Algebra Background

This background material for Algebra is divided up into 5 sections. These are What is Algebra?, A Recap and Some History, Pre-Algebra: A Foundation for Algebra, Teaching Pre-Algebra, Pattern and Developmental Stages. If you click on any of these you can go straight to that section.

At the start we want to say two things. First, algebra proper doesn’t really begin in the New Zealand Curriculum until Level 5. Levels 1 to 4, though, are important for algebra as they lay the foundation on which a good understanding of algebra can be built. Second, we have omitted any reference to inequalities at Level 2. We do not believe that the introduction of the inequality signs at this Level is appropriate and adds a valuable concept to children’s repertoires. Generally it is only introduced as an isolated piece of knowledge that is not linked to any other area and so has little relevance for the children. We believe that it is more appropriate to introduce it at a later point in their mathematical career when it fits in with other ideas. Hence we will not refer to Achievement Objective 3 of Level 2 here.

What is Algebra?

As we have said above, this site is not strictly about Algebra. It is really about Pre-Algebra. The work in the Algebra Strand of Mathematics in the New Zealand Curriculum in Levels 1 to 4 is meant to pave the way for the introduction of Algebra proper from Level 5 onwards. So let’s consider what Algebra is and then see how Levels 1 to 4 of the Algebra Strand fit into this branch of mathematics.

Put simply, Algebra is the area of mathematics that uses letters and symbols to represent numbers, points and other objects, as well as the relationships between them. It’s value is that it enables general statements to be made simply, it enables a wide range of problems to be solved efficiently and it even enables some problems to be solved that can’t yet be solved in any other way.

Consider the following example. It shows three things. First it shows that Algebra can be used to simplify communication. Second it shows that a calculation that can be done for particular numbers can be done for a general number and so we can save time calculating specific cases. Third it shows how Algebra can be used to solve equations.

You may find it better to skim through this example on a first reading. It’s not necessary to understand the details; it would be helpful though, to try to understand the ideas. So first look for simplification of communication. This is done by the introduction of letters to reduce the language required and to focus in on the particular aspects under consideration. We mark these places by [1]. The second idea of calculations performed for a general number can be seen when we redo the calculation for the sum of the numbers 1 to 100 using the numbers 1 to n, where n can stand for any whole number at all. This is marked by [2]. When we get to part where we solve equations we mark it with [3].

We are going to use the triangular numbers as an example. These are numbers of squares that form triangular shapes. We show a representation of the first five triangular numbers below.

square. 3 squares. 6 squares. 10 squares.
1 3 6 10

15 squares.

15

So the first triangular number is 1, the second is 3 (= 1 + 2), the third is 6 (= 1 + 2 + 3), the fourth is 10 (= 1 + 2 + 3 + 4) and the fifth is 15 (= 1 + 2 + 3 + 4 + 5).

Now suppose that we wanted to find the seventh triangular number. We could easily do that by constructing first the sixth triangular number and then the seventh. The sixth triangular number has 21 squares (it’s 15 + 6) and the seventh triangular number has 21 + 7 = 28.

The first simplification that Algebra can contribute to this discussion is a simplification of language. It gets cumbersome to have to say “sixth triangular number”, “seventh triangular number”, … But we can express this quite simply by using Algebra. [1]. Instead of “first triangular number”, let’s just put T(1). The T tells us the simplification is about Triangular numbers and the 1 in brackets tells us that this one is the first triangular number. So T(2) is the second triangular number, T(3) is the third and so on. [1]. Incidentally, we don’t have to use T for triangular number but it does help us to remember what the T stands for.

It should now be clear that T(6) = 21 and T(7) = 28. And what we have done is to produce a shorthand so that we can discuss any triangular number quite succinctly.

It is now easier to say that we want to find T(100) than it is to say that we want to find the “one-hundredth triangular number”. But at the moment, although it’s easier to say, it’s no easier to find. After all,

T(100) = 1 + 2 + 3 + 4 + … + 98 + 99 + 100.

That looks as if it is going to be a pain to calculate. Maybe your initial response is to get out a calculator and plug in all the numbers from 1 to 100 to get the answer. The famous German mathematician Carl Friedrich Gauss (1777–1855) was faced with this problem in school. His teacher had given it to other classes before he gave it to Gauss’ class. The teacher knew that it would keep them quiet for quite a while. But he wasn’t to know that he had a mathematical over-achiever in his class. This is what Gauss did.

