1.1: Mental Arithmetic and Algebra

1.1.1: Times Tables

Problem 1

Using only mental arithmetic:

(a) Compute for yourself, and learn by heart, the times tables up to \(9 \times 9\).

(b) Calculate instantly:

(i) \(0.004 \times 0.02\)

(ii) \(0.0008 \times 0.07\)

(iii) \(0.007 \times 0.12\)

(iv) \(1.08 \div 1.2\)

(v) \((0.08)^{2}\)

Multiplication tables are important for many reasons. They allow us to appreciate directly, at first hand, the efficiency of our miraculous place value system – in which representing any number, and implementing any operation, are reduced to a combined mastery of

(i) the arithmetical behaviour of the ten digits \(0–9\), and

(ii) the index laws for powers of \(10\).

Fluency in mental and written arithmetic then leaves the mind free to notice, and to appreciate, the deeper patterns and structures which may be lurking just beneath the surface.

1.1.2: Squares, Cubes, and Powers of 2

Algebra begins in earnest when we start to calculate with expressions involving powers. As one sees in the language we use for squares and cubes (i.e. 2nd and 3rd powers), these powers were interpreted geometrically for hundreds and thousands of years – so that higher powers, beyond the third power, were seen as being somehow unreal (like the 4th dimension). Our uniform algebraic notation covering all powers emerged in the 17th century (with Descartes (1596–1650)). But before one begins to work with algebraic powers, one should first aim to achieve complete fluency in working with numerical powers.

Problem 2

(a) Compute by mental arithmetic (using pencil only to record results), then learn by heart:

(i) the squares of positive integers: first up to \(12^{2}\); then to \(31^{2}\).

(ii) the cubes of positive integers up to \(11^{3}\).

(iii) the powers of \(2\) up to \(2^{10}\).

(b) How many squares are there: (i) \(\lt 1000\)? (ii) \(\lt 10,000\)? (iii) \(\lt 100,000\)?

(c) How many cubes are there: (i) \(\lt 1000\)? (ii) \(\lt 10,000\)? (iii) \(\lt 100,000\)?

(d) (i) Which powers of \(2\) are squares? (ii) Which powers of \(2\) are cubes?

(e) Find the smallest square greater than \(1\) that is also a cube. Find the next smallest.

Evaluating powers, and the associated index laws, constitute an example of a direct operation. For each direct operation, we need to think carefully about the corresponding inverse operation – here “extracting roots”. In particular, we need to be clear about the distinction between the fact that the equation \(x^{2} = 4\) has two different solutions, while \(\sqrt{4}\) has just one value (namely \(2\)).

Problem 3

(a) The operation of “squaring” is a function: it takes a single real number \(x\) as input, and delivers a definite real number \(x^{2}\) as output.

• Every positive number arises as an output (“is the square of something”).
• Since \(x^{2} = (-x)^{2}\), each output (other than \(0\)) arises from at least two different inputs.
• If \(a^{2} = b^{2}\), then \(0 = a^{2} - b^{2} = (a-b)(a+b)\), so either \(a=b\), or \(a= -b\). Hence no two positive inputs have the same square, so each output (other than \(0\)) arises from exactly two inputs (one positive and one negative).
• Hence each positive output \(y\) corresponds to just one positive input, called \(\sqrt{y}\).

Find:

(i) \(\sqrt{49}\)

(ii) \(\sqrt{144}\)

(iii) \(\sqrt{441}\)

(iv) \(\sqrt{169}\)

(v) \(\sqrt{196}\)

(vi) \(\sqrt{961}\)

(vii) \(\sqrt{96,100}\)

(b) Let \(a \lt 0\) and \(b \lt 0\). Then \(\sqrt{ab} \lt 0\), and \(\sqrt{a} \times \sqrt{b} \lt 0\), so both expressions are positive. Moreover, they have the same square, since

\((\sqrt{a b})^{2}=a b=(\sqrt{a})^{2} \cdot(\sqrt{b})^{2}=(\sqrt{a} \times \sqrt{b})^{2}\)

\(\therefore \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\)

Use this fact to simplify the following:

(i) \(\sqrt{8}\)

(ii) \(\sqrt{12}\)

(iii) \(\sqrt{50}\)

(iv) \(\sqrt{147}\)

(v) \(\sqrt{288}\)

(vi) \(\sqrt{882}\)

(c) [This part requires some written calculation.] Exact expressions involving square roots occur in many parts of elementary mathematics. We focus here on just one example – namely the regular pentagon. Suppose that a regular pentagon \(ABCDE\) has sides of length \(1\).

(i) Prove that the diagonal \(AC\) is parallel to the side \(ED\).

(ii) If \(AC\) and \(BD\) meet at \(X\), explain why \(AXDE\) is a rhombus.

(iii) Prove that triangles \(ADX\) and \(CBX\) are similar.

(iv) If \(AC\) has length \(x\), set up an equation and find the exact value of \(x\).

(v) Find the exact length of \(BX\).

