number of cows’ legs in a field and the number of cows would normally be in direct proportion.

The problems always involve two variables, which I have called variable A and variable B in Figure 18.1. For example, in problem (1) above, variable A would be the number of people and variable B would be the number of eggs. The numbers w and y are values of variable A, and the numbers x and z are values of variable B. So, for example, in recipe problem (1) above, w = 6 and y = 8, x = 12 and z is to be found. (See Figure 18.2(a).) In situations like the recipe problems above, we say that the two variables are ‘in direct proportion’ (or, often, just ‘in proportion’ or ‘proportional to each other’), meaning that the ratio of variable A to variable B (for example, the ratio of people to eggs) is always the same. This means that the ratio w:x must be equal to the ratio y:z. It is also true that the ratio w:y is equal to the ratio x:z. As a consequence, when it comes to solving problems of this kind you can work with either the left-to-right ratios or the top-to-bottom ratios, depending on the numbers involved. It is important to note, therefore, that the most efficient way of solving one of these problems will be determined by the numbers involved, as illustrated in Figure 18.2(a), (b), (c) and (d).

Recipe problem (1) (Figure 18.2a) – here I am attracted immediately by the simple relationship between 6 and 12. Double 6 gives me 12. So, I work from left to right, doubling the 8, to get 16. Answer: 16 eggs.

Recipe problem (2) (Figure 18.2b) – this time it’s the relationship between the 6 and the 9 that attracts me. Halving 6 and multiplying by 3 gives 9. So, I work from top to bottom, and do the same thing to the 4, halving it and multiplying by 3, to get 6. Answer: 6 eggs.

Recipe problem (3) (Figure 18.2c) – the left-to-right relationship is the easier to work with here: multiplying 6 by 20 gives 120. So, do the same to the 7, to get 140. Answer: 140 g of flour.

Recipe problem (4) (Figure 18.2d) – I think it’s easier to use the ratio of 8 to 6 than 8 to 500. So I’ll work from top to bottom, going from 8 people to 4 to 2, and then to 6. Now, 8 people need 500 g, so 4 people need 250 g, so 2 people need 125 g. Adding the results for 4 people and 2 people, 6 people need 375 g.

Use the four-cell diagram to make clear the structure of direct proportion problems and encourage children to use the most obvious relationships between the three given numbers to find the fourth number.

Recipe problem (5):

6 people require 140 g

So, 1 person requires 140 ÷ 6 = 23.333 g (using a calculator)

So, 14 people require 23.333 × 14 = 326.662 g (using a calculator)

Answer: approximately 327 g.

What does ‘per cent’ mean?

Per cent means ‘in each hundred’ or ‘for each hundred’. The Latin root cent, meaning ‘a hundred’, is used in many English words, such as ‘century’, ‘centurion’, ‘centigrade’ and ‘centipede’. We can use the concept of ‘per cent’ to describe a proportion of a quantity or of a set, just as we did with fractional notation.

So, for example, if there are 300 children in a school and 180 of them are girls, we might describe the proportion of girls as ‘sixty per cent’ (written as 60%) of the school population. This means simply that there are 60 girls for each 100 children. If on a car journey of 200 miles a total of 140 miles is single carriageway, we could say that 70% (seventy per cent) of the journey is single carriageway, meaning 70 miles for each 100 miles. In effect, we have here the structure shown in Figure 18.1 for a direct proportion problem, but where one of the numbers must be 100, as shown in Figure 18.3.

How do you use ad hoc methods to express a proportion as a percentage?

The concept of a percentage is simply a special case of a fraction, with 100 as the bottom number. So, 60% is an abbreviation for 60/100 and 70% for 70/100. In examples like these, it is fairly obvious how to express the proportions involved as ‘so many per hundred’. This is not always the case. The following examples demonstrate a number of approaches to expressing a proportion as a percentage, using ad hoc methods, when the numbers can be related easily to 100. The trick is to find an equivalent proportion for a population of 100, by multiplying or dividing by appropriate numbers.

1.   In a school population of 50, there are 30 girls. What percentage are girls? What percentage are boys?

30 girls out of 50 children is the same proportion as 60 out of 100.

So, 60% of the population are girls.

This means that 40% are boys (since the total population must be 100%).

