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GCSE/Mathematics/AQA

A6Distinguish equations, identities, formulae; argue with algebraic equivalence

Notes

Identities — proving algebraic equivalence

An identity is an equation that holds true for every value of the variable. To prove one, you transform one side until it matches the other — you never assume the result is true and "balance" both sides like an equation.

How identities differ from equations

  • Equation: 2x + 4 = 10 → solve to find x = 3.
  • Identity: 2(x + 2) ≡ 2x + 4 → no solving; both sides describe the same expression.

The symbol ("identically equal to") signals an identity. In examination questions you might also see "Show that..." or "Prove that...".

The proof technique

Pick the more complicated side (usually the LHS) and transform it step by step into the other side. Show every step. Never start from "what we are trying to prove".

Worked exampleWorked example 1 — straight expansion

Show that (x + 4)(x - 4) ≡ x² - 16.

LHS = (x + 4)(x - 4) = x² - 4x + 4x - 16 = x² - 16 = RHS. □

Worked exampleWorked example 2 — perfect square

Show that (2n + 1)² ≡ 4n² + 4n + 1.

LHS = (2n + 1)² = 4n² + 2n + 2n + 1 = 4n² + 4n + 1 = RHS. □

Worked exampleWorked example 3 — comparing coefficients

Given (x + a)(x + b) ≡ x² + 7x + 12 for all x, find a and b.

Expand: x² + (a + b)x + ab. Compare:

  • Coefficient of x: a + b = 7.
  • Constant: ab = 12.

So a and b are 3 and 4 (either way round).

Worked exampleWorked example 4 — proof with consecutive integers

Show that the sum of three consecutive integers is always a multiple of 3.

Let the integers be n - 1, n, n + 1. Sum = (n - 1) + n + (n + 1) = 3n, which is a multiple of 3. □

This consecutive-integer trick is examined every year. Pick a clever centre (n) so symmetry kills the constants.

Worked exampleWorked example 5 — algebraic divisibility

Show that (2n + 1)² - (2n - 1)² is always a multiple of 8.

Use difference of two squares: (2n + 1 + 2n - 1)(2n + 1 - 2n + 1) = (4n)(2) = 8n. So the expression is exactly 8n, a multiple of 8. □

Common mistakesCommon mistakes (examiner traps)

  1. Starting from the conclusion. Assuming what you need to prove and "balancing" both sides earns 0 marks. Always work from one side to the other.
  2. Sign errors during expansion. Especially with perfect squares: (a - b)² = a² - 2ab + b², not a² - b².
  3. Forgetting the symmetry trick for "any/all integers" proofs. Choose your variables wisely (e.g. n - 1, n, n + 1 for three consecutives).
  4. Comparing constants but forgetting to compare x-coefficients when matching identities.
  5. Stopping before the answer matches the RHS. Show the final simplification step.

Try thisQuick check

Show that (n + 3)² - (n + 1)² ≡ 4n + 8. LHS = (n² + 6n + 9) - (n² + 2n + 1) = 4n + 8. □

AI-generated · claude-opus-4-7 · v3-deep-algebra

Practice questions

Try each before peeking at the worked solution.

  1. Question 13 marks

    Verify a simple identity

    (F1) Show that (x + 2)(x + 3) ≡ x² + 5x + 6.

    [Foundation tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  2. Question 22 marks

    Identity with a square

    (F/H2) Show that (2x - 3)² ≡ 4x² - 12x + 9.

    [Foundation/Higher crossover]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  3. Question 34 marks

    Compare coefficients

    (H3) Given that (x + p)(x - 2) ≡ x² + 3x - 10 for all x, find p.

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  4. Question 43 marks

    Sum of three consecutive integers

    (H4) Prove that the sum of three consecutive integers is always a multiple of 3.

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  5. Question 53 marks

    Difference of squares proof

    (H5) Prove that (n + 5)² - (n + 3)² is always a multiple of 4.

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  6. Question 63 marks

    Proof involving even/odd numbers

    (H6) Prove that the sum of two consecutive even numbers is divisible by 2 but not necessarily 4.

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

  7. Question 74 marks

    Identity with multiple unknowns

    (H7) Find a and b so that (2x + a)(x + 3) ≡ 2x² + bx + 15 for all x.

    [Higher tier]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-deep-algebra

Flashcards

A6 — Identities and proof

10-card SR deck for AQA GCSE Maths topic A6

10 cards · spaced repetition (SM-2)