METHOD CONCEPT / EDITOR
Boom voortgezet onderwijs
AUTHORS
Benjamin del Canho
Maartje Elsinga
Jacqueline Kooiman
Gijs Langenkamp
Chantal Neijenhuis
Sibren Stienstra
Roosmarij Vanhommerig
Vera de Visser
KERN WISKUNDE
VWO
YEAR 3 PART A
BOOM VOORTGEZET ONDERWIJS
© 2022 Boom voortgezet onderwijs, Groningen, The Netherlands
Behoudens de in of krachtens de Auteurswet van 1912 gestelde uitzonderingen mag niets uit deze uitgave worden verveelvoudigd, opgeslagen in een geautomatiseerd gegevensbestand, of openbaar gemaakt, in enige vorm of op enige wijze, hetzij elek tronisch, mechanisch door fotokopieën, opnamen of enig andere manier, zonder voorafgaande schriftelijke toestemming van de uitgever.
Voor zover het maken van kopieën uit deze uitgave is toegestaan op grond van artikelen 16h t /m 16m Auteurswet 1912 jo. besluit van 27 november 2002, Stb 575, dient men de daarvoor wettelijk verschuldigde vergoeding te voldoen aan de Stichting Reprorecht te Hoofddorp (postbus 3060, 2130 kb , www.reprorecht.nl) of contact op te nemen met de uitgever voor het treffen van een rechtstreekse regeling in de zin van art. 16l, vijfde lid, Auteurswet 1912. Voor het overnemen van gedeelte(n) uit deze uitgave in bloemlezingen, readers en andere compilatiewerken (artikel 16, Auteurswet 1912) kan men zich wenden tot de Stichting PRO (Stichting Publicatie- en Reproductierechten, postbus 3060, 2130 kb Hoofddorp, www.stichting- pro.nl).
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, recording or otherwise without prior written permission of the publisher.
isbn 978 94 6442 007 4 www.boomvoortgezetonderwijs.nl
KERN Wiskunde is an RTTI-certified method, which consists of four types of questions:
r Reproduction questions
t1 Training-related application questions
t2 Transfer-related application questions
i Insight questions
See www.docentplus.nl for more information regarding the RTTI-certification.
Translation
Anne van Eeden Marieke Spijkstra Eline Wilhelm Luuk Wilhelm
Book design & cover René van der Vooren, Amsterdam
Layout & technical drawings PPMP, Wolvega
1 Numbers academy Sequences 8
1.1 Integers 12
1.2 Rational, irrational and real numbers 18
1.3 Roots 24
1.4 Number theory 30
1.5 Highest common factor and lowest common multiple 36
Test preparation 42 2 Trigonometry academy Drawing in perspective 46
2.1 Similarity 50
2.2 Right-angled triangles 56
2.3 Trigonometric ratios 62
2.4 Trigonometric calculations 68
2.5 Applications 74
Test preparation 80
3 Functions
academy Why are urban areas warmer than rural areas? 84
3.1 Functions 88
3.2 Quadratic functions 94
3.3 Different types of functions 100
3.4 Translating, stretching and compressing graphs 106
3.5 Symmetry 112
Test preparation 118 4 Statistics
academy Florence Nightingale, the lady with the lamp 122
4.1 Tables 126
4.2 Graphs, charts and diagrams 132
4.3 Grouping data 138
4.4
Measures of central tendency 144
4.5 Measures of spread 150
Test preparation 156
Appendix 160
Index 170
Numbers
In this chapter you will learn about different types of numbers and how to prove their properties. You will also learn what repeating decimals and nth roots are. In addition, you will learn how the prime factors of two numbers are related to the highest common factor and the lowest common multiple of these numbers. You will also be introduced to the Lichtenberg Ratio, the famous Euclidean algorithm and Gauss’s trick for adding up the integers from 1 to 100. Finally, you will learn that infinity plus 1 oddly enough equals infinity and how to prove that there are infinitely many prime numbers.
ACADEMY Sequences 8
1.1 Integers 12
1.2 Rational, irrational and real numbers 18
1.3 Roots 24
1.4 Number theory 30
1.5 Highest common factor and lowest common multiple 36 Test preparation 42
ACADEMY
GOAL You will learn what a sequence of integers is. You will also learn what an arithmetic sequence is and how you can calculate the sum of several terms in a sequence.
Sequences
Sequence A sequence of numbers is a list of several numbers in a specific order. The sequence (5, 10, 15, 20, 25, 30) consists of the first six positive multiples of 5. A sequence can consist of infinitely many numbers. For example, the sequence (2, 4, 6, 8, …) consists of all even positive integers. The ellipsis (the three dots) shows that the sequence continues indefinitely. Sequences are represented by a lowercase letter, for example (a n). The subscript n shows that the sequence consists of several elements or terms. a1 is the first term, a2 is the second term, and so on. If a sequence is regular, it is sometimes possible to set up a formula for the sequence. For example, the first term a1 of the sequence (a n) = (2, 4, 6, 8, …) is 2, the second term a2 is 4 and the n th-term a n is equal to 2n. The n th-term of the sequence (b n) = (1, 3, 9, 19, 33, …) is equal to b n = 2(n – 1)2 + 1. The use of a formula makes it easier to calculate the numbers of a sequence. However, determining the formula that corresponds to a given sequence is often difficult. Consider the sequence (b n) above. The formula that is used to calculate the numbers in the sequence cannot be determined easily.
Arithmetic sequence In the sequence of all even numbers, the difference between two consecutive terms is always equal to two. This type of sequence, where the difference between two consecutive terms is always the same, is called an arithmetic sequence. It is always possible to set up a formula to determine the terms of an arithmetic sequence. For example, the sequence (c n) = (5, 11, 17, 23, 29, …) is an arithmetic sequence, because the difference between two consecutive terms is always 6. The formula that corresponds to this sequence is c n = 5 + 6(n – 1). Note: the difference 6 is multiplied by n – 1 rather than n, because the sequence starts at n = 1 and not n = 0. The formula can be rearranged to c n = 6n – 1.
Summation formula The story goes that over 200 years ago, a German teacher wanted to keep his students busy for a while. To do this, he asked them to add up all the numbers from 1 to 100. To his surprise, one of his students, Carl Friedrich Gauss (1777–1855), gave the answer almost immediately: 5050. He did not find the answer by adding up all the numbers one by one, but he understood that adding up the ‘opposite’ numbers, always gives 101:
1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, …, 99 + 2 = 101, 100 + 1 = 101. This is the same as writing down the summation twice and reversing the order the second time:
1 + 2 + 3 + … + 99 + 100
100 + 99 + 98 + … + 2 + 1 +
101 + 101 + 101 + … + 101 + 101
The result of the summation below the line is 100 · 101 = 10 100. All the numbers from 1 to 100 are added up twice, so the result has to be divided by two. Therefore the result is 1 2 × 100 × 101 = 5050.