2.   In a school population of 250, there are 130 girls. What percentage are girls? What percentage are boys?

130 girls out of 250 children is the same proportion as 260 out of 500.

260 girls out of 500 children is the same as 52 per 100 (dividing 260 by 5).

So, 52% of the population are girls.

This means that 48% are boys (52% + 48% = 100%).

3.   In a school population of 75, there are 30 girls. What percentage are girls? What percentage are boys?

30 girls out of 75 children is the same proportion as 60 out of 150.

60 girls out of 150 children is the same proportion as 120 out of 300.

120 girls out of 300 children is the same as 40 per 100 (dividing 120 by 3).

So, 40% of the population are girls, and therefore 60% are boys.

4.   In a school population of 140, there are 77 girls. What percentage are girls? What percentage are boys?

77 girls out of 140 children is the same proportion as 11 out of 20 (dividing by 7).

11 girls out of 20 children is the same proportion as 55 per 100 (multiplying by 5).

So, 55% of the population are girls, and therefore 45% are boys.

When the numbers do not relate so easily to 100 as in the examples above, the procedure is more complicated and is best done with the aid of a calculator, as in the following example:

5.   In a school population of 140, there are 73 girls. What percentage are girls? What percentage are boys?

73 girls out of 140 children means that 73/140 are girls.

The equivalent proportion for a population of 100 children is 73/140 of 100.

Work this out on a calculator. (Key sequence: 73, ÷, 140, ×, 100, =.)

Interpret the display (52.142857): just over 52% of the population are girls.

This means that just under 48% are boys.

Give children the opportunity to use calculators to explore how to express more difficult proportions as percentages, showing them the various different ways of doing this. Be aware that not all calculators follow the same key sequences for finding percentages.

How do percentages relate to decimals?

a decimal such as 0.37 means 37 hundredths. Since 37 hundredths also means 37 per cent, we can see a direct relationship between decimals with two digits after the point and percentages. So, 0.37 and 37% are two ways of expressing the same thing. Here are some other examples: 0.50 is equivalent to 50%, 0.05 is equivalent to 5% and 0.42 is equivalent to 42%. It really is as easy as that: you just move the digits two places to the left, because effectively what we are doing is multiplying the decimal number by 100. It’s just like changing pennies into pounds and vice versa. This works even if there are more than two digits, for example 0.125 = 12.5% and 1.01 = 101%.

This means that we have effectively three ways of expressing proportions of a quantity or of a set: using a fraction, using a decimal or using a percentage. It is useful to learn by heart some of the most common equivalences, such as the following:

Other equivalences can be quickly deduced from these results. For example, we can use the fact that an eighth is half of a quarter (think of a quarter as 0.250) to work out that 1/8 written as a decimal is 0.125 and therefore equivalent to 12.5% as a percentage. We can then multiply by 3 to deduce 3/8 is equal to 0.375, which is equivalent to 37.5%. (See also self-assessment question 18.4 at the end of this chapter.)

Knowledge of these equivalences is useful for estimating percentages. For instance, if 130 out of 250 children in a school are girls, then, because this is just over half the population, I would expect the percentage of girls to be just a bit more than 50% (it is 52%). If the proportion of girls is 145 out of 450 children, then because this is a bit less than a third, I would expect the percentage of girls to be around 33% (to two decimal places it is 32.22%).

Also, because it is so easy to change a decimal into a percentage, and since we can convert a fraction to a decimal just by dividing the top number by the bottom number, this gives us another direct way of expressing a fraction or a proportion as a percentage. For example, 23 out of 37 corresponds to the fraction 23/37. On a calculator, enter: 23, ÷, 37, =. This gives the approximate decimal equivalent, 0.6216216. Since the first two decimal places correspond to the percentage, we can just read this straight off as ‘about 62%’. If we wish to be more precise, we could include a couple more digits, giving the result as about 62.16%. 

So, in summary, we now have these four ways of expressing a proportion of ‘A out of B’ as a percentage:

1.   Use ad hoc multiplication and division to change the proportion to an equivalent number out of 100.

2.   Work out A/B × 100, using a calculator, if necessary.

3.   Enter on a calculator: A, ÷, B, % (but note that calculators may vary in the precise key sequence to be used).

4.   Use a calculator to find A ÷ B and read off the decimal answer as a percentage, by shifting all the digits two places to the left.