Now                  T(100) =    1 +   2 +   3 + … + 98 + 99 + 100.

Writing this around the other way we get

                        T(100) = 100 + 99 + 98 + … +  3 +   2 +     1.

At this point, Gauss then added the two sets of numbers. At first sight this would appear to make things harder but wait, you need to group the numbers in the right way.

          T(100) + T(100) = (1 + 100) + (2 + 99) + (3 + 98) + … + (98 + 3) + (99 + 2) + (100 + 1).

From here it’s not too hard. All the numbers in brackets on the right hand side of the equation add to 101. 

          T(100) + T(100) = (101) + (101) + (101) + … + (101) + (101) + (101).

And there are 100 brackets, one for each of the numbers from one to 100. So the right hand side equals 101 ´ 100. 

          T(100) + T(100) = 101 ´ 100.

Well the left hand side is two T(100)s. So to get one T(100) all that Gauss needed to do was to divide 101 ´ 100 by 2. That gave him.

                        T(100) = (101 ´ 100) ¸ 2 = 5050.

Clearly this is one up to young Carl Friedrich. 

But Gauss may have seen this geometrically. Let’s try it with a smaller example. We’ll find T(5). Now T(5) = 1 + 2 + 3 + 4 + 5. Writing it the other way we get T(5) = 5 + 4 + 3 + 2 + 1. Geometrically we have T(5) represented in two ways. First as columns of blocks starting from one and going up to 5.

columns of blocks.

And then as these blocks inverted.

blocks inverted.

Adding these two lots of blocks together gives five columns of 6 blocks.

blocks together.

In this final picture there are 30 blocks. But this is twice as many blocks as we started with. So there were 15 blocks at the start. T(5) = 15.

But things gets better. The same two methods will work for any triangular number. Try it for yourself. Work out T(10) or T(200) or T(315). It certainly beats adding all of the numbers from 1 to 10 or from 1 to 200 or from 1 to 315. We’ll find T(8) using the first way.

T(8) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
T(8) = 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1

So     2T(8)  = (1 + 8) + (2 + 7) + (3 + 6) + (4 + 5) + (5 + 4) + (6 + 3) + (7 + 2) + (8 + 1) 
                  = 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9
                  = 9 x 8.

So       T(8) = (9 x 8) / 2 = 36.

But let’s say it again. The same method we have just used will work for any triangular number – no matter how big or how small. And now this is where the power of Algebra starts to kick in. This method will give us the value of any triangular number!

To see this, we first replace the 100 in T(100) by n. Look at T(n). [1, 2]. What does it mean? It’s the value of the n-th triangular number. That is, it’s the value we get if we add the numbers 1 + 2 + 3 + … + n, where n is any whole number we might like to think of. [2]. And we can calculate T(n) using Gauss’ method.  That means that we can get a formula for T(n). Read on.

Now      T(n) =  1 +  2 +  3 + …          + (n – 1) + n.    [2].

 Writing this around the other way we get

             T(n) = n +  (n – 1) + …          +  3 +  2 + 1.    [2].

Add the two sets of numbers and group them in the right way.

             T(n) + T(n) = (1 + n) + (2 + n – 1) + … + (n - 2 + 3) + (n - 1 + 2) + (n + 1)   [2].

 All the numbers in brackets on the right hand side of the equation add to n + 1.

             T(n) + T(n) = (n + 1) + (n + 1) + (n + 1) + … + (n + 1) + (n + 1) + (n + 1).

 And there are n brackets, one for each of the numbers from 1 to n. So the right hand side equals (n + 1) x n.

             T(n) + T(n) = (n + 1) x n.  [2].

 Well the left hand side is two T(n)s. So to get one T(n) all we have to do is to divide (n + 1) x n by 2. This gives

             T(n) = [(n + 1) x n] / 2.  [2].