The following is true for the sum of the first n positive integers:
The result of the summation below the line is n(n + 1). The sum has to be divided by two:
1 + 2 + 3 + … + n = 1 2 n(n + 1)
Gauss’s method can be used for any arithmetic sequence. The sum of the first n terms can be calculated with the formula: sum = 1 2 n(first term + n th term). This formula is called the summation formula for an arithmetic sequence.
Portrait of Carl Friedrich Gauss made in 1840 by the Danish portrait painter christian albrecht jensen . Gauss was a German mathematician and physicist. He is regarded as one of the greatest mathematicians since classical antiquity.
Questions about the text
1 The arithmetic sequence (a n) = (1, 4, 7, 10, 13, …) is given.
a What is an arithmetic sequence? R
b Write down the next two terms of this sequence. T1
c Set up a formula to calculate the n th term of this sequence. T2
d Write down the hundredth term of this sequence. T1
2 The following three sequences are given: T1
(a n) where a n = 4n − 7
(b n) where b n = 2n2 + 5
(c n) where c n = (–1) n · n + 2
a Calculate the first five terms of each sequence.
b Calculate a20, b7 and c100.
c For each sequence, explain whether it is an arithmetic sequence.
3 a Calculate the sum of all integers from 1 through 50. T1
b Calculate the sum of all integers from 1 through 999. T2
c Calculate the sum of all integers from 500 through 999. T2
4 Consider the arithmetic sequence (b n) = (6, 13, 20, 27, 34, 41, …).
a Set up a formula to calculate the n th term of this sequence. T2
b Calculate the sum of the first fifty terms of this sequence. T2
c Calculate the sum of the 42nd term through the 80 th term of this sequence. I
In–depth questions
5 Read the text The theatre of Epidaurus on the next page.
The first ring of the theatre seats 80 spectators. Every next ring seats approximately six people more. T2
a Approximately how many seats are there on the twentieth ring?
b Use a calculation to show that this theatre seats approximately thirteen thousand spectators.
c How many seats were added in the second century BC?
Research assignment
6 Three different sequences of figures are shown below. For every sequence, calculate the number of dots in the hundredth figure. I
The theatre of e pidaurus
In the Greek city Epidaurus, a famous 2000-years-old outdoor theatre can be found. Various arts were performed in the theatre, including music, dance and theatrical performances. At first, there were 34 rings with seats, but in the second century BC another 21 rings were added. 21 and 34 are consecutive numbers in Fibonacci’s famous sequence.
The ratio 34 : 21 ≈ 1.619 approximates the golden ratio φ quite well. This ratio can be found in unexpected places in mathematics and was already known to the Greeks. Therefore, the number of rings in the theatre may not have been randomly chosen.
The theatre was built in a semicircular shape and seats about 12 000 to 14 000 spectators. The theatre’s acoustics are amazing. Its construction allows for sounds from the stage to be strengthened, while sounds from outside the theatre are filtered away. If you drop a coin in the middle of the stage, it can be heard everywhere. Guides giving tours of the theatre love to demonstrate this phenomenon.
Have you reached your goal?
R I know what a(n) (arithmetic) sequence is and I know the summation formula for an arithmetic sequence.
T1 I can calculate the terms of a sequence using a formula. I can also recognise an arithmetic sequence.
T2 I can set up a formula for an arithmetic sequence. I can also calculate the sum of the terms of an arithmetic sequence.
I I can calculate the number of dots in figures with a regular pattern.
VOCA bu LARY sequence
term arithmetic sequence summation formula
The theatre of Epidaurus, Greece. Ancient tragedies and comedies are still performed at the theatre during summer.
Integers
GOAL You will learn what natural numbers and integers are and how to find the prime factorisation of a number.
Sets A set is a collection of numbers. Numbers that belong to a set often have special properties. A = {1, 2, 4, 8} is the set of all divisors of 8 and B = {5, 10, 15, …} is the set of all positive multiples of 5. The ellipsis (three dots) indicates that set B is infinite. The numbers that are part of a set are called the elements of the set. For example, the number 2 is an element of A and the number 7 is not. This is written as 2 ∈ A (2 is in A) and 7 ∉ A (7 is not in A).
Natural numbers and integers The numbers 0, 1, 2, 3, … are called natural numbers. ℕ is used to refer to the set of all natural numbers*.
The numbers …, –3, –2, –1, 0, 1, 2, 3, … are called integers. For example, –8 and 2 are integers. 2 is also a natural number, but –8 is not. ℤ is used to refer to the set of all integers. ℕ consists of the number 0 and all positive integers 1, 2, 3, … . Therefore, ℤ is equal to ℕ supplemented with all negative integers. As every element of ℕ is also an element of ℤ, ℕ is a subset of ℤ. This is written as ℕ ⊂ ℤ. The Venn diagram on the right shows the relationship between set ℕ and set ℤ. In this Venn diagram you can see that 4 ∈ ℕ, 4 ∈ ℤ, −8 ∉ ℕ and −8 ∈ ℤ.
e xamples
1 5 is an integer and a natural number.
2 –100 is an integer, but not a natural number.
3 (–1)4 gives a natural number, because (–1)4 = 1.
4 (–1)5 gives an integer, but not a natural number, because (–1)5 = –1.
5 11 2 is not an integer, nor is it a natural number, because 11 2 = 1.5.
ℕ = {0, 1, 2, 3, 4, … }
ℤ = {…, −3, −2, −1, 0, 1, 2, 3, …} 0 4 150 6 –8 –1 –100
* Sometimes 0 is excluded from the set of natural numbers. In this book 0 is always included in ℕ
Sets
7 Set A = {0, 3, 6, 9, 12} is given. T1
a How many elements does set A have?
b Copy and fill in ∈ or ∉ 10 A 6 A −3 A
8 Consider the set B of all positive multiples of 8 that are less than 100. T1
a Write down all elements of set B between curly brackets { }.
b How many elements does set B have?
c Copy and fill in ∈ or ∉.
−40 … B 42 … B 96 … B
9 Consider the set C of all multiples of 3 between –100 and 100. T1
a How many elements does set C have?
b Copy and fill in ∈ or ∉
−72 C 56 C 162 C
10 Write down all elements of the following sets between curly brackets. T1
a The set D of all positive divisors of 18.
b The set M of all positive multiples of 13 that are less than 100.
Natural numbers and integers
11 a What is a natural number? R
b Which symbols are used to refer to the set of all natural numbers and the set of all integers? R
c Draw a number line from 0 to 10 and indicate the location of the natural numbers. T1
12 Which of the following numbers are natural numbers? And which are integers? T1
13 Which elements of the set { √1 , √2 , √3 , √4 , …, √100 } are also elements of ℕ? T2
14 A Venn diagram is shown below.
a Explain how this diagram shows that ℕ ⊂ ℤ T1
b Copy the diagram and write the following numbers in the correct area. T2
Note: write the numbers that are not an element of ℕ and/or ℤ outside the diagram.