How do you use ad hoc methods to calculate a percentage of a quantity?

Examples: 

To find 25% of £48, simply change this to 1/4 of £48, which is £12.

Then, when the quantity in question is a nice multiple of 100, we can often find easy ways to work out percentages. First, let us note that, for example, 37% of 100 is 37. So, if I had to work out 37% of £600, we could reason that, since 37% of £100 is £37, we simply need to multiply £37 by 6 to get the answer required, namely £222.

It is also possible to develop ad hoc methods for building up a percentage, using easy components. One of the easiest percentages to find is 10% and most people intuitively start with this. (Note, however, that the fact that 10% is the same as a tenth makes it a very special case: 5% is not a fifth, 7% is not a seventh, and so on.) So, to find, say, 35% of £80, I could build up the answer like this:

    10% of £80 is £8

so  20% of £80 is £16  (doubling the 10%)

and  5% of £80 is £4  (halving the 10%)

Adding the 10%, 20% and 5% gives me the 35% required: £28.

What about percentage increases and decreases?

Percentage increase or percentage decrease - An increase or decrease expressed as a percentage of the original value.

One of the most common uses of percentages is to describe the size of a change in a given quantity, by expressing it as a proportion of the starting value in the form of a percentage increase or percentage decrease. We are familiar with percentage increases in salaries, for example. So, if your monthly salary of £1500 is increased by 5%, to find your new salary you would have to find 5% of £1500 (£75) and add this to the existing salary.

There is a more direct way of doing this: since your new salary is the existing salary (100%) plus 5%, it must be 105% of the existing salary. So, you could get your new salary by finding 105% of the existing salary, that is, by multiplying by 1.05 (remember that 105% = 1.05 as a decimal).

Similarly, if an article costing £200 is reduced by 15%, then to find the new price we have to find 15% of £200 (£30) and deduct this from the existing price, giving the new price as £170. More directly, we could reason that the new price is the existing price (100%) less 15%, so it must be 85% of the existing salary. Hence, we could just find 85% of £200, for example, by multiplying 200 by 0.85.

The trickiest problem is when you are told the price after a percentage increase or decrease and you have to work backwards to get the original price. For example, if the price of an article has been reduced by 20% and now costs £44, what was its original price? This problem is represented in Figure 18.4. The £44 must be 80% of the original price. We have to find what is 100% of that original price. It’s fairly easy now to get from 80% (£44) to 20% (£11) and then to 100% (£55).

There is an interesting phenomenon, related to percentage increases and decreases, that often puzzles people. If you apply a given percentage increase and then apply the same percentage decrease, you do not get back to where you started! For example, the price of an article is £200. The price is increased one month by 10%. The next month the price is decreased by 10%. What is the final price? Well, after the 10% increase, the price has gone up to £220. Now we apply the 10% decrease to this. This is a decrease of £22, not £20, because the percentage change always applies to the existing value. So the article finishes up costing £198.

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Calculations with Decimals

What are the procedures for addition and subtraction with decimals?

The procedures for addition and subtraction with decimals are effectively the same as for whole numbers. Difficulties would arise only if you were to forget about the principles of place value, regarding the decimal point. Provided you remember which digits are units, tens and hundreds, or tenths, hundredths, and so on, then the algorithms (and adhocorithms) employed for addition and subtraction of whole numbers work in an identical fashion for decimals

When adding or subtracting with decimals, show children how the principle of ‘one of these can be exchanged for ten of those’ works in the same way as when working with integers.

A useful tip with decimals is to ensure that the two numbers in an addition or a subtraction have the same number of digits after the decimal point. If one has fewer digits than the other then fill up the empty places with zeros, acting as ‘place holders’. So, for example: 1.45 + 1.8 would be written as 1.45 + 1.80; 1.5 – 1.28 would be written as 1.50 – 1.28; and 15 – 4.25 would be written as 15.00 – 4.25. This makes the standard algorithms for addition and subtraction look just the same as when working with whole numbers, but with the decimal points in the two numbers lined up, one above the other, as shown in Figure 17.1. The reader should note that if the decimal points are removed, these calculations now look exactly like 286 + 404, 145 + 180, 150 – 128 and 1500 – 425. There really is nothing new to learn here, apart from the need to line up the decimal points carefully and to fill up empty spaces with zeros.

