1.2: Direct and Inverse Procedures
We all learn to calculate – with numbers, with symbols, with functions, etc. But we may not notice that most calculating procedures come in pairs .
- First we learn a direct , deterministic, handle-turning technique, where answers are easy to churn out (as with addition, or multiplication, or working out powers, or multiplying brackets in algebra, or differentiating).
- Then we try to work backwards, or to “undo” this direct operation (as with subtraction, or division, or finding roots, or factorising, or integrating). This inverse procedure requires one to be completely fluent in the corresponding direct procedure; but it is much more demanding, in that one has to juggle possibilities as one goes, in order to home in on the required answer.
To master inverse procedures requires a surprising amount of time and effort. And because they are harder to master, they can easily get neglected. Even where they receive a lot of time, there are aspects of inverse procedures which tend to go unnoticed.
In how many different ways can the missing digits in this short multiplication be completed?
\(\begin{array}{r}{\square 6} \\ { \times \quad \square} \\ \hline \square 28 \\ \hline\end{array}\)
One would like students not only to master the direct operation of multiplying digits effectively, but also to notice that the inverse procedure of
“identifying the multiples of a given integer that give rise to a specified output”
depends on
the \(\text{HCF}\) of the multiplier and the base \((10)\) of the numeral system.
- Multiplying by \(1, 3, 7,\) or \(9\) induces a one-to-one mapping on the set of ten digits \(0–9\); so an inverse problem such as "\(7 \times \square\) ends in \(6\)" has just one digit-solution.
- Multiplying by \(2, 4, 6,\) or \(8\) induces a two-to-one mapping onto the set of even digits (multiples of \(2\)); so an inverse problem such as "\(6 \times \square\) ends in \(4\)" has two digit-solutions, and an inverse problem such as "\(6 \times \square\) ends in \(3\)" has no digit-solutions.
- Multiplying by \(5\) induces a five-to-one mapping onto the multiples (\(0\) and \(5\)) of \(5\), so an inverse problem such as "\(5 \times \square\) ends in \(0\)" has five digit-solutions and an inverse problem such as "\(5 \times \square\) ends in \(3\)" has no digit-solutions at all.
- Multiplying by \(0\) induces a ten-to-one mapping onto the multiples of \(0\) (namely \(0\)); so an inverse problem such as "\(0 \times \square\) ends in \(0\)" has ten digit-solutions and an inverse problem such as "\(0 \times \square\) ends in \(3\) (or any digit other than \(0\))” has no digit-solutions at all.
The next problem shows – in a very simple setting – how elusive inverse problems can be. Here, instead of being asked to perform a direct calculation, the rules and the answer are given, and we are simply asked to invent a calculation that gives the specified output.
(a) In the “\(24\) game” you are given four numbers. Your job is to use each number once, and to combine the four numbers using any three of the four basic arithmetical operations – using the same operation more than once if you wish, and as many brackets as you like (but never concatenating different numbers, such as “\(3\)” and “\(4\)” to make “\(34\)”). If the given numbers are \(3, 3, 4, 4\), then one immediately sees \(3 \times 4 + 3 \times 4 = 24\). With \(3, 3, 5, 5\) it may take a little longer, but is still fairly straightforward. However, you may find it more challenging to make \(24\) in this way:
(i) using the four numbers \(3, 3, 6, 6\)
(ii) using the four numbers \(3, 3, 7, 7\)
(iii) using the four numbers \(3, 3, 8, 8\)
(b) Suppose we restrict the numbers to be used each time to “four \(4\)s” \((4,4,4,4)\), and change the goal from “make \(24\)”, to “make each answer from\(0–10\) using exactly four \(4\)s”.
(i) Which of the numbers \(0–10\) cannot be made?
(ii) What if one is allowed to use squaring and square roots as well as the four basic operations? What is the first inaccessible integer?
Calculating by turning the handle deterministically (as with addition, or multiplication, or multiplying out brackets, or differentiating) is a valuable skill. But such direct procedures are usually only the beginning. Using mathematics and solving problems generally depend on the corresponding inverse procedures – where a certain amount of juggling and insight is needed in order to work backwards (as with subtraction, or division, or factorisation, or integration). For example, in applications of calculus, the main challenge is to solve differential equations (an inverse problem) rather than to differentiate known functions.
Problem 14 captures the spirit of this idea in the simplest possible context of arithmetic: the required answer is given, and we have to find how (or whether) that answer can be generated. We will meet more interesting examples of this kind throughout the rest of the collection.
1.2.1: Factorisation
(a) (i) Expand \((a + b)^{2}\) and \((a + b)^{3}\).
(ii) Without doing any more work, write out the expanded forms of \((a + b)^{2}\) and \((a + b)^{3}\).
(b) Factorise
(i) \(x^2 + 2x + 1\)
(ii) \(x^{4} -2x^{2} + 1\)
(iii) \(x^{6} -3x^{4} + 3x^{2} -1\)
(c) (i) Expand \((a-b)(a+b)\).
(ii) Use (c)(i) and (a)(i) to write down (with no extra work) the expanded form of
\((a-b-c)(a+b+c)\)
and of
\((a-b+c)(a+b-c)\)
(d) Factorise \(3x^{2} + 2x - 1\)