WJEC A-Level Mathematics

What a proof is

A proof is a chain of logical steps that starts from given assumptions, or from results already established, and ends at the statement to be shown. Each step must follow from the one before, and the conclusion has to hold in every case covered, not just in the cases tried. A few examples that work prove nothing on their own; a single example that fails is enough to disprove a general claim.

Three methods are used at this level.

Proof by deduction

Start from known facts and use algebra to reach the result. To prove something about a family of integers, represent them algebraically: an even number is $2n$, an odd number is $2n+1$, and consecutive integers are $n$ and $n+1$, where $n$ is an integer.

Claim: the sum of two consecutive odd numbers is a multiple of 4.

Since $n+1$ is an integer, the sum is a multiple of 4.

Claim: the square of any odd number is one more than a multiple of 8. Write the odd number as $2n+1$. Then $(2n+1)^2 = 4n^2 + 4n + 1 = 4n(n+1) + 1$. One of $n$ and $n+1$ is even, so $n(n+1)$ is even and $4n(n+1)$ is a multiple of 8.

The laws of logarithms are also proved by deduction, from the definition that $x = a^n$ means $n = \log_a x$ (for $a > 0$, $x > 0$). Let $x = a^m$ and $y = a^n$, so that $\log_a x = m$ and $\log_a y = n$.

• The product: $xy = a^m \times a^n = a^{m+n}$, so $\log_a(xy) = m + n = \log_a x + \log_a y$.
• The quotient: $\dfrac{x}{y} = \dfrac{a^m}{a^n} = a^{m-n}$, so $\log_a\!\left(\dfrac{x}{y}\right) = m - n = \log_a x - \log_a y$.
• The power: $x^k = (a^m)^k = a^{mk}$, so $\log_a(x^k) = mk = k\log_a x$, for any $k$, including $k=-1$ and $k = -\tfrac12$.

Proof by exhaustion

Split the claim into a finite number of cases and check every one. It works only when the cases can genuinely be listed.

Claim: $n^2 + n + 11$ is prime for $n = 1, 2, 3, 4, 5$. The values are 13, 17, 23, 31 and 41, and each is prime, so the claim is proved for those five values. The same technique proves that the square of an integer leaves remainder 0 or 1 when divided by 4: an integer is either even, $2n$, with square $4n^2$, or odd, $2n+1$, with square $4n^2+4n+1$, and these two cases cover every integer.

Disproof by counter-example

One case that breaks a general statement disproves it. The claim "$n^2 + n + 11$ is prime for every positive integer $n$" is false, because $n = 10$ gives $100 + 10 + 11 = 121 = 11^2$. The claim "if $a^2 > b^2$ then $a > b$" is false, because $a = -3$, $b = 2$ gives $9 > 4$ but $-3 < 2$.

Showing that a statement holds for many values is not a proof. The statement above about $n^2+n+11$ holds for $n = 1$ to $9$ and still fails.

Implication symbols

An identity such as $(x+1)^2 \equiv x^2 + 2x + 1$ holds for all $x$, whereas an equation such as $x^2 = 4$ holds only for particular values. In solving an equation, two lines are connected by $\Leftrightarrow$ only if the step can be reversed. Squaring both sides of $x = 3$ gives $x^2 = 9$, which is an implication ($\Rightarrow$) and not an equivalence, because $x^2 = 9$ also allows $x = -3$.

Sets and their notation

A set is a collection of distinct elements.

The standard number sets are $\mathbb{N} = \{1, 2, 3, \dots\}$, $\mathbb{Z}$ for the integers, $\mathbb{Z}^+$ for the positive integers, $\mathbb{Q}$ for the rational numbers (those that can be written as a fraction $\frac{p}{q}$ of integers) and $\mathbb{R}$ for the real numbers.

Intervals of real numbers are written with brackets, where a square bracket includes the end point and a round one excludes it: $[a,b] = \{x \in \mathbb{R} : a \le x \le b\}$, $[a,b) = \{x \in \mathbb{R} : a \le x < b\}$, $(a,b]$ likewise, and $(a,b)$ with neither end point.

Example. With $\varepsilon = \{1, 2, \dots, 10\}$, $A = \{2, 4, 6, 8, 10\}$ and $B = \{3, 6, 9\}$:

• $A \cap B = \{6\}$ and $A \cup B = \{2, 3, 4, 6, 8, 9, 10\}$, so $n(A \cup B) = 7$;
• $A' = \{1, 3, 5, 7, 9\}$;
• $A' \cap B = \{3, 9\}$.

The same language describes solutions to inequalities. The solution of $x^2 - x - 6 > 0$ is $x < -2$ or $x > 3$, which is the union $\{x : x < -2\} \cup \{x : x > 3\}$. A solution that has to satisfy two conditions at once, such as $x > 2$ and $x < 5$, is an intersection, and is written $2 < x < 5$. The words "and" and "or" correspond to intersection and union.