Examples of real life situations in which addition and subtraction of decimals are applied: 

a.   If I save £2.86 one month and £4.04 the next month, how much have I saved altogether?

b.   Find the total length of wall space taken up by a cupboard that is 1.45 m wide and a bookshelf that is 1.80 m wide.

c.   What is the difference in height between a girl who is 1.50 m tall and a boy who is 1.28 m tall?

d.   What is the change from £15.00 if you spend £4.25?

Note that example (d) here is most easily done by adding on from £4.25 to get to £15: first £0.75 to get to £5, then a further £10 to get to £15, giving the result £10.75. This illustrates again that often an informal calculation strategy will be easier to handle than formal vertical layout (as in Figure 17.1(d)), particularly when the calculation is in a real-life context.

How can you check that the answer to an addition or a subtraction is reasonable?

Always check the reasonableness of your answer to a calculation involving decimals by making a mental estimate based on simple approximations. In example (a) above, the amounts of money could be approximated to £3 and £4 to the nearest pound, so we would expect an answer around £7. In example (b), the lengths are about 1 m and 2 m, so we would expect an answer around 3 m. In example (c), we might approximate the heights to 150 cm and 130 cm, so an answer around 20 cm would be expected. And in example (d), approximating this to £15 – £4, we would expect the change to be around £11.

An addition example:

 6.47 + 7.39

Round both up: 17 + 8 = 25 Round both down: 6 + 7 = 23 So, the answer lies between 23 and 25.

A subtraction example:

16.47 – 7.39

Round first number up, second one down: 17 – 7 = 10 Round second number up, first one down: 16 – 8 = 8 So, the answer lies between 8 and 1

What about multiplications and divisions involving decimals in a real-life context?

Example: find the cost of 12 rolls of sticky tape at £1.35 per roll. The mathematical model of this real-world problem is the multiplication, 1.35 × 12. On a calculator, we enter 1.35, ×, 12, =, read off the mathematical solution (16.2) and then interpret this as a total cost of £16.20.

There is some potential to get in a muddle with the decimal point when doing this kind of calculation by non-calculator methods. So, a useful tip is, if you can, avoid multiplying the decimal numbers altogether! In the example above, this is easily achieved, simply by rephrasing the situation as 12 rolls at 135p per roll, hence writing the cost in pence rather than in pounds. We then multiply 135 by 12, by whatever methods we prefer, to get the answer 1620, interpret this as 1620p and, finally, write the answer as £16.20.

Almost all the multiplications involving decimals we have to do in practice can be tackled like this. Here’s another example: find the length of wall space required to display eight posters, each 1.19 m wide. Rather than tackle 1.19 × 8, rewrite the length as 119 cm, calculate 119 × 8 and convert the result (952 cm) back to metres (9.52 m). So, 1.19 × 8 = 9.52.

Realistic multiplication and division problems with decimals involving money or measurements can often be recast into calculations with whole numbers by changing the units (for example, pounds to pence, metres to centimetres). Teach children how to do this.

The same tip applies when dividing a decimal number by a whole number. The context from which the calculation has arisen will suggest a way of handling it without the use of decimals. For example, a calculation such as 4.35 ÷ 3 could have arisen from a problem about sharing £4.35 between 3 people. We can simply rewrite this as the problem of sharing 435p between 3 people and deal with it by whatever division process is appropriate (see Chapters 11 and 12), concluding that each person gets 145p each. The final step is to put this back into pounds notation, as £1.45. So 4.35 ÷ 3 = 1.45.

How do you explain the business of moving the decimal point when you multiply and divide by 10 or 100, and so on?

Because of this phenomenon (see below) we tend to think of the effect of multiplying a decimal number by 10 to be to move the decimal point one place to the right, and the effect of dividing by 10 to be to move the decimal point one place to the left. Since multiplying (dividing) by 100 is equivalent to multiplying (dividing) by 10 and by 10 again, this results in the point moving two places. Similarly, multiplying or dividing by 1000 will shift it three places, and so on for other powers of 10.