INTEGERS
Prime factorisation An integer greater than one with exactly two positive divisors, 1 and itself, is called a prime number. There is an infinite number of prime numbers. The first five prime numbers are 2, 3, 5, 7 and 11.
Every integer greater than 1 can be written as a product of prime numbers. For instance, 15 can be written as 3 5. Finding these prime numbers is called prime factorisation. In this case, 3 and 5 are the prime factors. When a number is a prime number, there is only one prime factor. You can find a prime factorisation by systematically finding larger prime factors of a number. An example for the number 60 is shown on the right.
The fundamental theorem of arithmetic (also called the unique factorisation theorem) states that every integer greater than 1 can be written as a product of prime factors in just one way, not taking into account the order of the factors.
e xamples
Give the prime factorisation of the numbers 40 and 99.
Prime factorisation
15 a What is the definition of a prime number? R
b Give the first five prime numbers. R
c Why is 25 not a prime number? T1
d Give all prime numbers between 20 and 30. T1
16 In this exercise you will find the prime factorisation of 150. T1
a Copy and fill in the diagram shown below. 150
17 a Write down the set of all positive divisors of 24. T1
b Give the prime factorisation of 24. T1
18 Give the prime factorisation of the following numbers. T1
a 38 c 256
b 99 d 700
19 Simplify the fraction 525 6000 by first finding the prime factorisation of both its numerator and its denominator. T2
b Copy and fill in the blanks.
150 = 2
EXERCISES — EXPLORING & INVESTIGATING
exploring
20 The sets A = {1, 3, 5, 7, …, 99} and B = {0, 5, 10, 15, …} are given. T2
a How many elements does set A have?
How about set B?
b Find the intersection of A and B
The intersection is the set of all elements that are present in both A and B.
c How many elements does the union of A and B have? The union is the set of all elements that are present in A and/or B.
d How many elements are in the set that consists of the numbers that are not an element of B, but are an element of A?
21 a Give the prime factorisation of 210. T1
b Determine the number of positive divisors of 210 by using its prime factorisation. T2
22 a Give the fundamental theorem of arithmetic. R
b One of the reasons that the number 1 is not considered a prime number, is that the fundamental theorem of arithmetic would no longer be valid if it were. Show this with a concrete example. I
Investigating
23 Read the text Hilbert’s hotel on the next page. T2
a Suppose a bus with one hundred new guests arrives at Hilbert’s Hotel. The receptionist makes a request to the current guests. If they respond positively, each new guest can be accommodated. What is the receptionist’s request?
b Suppose a bus with an infinite number of new guests arrives at Hilbert’s Hotel. The receptionist makes a request to the current guests. If they respond positively, each new guest can be accommodated. What is the receptionist’s request?
24 Use Hilbert’s Hotel and a bus with new guests to explain why the sets below contain the same number of elements. I
A ℕ and {1, 2, 3, 4, …}
B ℕ and {1, 3, 5, 7, …}
C ℕ and ℤ
Hint : One of the two sets is Hilbert’s Hotel before the bus arrives, while the other set is Hilbert’s Hotel after the bus arrives.
CO m P u TATIONAL SKILLS
C1 There are 7351 rooms in the First World Hotel in Pahang, Malaysia.
I Round the number of rooms to the nearest hundred.
II Round the number of rooms to the nearest thousand.
III Another hotel has 4700 rooms, rounded to the nearest hundred. So there are fewer rooms in this hotel than in the First World Hotel. How many fewer? Give the minimum as well as the maximum number.
Hilbert’s Hotel
In 2008 the First World Hotel in Pahang, Malaysia, lost its title of largest hotel in the world. To regain the title, a new hotel wing was built. Since 2015, the First World Hotel is once again the largest in the world, with 7351 rooms. However, not for much longer; at the time this book was written a larger hotel, called the Abraj Kudai, was under construction in Mecca, Saudi Arabia.
In 1924, the German mathematician David Hilbert (1862–1943) gave a famous speech in which he described a thought experiment about a hotel with an infinite number of rooms. He wondered what would happen if a new guest were to arrive, while all the rooms were occupied. At first you might think that this new guest could not be accommodated. However, mathematics with infinite numbers is not the same as mathematics with normal numbers. If the receptionist of this hotel with infinite rooms asked every single guest to move from their current room to the next room, the first room would be free to use. Therefore, there are always free rooms in Hilbert’s hotel, even when there are no available rooms.
With this thought experiment, Hilbert sought to create a better understanding of the question whether infinity plus 1 is more than infinity Everyone understands that a hotel with 7352 rooms has more rooms than a hotel with 7351. But does that mean that a hotel with an infinite number of rooms plus 1 has more rooms than a hotel with an infinite number of rooms? Or is infinity plus 1 equal to infinity?
Have you reached your goal?
R I know what natural numbers and integers are. I also know what prime numbers are and I know the fundamental theorem of arithmetic.
T1 I can check if an element belongs to a specific set in simple cases. I can find the prime factorisation of a natural number.
T2 I can check if an element belongs to a specific set in more complex cases.
I I can think about and give reasoned responses to questions about sets.
VOCA bu LARY set element natural number integer subset Venn diagram prime number prime factorisation prime factor fundamental theorem of arithmetic
The First World Hotel in Pahang, Malaysia
Rational, irrational and real numbers
GOAL You will learn what rational, irrational and real numbers are.
Rational numbers A number that can be expressed as a division or a quotient of two integers is called a rational number. Therefore a rational number can be written as p q for all p, q ∈ ℤ and q ≠ 0. ℚ is used to refer to the set of all rational numbers. Every integer n can be written as a quotient n 1, so integers are also rational numbers. This means that the set of integers is a subset of the set of rational numbers: ℤ ⊂ ℚ
When you write a rational number p q as a decimal number, there are two possibilities (see also page 23):
1 The number of decimals is finite. Examples of this are 1 10 = 0.1 and 3 4 = 0.75. In this case, the denominator q is a power of 10 or p q can be rewritten as a fraction with a denominator which is a power of 10. These fractions are called terminating decimals. For example, 345 1000 = 0.345 and 11 8 = 11 125 8 125 = 1375 1000 = 1.375.