(vi) Prove that triangles \(ABD\) and \(BXA\) are similar.

(vii) Find the exact values of \(\cos 36^{\circ}\), \(\cos 72^{\circ}\).

(viii) Find the exact values of \(\sin 36^{\circ}\), \(\sin 72^{\circ}\).

Every calculation with square roots depends on the fact that “\(\sqrt{\text{ }}\) is a function”. That is: given \(y \gt 0\),

\(\sqrt{y}\) denotes a single value – the positive number whose square is \(y\).

The equation \(x^{2} = y\) has two roots, namely \(x = \pm \sqrt{y}\); however, \(\sqrt{y}\) has just one value (which is positive).

The mathematics of the regular pentagon is important – and generally neglected. It is included here to underline the way exact expressions involving square roots arise naturally.

In Problem 3(c), parts (iii) and (vi) require one to identify similar triangles using angles. The fact that “corresponding sides are then proportional” leads to a quadratic equation – and hence to square roots.

Parts (vii) and (viii) illustrate the fact that basic tools, such as

• the trigonometric identity \(\cos^{2} \theta + \sin2 \theta = 1\)
• the Cosine Rule, and
• the Sine Rule

should be part of one’s stock-in-trade. Notice that the exact values for

\(\cos 36^{\circ}\), \(\cos 72^{\circ}\), \(\sin 36^{\circ}\), and \(\sin 72^{\circ}\)

also determine the exact values of

\(\sin{54^{\circ}} = \cos{36^{\circ}}, \sin{18^{\circ}} = \cos{72^{\circ}}, \cos{54^{\circ}} = \sin{36^{\circ}}, \cos{18^{\circ}} = \sin{72^{\circ}}\)

1.1.3: Primes

Problem 4

(a) Factorise \(12,345\) as a product of primes.

(b) Using only mental arithmetic, make a list of all prime numbers up to \(100\).

(c) (i) Find a prime number which is one less than a square.
(ii) Find another such prime.

There are \(4\) prime numbers less than \(10\); \(25\) prime numbers less than \(100\); and \(168\) prime numbers less than \(1000\).

Problem 4(c) is included to emphasize a frequently neglected message:

Words and images are part of the way we communicate.

But most of us cannot calculate with words and images.

To make use of mathematics, we must routinely translate words into symbols. For example, unknown numbers need to be represented by symbols, and points in a geometric diagram need to be properly labeled, before we can begin to calculate, and to reason, effectively.

1.1.4: Common Factors and Common Multiples

To add two fractions we need to find a common multiple, or the \(\text{LCM}\), of the two given denominators. To cancel fractions, or to simplify ratios, we need to be able to spot common factors and to find \(\text{HCF}\)s. Two positive integers \(a\), \(b\) which have no (positive) common factors other than \(1\) (that is, with \(HCF\left(a,b\right) = 1\)) are said to be relatively prime, or coprime.

Problem 5

[This problem requires a mixture of serious thought and written proof.]

(a) I choose six integers between \(10\) and \(19\) (inclusive).

(i) Prove that some pair of integers among my chosen six must be relatively prime.

(ii) Is it also true that some pair must have a common factor?

(b) I choose six integers in the nineties (from \(90–99\) inclusive).

(i) Prove that some pair among my chosen integers must be relatively prime.

(ii) Is it also true that some pair must have a common factor?

(c) I choose \(n + 1\) integers from a run of \(2n\) consecutive integers.

(i) Prove that some pair among the chosen integers must be relatively prime.

(ii) Is it also true that some pair must have a common factor?

1.1.5: The Euclidean Algorithm

School mathematics gives the impression that to find the \(\text{HCF}\) of two integers \(m\) and \(n\), one must first obtain the prime power factorizations of \(m\) and of \(n\), and can then extract the \(\text{HCF}\) from these two expressions. This is fine for beginners. But arithmetic involves unexpected subtleties. It turns out that, as the numbers get larger, factorizing integers quickly becomes extremely difficult – a difficulty that is exploited in modern encryption systems. (The limitations of any method that depends on first finding the prime power factorisation of an integer should have become clear in Problem 4(b), where it is all too easy to imagine that \(91\) is prime, and in Problem 4(c)(ii), where students regularly think that \(143\), or that \(323\) are prime.)

Hence we would like to have a simple way of finding the \(\text{HCF}\) of two integers without having to factorize each of them first. That is what the Euclidean algorithm provides. We will look at this in more detail later. Meanwhile here is a first taste.

Problem 6

(a) (i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m – n\), and of their sum \(m + n\).

(ii) Prove that

\(HCF\left(m, n\right) = HCF\left(m - n, n\right)\).

(iii) Use this to calculate in your head \(HCF\left(1001,91\right)\) without factorizing either number.

(b) (i) Prove that: \(HCF\left(m,m + 1\right) = 1\).

(ii) Find \(HCF\left(m, 2m + 1\right)\).

