amazon.co.ukProof by deduction, exhaustion, counter-example and contradiction, algebraic identities and partial fractions - A-Level pure maths proof (Edexcel 9MA0 Topics 1-2, AQA 7357 A1/B8) revised with full worked proofs and mark-scheme logic.
A proof is a chain of logical steps from given assumptions to a conclusion, and examiners mark the chain, not just the destination. Four techniques appear at A-level (Edexcel 9MA0 Topic 1; AQA 7357 A1; OCR A H640 1.01; WJEC/Eduqas and CCEA AS/A2 Pure Mathematics), and the first skill is choosing the right one. Proof by deduction is the default: start from algebra you know, manipulate validly, land on the result - used for statements that are genuinely always true. Proof by exhaustion applies when the domain splits into a small finite number of cases. Disproof by counter-example applies to universal claims that are false: one failing value is enough. Proof by contradiction applies when direct deduction is awkward but assuming the opposite produces something impossible.
Distinguish an identity from an equation. An equation such as 2x = 6 is true for particular values of x and is solved. An identity such as (x + 1)^2 = x^2 + 2x + 1 is true for all values and is proved, with each side simplified independently - you never subtract one side from the other as if solving. Identities are proved by expanding or factorising one side into the other, and the equivalence sign should be used once the equivalence is established.
Mark schemes award specific marks for specific structural moves: M1 for stating the assumption or setting up the cases, M1 for valid manipulation, A1 for the contradiction or conclusion, often with a final B1 requiring the conclusion to be stated in words. A proof that ends in algebra with no concluding sentence routinely drops the last mark.
Deduction proofs at this level are short algebraic arguments. Two standard patterns cover most questions: factorisation patterns (showing an expression is a multiple of something) and parity patterns (showing an expression is even or odd).
Prove that n^3 - n is divisible by 3 for all positive integers n.
Line 1: n^3 - n = n(n^2 - 1) (M1: factor out n).
Line 2: n^2 - 1 = (n - 1)(n + 1), so n^3 - n = (n - 1)n(n + 1) (M1: difference of two squares).
Line 3: This is a product of three consecutive integers, so exactly one of the factors is a multiple of 3 (A1: valid argument).
Line 4: Hence the product is divisible by 3 for every positive integer n (A1: conclusion stated).
Prove that the sum of the squares of two consecutive integers is always odd.
Line 1: Let the integers be n and n + 1. Sum of squares: n^2 + (n + 1)^2 = n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1 (M1).
Line 2: 2n^2 + 2n + 1 = 2n(n + 1) + 1 (M1: factor out 2).
Line 3: 2n(n + 1) is even, so the sum is one more than an even number, hence odd (A1: conclusion).
Identity proofs follow the same algebra but work side-by-side. The classic A* identity: prove that (x + 1)(x + 2)(x + 3)(x + 4) + 1 = (x^2 + 5x + 5)^2 for all real x. Group (x + 1)(x + 4) = x^2 + 5x + 4 and (x + 2)(x + 3) = x^2 + 5x + 6; writing u = x^2 + 5x, the product is (u + 4)(u + 6) = u^2 + 10u + 24, so adding 1 gives u^2 + 10u + 25 = (u + 5)^2 = (x^2 + 5x + 5)^2. The substitution u = x^2 + 5x is the A* step that keeps the algebra short.
To disprove a universal claim, you do not need an argument - you need one example. The value must satisfy the conditions of the statement but fail the conclusion. Two classics: the claim that n^2 + n + 41 is prime for all positive integers n holds for n = 1 to 39 but fails at n = 40, since 40^2 + 40 + 41 = 1681 = 41^2; and the claim that x^2 > x for all real x fails at x = 1/2, where x^2 = 1/4 < 1/2. In an exam, state the counter-example, verify it satisfies the conditions, and evaluate it to show the claim fails - one line each.
Proof by exhaustion is the correct technique when the domain splits into a small number of cases. The standard example: prove that every perfect square ends in the digit 0, 1, 4, 5, 6 or 9. Every integer has one of ten possible last digits, so there are ten cases, checked in a table:
| Last digit n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Last digit of n^2 | 0 | 1 | 4 | 9 | 6 | 5 | 6 | 9 | 4 | 1 |
The ten cases are exhausted: no last digit produces a square ending in 2, 3, 7 or 8, so the claim is proved. Another exhaustion pattern: for 1 <= n <= 6, check each value of n against the claim individually - legitimate because the list is finite and complete.