2 The number of decimals is infinite with a repeating pattern. Examples of this are 1 3 = 0.333… and 3 11 = 0.272727… . This is the case if p q cannot be rewritten as a fraction with a denominator which is a power of 10. These fractions are called repeating decimals. To make repeating decimals easier to write, a bar or dot can be placed over the repeating pattern: 1 3 = 0.3 (bar notation) or 0.3 (dot notation) and 3 11 = 0.27 or 0.27 ˙ Note: the repeating pattern does not always start directly after the decimal point. For example, 5 6 = 0.83333333… = 0.83
In reverse, any decimal number with a finite number of decimals or with an infinitely repeating pattern after the decimal point is a rational number.
e xamples
1 0.125 is a rational number, because 0.125 = 1 8
2 8.231 is a rational number, because 8.231 = 8231 1000
3 2 is a rational number, because 2 = 2 1
4 –61 7 is a rational number, because –61 7 = 43 7
5 √1 4 = 1 2, so √1 4 is a rational number. 1 5 = 1 2 5 2 = 2 10 = 0.2
1 7 = 0.142857142857… = 0.142857
22 15 = 1.4666… = 1.46
EXERCISES — PRACTICE
Rational numbers
25 a What is the definition of a rational number? R
b What symbol is used to refer to the set of all rational numbers? R
c How do you know if a decimal number with one or more decimals is a rational number? R
26 Which of the following numbers are elements of ℚ? T1
a 0.5 e π = 3.1415926535…
b 0.3333333… f −11 2
c 1.25 g 3.1415926535
d 5 7 h √2 = 1.4142135623…
27 Perform the calculations below and indicate which of the results is an element of ℚ but not of ℤ. T2
a 2 1 3 − 6 2 3 c 5 7 · 2 1 6
b 13 8 + 5 1 4 − 8 5 8 d 1 5 : 1 25
28 Copy and fill in the blanks. T1
a 5.9 = 10
b 1.12 = 1 100 =
c 0.105 = 1000 =
d 6.0082 = 6 10 000 =
29 Copy and fill in the blanks. T1
a 3 5 = 10 = 0.
b 17 20 = 100 = 0.
c 69 250 = 1000 = 0.
d 1554 5000 = 10 000 = 0.
30 a Write the following numbers as a fraction without using a calculator. T1
1.5 / 0.25 / −6.9 / 0.375
b Write the following fractions as a decimal number without using a calculator. T1
5 8 / 4 25 / 339 500 / 199 1250
31 Give the first eight decimals of the number 0.17828. T1
32 Which of the following fractions is equal to the number 3.416? T2
Irrational numbers and real numbers Some numbers, such as √2 and π, cannot be written as a quotient of two integers (see exercise 90). Thus, there is no quotient of two integers whose outcome is equal to √2 or π. These numbers are called irrational numbers. Irrational numbers can be written as decimal numbers. For example, √2 = 1.414213… and π = 3.141592… . The difference with rational numbers is that there is no infinitely repeating pattern after the decimal point, no matter how many decimals you consider. Therefore, when written as a decimal number, irrational numbers always have an ellipsis at the end or they are rounded. There are many irrational numbers. For example, every square root of a natural number that is not a perfect square is an irrational number. Together, the rational and irrational numbers are called the real numbers. Every number that can be indicated on a number line is a real number.
Rational numbers are numbers that can be written as a fraction. Irrational numbers are numbers that cannot be written as a fraction.
The orange numbers are the irrational numbers.
ℝ is used to refer to the set of all real numbers. The set of rational numbers is a subset of the set of real numbers: ℚ ⊂ ℝ
Irrational numbers and real numbers
33 a What is the definition of an irrational number? R
b Which symbol is used to refer to the set of all real numbers? R
c How do you know if a decimal number is an irrational number? R
34 a What is the definition of a real number? R
b Draw a number line from –1 to 2 and use arrows to indicate five different rational numbers. T1
c Use arrows to approximately indicate three irrational numbers on the same number line. Write the numbers next to the arrows. T1
35 For each of the numbers below, explain whether it is a rational or an irrational number. T1
37 Copy and fill in ∈ or ∉ T1 a b c −2 ℚ 0 ℤ 0.78 ℤ
38 Copy the Venn diagram and write the following numbers in the correct area. T2
36 Copy the table. In each of the cells, put a tick if the number belongs to the set on the left and a cross if the number does not belong to the set on the left. T1
39 For each of the following square roots, explain whether it is a rational or an irrational number. T2 a √49 d √144 b √68 e √250 c √94 f √10 000
exploring
40 Give the hundredth decimal of the number 0.82365 T2
41 The line that corresponds to the formula y = ax is given, where a ∈ ℝ. Depending on the value of a , the line passes through one or more gridpoints.
a The line passes through one specific gridpoint for any value of a. Give the coordinates of this gridpoint. T2
b Find two other gridpoints the line passes through if a = 2 5. T2
c Explain how many gridpoints the line passes through if a = √2 . I
42 Which of the following numbers are rational and which are irrational? Explain your answer. I
a 0.123456789101112131415…
b 8.76123123123123123123… c 1.121221222122221222221…
Investigating
43 Read the text Repeating decimals on the next page. Perform the following calculations using long division. T1
a 204 : 12 c 795 : 6
b 966 : 7 d 100 : 3
44 Write the following fractions as a decimal number without using a calculator. T2
a 4 15 c 5 22
b 10 37 d 22 41
45 Consider the number A = 0.18. T2
a Prove that 100A – A = 18.
b Show that from this, it follows that A = 2 11.
c All decimal numbers with a repeating pattern after the decimal point are rational numbers. Explain why.
46 Write the following numbers as a fraction. I
a 0.76 c 0.9
b 0.723 d 0.231
CO m P u TATIONAL SKILLS
C2 What is the total amount owed for the products below?
product quantity price per product wall paint per 10 litres 1 ₤ 28.69 laminate per package 12 ₤ 12.95
skirting boards per 250 cm 9 ₤ 4.52
Repeating decimals
A fraction such as 23 54 can be converted to a decimal number by performing the division 23 : 54. The image on the right shows how to use long division to divide these numbers.
230 : 54 = 4 r14 (tenths)
140 : 54 = 2 r32 (hundredths)
320 : 54 = 5 r50 (thousandths)
500 : 54 = 9 r14 (ten thousandths)
140 : 54 = 2 r32 (hundred thousandths)
320 : 54 = 5 r50 (millionths)
500 : 54 = 9 r14 (ten millionths) And so on.
The calculation above repeats itself every time the remainder is 14. Therefore the decimals 2, 5 and 9 also repeat themselves. Thus, 23 54 = 0.4259.
A similar calculation can be performed for every fraction. There are two possible outcomes:
The division eventually leads to a result without a remainder. In this case, the fraction can be written as a terminating decimal.
The division leads to a remainder that has previously appeared in the calculation. A division of integers has a limited number of possible remainders, so this will happen sooner or later. In this case the fraction can be written as a repeating decimal.
54/2 3.000000\0. 42592 ...
Have you reached your goal?
R I know what rational, irrational and real numbers are. I also know what terminating and repeating decimals are.
T1 I can identify to which set a number belongs. I can also write fractions as decimal numbers.
T2 I can put numbers in a Venn diagram and I can think about and give reasoned responses to questions about rational and real numbers.
I I can write decimal numbers as fractions.