 The importance of this is that it gives us a formula for T(n). We can find the n-th triangular number for any value of n we like. So you want T(10)? Well that’s just 11 x 10 / 2 = 55. How about T(200)? Well that’s just 201 x 200 / 2 = 201 x 100 = 20100. And what about T(315)? Well that’s just (316) x 315 / 2 = 49770.

 The important thing that we’ve managed to do, using Algebra, is to find an expression for every triangular number that there is. No longer do we have to draw the n-th triangular number for a particular value of n or write out 1 + 2 + 3 + … and add them all up. All we have to do is to use the formula (n + 1) x n / 2.

 Can you see the power of Algebra here? All possible triangular numbers, an infinite number of them, are tied up in (n + 1) x n / 2.

 But more than that, perhaps we could have been happy using Gauss’ method every time that we came across a new triangular number. It would have worked. But the same method works for every triangular number. And Algebra allows us to encapsulate this method in the most general case.

Oh but what if we wanted to know what triangular number had the value 2080? Then we could put (n + 1) x n / 2 equal to 2080 and see if we could find n. Let’s do it.

 Put                       (n + 1) x n / 2 = 2080.  [3].

 Now there are many ways to solve this equation. For instance, you could use trial and error. If you drop a few values in you should eventually be able to find a number that would satisfy this equation. Perhaps it might be better to use a table and approach the problem more systematically. We can even look at the factors of 2080 and see if any of them fit the pattern of the left hand side of the equation.

 However, Algebra has made a living out of solving equations. It’s one of the things that it does really well. So we could rearrange the equation so that it looks like ax2 + bx + c = 0 and then use the formula that exists to solve quadratic equations like this. As a result we are quickly able to discover that T(64) = 2080.

A Recap and Some History

Let’s just highlight what we have done. First, we have used letters to make communication more efficient and less prone to error. By using T(n), especially, we have saved ourselves a lot of words. These words may well have got in the way of any calculations that we might have been trying to use. Once defined, the symbol is easily recognisable and saves us saying “the n-th triangular number”. While this may not appear to be much of a saving, try working the above example through without using any symbols at all. Not only will the lack of symbols make it hard to follow but it will also make it tedious too.

Second, we were able to take a calculation that we could do for 100 and show that it could be done for any whole number. The process we had carried out with 100, we were able to perform with the variable n.

Third, we were able to solve an equation for an unknown, when we found which triangular number was equal to 2080. Now being able to solve equations is extremely important. In fact given an equation, we’d like to know if it has a solution or not and how many solutions it has.

Historically, solving equations has led to new types of numbers. Now initially you might want to solve an equation like x – 2 = 0. This clearly gives x = 2. But at one time people were perplexed by x + 2 = 0, because they didn’t think that negative numbers existed. By wanting to solve x + 2 = 0 and similar equations, the negative numbers were invented.

In a similar way we know that there are no whole numbers that give a solution to x2 – 2 = 0. It was equations like this that led to the invention of surds, numbers like square root. 2. Of course x2 + 1 has no solution using real numbers, so the invention of the complex numbers came about. The surprising thing is that all of these ‘new’ numbers have valuable uses now and we couldn’t do without them.

But history pushed further in this direction by inventing abstract number-like systems. These were able to operate on objects that are not numbers. Things like rotations and symmetries can now be manipulated in ways that are similar to the way that we manipulate numbers. Even more surprisingly, these abstract systems have useful applications too.

Let’s take the above discussion up in the setting of history. How did Algebra develop historically? Well initially, problems that we might think of being arithmetical or even algebraic, were solved without the use of any special signs for unknowns or for arithmetic operations. As we have suggested above, this must have been fairly tedious. Even as late as Sir Isaac Newton (17th Century), much more language was used than is used today. For instance, Fermat, when he conjectured that there were no integer solutions of xn + yn = zn for n a whole number greater than 2, expressed it in the following cumbersome way:

It is impossible for a cube to be written as the sum of two cubes or a fourth power to be written as the sum of two fourth powers or, in general, for any number which is a power greater than the second to be written as the sum of two like powers.

This makes it very difficult to understand. It lacks the clarity brought on by the succinctness of notation like the T(100) that we used in the triangular numbers example or the xn + yn = zn that we just used above.