However, to understand this phenomenon, rather than just observing it, it is more helpful to suggest that it is not the decimal point that is moving, but the digits. This is shown by the results displayed on the basis of place value, as shown in Figure 17.2(b). Each time we multiply by 10, the digits all move one place to the left. The decimal point stays where it is! To understand why this happens, trace the progress of one of the digits, for example the 3. In the original number, it represents 3 hundredths. When we multiply the number by 10, each hundredth becomes a tenth, because ten hundredths can be exchanged for a tenth. This is, once again, the principle that ‘ten of these can be exchanged for one of those’, as we move right to left. So, the 3 hundredths become 3 tenths and the digit 3 moves from the hundredths position to the tenths position. Next time we multiply by 10, these 3 tenths become three whole units, and the 3 shifts to the units position. Next time we multiply by 10, these 3 units become 3 tens, and so on. Because the principle that ‘ten of these can be exchanged for one of those’, as you move from right to left, applies to any position, each digit moves one place to the left every time we multiply by 10.

Since dividing by 10 is the inverse of multiplying by 10 (in other words, one operation undoes the effect of the other), clearly the effect of dividing by 10 is to move each digit one place to the right.

Base your explanation of multiplication and division of decimal numbers by 10 (and 100 and 1000) on the principle of place value. Talk about the digits moving, rather than the decimal point. Allow children to explore repeated multiplications and divisions by 10 with a calculator, making use of the constant facility.

How do you use this when multiplying by a decimal number?

Older children in primary school should be able to handle multiplications such as (a) 7 × 0.4, (b) 6 × 0.03, (c) 24 × 0.6 and (d) 75 × 0.02. These calculations can, of course, be done by reinterpreting them in the context of money, as explained above (7 items at 40p each, 6 items at 3p each, and so on). But, in preparation for more complex multiplications involving decimals that they will have to learn in secondary school, it is helpful even at this level of difficulty to learn how to manage these as abstract calculations.

The key principle is that a multiplication by a decimal number can always be done as a multiplication by a whole number and then division by a power of 10. I will show how this principle works with the four examples here:

(a)   Because 0.4 = 4 ÷ 10

to multiply by 0.4, first multiply by 4 and then divide by 10

7 × 4 = 28, so 7 × 0.4 = 2.8

(b)   Because 0.03 = 3 ÷ 100

to multiply by 0.03, first multiply by 3 and then divide by 100

6 × 3 = 18, so 6 × 0.03 = 0.18

(c)   Because 0.6 = 6 ÷ 10

to multiply by 0.6, first multiply by 6 and then divide by 10

24 × 6 = 144, so 24 × 0.6 = 14.4

(d)   Because 0.02 = 2 ÷ 100

to multiply by 0.02, first multiply by 2 and then divide by 100

75 × 2 = 150, so 75 × 0.02 = 1.50

In (d) here, because there is no context for the calculation, the 0 at the end of 1.50 tells us simply that there are no hundredths. But nor are there any thousandths, or ten thousandths, and so on! This ‘trailing zero’ can therefore be dropped and the answer given just as 1.5. In terms of abstract numbers, 1.50, 1.500, 1.5000, and so on are all equal to 1.5. In some contexts, these trailing zeros are retained by convention (as in £1.50) or as an indication that a quantity has been rounded (for example, 1.498 rounded to two decimal places is 1.50).

A prerequisite for understanding how to multiply and divide with decimal numbers is to be confident in multiplying and dividing by 10 and 100 (and other powers of 10).

How can the remainder in a written division calculation be expressed as a decimal?

We know how dividing the numerator of a fraction by the denominator gives a decimal number equivalent to the fraction. For example, by dividing 21 by 40 we find that 21/40 was equal to 0.525. This means that in any division calculation that does not work out exactly as a whole number answer we can express the remainder as a decimal.

All that is involved is to continue the division process beyond the decimal point. Figure 17.3 provides two examples. In (a) 647 is divided by 5 using the method of short division. Instead of stopping at the point where the result would be 129 remainder 2, note that this 2 has been changed into 20 tenths. This is facilitated by writing in an additional zero in the tenths position. Then this 20 divided by 5 gives 4 in the tenths position in the answer. Hence, we get 647 ÷ 5 = 129.4. Again, we should remember that all genuine calculations arise from some context and the context will determine whether it is appropriate to give the result in the form of a decimal. For example, if the problem had been to divide a length of 647 m of rope into 5 equal parts, then to say that the length of each part would be 129.4 m is a very appropriate result. But if the problem had been to find out how many lengths of 5 m could be cut from a length of 647 m, then the appropriate solution is 129 lengths with 2 m of rope unused (the remainder).

