Mathematical Misconceptions: A Guide for Primary Teachers

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Edited by: Anne D. Cockburn & Graham Littler

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  • Appendix: Division by Zero and One: A More Mathematical Explanation

    CarloMarchini and PaolaVighi

    In this final part of the book we discuss zero, one and division. What follows may appear to be fairly technical, but we have endeavoured to make it as accessible as possible by including numerous examples to illustrate key points. Although it is not crucial that you, as a primary teacher, have a full understanding of the complexities of these topics, our view is that knowledge of more advanced mathematics is never wasted in the primary school classroom.

    Some might describe division by 0 and 1 as a sort of mathematical nightmare, a ghost often found haunting high schools and universities. We can consider this ghost as a typical misconception which, as you will see, might well have arisen in primary school.

    The Primary Framework document (Department for Education and Skills, 2006) refers to division across the primary years as set out in Table A.1. If you stop to think about it, some of the words related to division are quite tricky. For example, one does not immediately discern the presence of 2 in ‘half’ or the presence of 4 in ‘quarter’. It is also interesting to note that the specific words ‘dividend’ and ‘divisor’ are not present in DfES vocabulary.

    Table A.1 Division across the primary years as set out in the Primary Framework (Department for Education and Skills, 2006)
    Reception‘Share objects into equal groups and count how many in each group’ (p. 70)
    Year 1‘Solve practical problems that involve … sharing into equal groups’ (p. 72)
    Year 2‘Represent … sharing and repeated subtraction (grouping) as division; use practical and informal written methods and related vocabulary to support … division, including calculations with remainders.’ (p. 74)
    Year 3‘Use practical and informal written methods to … divide two-digit numbers (e.g. … 50 ÷ 4); round remainders up or down, depending on the context.
    Understand that division is the inverse of multiplication and vice versa; use this to derive and record related multiplication and division number sentences.’ (p. 76)
    Year 4‘Multiply and divide numbers to 1000 by 10 and then 100 (whole-number answers), understanding the effect; relate to scaling up or down.
    Develop and use written methods to record, support and explain … division of two-digit numbers by a one-digit number, including division with remainders (e.g. … 98 ÷ 6)’ (p. 78)
    Year 5‘Use understanding of place value to … divide whole numbers and decimals by 10, 100 or 1000.
    Refine and use efficient written methods to … divide … HTU ÷ U. Find fractions using division (e.g. 1/100 of 5 kg)’ (p. 80)
    Year 6‘Calculate mentally with integers and decimals … TU U … U.t ÷ U Use efficient written methods to … divide integers and decimals by a one-digit integer.
    Relate fractions to multiplication and division (e.g. 6 ÷ 2 = ½ of 6 = 6 × ½; express a quotient as a fraction or decimal (e.g. 67 ÷ 5 = 13.4 or (p. 82)

    The idea of ‘sharing equally’ looses its meaning if the divisor is greater than the dividend (e.g. 4 ÷ 6 = □) but also if it is 1 (e.g. 4 ÷ 1 = □), therefore the early encounters with division could result in misunderstanding.

    There is another problem connected with division. In Chapter 7 we pointed out that it was not always possible to use division within the natural number system. If we ask ‘what is the result of 86 ÷ 25?’ it may not be clear what answer is required: is it 3.44 or 3 with remainder 11, or simply 3? And what is the result of 85 ÷ 26? Is it 3 with remainder 7 or 3 or 3.2692307692307692307692307…, with an infinite number of decimal digits?

    Division with a remainder, which is also called Euclidean division, can be performed for every pair of natural numbers (dividend, divisor), with the proviso that divisor is different from 0, and its result is a pair of natural numbers (quotient, remainder). These pairs (dividend, divisor) and (quotient, remainder) are ordered, since if the numbers are reversed the results are different: the division of 8 by 3 is different from the division of 3 by 8 (division is not commutative). In what follows we consider the ‘result’ of a (Euclidean) division of Relate fractions to multiplication and division (e.g. 6 ÷ 2 = ½ of 6 = 6 × ½; express a quotient as a fraction or decimal (e.g. 67 ÷ 5 = 13.4 or 13 2/5′ (p. 82) an ordered pair (quotient, remainder). In Italian and Israeli schools the customary notation 8 ÷ 3 = 2 (2) or 8 ÷ 3 = 2 (r. 2), in which it is clear that the two instances of 2 have different meanings, is used.