VOCA bu LARY
quotient rational number terminating decimal repeating decimal irrational number real number long division remainder
GOAL You will learn how to perform calculations with square roots. You will also learn what nth roots are and how to perform calculations with them.
Performing calculations with square roots The square root of a number a is equal to the non–negative number whose square is a For example, √36 = 6, because 62 = 36. The square root is also called the second root or root of degree 2. Taking a square root is the opposite of squaring. Like radical terms are terms with the same radicals. Like radical terms can be added or subtracted. For example,
2 √3 + 4 √3 = 6 √3
Square roots can also be multiplied or divided. The following rules apply: √a √b = √ab and √a √b = √a b .
e xamples
1
2
e xamples 1
Simplifying square roots Some square roots can be simplified.
When the radicand is divisible by a perfect square, you can take out the square root of that perfect square and place it in front of the radical.
When the radicand is a fraction, you can use a multiplication to place the modified fraction in front of the radical.
When the denominator of a fraction contains a square root, you can use a multiplication to move the square root to the numerator. Always simplify square roots as much as possible.
e xamples
1
EXERCISES — PRACTICE
Performing calculations with square roots
47 Calculate without using a calculator. T1
a √49 c √144 b √81 d √400
48 Copy and fill in the blanks. T1
Simplifying square roots
52 Simplify. Do not use a calculator. T1 a √20 c √176 b √99 d √1000
53 Calculate the lengths of sides AC and EF of the two triangles shown below. Simplify as much as possible. T2
49 a What is the length of the sides of a square with an area of 196 cm 2? T2
b What is the length of the sides of a square with an area of 200 cm 2? Round your answer to the nearest hundredth of a cm. T2
50 Calculate without using a calculator. T1
a 8 √5 + 7 √5 − 6 √5 b 10 √10 − 4 √10 + 10
c 3 − 4 √2 − 1 + 5 √2
d √7 + 3 √3 + 9 √7 − 4 √3
51 Calculate without using a calculator. T1 a √2 √5 d √15 √3
b 2 √5 8 √7 e 9 √14 3 √2
c 6 √3 3 √3 f 4 √8 √2
54 Simplify. Do not use a calculator. T2
55 Simplify. Make sure there is no square root in the denominator. Do not use a calculator. T2
n th roots It is also possible to find the nth root of a number. The n th root is also called the root of degree n. The n th root of a number raised to the power of n equals that number. For example, if you want to determine the length of the edges of a cube with a volume of 8, you want to find the number that equals 8 when raised to the power of three. This number is the third root, or cube root, of 8 and is written as 3√8 . Since 23 = 8, 3√8 = 2.
The n in n th root can be any number. For example, 4√81 is the number that equals 81 when raised to the power of four. 3 4 = 81, so 4√81 = 3.
Make sure to note the following when working with n th roots:
It is not possible to extract an even root from a negative number. For instance, 4√ 16 does not exist, because a number raised to the power of four is always greater than or equal to 0.
It is possible to extract an odd root from a negative number. For instance, 3√ 8 = –2, because (–2)3 = 8.
Even roots are always positive. For instance, 4√81 = 3, even though 3 4 = 81 and (–3) 4 = 81. This agreement assures that it is clear which number 4√81 equals.
e xamples
1 3√27 = 3, because 33 = 27
2 4√256 = 4, because 44 = 256
Approximating n th roots Many square roots cannot be written as a fraction and are therefore irrational. This also applies to many n th roots, such as 3√100 . The number 3√100 is greater than 3√64 = 4 and smaller than 3√125 = 5. Therefore 4 < 3√100 < 5. You can find a more accurate approximation using your calculator: 3√100 = 4.6415…, so 3√100 ≈ 4.64.
EXERCISES
nth roots
56 Copy and fill in the blanks. T1
a 3√27 = 3, because 3 =
b 4√625 = , because 4 =
c 3√ = −4, because 3 =
57 Calculate without using a calculator. T1
a 4√81 d 3√1 8
b 5√32 e 3√0.001
c 3√−125 f 999√−1
58 a Explain why 3√ 64 exists, but 4√ 256 does not. T2
b Explain why 4√256 = 4, but 4√256 ≠ − 4, even though 44 and ( 4)4 are both equal to 256. T2
59 What is the length of the edges of a cube with a volume of 216 cm 3? T1
60 Calculate if possible. If not possible, explain why. Do not use a calculator. T2
a 8 − 5 3√27 e 5√95 − 8√(−4)8
b 6 4√81 − √81 f 3 4√19 − 102
c 4√−256 + 2 √100 g 3 − 2 6√82 d 3
Approximating nth roots
61 Calculate. Round your answer to the nearest thousandth. T1
a 3√100 d 3√−35
b 4√2 e 10√1002
c 7√−98 f 3√6.89
62 Order the following numbers from smallest to largest. Do not use a calculator. T2 3√30 / 5√1200 / 4√40 / √64 / 3√−20 / 7√−100
63 What is the length of the edges of a cube with a volume of 5000 cm 3? Round your answer to the nearest cm. T1
64 The cube of a number is that number raised to the power of three. T2
a Explain why the third power of a number is called its cube.
b Give the first five cubes of natural numbers.
c Which of the following numbers are cubes of natural numbers? Do not use a calculator.
243 / 1000 / 2197 / 10 000
exploring
65 A house has a volume of 536 m 3. A scale model of this house has a volume of 67 m 3 What scale is used for this model? T2
66 Do not use a calculator.
a Show that √√2 = 4√2 . T2
b Show that 6√102 = 3√10 . T2
c Which number is larger: √2 or (5√2 )2 ? I
67 A stack of 25 cubes is shown below. Each cube has a volume of 8 cm 3 .
a Draw the full-scale front elevation and the full-scale right elevation of the stack of cubes. T2
b Explain whether 25 cubes can be stacked to form a larger cube. I
c Explain whether 343 cubes can be stacked to form a larger cube. I
Investigating
68 Read the text Paper sizes on the next page. T2
a The dimensions of an A0 sheet of paper are shown on the next page. Show that these dimensions conform to the Lichtenberg Ratio.
b Calculate the dimensions of an A5 sheet of paper.
c Calculate the area of an A11 stamp. Round your answer to the nearest cm 2 .
69 A sheet of paper conforms to the Lichtenberg Ratio. When you cut this sheet of paper into two equal parts along its width, you get two sheets of paper where the ratio of length to width is 1 : 1 2 √2 . Explain why, and show that these sheets of paper also conform to the Lichtenberg Ratio. I
CO m P u TATIONAL SKILLS
C3 A motorcycle racer completed 20 rounds on a 3.8 km circuit in 24 minutes.
I What was the motorcycle racer’s average speed in km/h?
II A race car driver completed 15 rounds on the same circuit with an average speed of 210 km/h. How long did it take them to complete those 15 rounds? Give your answer in minutes and seconds.
