Cambridge IGCSE Mathematics

Kinds of number

The natural numbers are the counting numbers $1, 2, 3, \ldots$. The integers are the whole numbers, positive, zero and negative: $\ldots, -2, -1, 0, 1, 2, \ldots$. A number is rational if it can be written as a fraction $\dfrac{a}{b}$ of two integers with $b \ne 0$. That covers every integer, every terminating decimal such as $0.25 = \tfrac{1}{4}$ and every recurring decimal such as $0.\dot{3} = \tfrac{1}{3}$. An irrational number cannot be written that way, and its decimal neither ends nor repeats. The square root of a number that is not a perfect square is irrational, so $\sqrt{2}$ and $\sqrt{17}$ are irrational while $\sqrt{49} = 7$ and $\sqrt{\tfrac{9}{16}} = \tfrac{3}{4}$ are rational. The number $\pi$ is irrational too, which is why $\pi$ is left in an answer when a question asks for it in terms of $\pi$.

Large numbers are read in groups of three digits, so 10 007 is ten thousand and seven, and six billion is 6 000 000 000 (a billion is $10^9$).

Square numbers, cube numbers and reciprocals

The square numbers are $1, 4, 9, 16, 25, \ldots$ and the cube numbers are $1, 8, 27, 64, 125, \ldots$. The squares of the integers from 1 to 15 and the cubes of 1, 2, 3, 4, 5 and 10 are worth knowing without a calculator, because they make roots, factors and estimates quick.

The reciprocal of a number $x$ is $\dfrac{1}{x}$, the number that multiplies with $x$ to give 1. The reciprocal of 5 is $\tfrac{1}{5}$, the reciprocal of $\tfrac{3}{4}$ is $\tfrac{4}{3}$ (turn the fraction over), and the reciprocal of $0.2$ is $5$. Zero has no reciprocal, because nothing multiplied by 0 gives 1.

Factors, multiples and primes

A factor of a number divides it exactly, and a multiple is what you get by multiplying it by an integer. The factors of 18 are $1, 2, 3, 6, 9, 18$, and the multiples of 18 are $18, 36, 54, \ldots$. A common factor of two numbers is a factor of both, and a common multiple is a multiple of both.

A prime number has exactly two factors, 1 and itself, so the primes begin $2, 3, 5, 7, 11, 13, \ldots$. The number 1 is not prime because it has only one factor, and 2 is the only even prime.

Every integer greater than 1 is either prime or a product of primes in exactly one way, apart from the order. Writing a number as that product is prime factorisation, and it is found by dividing repeatedly by primes. For 72:

Highest common factor and lowest common multiple

The highest common factor (HCF) of two numbers is the largest number that divides both, and the lowest common multiple (LCM) is the smallest number that both divide into. Both come from the prime factorisations. The HCF takes each prime to the lower of its two powers, and the LCM takes each prime to the higher power (a prime missing from one number counts as power 0).

For 84 and 90:

As a check, $\text{HCF} \times \text{LCM} = 6 \times 1260 = 7560 = 84 \times 90$, which holds for any pair of numbers. The LCM is what you use to find when two repeating events coincide. Two lights that flash every 12 seconds and every 18 seconds flash together again after $\text{LCM}(12, 18) = 36$ seconds. The HCF is what you use to split things into the largest equal groups: 84 red and 90 blue counters can be shared into 6 identical bags.

Ordering and inequality symbols

The symbols are $=$ (equal to), $\ne$ (not equal to), $>$ (greater than), $<$ (less than), $\geqslant$ (greater than or equal to) and $\leqslant$ (less than or equal to). The open end of the symbol points at the larger quantity, so $-7 < -2$ is read "minus 7 is less than minus 2".

To order numbers written in different forms, convert them to the same form first, usually decimals. To order $0.6$, $\tfrac{5}{8}$ and $62\%$, write them as $0.6$, $0.625$ and $0.62$, which gives $0.6 < 62\% < \tfrac{5}{8}$. Negative numbers reverse the pattern: the further a number is below zero, the smaller it is.

Sets

A set is a collection of objects, called its elements, written in curly brackets. Sets can be listed, $B = \{2, 3, 5, 7\}$, or described by a rule with a colon meaning "such that": $A = \{x : x \text{ is a natural number}\}$ and $C = \{x : 2 < x \leqslant 7,\ x \text{ an integer}\} = \{3, 4, 5, 6, 7\}$. Extended tier only: sets can also hold ordered pairs, such as $\{(x, y) : y = mx + c\}$, the points of a line.

Take $\xi = \{1, 2, \ldots, 12\}$, $A = \{2, 4, 6, 8, 10, 12\}$ (the even numbers) and $B = \{3, 6, 9, 12\}$ (the multiples of 3). Then $A \cap B = \{6, 12\}$, $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$ so $n(A \cup B) = 8$, $A' = \{1, 3, 5, 7, 9, 11\}$ and $(A \cup B)' = \{1, 5, 7, 11\}$. The set $\{6, 12\}$ is a subset of both $A$ and $B$.

Venn diagrams

A Venn diagram draws $\xi$ as a rectangle and each set as a loop, with overlapping loops for elements that can be in both. The regions are read as unions, intersections and complements. When numbers of elements are given, fill the diagram from the middle outwards: start with the overlap, then subtract it from each loop to get the parts that belong to one set only, and finally find the outside from the total.

In a group of 40 students, 25 study French, 18 study Spanish and 7 study both.

• Both: 7.
• French only: $25 - 7 = 18$.
• Spanish only: $18 - 7 = 11$.
• Neither: $40 - (18 + 7 + 11) = 4$.

The same idea gives $n(F \cup S) = n(F) + n(S) - n(F \cap S) = 25 + 18 - 7 = 36$, and the probability that a student chosen at random studies neither is $\tfrac{4}{40}$.

Three sets

Extended tier only. Venn diagrams may have three overlapping loops, which make eight regions. Start with the region where all three overlap, then the regions where exactly two overlap, then the parts of one loop only, and finish with the outside.

Suppose $n(\xi) = 50$, $n(A) = 20$, $n(B) = 18$, $n(C) = 15$, $n(A \cap B) = 6$, $n(B \cap C) = 5$, $n(A \cap C) = 4$ and $n(A \cap B \cap C) = 2$.

• All three: 2.
• $A$ and $B$ only: $6 - 2 = 4$; $B$ and $C$ only: $5 - 2 = 3$; $A$ and $C$ only: $4 - 2 = 2$.
• $A$ only: $20 - 4 - 2 - 2 = 12$; $B$ only: $18 - 4 - 3 - 2 = 9$; $C$ only: $15 - 3 - 2 - 2 = 8$.
• Inside at least one loop: $12 + 9 + 8 + 4 + 3 + 2 + 2 = 40$, so the outside holds $50 - 40 = 10$.