    The fundamental property of this kind of division is as follows. Consider 86 ÷ 25 = 3 (r. 11). The dividend 86 is equal to the sum of the divisor 25 multiplied by the quotient 3, added to remainder 11, that is, 86 = 25×3 + 11. Note that the remainder satisfies the condition 0 ≤ 11 < 25. This chain of inequalities ensures the uniqueness of the quotient and remainder, given the dividend 86 and the divisor 25. To record this in a more abstract mathematical manner, we can say that, in a (Euclidean) division – where the divisor is never 0 – the result is a (unique) ordered pair of natural numbers, given by the quotient and the remainder. These two numbers are such that the product of the divisor and the quotient added to the remainder is equal to the dividend and the remainder is greater than, or equal, to 0 and less than the divisor. In short,

    dividend ÷ divisor = quotient (remainder)

    under the following conditions:

    • divisor different from 0;
    • dividend = divisor × quotient + remainder;
    • 0 ≤ remainder < divisor.

    Thus, given the natural numbers 9 and 4, where 9 is the dividend and 4 is the divisor, we can compute 9 ÷ 4 (since 4 is different from 0) resulting in a quotient 2 remainder 1, 9 ÷ 4 = 2 (r. 1). The main property of division is satisfied, that is 9 = 4 × 2 + 1. In other words, the dividend (9) is equal to the sum of: the product of divisor (4) and the quotient (2) plus the remainder (1) and the remainder (1) is greater than or equal to 0 and less than the divisor (4). Note that we also can write 9 = 2 × 4 + 1, thanks to commutative property of multiplication and 0 ≤ 1 < 2. Therefore we can also state that 9 ÷ 2 = 4 (r. 1).

    Warning: 11 ÷ 4 = 2 (r. 3) is equivalent to 11 = 4 × 2 + 3 but, contrary to expectations given the above, 11 ÷ 2 = 5 (r. 1) as it is not the case that 3 < 2!

    This is not a trick question and is, in fact, quite simple: consider sharing 9 marbles equally among 9 children, for example, or sharing 21 sweets equally in a class of 21 pupils. The answer is one each, in both cases. But this answer hides the fact that we are in the context of Euclidean division with remainder. Of course the remainder is 0, nevertheless this remainder must be mentioned, and therefore the answer is: the result of a division in which divisor and dividend are equal is 1 with remainder 0. There is the temptation – to which we often fall prey – to give only the quotient as the result of such a division problem, since 0 ‘is nothing’ and when the remainder is 0 there is nothing to say! But in this case 1 and 0 are strictly tied. And remember that the dividend must be equal to a sum, of a suitable product together with a remainder. Therefore answering 1 to the challenge is not a complete answer as only the product – divisor × quotient – is presented, and thus we have one addend of the addition – whose sum is the dividend – but not the other.

    This question points out one of the main misconceptions regarding division: that division with natural numbers is possible only when the dividend is greater than the divisor. This is a typical misconception almost certainly originating from the fact that in school teaching and in textbooks most examples of division are with dividend greater than divisor, since the aim is to teach the division algorithm. For example, you almost certainly would not see 3 ÷ 7 = □ in such a book. The model of division of cakes is sometimes misleading: the slice is not a cake; moreover, it is smaller than the whole cake. Furthermore, a cake is something of a continuous thing in that, in theory, it can be cut into any number of slices of varying size. In contrast, Euclidean division works for natural numbers (a discrete set) and the ‘result’ is the (ordered) pair: quotient and remainder. The only requirement for a (Euclidean) division is the fact that the divisor must be different from 0, and no other condition is necessary. Thus, for example, 1 ÷ 132 = 0 (r. 1), or 1 ÷ 3 = 0 (r. 1). But 1 ÷ 1 = 1 (r. 0) with a divisor of 1 is a unique case in which the dividend 1 does not give quotient 0 and remainder 1. In other words, the answer to the above challenge is simple. Put more abstractly, the dividend is the sum of the product of divisor times the quotient (0) added to the remainder (1). Since a factor of the product is 0, the product itself is 0 and the sum is 1, since the remainder is 1. But the condition for the remainder is 0 ≤ 1 < divisor, hence the divisor is any natural number greater than 1. Note that 3 ÷ 7 = 0 (r. 3) and that in every case in which the dividend is less than the divisor the quotient is 0 and the remainder equals the dividend.

    This challenge is quite complex, but with patience we can unwind the skein. Take, for example, the number 18 as a dividend in a division with quotient 0. For the main property of division we have 18 = divisor × 0 + remainder. Now it is evident that whatever the divisor – given that we want the quotient to be 0 – the product divisor × 0 = 0, therefore 18 = remainder. But in every case, the divisor can never be 0 and 0 must be less than, or equal to, the remainder (18) which, in turn, is less than the divisor. Therefore we get 18 < divisor. In the above example, the divisor happened to be 18 but it could have been any other natural number such as 23, or 19, or 156. We can state that when dividing, if the quotient is 0, the dividend is equal to the remainder and the dividend is less than the divisor. Conversely, if dividend and remainder are equal, then divisor is greater than the dividend and, therefore the quotient is zero: 18 = divisor × quotient + 18, therefore divisor × quotient = 0. But divisor is different from 0 and hence quotient must be 0.