There was a change for the better in about 250 AD, when Diophantus came on the scene. He used letters to stand for unknowns. But although this was progress there was still a long way to go. The concern of algebraists from Diophantus, right through to the 16th Century, was to find unknowns – to solve equations. So although they developed many useful techniques these were not encapsulated using unknowns as we did in the previous section when we added the first n numbers. However, mathematicians did work hard throughout this period to develop ways to solve equations.

Viète (1540 – 1603), was the man who introduced the idea of generalising techniques by using unknowns. He was responsible for us being able to add the first n integers in a neat way. So now we have letters being used for two types of variable. The first was for unknowns. These are generally things that we are trying to find. The n in the equation (n + 1) x n / 2 = 2080 is an unknown. So is the x in the equation ax2 + bx + c = 0.

The second type of variable that Viète produced was for givens or parameters. These are numbers that we can insert when we want to. For instance, in the secondary school when students want to solve quadratic equations, they are thinking about equations like ax2 + bx + c = 0. Here we can replace the a, b and c by whatever numbers we have in the quadratic we are trying to solve. There are a lot of quadratic equations like ax2 + bx + c = 0. Using a, b and c allows us to group them altogether and give their solution in one formula. If we are interested in, say, x2 + x – 4160 = 0, then the givens are a = 1, b = 1 and c = -4160. To solve x2 + x – 4160 = 0, we just replace a, b, and c in the formula by 1, 1, and –4160.

But the x in the last paragraph cannot be replaced by anything. We just have to find its value. It is a true unknown and it remains unknown right up to the moment that we find its value.

Pre-Algebra:A Foundation for Algebra

In a sense we have been discussing things that are not the immediate concern of Algebra, Levels 1 to 4. The full power of Viète’s revolution doesn’t really come into being until Level 5. But the importance of the Levels 1 to 4 work is similar to that of Algebra prior to the 17th Century. There is a sense in which that whole of the pre 17th Century period was building up to Viète’s innovations. Only after history had produced a solid foundation for him was he able to move forward. It was a bit like Newton who said “If I have seen further than other men it is because I have stood on the shoulders of giants.”

So Levels 1 to 4 are about providing a foundation. Since this foundation took the human race many centuries to construct it is unlikely that it can be done overnight for our children. The essential ideas will take time to develop. But what are those essential ideas? What are the Foundations of Algebra?

A good platform for children to move forward to algebra is provided by a sound knowledge of the properties of number and of the four basic operations. We list these properties under the following headings:

Equality: One of the first difficulties that children have with algebra may arise because of an incomplete knowledge of what the equals sign signifies. When they initially meet equations they are learning arithmetic and come across equations like 7 + 9 = 16. Here the equals sign indicates that an answer has to be found. (The question is often presented in the form 7 + 9 =.) So they come to think of the equals sign as meaning ‘find the answer’.

SUGGESTION  When they initially meet equations they are learning arithmetic and come across equations like 7 + 9 = 16. This means that 7 + 9 is equal to 16, 16 is equal to 7 + 9 and wherever 16 is used, 7 + 9 could be used instead. However, the question is often presented in the form 7 + 9 =   , so children come to think of the = sign as meaning ‘find the answer’, rather than “is equal to”.   They then are reluctant to write that 16 = 7 + 9, which is just as true a statement.

In algebra they may see equations like 3(x – 1) = 3x – 3. Here the right hand side doesn’t look like an answer. In fact we have a statement about equality. The two sides are equal. When we want to, we can replace one side of the equation by the other, just like we can use 7 + 9 instead of 16.

Children need to know that equals signs represent equalities.

This problem is underlined by statements such as 4 + 8 = 12 ´ 3 = 36. Here they may well have used their calculators correctly to find (4 + 8) x 3. They have certainly carried out the calculations correctly. However, it is certainly not true that 4 + 8 = 36. In such situations, the children should be led to use the equals sign correctly. It may be a little tedious but it will lead to habits that will be valuable later, if they write

(4 + 8) x 3 = 12 x 3 = 36.

Here it is true that quantities on either side of the equals signs are equal

The next thing that children have to handle is a string of logically equivalent equalities. Things like the following are common.

3x + 7
3x
 x 
= 10
= 3
= 1.

Algebraic beginners find it hard to deal with this kind of logical reasoning .