The mark-scheme discipline: a counter-example must be specific (n = 40, not some large n) and exhaustion must be complete (all cases, explicitly listed). Partial exhaustion or a vague counter-example earns zero on the disproof mark.
Contradiction proofs have a fixed skeleton: (1) assume the statement is false; (2) deduce consequences validly until something impossible appears; (3) conclude the assumption was wrong, so the statement is true. AQA 7357 A1 requires this technique explicitly, and it appears in OCR A, WJEC and CCEA specifications; Edexcel 9MA0 centres on deduction, exhaustion and counter-example, but contradiction-style arguments appear in past paper extensions.
Line 1: Assume sqrt(2) is rational, so sqrt(2) = a/b where a, b are integers with no common factor (fraction in lowest terms) (M1: assumption stated).
Line 2: Squaring: 2 = a^2/b^2, so a^2 = 2b^2 (M1: valid algebra).
Line 3: a^2 is even, so a is even; write a = 2k (A1: justified - if a were odd, a^2 would be odd).
Line 4: Substituting: 2b^2 = 4k^2, so b^2 = 2k^2, hence b is also even (A1).
Line 5: a and b are both even, so they share the factor 2 - contradicting the choice of a/b in lowest terms (A1: contradiction identified).
Line 6: The assumption is false, so sqrt(2) is irrational (A1: conclusion stated).
Line 1: Assume there are finitely many primes, listed p_1, p_2, ..., p_n (M1).
Line 2: Let N = p_1 p_2 ... p_n + 1 (M1: construction).
Line 3: N leaves remainder 1 when divided by each p_i, so no listed prime divides N (A1).
Line 4: So N is either itself prime or has a prime factor outside the list - either way a prime exists beyond the finite list, contradicting the assumption (A1: contradiction).
Line 5: Hence there are infinitely many primes (A1: conclusion).
Two further contradiction patterns to know: to prove that if n^2 is even then n is even, assume n is odd, write n = 2k + 1, so n^2 = 4k^2 + 4k + 1 is odd - contradicting n^2 even. And to prove that the sum of a rational and an irrational number is irrational, assume it is rational and subtract to force the irrational number to be rational. Each follows the same skeleton: assumption, valid steps, impossible consequence, conclusion.
Partial fractions (Edexcel 9MA0 2.5; AQA 7357 B8; OCR A H640 6.01) reverse the process of adding fractions, and they matter because Year 2 integration requires them. First, classify the fraction. A proper fraction (numerator degree below denominator degree) is split directly. An improper fraction (numerator degree at or above denominator degree) must be divided first, leaving a polynomial plus a proper remainder.
Forms: for distinct linear factors (x - a)(x + b), write A/(x - a) + B/(x + b). For a repeated linear factor (x + c)^2, write A/(x + c) + B/(x + c)^2. The cover-up and substitution method finds constants quickly: multiply through by the denominator and substitute the root of each factor in turn.
Express (5x - 2)/((x - 2)(x + 1)) in partial fractions.
Line 1: 5x - 2 = A(x + 1) + B(x - 2) (M1: correct form).
Line 2: x = 2: 8 = 3A, so A = 8/3 (A1).
Line 3: x = -1: -7 = -3B, so B = 7/3 (A1).
Line 4: (5x - 2)/((x - 2)(x + 1)) = 8/(3(x - 2)) + 7/(3(x + 1)) (A1).
Express (2x^2 + 3x + 5)/(x + 1)^2 in partial fractions.
Line 1: Improper (both degrees 2), so divide: 2(x + 1)^2 = 2x^2 + 4x + 2, and (2x^2 + 3x + 5) - (2x^2 + 4x + 2) = -x + 3 (M1).
Line 2: So the fraction = 2 + (-x + 3)/(x + 1)^2 (A1).
Line 3: -x + 3 = A(x + 1) + B. Setting x = -1: 4 = B (A1). Comparing coefficients of x: -1 = A.
Line 4: Answer: 2 - 1/(x + 1) + 4/(x + 1)^2 (A1).
Always check by substituting one value of x into both sides - if they disagree, a constant is wrong. Note the sign trap: when a substitution gives a negative product, solve carefully (a frequent source of wrong constants). AQA and OCR also expect the form (Ax + B)/(irreducible quadratic) for quadratic denominators; Edexcel 9MA0 restricts to linear and repeated linear factors.