Proof by deduction, proof by exhaustion, and disproof by counterexample, with the language and structure WJEC rewards.

What this dot point is asking

WJEC wants you to construct a rigorous argument using proof by deduction and proof by exhaustion, and to disprove a false statement using a counterexample. Proof is woven through the whole course: the same logical discipline appears when you justify a coordinate-geometry result, verify an identity, or argue that a turning point is a minimum. Examiners reward precise structure and correct logical language as much as the algebra.

The answer

Proof by deduction

A direct or deductive proof chains together known facts and definitions in logically valid steps until you reach the required conclusion. The key skill at this level is to represent the objects generally so that the argument covers every case, not just one example.

The standard representations you should know:

• An even integer is $2n$; an odd integer is $2n+1$ (or $2n-1$).
• A multiple of three is $3n$; consecutive integers are $n,\ n+1,\ n+2$.
• A rational number is $\dfrac{a}{b}$ with $a$ and $b$ integers and $b \neq 0$.

Proof by exhaustion

When a statement can only involve a finite number of cases, you can prove it by checking every case. This is legitimate provided you genuinely cover all possibilities and show the working for each.

For example, to show that $n^2 + n$ is even for every integer $n$, split into two cases: if $n$ is even then $n^2+n$ is even (even plus even); if $n$ is odd then $n^2$ is odd and $n$ is odd, so their sum is even. Both cases give an even result, so the statement holds for all integers.

Disproof by counterexample

A statement of the form "for all $x$, property $P$ holds" is false as soon as you find one value of $x$ for which $P$ fails. That single value is a counterexample.

To disprove "the product of two irrational numbers is always irrational", take $\sqrt{2} \times \sqrt{2} = 2$, which is rational. One counterexample is enough.

Examples in context

Example 1. A divisibility claim. Prove that $n^3 - n$ is divisible by $6$ for every integer $n$. Factor: $n^3 - n = n(n-1)(n+1)$, a product of three consecutive integers. Among any three consecutive integers one is divisible by $3$ and at least one is even, so the product is divisible by $3 \times 2 = 6$. This blends deduction with a short exhaustion-style argument about residues.

Example 2. Spotting a false identity. A student claims $(a+b)^2 = a^2 + b^2$ for all real $a, b$. Take $a = b = 1$: the left side is $4$, the right side is $2$. The counterexample disproves it. The correct identity, $(a+b)^2 = a^2 + 2ab + b^2$, then explains exactly which term was missing.

Try this

Q1. Prove by deduction that the sum of three consecutive integers is divisible by $3$. [3 marks]

• Cue. Take $n + (n+1) + (n+2) = 3n+3 = 3(n+1)$, a multiple of $3$.

Q2. Disprove: "for all real $x$, $x^2 > x$". [2 marks]

• Cue. Try $x = \tfrac{1}{2}$: $x^2 = \tfrac{1}{4} < \tfrac{1}{2}$, so the statement is false.

Q3. Use exhaustion to show that no perfect square ends in the digit $7$. [3 marks]

• Cue. Check the last digit of $n^2$ for $n$ ending in $0$ to $9$: the only possible last digits are $0,1,4,5,6,9$, never $7$.