To assist the children when they are asked to attempt problems like 7 + 9 =, sometimes write the problem the other way round as  = 7 + 9. You could also ask them to write as many equivalent statements as possible so that they are encouraged to write things like 7 + 9 = 8 + 8; 7 + 9 = 3 + 3 + 4 + 6; 12 + 2 = 7 + 9.

Operations: The answers to a number of problems can be found by using different operations. For instance, suppose that Garry was making trolleys that needed 4 wheels each. If he had 96 wheels, how many trolleys could he make?

This could be done by using addition, taking 4 + 4 + 4 + … 4 = 96 and noticing that it would need 24 fours. Or we could use subtraction: 96 – 4 – 4 – 4 –  …  – 4 = 0. Alternatively we could divide 96 by 4 to get 24.

On the other hand, if Garry had w wheels the only way to determine the number of trolleys he could make would be to divide by 4. So the number of trolleys is w/4.

Difficulties arise for children who are only able to see ‘multiplication’ in terms of repeated addition. Multiplication needs to be understood as an operation in its own right. At the same time, children need to know the properties of multiplication, especially the fact that the order of multiplication is unimportant, that 12 x 3 = 3 x 12.

In the same way, division needs to be understood as an operation in its own right and not just as repeated subtraction. Knowing the properties of division is vital too. The order of division is important (12 / 3 is not equal to 12 / 3) and division is the inverse of multiplication (dividing by 3 ‘cancels out’ multiplication by 3).

One of the processes that is frequently needed in algebra is renaming of expressions. Equalities such as

a – (b + c) = (a – b) – c

are needed quite often. In order to understand what is going on here children need to know the properties of subtraction. Clearly this is based on what happens when they are subtracting numbers. So they can always check that they are using the algebraic expressions correctly if they have a sound knowledge of subtraction with numbers.

The interaction of operations is also important. We have mentioned the inverse relation between multiplication and subtraction and a similar relation holds between addition and subtraction. But perhaps of even more importance is the hierarchy of the operations in an expression. Children need to know how to calculate 5 + 4 x 7 / 3. Children need ample examples of this type in order to use algebraic expressions with confidence.

Range of Numbers: Many children go through the early years of school only having experienced relatively small whole numbers. They are not sure whether or not the basic operations apply equally as well to large numbers, fractions and decimals. This can easily be overcome by using a range of numbers in their problems. Make sure that they regularly come across problems that involve numbers in the hundreds and thousands, fractions and decimals.

Patterns: Patterns are essential to mathematics so substantial practice with patterns is important for general purposes. However, children need to be able to express the relationships between objects in patterns in mathematical ways if they are to be able to make progress in the algebraic side of the topic. An important step towards this is to allow children the chance to tell the class what pattern they see. Then they might be guided towards more accurate statements of a relationship.

We should point out here that much of what we have said above occurs in the Number Strand and we won’t always directly produce units to cover that material on the Algebra web site. It may come in here incidentally when does occur. However, more may appear on the Number web site when that is established. Having said that though, this site is divided up into two themes, Patterns and Number Operations. Much of this site is currently about patterns though consideration has also been given to number operations. Hence there are some units that cover some of the ideas mentioned above.

Teaching Pre-Algebra

The difficulty in presenting a well-considered plan for the teaching of pre-algebra in Levels 1 to 4 is that the research has still to be done. Currently we know what difficulties children have with this material but we don’t really know how to overcome these difficulties. In this section then, we will discuss some of the difficulties followed by some of the suggested solutions. In the next section we will consider learning sequences that we believe will be useful.

Researchers refer to things as being a process or operational versus being an object or structural. This dichotomy appears in many areas of mathematics education. Actually in many areas of mathematical education research the terminology extends further than this but we won’t worry about this here.

By operational or process understanding we mean that a child can treat something in a procedural way, as a process. For instance, a child who was working at a procedural level in arithmetic could add two single digit numbers together. In algebra, a child working at  an operational level will be able to substitute numbers into an algebraic expression. 

On the other hand, a structural or object view of a topic means that a child is able to work with something as if it were an object. In this case in number, dividing something by 8 is a process while an eighth is an object. Further, the child who calculates 27 + 45 – 27 directly is working at an operational level, while the one who immediately sees this as 45, is working with the expression as an object.

