AQA A-Level Mathematics

Notation and language

A mathematical argument is built from precise statements joined by connecting words, so the symbols need to be used exactly. The symbol $\Rightarrow$ means "implies": $x = 3 \Rightarrow x^2 = 9$ is true, but the reverse is not, because $x$ could be $-3$. The symbol $\Leftrightarrow$ means "implies and is implied by", so that $2x = 6 \Leftrightarrow x = 3$. The symbol $\equiv$ marks an identity, which is true for every value of the variable, while $=$ in an equation is true only for particular values. The symbol $\therefore$ means "therefore".

Set notation is used for solutions, inequalities and probability.

The solution of $x^2 > 4$ can be written $x < -2$ or $x > 2$, or as the set $\{x : x < -2\} \cup \{x : x > 2\}$. The solution of $2 < x < 5$ is the intersection $\{x : x > 2\} \cap \{x : x < 5\}$. "And" corresponds to an intersection and "or" to a union.

A rational number can be written as $\frac{p}{q}$ with $p$ and $q$ integers and $q \neq 0$. An irrational number cannot. An even integer has the form $2k$ and an odd integer has the form $2k + 1$, where $k$ is an integer.

Proof by deduction

A proof by deduction starts from known facts or definitions and moves by logical steps to the result, so the result is true for every case at once. Algebra is the usual tool, because a letter stands for any number of the right type.

Show that the sum of two consecutive odd numbers is a multiple of 4. Write the odd numbers as $2k + 1$ and $2k + 3$, where $k$ is an integer. Their sum is

and since $k + 1$ is an integer, the sum is a multiple of 4.

Show that $n^3 - n$ is divisible by 6 for every integer $n$. Factorising gives $n^3 - n = n(n-1)(n+1) = (n-1)\,n\,(n+1)$, which is the product of three consecutive integers. Among any three consecutive integers at least one is even and exactly one is a multiple of 3, so the product is divisible by 2 and by 3, and therefore by 6.

A proof needs a final statement that says what has been shown, and every step must follow from the one before. Checking a few numerical cases does not prove a general result.

Proof by exhaustion

Proof by exhaustion splits the possibilities into a finite number of cases and checks each one. It only works when the cases really do cover everything.

To show that the square of an integer is never one less than a multiple of 3, let $n$ be any integer. Then $n$ is of the form $3k$, $3k + 1$ or $3k + 2$, and

So every square leaves remainder 0 or 1 when divided by 3, and never remainder 2, which is the remainder of a number one less than a multiple of 3.

Disproof by counter-example

To show a statement is false it is enough to find one case where it fails. This is a counter-example, and it should be checked and stated clearly.

The statement "$n^2 + n + 11$ is prime for every positive integer $n$" is false, because for $n = 11$ the value is $121 + 11 + 11 = 143 = 11 \times 13$. The statement "if $a^2 > b^2$ then $a > b$" is false, because $a = -5$ and $b = 3$ give $25 > 9$ but $-5 < 3$.

Proof by contradiction

In a proof by contradiction, assume the statement is false, follow the consequences logically, and reach something impossible. The assumption must then be wrong, so the statement is true.

The square root of 2 is irrational

Assume $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are integers with no common factor (the fraction is in its lowest terms). Squaring gives $2 = \frac{p^2}{q^2}$, so $p^2 = 2q^2$. Then $p^2$ is even, so $p$ is even, because the square of an odd number is odd. Write $p = 2m$. Then $4m^2 = 2q^2$, so $q^2 = 2m^2$, and $q^2$ is even, so $q$ is even. Now $p$ and $q$ are both even, which contradicts the assumption that they have no common factor. So $\sqrt{2}$ is irrational.

There are infinitely many primes

Assume there are finitely many primes, and list them all as $p_1, p_2, \ldots, p_n$. Form the number $N = p_1 p_2 \cdots p_n + 1$. When $N$ is divided by any $p_i$ the remainder is 1, so no prime in the list divides $N$. But $N$ is either prime or has a prime factor, and in both cases there is a prime that is not in the list. This contradicts the assumption, so there are infinitely many primes.

Unfamiliar proofs

The same pattern can be applied to a new statement. To show that there is no smallest positive rational number, assume $r$ is the smallest. Then $\frac{r}{2}$ is a smaller positive rational number, which contradicts the assumption.

To show that there are no integers $a$ and $b$ with $a^2 - b^2 = 2$, assume there are. The left side factorises as $(a - b)(a + b)$. The two factors $a - b$ and $a + b$ differ by $2b$, which is even, so they are both even or both odd. If both are odd their product is odd, and if both are even their product is a multiple of 4. Neither can equal 2, so there are no such integers.

Choosing a method

Deduction is the default for a general result about all integers or all values of a variable. Exhaustion suits a statement that splits into a small number of cases, such as remainders or signs. A single counter-example settles a claim that something is always true. Contradiction suits a statement of the form "there is no…" or "this number is irrational", where assuming the opposite gives something concrete to work with.