Note that the 0.4 in the result for 647 ÷ 5 is the decimal equivalent of the fraction 2/5. In Chapter 15 we saw how the remainder of a division problem could be interpreted as a fraction. So we now have three possible exact results for this division, which would be used as appropriate to the context for the calculation:

647 ÷ 5 = 129 remainder 2

647 ÷ 5 = 1292/5

647 ÷ 5 = 129.4

Figure 17.3(b) shows the written calculation of 346 ÷ 8, using the long division format, again extending the process beyond the decimal point. This time we need to write in two additional zeros in order to be able to complete the division and to get the answer 43.25. Once again, it is helpful to note the three different exact outcomes here:

346 ÷ 8 = 43 remainder 2

346 ÷ 8 = 432/8 (which can be simplified to 431/4)

346 ÷ 8 = 43.25

How do the digits after the decimal point in the result of a division calculation relate to the remainder?

Of the three outcomes for the division 346 ÷ 8 obtained above, note that the first of these would be the appropriate solution if calculating how many teams of 8 could be made from 346 children (43 teams, 2 children remaining); the second, if calculating how many cakes each person gets if 346 cakes are shared equally between 8 people (43 cakes and a quarter of a cake each); the third, if sharing £346 equally between 8 people (£43.25 each). The different solutions are appropriate for different kinds of problems. We explore these ideas further by considering three problems (A–C below), all involving division, but with different structure.

A.   How many coaches seating 60 children each are needed to transport 250 children?

Using the process of mathematical modelling (see Chapter 5), we could represent this problem with the division, 250 ÷ 60. If we divide 250 by 60, using whatever mental or written procedure we are confident with, we could get to the result ‘4 remainder 10’. To interpret this result, we should note that the ‘remainder 10’ does not stand for 10 coaches; it represents the 10 children who would have no seats if you ordered only 4 coaches. When we do the calculation, either by short division extended beyond the decimal point or on a calculator, we get the recurring decimal answer ‘4.1666…’ or, say, 4.167 rounded to three decimal places (see Chapter 16). The ‘.167’ represents a bit of a coach, not a bit of a child. So, in this example, the digits after the decimal point represent a bit of a coach, whereas the remainder represents a number of children. They are certainly not the same thing. This is a very significant observation, requiring careful explanation to children, through discussion of a variety of examples.

B.   If I share 150 bananas equally between 18 people, how many bananas do they each get?

The model for this problem is the division, 150 ÷ 18. The result of this division is either ‘8 remainder 6’ or the recurring decimal answer ‘8.3333333’, or, say, 8.333 to three decimal places. In interpreting the answer ‘8 remainder 6’, the remainder represents the 6 bananas left over after you have done the sharing, giving 8 bananas to each person. But the digits after the decimal point in the calculator answer (the ‘.333’ part of 8.333) represent the portion of a banana that each person would get if you were able to share out the remainder equally. This would, of course, involve cutting up the bananas into bits. So, in this example, both the remainder and the digits after the decimal point refer to bananas: the first refers to bananas left over, and the second to portions of banana received if the left-overs are shared out. (Whether or not you can actually share out the remainder in practice will, of course, depend on what you are dealing with. For example, people are not usually cut up into smaller bits, but lengths, areas, weights and volumes can usually be subdivided further into smaller units.)

C.   How many boxes that hold 18 pencils each do you need to store 150 pencils?

The model for this problem is the same division as in B: 150 ÷ 18. The remainder 6 now represents the surplus of pencils after you have filled up 8 boxes of 18. The digits after the decimal point in the calculator answer (the ‘.333’ part of 8.333) represent what fraction this surplus is of a full set of 18. In our example, the ‘.333’ means the fraction of a box that would be taken up by the 6 remaining pencils. So, the remainder in this division problem represents surplus pencils, but the digits after the decimal point represent ‘a bit of a box’. In simple language, we might say either that ‘we need eight boxes and there will be six pencils left over’, referring to the answer with the remainder, or ‘we need eight boxes and a bit of a box’, referring to the answer with digits after the decimal point. Of course, in practice we need 9 boxes, since you cannot purchase a bit of a box.