Given an unknown number, x, children can happily deal with the result of adding 31 to it.  And most can call it x + 31. They can also easily work out what will happen when the unknown is 10 etc. However, the difference between procedural and structural understanding seems to come when they have to do something to x + 31, such as adding another x.  Some students cannot do this because they want to find out the answer to x + 31 first – they see it as a process (add 31 to the unknown number), not the product of having done the addition.  Students who have moved beyond a procedural understanding can deal with x + 31 as if it is a thing, not an instruction to do some adding.  

Clearly being able to handle something as an object shows a deeper cognitive ability than if it can only be handled as a process. In studying algebra, and indeed most of mathematics, some of the difficulties appear to lie in moving the child from a procedural to a structural understanding. This requires a series of what are often painful steps as the child develops from one level to another in each topic. In algebra, these adjustments appear to be relatively difficult. Children who come through this section of their schoolwork with confidence generally appear to be able to proceed successfully at least to senior secondary mathematics. Research suggests that children who do not develop this fluency with algebra have only an operational understanding of the properties of number, of expressions, and of operations.

So far we have talked in generalities. Now we consider some particular difficulties that children encounter in algebra.

One of the first problems that children encounter with algebra is its syntax. For instance, they are happy that a + 3a + b = 4a + b but also believe that c + c + c x 5 should equal 3c x 5. This is because they apparently work from the left and think of the letters as some strange concrete object rather that as generalised numbers. This difficulty might be lessened if children were confident with numbers and saw the link between the expressions with unknowns and work that they had already done in number.

When word problems are introduced again we get the left to right difficulty appearing. Children may tend to change information into letters and symbols in the order in which they appear in the problem. The famous example from the literature here is the one containing the following sentence. “There are six times as many students as there are professors.” This statement is frequently translated into algebra as 6s = p (s is the number of students and p the number of professors) instead of s = 6p. This trap of order is not uncommon and is easy for the unwary to slip in to. 

This difficulty, though, could have be overcome by the children if they think about substituting particular values for s and p. Clearly if s = 2, p = 6. If there are six times as many students, then there are more students than there are professors. A simple check should sound warning bells. When patterns are found, even in verbal form, it is good practice to check them out with particular values to see if they make any sense. If not, then it is time to go back to the drawing board.

Pattern

In much of the early pattern work, examples can be compiled in tables like the one below (see MiNZC, Level 5, p. 150).

length of side of garden

1

2

3

4

5

6

number of paving stones

8

12

16

20

In problems like this there are two directions that one can go. One of these ways is to work along the number of paving stones. This gives a relation between consecutive numbers in the second row. It would appear that adding 4 to 8 gives 12, the second number; adding 4 to 12 gives 16; the third number, and so on. This is often referred to as a recurrence relation. It allows us to calculate the next number from the one before or from previous numbers. The relation occurs along the second row.

The other relation is between the numbers in the top row and the numbers in the bottom row. In the table above, this is the relation between the length of the side of the garden and the number of paving stones required. This relation is an equation linking the letter n that we will use to represent the length, and the quantity N(n) that will represent the number of paving stones. (Note that the N and the n have different roles here. They do not represent the same thing.) The equation here is N(n) = 4n + 4. It is a functional relationship or functional equation that gives us the number of paving stones as an output that is dependent upon the input number of stones.

In practice, students find recurrence relationships easier to determine than functional ones.

Developmental Stages

In this section we look at the stages that children go through from first meeting patterns to the algebraic formulation of a pattern. According to Wright there are five stages. We represent them in the diagram below. They approximately correspond to the first five Levels of Algebra in the Curriculum.

Copy a pattern and create the next element

Predict relationship values by continuing the pattern with systematic counting

Predict relationship values using recursive methods e.g. table of values, numeric expression

Predict relationship values using direct rules e.g.   ? x 3 + 1

Express a relationship using algebraic symbols with structural understanding e.g. m = 6f + 2 or m = 8 + 6(f – 1)

Level 1:Copy a Pattern And create the next element.