Paper sizes
Worldwide, there are many different standards for paper sizes. In the Netherlands, the A series is generally used. The A4 size paper is part of the A series.
The largest A series size is A0. An A0 sheet of paper has dimensions of 1189 mm by 841 mm. The area of such a sheet of paper is exactly 1 m 2 . Each subsequent size in the A series is made by folding the larger sheet in half with the crease parallel to the shortest sides (see the figure on the right). If necessary, the length in mm is rounded down. For example, an A1 sheet of paper is 841 mm by 594 mm. The area of this sheet of paper is 0.5 m 2. The A11 size is the smallest sheet of paper in use. Some stamps are this size.
A series sheets of paper always have a length to width ratio of √2 : 1. This ratio is called the Lichtenberg Ratio. As the ratio of length to width is always the same, images can be enlarged or reduced to a different size in the A series without parts being cut off; nor is it necessary to stretch or compress the image.
VOCA bu LARY
square root
second root taking the square root
extracting the square root like radical terms
nth root root of degree n
third root
cube root
Lichtenberg Ratio
The A series paper sizes
A1 A0 A2
Have you reached your goal?
R I know what like radical terms are and I know what nth roots are.
T1 I can add, subtract, multiply and divide square roots. I can also calculate nth roots.
T2 I can perform calculations with nth roots. I can also order nth roots from smallest to largest without using my calculator.
I I can show that the dimensions of paper sizes in the A series conform to the Lichtenberg Ratio.
Number theory
GOAL You will learn how you can prove properties of integers and how you can use a counterexample to disprove a statement.
Proofs Suppose someone claims that the sum of two odd numbers is always even. This statement is easy to check using any two odd numbers as an example, such as 3 + 5 = 8 and 19 + 27 = 46. However, just because the statement is true for these pairs of odd numbers does not mean it is true for every pair of odd numbers. You can only know this for certain if you can find an argument that shows the statement is always true. Such an argument is called a proof. A proof is based on other statements that have been proved previously or are true by definition.
e xample
To prove that the sum of two odd numbers is always even, you use the following properties of even and odd numbers. Even numbers are divisible by 2 and can therefore be written as 2n, where n ∈ ℕ. Odd numbers are not divisible by 2 and can therefore be written as 2m + 1, where m ∈ ℕ. For example, 8 = 4 2 and 9 = 4 2 + 1.
The sum of two odd numbers can be written as 2m1 + 1 + 2m2 + 1, where m1, m2 ∈ ℕ. This is equal to 2m1 + 2m2 + 2 = 2(m1 + m2 + 1). The resulting number can be written as 2n, where n = m1 + m2 + 1, which means it is an even number.
In the example above, some of the properties of even and odd numbers are used. By the same logic, numbers which are divisible by 3 can be written as 3n. Numbers which are not divisible by 3 will have a remainder when divided by 3. These numbers can be written as 2n + 1 if the remainder is 1 and 2n + 2 if the remainder is 2. Numbers which are divisible by 4 can be written as 4n, and so on. These properties follow directly from the definition of divisibility and are useful when proving statements about integers.
Number theory is a branch of mathematics that studies the properties of natural numbers and integers. In this section, you may assume that the word ‘number’ always refers to an integer.
EXERCISES — PRACTICE
Proofs
70 a Prove that the square of an even number is always even. T1
b Prove that the square of an odd number is always odd. T1
c Prove that the product of two odd numbers is always odd. T1
71 It is given that 98 23 251 = 2 278 598. Show that 7 is a divisor of 2 278 598 without using a calculator. T1
72 Calculate. Include the remainder in your answer. Do not use a calculator. T1
a 28 : 3 c 225 : 13 b 127 : 5 d 79 813 : 223
73 A natural number that is divisible by 5 can be written as 5n, where n ∈ ℕ. A natural number that is not divisible by 5 can be written as either 5n + 1 or 5n + 2 or 5n + 3 or 5n + 4, where n ∈ ℕ. Write the following numbers in one of these five ways. T1
a 75 c 378 b 87 d 629
74 Consider a natural number n that is divisible by 3. T1
a Explain that n can be written as 3m Which set is m in?
b Prove that n2 is divisible by 9.
c Prove that n 3 is divisible by 27.
75 Consider a natural number n which is not divisible by 3.
Therefore, n ∈ {1, 2, 4, 5, 7, 8, …}. T2
a Explain that n can be written as n = 3q + 1 or as n = 3q + 2, where q ∈ ℕ.
b Can n2 be written as 3q + 1 or 3q + 2?
c Prove that n2 – 1 is divisible by 3.
76 a Assume n is an odd number. Prove that n2 – 2 is odd. T2
b Assume n is a natural number. Prove that n 3 + n is even. T2
Hint : start with a proof for even values of n Then do the same for odd values of n
77 Assume a is an odd number. T2
a Prove that either a + 1 or a – 1 is a multiple of 4.
b Prove that a2 – 1 is divisible by 8. Hint : factorise a2 – 1.
Counterexample Not all statements are true. It is not possible to prove that a statement is false. However, you can disprove a statement by giving a counterexample. A counterexample is an example which shows that the statement is false. When you give a counterexample, you contradict the statement.
e xamples
1 Somebody claims that the result of the formula P = n2 − n + 41 is a prime number for every n ∈ ℕ. n = 0 and n = 1 give P = 41, n = 2 gives P = 43 and n = 3 gives P = 47, which are all prime numbers. Some larger values of n also give prime numbers. However, n = 41 gives a result of P = 412 − 41 + 41 = 412 = 1681, which is not a prime number, because 412 is divisible by 41. Therefore n = 41 is a counterexample, which proves that the result of the formula is not always a prime number.
2 Prove or give a counterexample.
The sum of three odd numbers is always divisible by 3.
The sum of 3, 5 and 9 is 17, which is not divisible by 3. The three odd numbers 3, 5 and 9 are a counterexample. Therefore the statement is false.
EXERCISES — PRACTICE
Counterexample
78 Prove or give a counterexample. T1
a An even number divided by an even number is always odd.
b The sum of two natural numbers, neither of which is a prime number, is never a prime number.
c The only even prime number is 2.
79 Prove or give a counterexample.
a If 4 and 6 are divisors of an integer, then 24 is also a divisor of that integer. T1
b If 15 and 8 are divisors of an integer, then 120 is also a divisor of that integer. T2
80 Prove or give a counterexample. T2
a An integer is divisible by 4 if its last digit is divisible by 4.
b An integer is divisible by 4 if its last two digits are both divisible by 4.
c An integer n is divisible by 4 if the number m formed by the last two digits of n is divisible by 4.
81 Prove or give a counterexample. T2
The square of a number that is not equal to 0 is always greater than that number.
82 Prove or give a counterexample. T2
Every prime number can be written as 2 n – 1, where n ∈ ℕ.