(iii) Find \(HCF\left(m^{2}+1, m - 1\right)\).

1.1.6: Fractions and Ratio

Problem 7

Which is bigger: \(17\%\) of nineteen million, or \(19\%\) of seventeen million?

Problem 8

(a) Evaluate

\(\left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3} \right) \left(1 + \frac{1}{4} \right) \left(1 + \frac{1}{5} \right)\)

(b) Evaluate

\(\sqrt{1 + \frac{1}{2}} \times \sqrt{1 + \frac{1}{3}} \times \sqrt{1 + \frac{1}{4}} \times \sqrt{1 + \frac{1}{5}} \times \sqrt{1 + \frac{1}{6}} \times \sqrt{1 + \frac{1}{7}}\)

(c) We write the product "\(4 \times 3 \times 2 \times 1\)" as "\(4!\)" (and we read this as "\(4 \text{ factorial}\)"). Using only pencil and paper, how quickly can you work out the number of weeks in \(10!\) seconds?

Problem 9

The “DIN A” series of paper sizes is determined by two conditions. The basic requirement is that all the DIN A rectangles are similar; the second condition is that when we fold a given size exactly in half, we get the next smaller size. Hence

• a sheet of paper of size A3 folds in half to give a sheet of size A4 – which is similar to A3; and
• a sheet of size A4 folds in half to give a sheet of size A5; etc..

(a) Find the constant ratio

\(r = “(\text{longer side length}) : (\text{shorter side length})”\)

for all DIN A paper sizes.

(b) (i) To enlarge A4 size to A3 size (e.g. on a photocopier), each length is enlarged by a factor of \(r\). What is the “enlargement factor” to get from A3 size back to A4 size?

(ii) To “enlarge” A4 size to A5 size (e.g. on a photocopier), each length is “enlarged” by a factor of \(\frac{1}{r}\). What is the enlargement factor to get from A5 size back to A4 size?

Problem 10

(a) In a sale which offers “\(15\%\) discount on all marked prices” I buy three articles: a pair of trainers priced at \(£ 57.74\), a T-shirt priced at \(£ 17.28\), and a yo-yo priced at \(£ 4.98\). Using only mental arithmetic, work out how much I should expect to pay altogether.

(b) Some retailers display prices without adding VAT – or “sales tax” – at \(20\%\) (because their main customers need to know the pre-VAT price). Suppose the prices in part (a) are the prices before adding VAT. Each price then needs to be adjusted in two ways – adding VAT and subtracting the discount. Should I add the VAT first and then work out the discount? Or should I apply the discount first and then add the VAT?

(c) Suppose the discount in part (b) is no longer \(15\%\). What level of discount would exactly cancel out the addition of VAT at \(20\%\)?

Problem 11

(a) Using only mental arithmetic:

(i) Determine which is bigger:

\(\frac{1}{2} + \frac{1}{5}\) or \(\frac{1}{3} + \frac{1}{4}\)

(ii) How is this question related to the observation that \(10 \lt 12\)?

(b) [This part will require some written calculation and analysis.]

(i) For positive real numbers \(x\), compare

\(\frac{1}{x+2} + \frac{1}{x+5}\) and \(\frac{1}{x+3} + \frac{1}{x+4}\)

(ii) What happens in part (i) if \(x\) is negative?

1.1.7: Surds

Problem 12

(a) Expand and simplify in your head:

(i) \(\left(\sqrt{2} + 1\right)^{2}\)

(ii) \(\left(\sqrt{2} - 1\right)^{2}\)

(iii) \(\left(\sqrt{2} + 1\right)^{3}\)

(b) Simplify:

(i) \(\sqrt{10 + 4\sqrt{6}}\)

(ii) \(\sqrt{5 + 2\sqrt{6}}\)

(iii) \(\sqrt{\frac{3 + \sqrt{5}}{2}}\)

(iv) \(\sqrt{10 - 2\sqrt{5}}\)

The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surds arises naturally: for example, some of the expressions in the previous problem have already featured in Problem 3(c). In particular, surds will feature whenever Pythagoras’ Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example:

• A regular octagon with side length \(1\) can be surrounded by a square of side \(\sqrt{2} + 1\) (which is also the diameter of its incircle); so the area of the regular octagon equals \(\left(\sqrt{2} + 1 \right)^{2} - 1\) (the square minus the four corners).
• \(\sqrt{2} - 1\) features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express \(\sqrt{2}\) as a “continued fraction”.
• \(\sqrt{10 - 2\sqrt{5}}\) may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting and has already featured as \(4\sin 36^{\circ}\) in Problem 3(c).
• One of the simplest ruler and compasses constructions for a regular pentagon \(ABCDE\) (see Problem 185) starts with a circle of radius \(2\), centre \(O\), and a point \(A\) on the circle, and in three steps constructs the next point \(B\) on the circle, where \(\underline{A B}\) is an edge of the inscribed regular pentagon, and

\(\underline{A B}=\sqrt{10-2 \sqrt{5}}\)