Proof by Contradiction, Algebraic Division & Partial Fractions
Mathematical proof and algebraic manipulation underpin every topic in Pure Mathematics across AQA, Edexcel, and OCR A-Level Maths. In Year 2 Pure, you must construct rigorous proofs by contradiction, including proving the irrationality of √2 and Euclid's proof that there are infinitely many prime numbers. Alongside formal proof, decomposing rational functions into partial fractions is an essential gateway skill: you cannot integrate complex rational expressions or expand them using the general binomial theorem until you have split them into simpler fractions, including cases with repeated linear factors and improper algebraic fractions.
- Proof by Exhaustion and Counter-Example
- To prove a statement by exhaustion, divide all possible cases into a finite set (such as testing even n = 2k and odd n = 2k + 1, or checking every integer in a given interval) and show the statement holds for every case. To disprove a universal claim, you only need a single counter-example: substitute one specific value that satisfies the condition's hypothesis but fails its conclusion, and explicitly state that the claim is false.
- Proof by Contradiction: Irrationality of √2
- Assume the negation of the statement is true: suppose √2 is rational, so √2 = a / b where a and b are integers with no common factors (coprime, b ≠ 0). Squaring both sides gives 2 = a² / b², so a² = 2b². Since a² is a multiple of 2, a must also be even, so let a = 2k for some integer k. Substituting back gives (2k)² = 2b² → 4k² = 2b² → b² = 2k². Hence b² is even, so b is also even. Therefore both a and b share a common factor of 2, contradicting the assumption that a / b is in its simplest form. Hence √2 is irrational.
- Proof by Contradiction: Infinitely Many Primes
- Assume there is a finite number of prime numbers, and list all of them as p₁, p₂, p₃, ., p_n. Construct the number N = (p₁ × p₂ × p₃ × . × p_n) + 1. When N is divided by any prime p_i from our complete list, it leaves a remainder of 1, so N is not divisible by any of p₁, p₂, ., p_n. Therefore, either N is itself a new prime number not on our list, or N is composite and divisible by some prime factor not in our list. Either case contradicts the assumption that p₁, ., p_n is the complete finite list of primes. Thus there are infinitely many primes.
Quadratic Parabola Plotter, Roots, Turning Point & Quadratic Formula
Adjust coefficients a, b, and c in y = ax² + bx + c to see how the parabola, roots (x-intercepts), and turning point move on the graph.
y = x² - 2x - 3
(-2)² − 4(1)(-3) = 16 → 2 real roots (crosses x-axis twice)x = 1.00b in brackets when squaring: (-2)² = 4 (always positive!). Typing -2² into a calculator without brackets gives -4 and loses the accuracy mark.Key Concepts & Exam Definitions (6)
- Proof by Exhaustion and Counter-Example
- To prove a statement by exhaustion, divide all possible cases into a finite set (such as testing even n = 2k and odd n = 2k + 1, or checking every integer in a given interval) and show the statement holds for every case. To disprove a universal claim, you only need a single counter-example: substitute one specific value that satisfies the condition's hypothesis but fails its conclusion, and explicitly state that the claim is false.
- Proof by Contradiction: Irrationality of √2
- Assume the negation of the statement is true: suppose √2 is rational, so √2 = a / b where a and b are integers with no common factors (coprime, b ≠ 0). Squaring both sides gives 2 = a² / b², so a² = 2b². Since a² is a multiple of 2, a must also be even, so let a = 2k for some integer k. Substituting back gives (2k)² = 2b² → 4k² = 2b² → b² = 2k². Hence b² is even, so b is also even. Therefore both a and b share a common factor of 2, contradicting the assumption that a / b is in its simplest form. Hence √2 is irrational.
- Proof by Contradiction: Infinitely Many Primes
- Assume there is a finite number of prime numbers, and list all of them as p₁, p₂, p₃, ., p_n. Construct the number N = (p₁ × p₂ × p₃ × . × p_n) + 1. When N is divided by any prime p_i from our complete list, it leaves a remainder of 1, so N is not divisible by any of p₁, p₂, ., p_n. Therefore, either N is itself a new prime number not on our list, or N is composite and divisible by some prime factor not in our list. Either case contradicts the assumption that p₁, ., p_n is the complete finite list of primes. Thus there are infinitely many primes.
- Algebraic Division and the Factor Theorem
- A rational fraction P(x) / Q(x) is improper if the degree of the numerator P(x) is greater than or equal to the degree of the denominator Q(x). Dividing P(x) by Q(x) gives the identity P(x) ≡ Q(x) × Quotient(x) + Remainder(x), where the degree of Remainder(x) is strictly less than the degree of Q(x). By the Factor Theorem, if (ax - b) is a factor of P(x), then P(b / a) = 0.
- Partial Fractions: Distinct and Repeated Linear Factors
- For a proper fraction with distinct linear factors in the denominator, use the identity f(x) / [(x - a)(x - b)(x - c)] ≡ A / (x - a) + B / (x - b) + C / (x - c). If the denominator contains a repeated linear factor (x - b)², you must include both powers of the repeated factor in the decomposition: f(x) / [(x - a)(x - b)²] ≡ A / (x - a) + B / (x - b) + C / (x - b)². Multiply through by the full denominator and use strategic x-values (cover-up substitution) combined with equating coefficients of x² or constants.
- Improper Fractions into Partial Fractions
- When the degree of the numerator is greater than or equal to the degree of the denominator, you must either perform polynomial long division first or include polynomial terms (such as A + B/(x - p) + C/(x - q) when degrees are equal, or Ax + B + C/(x - p) + D/(x - q) when the numerator degree is 1 higher than the denominator) before finding the partial fraction constants.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
How do you express an odd integer and an even integer algebraically in a formal proof?