83 Two integers n and m are given. Prove or give a counterexample. T2
a If mn < 0, then m < 0 or n < 0.
b If mn > 0, then m > 0 and n > 0.
c If m2 < n2, then m < n.
d If m < n, then m n < 1.
84 Prove or give a counterexample. T2
If p is a prime number, then p – 2 or p + 2 is also a prime number.
85 Prove or give a counterexample. T2
The number of integers between two consecutive prime numbers equals zero only if the prime numbers are 2 and 3.
exploring
86 Prove or give a counterexample. T2
a If d is a divisor of a and b, then d is also a divisor of a + b
b If d is a divisor of a and a is a divisor of b, then d is also a divisor of b
c If d is a divisor of ab, then d is a divisor of a and/or b.
87 a How many prime factors does the number 130 have? How about the number 1302 = 16 900? T1
Hint : The number 72 = 23 · 32 has five prime factors, not two.
b Prove that the number a2, where a ∈ ℕ, has an even number of prime factors. T2
88 Prove or give a counterexample. I
The sum of three consecutive numbers is always divisible by 3.
Investigating
89 Read the text Prime numbers on the next page. Consider the number 5041 = 7 6 5 4 3 2 1 + 1. T2
a Explain why 5041 is divisible by a prime number greater than 7.
b Show that 5041 is divisible by 71.
c Give the prime factorisation of 5041.
90 In this exercise you will prove that √2 is an irrational number. To prove this you start with stating the opposite: assume √2 is a rational number. This means that there are two integers a and b, for which √2 = a b. I
a Show that this leads to 2b2 = a2 .
b Why are there no numbers a and b that satisfy this equation?
Hint : take another look at exercise 87b.
c Explain that from part b it follows that √2 cannot be written as a fraction, which means that √2 is an irrational number.
d Prove in a similar way tat √3 and 4√5 are irrational numbers.
CO m P u TATIONAL SKILLS
C4 A unit fraction is a fraction where the numerator is 1 and the denominator is a positive integer. For example, 1 5 and 1 7 are unit fractions.
I Write 3 4 as the sum of different unit fractions.
II Write 1 2 as the sum of different unit fractions.
III Give the largest possible unit fraction that is smaller than 4 27
Prime numbers
In December 2018 a prime number with almost 25 million digits was found. In 2021, this was the largest known prime number, found by a collaborative project of thousands of volunteers who made the computing power of their computers available to search for large prime numbers. It is only a matter of time before this record will be broken. Around 2300 years ago the Greek mathematician Euclid already proved that there is an infinite number of prime numbers. His proof is a so-called proof by contradiction. To prove something by contradiction, you first assume the opposite of the statement that you want to prove is true. If you can disprove this opposite statement by showing it leads to a contradiction, your original statement must be true.
Assume there is a finite number of prime numbers. Then there must be a largest prime number p
Consider the number N = p 6 5 4 3 2 1 + 1. This number is not divisible by 2, because it can be written as N = 2n + 1 (where n = p 6 5 4 3 1).
N is not divisible by 3 either, because it can be written as N = 3n + 1 (where n = p 6 5 4 2 1).
In this way you can show that n is not divisible by any number from 2 through p. This means that n is a prime number or that n is divisible by a prime number greater than p and less than N.
However this is in contradiction with the assumption that p is the largest prime number. From this contradiction it follows that the assumption cannot be true. Therefore there is an infinite number of prime numbers.
LARY proof counterexample proof by contradiction
Have you reached your goal?
R I know what a proof and a counterexample are. I also know the general expressions for numbers that are (not) divisible by a certain number.
T1 I can prove simple statements or give a counterexample.
T2 I can prove more complex statements or give a counterexample.
I I can prove that √2 is an irrational number.
VOCA bu
Highest common factor and lowest common multiple
GOAL You will learn different methods to find the highest common factor and the lowest common multiple.
HCF and LCm The highest common factor of two natural numbers is the largest natural number that is a factor of both numbers. The factors of 10 are 1, 2, 5 and 10. The factors of 35 are 1, 5, 7 and 35. Therefore HCF(10, 35) = 5.
The lowest common multiple of two natural numbers is the smallest natural number that is a multiple of both numbers. The positive multiples of 6 are 6, 12, 18, … . The positive multiples of 9 are 9, 18, 27, … . Therefore LCM(6, 9) = 18.
using prime factorisation If you know the prime factorisation of two numbers, the highest common factor and the lowest common multiple of these two numbers are easy to find. For example, consider the numbers 24 and 180. Prime factorisation gives 24 = 2 2 2 3 = 23 3 and 180 = 2 2 3 3 5 = 22 32 5.
The highest common factor is equal to the product of the common prime factors. The common prime factors of 24 and 180 are 2, 2 and 3. So HCF(24, 180) = 2 · 2 · 3 = 22 · 3 = 12.
The lowest common multiple is equal to the product of the combined prime factors. The common prime factors are only included once. The common prime factors of 24 and 180 are 2, 2 and 3. 24 has an additional prime factor 2 and 180 has additional prime factors 3 and 5. Therefore LCM(24, 180) = 2 · 2 · 2 · 3 · 3 · 5 = 23 · 32 · 5 = 8 · 9 · 5 = 360.
e xample
Calculate HCF(40, 420) and LCM(40, 420).
First, give the prime factorisation of 40 and 420.
40 = 23 5 and 420 = 22 3 5 7
The common prime factors are 2, 2 and 5. 40 has an additional prime factor 2 and 120 has additional prime factors 3 and 7.
HCF(40, 420) = 22 5 = 4 5 = 20
LCM(40, 420) = 23 3 5 7 = 840
EXERCISES — PRACTICE
HCF and LCm
91 Give the definition of the highest common factor of two numbers. R
92 a Write down all of the factors of 28. T1
b Write down all of the factors of 72. T1
c What is the highest common factor of 28 and 72? T1
93 Calculate. T1
a HCF(36, 54)
b HCF(75, 250)
94 a Write down the first ten positive multiples of 6. T1
b Write down the first ten positive multiples of 8. T1
c What is the lowest common multiple of 6 and 8? T1
95 Calculate. T1
a LCM(36, 54)
b LCM(75, 250)
using prime factorisation
96 Calculate by using prime factorisation. T1
a HCF(340, 835)
b HCF(198, 504)
97 1260 = 22 32 5 7 and 52 000 = 25 53 13. Calculate. T1
a HCF(1260, 52 000)
b LCM(1260, 52 000)
98 A clock is 14 minutes per hour slower than a clock that runs well. At a certain moment, both minute hands point at 12. How long will it take before both minute hands point at 12 at the same time again? T2
The e uclidean algorithm An algorithm is a sequence of instructions used to solve a problem step by step. The Euclidean algorithm is a special method used to find the highest common factor of two integers (see exercise 105):
Determine the number of times the smallest integer fits into the largest integer. Write down the remainder. If there is no remainder, the smallest integer is the highest common factor. If there is a remainder, then divide the smallest of the first two integers by the remainder. Continue until there is no remainder. The last non-zero remainder is the highest common factor.
e xamples
1 Calculate HCF(350, 500).
350 fits into 500 once with remainder 150. Continue with the integers 350 and 150.
