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IGCSE Maths · Edexcel 4XMA1 · Unit 2 Higher · Algebra

Algebraic Proof — Worked Solutions

Full worked solutions to all 12 questions (40 marks), set out the way an Edexcel examiner wants to see them.

Section A — Warm-up (grades 4–5)
  1. (2 marks)Give a counterexample to show that this statement is false: “The sum of two prime numbers is always even.”
    Solution
    1. Choose 2 (the only even prime) and an odd prime, e.g. 2 + 3 = 5.
    2. 5 is odd, so the statement is false.
    e.g. 2 + 3 = 5
  2. (3 marks)Prove that the sum of any three consecutive integers is a multiple of 3.
    Solution
    1. Let the integers be n, n + 1, n + 2.
    2. Sum = 3n + 3 = 3(n + 1).
    3. n + 1 is an integer, so the sum is a multiple of 3.
    3(n + 1)
  3. (3 marks)Prove that the difference between the squares of two consecutive integers is always odd.
    Solution
    1. Let the integers be n and n + 1.
    2. (n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1.
    3. 2n is even, so 2n + 1 is odd.
    2n + 1, which is odd
Section B — Core (grades 5–7)
  1. (3 marks)Prove that the product of any two odd numbers is odd.
    Solution
    1. Let the odd numbers be 2n + 1 and 2m + 1, where n and m are integers.
    2. (2n + 1)(2m + 1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) + 1.
    3. 2nm + n + m is an integer, so the product is of the form 2k + 1, which is odd.
    2(2nm + n + m) + 1
  2. (3 marks)Prove that the sum of the squares of two consecutive even numbers is a multiple of 4.
    Solution
    1. Let the even numbers be 2n and 2n + 2.
    2. (2n)² + (2n + 2)² = 4n² + 4n² + 8n + 4 = 8n² + 8n + 4.
    3. = 4(2n² + 2n + 1), which is a multiple of 4.
    4(2n² + 2n + 1)
  3. (3 marks)Prove that (x + 2)² − (x − 2)² ≡ 8x.
    Solution
    1. LHS = (x² + 4x + 4) − (x² − 4x + 4)
    2. = 8x = RHS.
    LHS ≡ RHS
  4. (3 marks)Prove that x² − 8x + 19 > 0 for all values of x.
    Solution
    1. x² − 8x + 19 = (x − 4)² − 16 + 19 = (x − 4)² + 3.
    2. (x − 4)² ≥ 0, so (x − 4)² + 3 ≥ 3 > 0 for all x.
    (x − 4)² + 3 > 0
  5. (4 marks)n is a positive integer. Prove that (n + 1)² + (n − 1)² is always even, and is 2 more than twice a square number.
    Solution
    1. (n + 1)² + (n − 1)² = n² + 2n + 1 + n² − 2n + 1 = 2n² + 2.
    2. 2n² + 2 = 2(n² + 1), so it is even.
    3. 2n² + 2 = 2 × n² + 2, which is 2 more than twice the square number n².
    2n² + 2
Section C — Stretch (grades 8–9)
  1. (4 marks)Prove that the difference between the squares of any two consecutive odd numbers is a multiple of 8.
    Solution
    1. Let the odd numbers be 2n + 1 and 2n + 3.
    2. (2n + 3)² − (2n + 1)² = (4n² + 12n + 9) − (4n² + 4n + 1) = 8n + 8.
    3. = 8(n + 1), and n + 1 is an integer, so it is a multiple of 8.
    8(n + 1)
  2. (4 marks)Prove that the square of any odd number is one more than a multiple of 8.
    Solution
    1. (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1.
    2. n(n + 1) is the product of consecutive integers, so it is even: n(n + 1) = 2k for some integer k.
    3. So (2n + 1)² = 4(2k) + 1 = 8k + 1, which is one more than a multiple of 8.
    8k + 1
  3. (4 marks)Prove that n³ − n is a multiple of 6 for all integers n.
    Solution
    1. n³ − n = n(n² − 1) = (n − 1)n(n + 1).
    2. This is the product of three consecutive integers.
    3. Among any three consecutive integers, at least one is even and exactly one is a multiple of 3.
    4. So the product is divisible by 2 and by 3, so it is a multiple of 6.
    (n − 1)n(n + 1) is divisible by 2 and 3
  4. (4 marks)(a) Write 2x² + 4x + 7 in the form a(x + b)² + c.
    (b) Hence prove that 2x² + 4x + 7 is always greater than 4.
    Solution
    1. (a) 2x² + 4x + 7 = 2(x² + 2x) + 7 = 2[(x + 1)² − 1] + 7 = 2(x + 1)² + 5.
    2. (b) 2(x + 1)² ≥ 0, so 2(x + 1)² + 5 ≥ 5 > 4 for all x.
    (a) 2(x + 1)² + 5 (b) minimum value 5 > 4
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