What is the correct partial fraction form for (5x² + 2) / [(x - 1)(2x + 3)²]?
What is the correct partial fraction form for the improper fraction (3x² + 4x - 5) / [(x - 1)(x + 2)]?
Why does if a² is divisible by 3, then a is divisible by 3 hold true in a proof that √3 is irrational?
What is the degree of the remainder R(x) when a polynomial P(x) of degree 4 is divided by a quadratic divisor Q(x) of degree 2?
How do you simplify (2x² - 5x - 3) / (4x² - 1) before integrating or differentiating?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Use proof by contradiction to prove that √3 is an irrational number. State every logical step clearly. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Assume the contrary: suppose √3 is rational, so √3 = a / b where a and b are integers (b ≠ 0) in their simplest form with no common factors (coprime) [1 mark]. 2. Square both sides and rearrange: 3 = a² / b² → a² = 3b² [1 mark]. 3. Since a² is a multiple of 3 (and 3 is prime), a must also be a multiple of 3; let a = 3k for some integer k ∈ ℤ [1 mark]. 4. Substitute a = 3k into a² = 3b²: (3k)² = 3b² → 9k² = 3b² → b² = 3k², which implies b² is a multiple of 3 and therefore b is also a multiple of 3 [1 mark]. 5. Since both a and b are multiples of 3, they share a common factor of 3, which contradicts our initial assumption that a and b have no common factors; therefore √3 is irrational [1 mark].
- Q2
Express f(x) = (11x² + 14x + 5) / [(x + 1)²(2x + 1)] in partial fractions. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Set up the correct identity with repeated linear factor: (11x² + 14x + 5) / [(x + 1)²(2x + 1)] ≡ A / (x + 1) + B / (x + 1)² + C / (2x + 1) [1 mark]. 2. Multiply through by the denominator (x + 1)²(2x + 1): 11x² + 14x + 5 ≡ A(x + 1)(2x + 1) + B(2x + 1) + C(x + 1)² [1 mark]. 3. Substitute x = -1: 11(-1)² + 14(-1) + 5 = 2 = B(-1) → B = -2 [1 mark]. 4. Substitute x = -1/2: 11(1/4) + 14(-1/2) + 5 = 3/4 = C(1/4) → C = 3 [1 mark]. 5. Equate coefficients of x²: 11 = 2A + C → 11 = 2A + 3 → 2A = 8 → A = 4; final expression: 4 / (x + 1) - 2 / (x + 1)² + 3 / (2x + 1) [1 mark].
- Q3
Given that (3x³ + 5x² - 7x + 6) ÷ (x² + 2x - 3) ≡ Ax + B + C / (x - 1) + D / (x + 3), find the values of the constants A, B, C, and D. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Perform polynomial long division of (3x³ + 5x² - 7x + 6) by (x² + 2x - 3): first quotient term is 3x; subtracting 3x(x² + 2x - 3) = 3x³ + 6x² - 9x leaves -x² + 2x + 6 [1 mark]. 2. Next quotient term is -1; subtracting -1(x² + 2x - 3) = -x² - 2x + 3 leaves remainder 4x + 3, so A = 3 and B = -1 [1 mark]. 3. Factorise the denominator x² + 2x - 3 = (x - 1)(x + 3) and decompose the remainder fraction: (4x + 3) / [(x - 1)(x + 3)] ≡ C / (x - 1) + D / (x + 3) → 4x + 3 ≡ C(x + 3) + D(x - 1) [1 mark]. 4. Substitute x = 1: 4(1) + 3 = 7 = 4C → C = 7/4 [1 mark]. 5. Substitute x = -3: 4(-3) + 3 = -9 = -4D → D = 9/4; thus A = 3, B = -1, C = 7/4, D = 9/4 [1 mark].
- Q4
Prove algebraically that for all positive integers n, the expression n³ - n is divisible by 6. (3 marks)
Click “Check Mark Scheme” to compare your answer, 1. Factorise n³ - n completely: n³ - n = n(n² - 1) = (n - 1)n(n + 1) [1 mark]. 2. Observe that (n - 1), n, and (n + 1) are three consecutive integers, so at least one of them is an even number (divisible by 2) and exactly one of them is divisible by 3 [1 mark]. 3. Since 2 and 3 are coprime factors of (n - 1)n(n + 1), their product is divisible by 2 × 3 = 6 for all positive integers n [1 mark].
Common Exam Questions (4)
What is the most common reason students lose the final mark in a proof by contradiction?
Failing to state at the very start that a and b are integers with NO common factors (or that the fraction a / b is in its simplest form), or forgetting to write a concluding sentence explaining the contradiction and restating the proven theorem.
Why do we need both B / (x - a) and C / (x - a)² for a repeated root in partial fractions?
A proper rational function whose denominator has degree 3 has a general numerator of degree at most 2 (px² + qx + r), which has 3 independent coefficients. Including both B / (x - a) and C / (x - a)² ensures 3 unknown constants (A, B, C) so the identity can match any quadratic numerator.
How can I check my partial fractions answer in 15 seconds during an exam?
Pick a simple random number not equal to any root of the denominator (such as x = 2 or x = 0) and substitute it into both your original fraction and your partial fractions answer on your calculator. If the two decimals match, your algebra is almost certainly right.
When should I use polynomial long division vs equating coefficients for improper fractions?
When the numerator degree is higher than the denominator degree (e.g., cubic over quadratic), polynomial long division is usually faster and less error-prone. When both degrees are equal (quadratic over quadratic), writing A + B/(x - p) + C/(x - q) and multiplying through is faster.
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