150 fits into 350 twice with remainder 50. Continue with the integers 150 and 50.
50 fits into 150 three times with no remainder. Therefore, the highest common factor of 350 and 150 is 50.
2 Calculate HCF(400, 1440).
400 fits into 1440 three times with remainder 240. Continue with the integers 400 and 240.
240 fits into 400 once with remainder 160. Continue with the integers 240 and 160.
160 fits into 240 once with remainder 80. Continue with 160 and 80.
80 fits into 160 twice with no remainder. Therefore, the highest common factor of 400 and 1440 is 80.
The euclidean algorithm
99 Use the Euclidean algorithm and the table below to find the highest common factor of 122 and 142. T1
142 and 22
− 22 = 10 22 and 10
− 2 10 = 10 and 10 − = 0
100 Use the Euclidean algorithm and the table below to find the highest common factor of 72 and 246. T1
246 and 72
101 Calculate using the Euclidean algorithm. T2
a HCM(60, 150)
b HCM(108, 891)
c HCM(1850, 4850)
102 The integers 77 089 and 44 831 are given. T2
a Use the Euclidean algorithm to find the highest common factor of these two integers.
b The Euclidean algorithm is a faster method to find the highest common factor than prime factorisation. Explain why.
103 Prove that the highest common factor of two consecutive integers is always greater than 1. T2
exploring
104 Two natural numbers are coprime if the highest common factor of these numbers is equal to 1. T2
a How many natural numbers less than 20 are coprime with 20?
b Assume p is a prime number. How many natural numbers less than p are coprime with p?
105 A schematic drawing of how to calculate the highest common factor of the numbers 350 and 500 using the Euclidean algorithm is shown below. I
Investigating
106 Read the text Algorithms and computers on the next page. Execute the following algorithm. Which shape do you get? T1
Go 5 cm to the right.
Go 4 cm up.
Go 5 cm to the left.
Go 4 cm down.
107 Design an algorithm that can be used to draw the figure shown below. T2
a Explain how the steps of the calculation are shown in the schematic drawing.
b Make a similar drawing for the calculation of HCF(45, 57) = 3.
c Use your drawing in part b to explain why the Euclidean algorithm produces the highest common factor.
108 Execute the following algorithm. What type of figure do you get? T2
Repeat 8 times:
Go 3 cm forward.
Turn 45° to the right.
109 Design your own algorithm and ask a classmate to execute it. I CO m P u TATIONAL SKILLS
C5 Calculate in an efficient way. I 4 37 25 IV 99 99 II 4 67 + 6 67 V 23 38 − 3 38
Algorithm and computers
An algorithm gives a very specific description of what you are supposed to do. You do not actually have to think, you just have to follow the instructions precisely. Therefore, computers are excellent at following algorithms. After all, computers do not think. They simple follow a sequence of instructions quickly and accurately.
Using the programming language Logo, you can make a turtle draw figures on a screen. You have to describe the moves the turtle has to make. For example, the algorithm below makes the turtle draw a square.
Go n cm forward.
Go n cm to the right.
Go n cm back.
Go n cm to the left.
2
1
3
4
By making the turtle repeat certain steps, you can use a loop to shorten the algorithm:
Repeat 4 times:
Go n cm forward.
Turn 90° to the right.
This usage of repetition is especially useful when you want complex figures to be drawn.
Have you reached your goal?
R I know what the highest common factor and the lowest common multiple of two numbers are. I also know what an algorithm is.
T1 I can find the highest common factor and the lowest common multiple of two numbers using prime factorisation.
T2 I can find the highest common factor of two numbers using the Euclidean algorithm.
I I can explain why the Euclidean algorithm produces the highest common factor.
VOCA bu LARY
highest common factor lowest common multiple algorithm Euclidean algorithm
Test preparation
At the end of every section, check whether you understand the vocabulary and have reached your learning goal. If not, reread the explanation or take another look at the videos. Then do the following exercises.
Academy
110 The arithmetic sequence (a n) = (5, 9, 13, 17, 21, 25, …) is given. T2
a Find out if 1223 is a term of this sequence.
b Calculate the sum of the first hundred terms of this sequence.
§ 1.1
111 Set A = {−2, −112, −1, 0, 1 2, 1, 112, 2} is given. T1
a How many elements does set A have?
b Which of the elements of this set are natural numbers?
c Explain if A ⊂ ℤ.
112 Give the prime factorisation of 1200. T1
113 A group of rectangles has a length of a and a width of b, where a , b ∈ ℕ and a > b T2
a How many of these rectangles have a perimeter of 20?
b How many of these rectangles have an area of 20?
§ 1.2
114 Copy and fill in ℚ or ℝ R
a is used to refer to the set of all real numbers.
b An irrational number is an element of , but not of c is a subset of
115 Write the following fractions as a decimal number without using a calculator. T1
7 20 / 43 250 / 3007 5000 / 2 3
116 Consider the number 0.12331
a Explain whether this is a rational or an irrational number. T1
b Write down the first eight decimals of this number. T1
c Find the fiftieth decimal of this number. I
§ 1.3
117 Calculate if possible. If not possible, explain why. Do not use a calculator. T2
118 A figure consists of three squares with areas of 25, 17 and 3. What is the perimeter of this figure? Simplify your answer as much as possible. I 25
§ 1.4
119 Prove the following statements or give a counterexample. T1
a The product of an even number and an odd number is always even.
b The product of two prime numbers always has exactly four divisors.
c If x2 ∈ ℕ, then x ∈ ℕ.
120 The set W consists of all numbers that can be written as 4n + 3, where n ∈ ℕ. T2
a Give four elements of this set.
b Explain whether the sum of two numbers that are elements of W can be a number which is also an element of W.
c Prove that the product of two numbers that are elements of W always has remainder 1 when it is divided by 4.
§ 1.5
121 Give the definition of the lowest common multiple of two numbers. R
122 8712 = 23 32 112 and 146 410 = 2 5 114
Calculate the following. T1
a HCF(8712, 146 410)
b LCM(8712, 146 410)
123 Use the Euclidean algorithm to find the highest common factor of 140 and 180. T2
Chapter 1
124 Copy and fill in ∈ or ∉ T2
125 Suppose you can only jump 33 units to the right or 75 units to the left on a number line. You start at 0. The number line below shows that you will end up at –9 after two jumps to the right and one to the left. I
a You start at 0 and jump three times to the left and twice to the right. At which number do you end up?
b Explain whether you can end up at any possible integer if you start at 0.
c You start at 0 and jump once to the right. Which jumps will get you back to 0 as quickly as possible?