In the initial stages of pattern, children need to spend a great deal of time exploring the idea of pattern. Here they are introduced to the concept of a pattern being something that is repeated. They learn to add the next object or number to a sequence in order to extend the pattern; they learn to see that there is a limitless variety of patterns; they learn to invent their own patterns; and they begin to be able to describe what the next object in the pattern will be.

At the same time, children begin to see relationships between things. This object is heavier than this object. This object is lighter than this object. These two objects are the same length. Two 5c coins are worth the same amount of money as a 10c coin. There are as many bottle tops as there are pencils. There are four more chairs than there are tables.

During this stage, children should be encouraged to work with a range of materials and try to discover as many patterns and relations as they can. An important idea here is the role of ordinal number in the development of relations with spatial patterns.

Activities:

make and describe patterns: using a variety of materials including money; using numbers; using geometric objects.

Level 2:Continue a pattern using systematic counting

Now number takes a more predominant part in pattern. Patterns have a first term; a fifth term; a tenth term. Patterns are made with numbers. The usual counting sequence 1, 2, 3, 4, …, has by now become well known. Children are now able to handle even numbers and odd numbers in order. Then they can manage multiples of 10, 5, and so on.

Along the way, they begin to realise that they can predict what number is going to come next. This may be done by recognising the number pattern or by counting on by the number required by the pattern.

At this stage, graphs can be introduced to illustrate loosely expressed relationships such as the one between energy levels and times of the day shown on p. 135 of MiNZC.

Activities:

play with number patterns that involve multiples of numbers;
predict the next number in a sequence;
produce graphs showing every day relationships.

Level 3:Predict values using relationships between successive terms

Sequences involving numbers now get more complicated again. Number patterns like 3, 8, 13, 18, … can now be handled relatively easily. Children know that they can make this pattern by recursively adding 5. With more complicated numbers a table of values will be necessary. Fibonacci’s sequence, 1, 1, 2, 3, 5, 8, …, where the next term is the sum of the two previous terms can now be comprehended.

As well as understanding what is happening, the children can describe in words the rules for given sequences. They can also make up and use their own recursive rules. Finding the number that comes before a given number in a sequence leads to the need to solve equations like value. + 7 = 39 ­­what is the number before 39 in the sequence that starts 4, 11, …?

But at this point too, spatial patterns, such as the triangle one on p 140 of MiNZC, can be handled.

Activities:

play with more complicated patterns; find the rule for a given pattern and find the pattern for a given rule; use tables to produce and describe patterns;
make equations that show how the next term is found recursively;
find the number before a given number in a number pattern;
play with spatial patterns;
sketch graphs showing everyday relationships where one variable involves a number.

Level 4:Predict values using rules

Now children are able to predict values of terms in a sequence using the direct rules involved. They can also produce the rules of a given number pattern and use these rules to predict future terms. Children can cope with linear relationship here but simple quadratics and exponentials can be used.

In the sequence 6, 8, 10, … what number comes before 26? This leads to the need to solve equations like  2 value. + 4 = 26.

I think this one is too easy to be a good example:  counting by even numbers and the one before 26 is 24.           

Activities:

play with patterns using direct rules; what is the direct rule of this relationship; 
what is the pattern described by this rule; what numbered term in the pattern is a given one?
sketch graphs of everyday situations that involve one or two numerical variables.

Level 5:Find and use an algebraic expression for a relation

The fifth box in Wright’s developmental sequence shows the beginnings of  “real” algebra. This stage involves the generalisation of processes that are now well understood. For instance, in solving 2 n + 7 = 35, they should know

the idea of n as a variable that can take a particular value when constrained;
understanding the equality relation implied by the equals sign;
‘undoing’ the + 7 by subtracting 7 from both sides; and
‘undoing’ the 2n by dividing both  sides by 2.

References

Kaye Stacey and Mollie MacGregor , Building Foundations for Algebra, Mathematics Teaching in the Middle School, February 1997, 252-257.
Mollie MacGregor and Kaye Stacey, A Flying Start with Algebra, Teaching Children Mathematics, October 1999, 79-85.
Wright, V. (1998). The Learning and Teaching of Algebra: Patterns, problems and Possibilities.In Exploring Issues in Mathematics Education. Proceedings of a Research Seminar on Mathematics Education. Wellington: Ministry of Education.