CCEA A-Level Mathematics

Symbols and connecting language

Mathematical arguments are written with a small set of symbols, and each one has a precise meaning that a sentence of working has to respect.

The difference between $=$ and $\equiv$ matters. The statement $x^2 - 1 = 0$ is true only for $x = 1$ and $x = -1$, whereas $x^2 - 1 \equiv (x - 1)(x + 1)$ is true whatever $x$ is. The arrow $\Rightarrow$ goes in one direction only: $x = 3 \Rightarrow x^2 = 9$ is true, but $x^2 = 9 \Rightarrow x = 3$ is false because $x$ could be $-3$. Since $x^2 = 9 \Leftrightarrow x = 3 \text{ or } x = -3$, the equivalence sign can be used there.

A condition is necessary for a statement if the statement cannot be true without it, and sufficient if it guarantees the statement. Being a multiple of $4$ is sufficient for a whole number to be even but not necessary, because $6$ is even and is not a multiple of $4$. Being even is necessary for being a multiple of $4$ but not sufficient, since $6$ is even and is not a multiple of $4$. When a condition is both, the two statements are equivalent and $\Leftrightarrow$ applies: a quadratic has equal roots if and only if its discriminant is zero.

Several everyday words have a fixed meaning in mathematical writing: a constant never changes, a variable can take different values, an index is a power, and a function gives one output for each input. In $3x^2 + 5x - 7$ there are three terms, the coefficient of $x^2$ is $3$, the coefficient of $x$ is $5$, the constant term is $-7$, and $x$ is the variable. An expression has no equals sign, an equation does, and an identity is an equation true for all values of its variable.

Sets

A set is a collection of objects, called its elements, written in curly brackets. The notation $x \in A$ says that $x$ is an element of $A$ and $x \notin A$ says it is not. Some sets are used so often that they have their own letters.

The other symbols for sets are these.

Sets can also be described by a rule. The set $\{x \in \mathbb{R} : x > 2\}$ is read as the set of real numbers $x$ such that $x$ is greater than $2$. Solutions to inequalities are often given this way, or as an interval, and a Venn diagram shows the relations between sets with overlapping regions drawn inside a rectangle that represents $\varepsilon$.

If $\varepsilon = \{1, 2, \dots, 10\}$, $A = \{2, 4, 6, 8, 10\}$ and $B = \{3, 6, 9\}$, then $A \cap B = \{6\}$, $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$ and $A' = \{1, 3, 5, 7, 9\}$. The same language is used in probability, where $A \cap B$ is the event that both happen and $A \cup B$ the event that at least one happens.

What a proof is

A proof starts from assumptions that are given or already established and moves by logical steps to a conclusion, so that the conclusion has to be true whenever the assumptions are. Checking a statement on a few examples is never a proof, however many cases work, because the next case might fail. Four methods are used at AS level and one more is added at A2 (proof by contradiction, in its own note).

Proof by deduction

A deductive proof works with general symbols, so that it covers every case at once. To prove a statement about all even numbers, write an even number as $2n$ with $n$ an integer; to cover odd numbers use $2n + 1$; and to cover consecutive integers use $n$ and $n + 1$.

Prove that the sum of the squares of two consecutive integers is always odd. Let the integers be $n$ and $n + 1$. Then

Since $n^2 + n$ is an integer, this has the form $2k + 1$ and so it is odd.

Prove that $(2a + 3)^2 - (2a - 3)^2$ is a multiple of $24$ when $a$ is an integer. Expanding, it equals $(4a^2 + 12a + 9) - (4a^2 - 12a + 9) = 24a$. The difference is $24$ times an integer, so it is a multiple of $24$.

An algebraic statement can be proved in the same way. To show that $x^2 - 6x + 11 > 0$ for all real $x$, complete the square to get $(x - 3)^2 + 2$. A square is never negative, so the expression is at least $2$ and therefore positive.

Proof by exhaustion

Some statements have a finite number of cases, and proof by exhaustion checks every one. To prove that the square of any integer leaves remainder $0$ or $1$ when divided by $4$, there are two cases, since an integer is either $2n$ or $2n + 1$. In the first, $(2n)^2 = 4n^2$ leaves remainder $0$. In the second, $(2n + 1)^2 = 4n^2 + 4n + 1 = 4(n^2 + n) + 1$ leaves remainder $1$. There are no other cases, so the statement is proved.

Exhaustion is a valid proof only when the cases are listed completely and each is actually checked. A small finite set, such as the integers from $1$ to $20$, can be done by working through them all. Often an infinite set is split into a finite number of types, as above with even and odd.

Disproof by counter-example

To show that a statement is false it is enough to find one case where it fails. The claim that $n^2 + n + 11$ is prime for every positive integer $n$ holds for $n = 1, 2, \dots, 9$ (giving $13, 17, 23, 31, 41, 53, 67, 83, 101$), yet at $n = 10$ it gives $121 = 11^2$, which is not prime. One failure settles it. The claim that $x^2 > x$ for every real $x$ fails at $x = \tfrac{1}{2}$, where $x^2 = \tfrac{1}{4}$.

A counter-example has to satisfy the conditions of the statement it attacks. To disprove "the sum of two irrational numbers is irrational" use $\sqrt{2}$ and $-\sqrt{2}$, each irrational, whose sum is $0$, which is rational.

Setting out a proof

A proof reads as a chain in which each line follows from the ones before, and the connecting words are part of it. State what is being assumed at the start (for example "let $n$ be an integer, so the integers are $n$ and $n + 1$"), show the steps, and finish with a sentence that says what has been proved and why, for example "this has the form $2k + 1$, so it is odd". Give the reason for any step that is not plain algebra. A statement of the form "if $P$ then $Q$" is proved by starting from $P$ and reaching $Q$; starting from $Q$ and reaching $P$ proves the converse, which is a different statement, and when both directions are needed each has to be shown.