    Therefore the conditions: dividend < divisor, quotient 0, dividend = remainder are equivalent.

    Now the answer to the third question is easy to find. Using the main property of (Euclidean) division, the given data are enough to assert that the dividend is 0 since divisor × 0 + 0 = 0. We have only to choose the divisor, but, as every natural number is different from 0 is a suitable divisor: e.g. 0 ÷ 5 = 0 (r. 0); 0 ÷ 1 = 0 (r. 0); 0 ÷ 1234 = 0 (r. 0).

    When the remainder is 0, the dividend is divisible by the divisor, since dividend = divisor × quotient + 0. Let us experiment by having a dividend 0; hence, for the above challenge, the quotient would be 0 and remainder would also be 0, for every divisor. Now let us try with a dividend 1, thus 1 = divisor × quotient + 0. But the product of two natural numbers is 1 if and only if both factors are 1, which means that the divisor is 1 and the quotient is 1: in other words, 1 ÷ 1 = 1 (r. 0). Is it the case that a divisor of 1 fulfils the condition of the challenge? By way of response, consider the following:

    • We know when the remainder is 0, whatever the dividend, 0 ≤ remainder < divisor.
    • We also know that if the divisor is 1, then 0 ≤ remainder <1.
    • Using the fact that the set of natural numbers is discrete, we can conclude that the remainder is 0.
    • Therefore, in this case dividend and quotient are equal: for example, 8 ÷ 1 = 8 (r. 0), 13 ÷ 1 = 13 (r. 0), 0 ÷ 1 = (r. 0).

    We can also deduce that division by 1 gives a dividend which is equal to the quotient and a remainder of 0.

    In the light of the above, consider the Table A.2, bearing in mind that we have stressed that 0 can never be a divisor. Notice how:

    Table A.2 Some division calculations with 36 as the dividend

    • In the first three rows a divisor greater than the dividend gives a quotient 0 and a remainder equal to the divisor. The table considers just two examples of the infinite number in which the divisor is greater than 36.
    • In the fourth row, if dividend and divisor are equal the quotient is 1 and the remainder 0.
    • From the fifth row on, the dividend is greater than divisor: the remainder oscillates, while the quotient grows ‘regularly’ until the last four rows.
    • From the fifth row onwards there are several occasions – in total 36 – where, despite variations in the divisor, the same quotient (other than 0) results. More specifically, had we included all the divisors from 38 down to 1 we would have had 18 quotients of 1, 6 of 2, 3 of 3 and 2 of 4.
    • If you cast your eye down the quotient and remainder columns from the fifth row onwards you will see that if the quotient is 1 the remainders increase by 1, if the quotient is 2 they increase by 2, and so on.
    • In the last four lines the remainder is 0 on the basis of the divisibility condition, since all the natural numbers 1, 2, 3, 4, are divisors of 36.
    • In the last line where the divisor is 1, the quotient equals the dividend and the remainder is 0.

    Note that the table does not have a row in which the divisor is 0 but we can imagine what might happen with such an (impossible) divisor.

    • The regularity in the table could suggest that as long as the divisor decreases, the quotient increases. Thus we might deduce that if we divide 36 by 0 the quotient would be greater than 36 (continuity argument).
    • Or we could consider what would happen if the remainder were greater than or equal to 0 and less than the divisor. If the divisor were 0, the remainder would have to be a natural number greater than or equal to 0 and, at the same time, less than 0. But there is no natural number less than 0, since 0 is the smallest natural number (discreteness argument).

    Here lies the conundrum we face if we want to divide by 0: from one point of view we must have a quotient greater than the dividend, and from the other we cannot have a remainder! In other words, the concepts of continuity and discreteness are incompatible.

    Remembering the trend of a growing quotient as the divisor decreases is useful when dividing by decimals or by fractions less than 1, as in such cases the intuitive view (see Chapter 3) that division makes smaller might be violated.

    The peculiarities of division by 0 and 1 are often not considered or, as we discovered in some textbooks, inadequately discussed. Indeed, the question of having a divisor equal to 0 is often passed over in silence and yet it is important later in school life when the fractions are used and much later on, when the incremental ratio is used in derivative calculus. It is tempting to provide you with the following conclusion, ‘Division by 0 is impossible. From now on, please remember: you can never divide by 0.’ Such a generalisation, however, is not strictly accurate and is, indeed, contrary to the spirit of this book where we countenance against making statements which do not universally apply. It is more accurate to say: ‘When you are working with real numbers (natural, integer, rational and irrational numbers) please refrain from division by 0. Dividing by 0 is not allowed when you are working with these numbers.’ Thus when the more mathematically minded members of your class encounter concepts such as infinitesimals, they will be prepared rather than having to confront a previously held misconception regarding division and zero.

    References
    Department for Education, and Skills (2006) Primary Framework for Literacy and Mathematics. London: Department for Education and Skills.

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