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TMUA Proof by Cases, Exhaustion and Counterexample

A 'for all' claim can be proved by splitting into cases that cover everything, or disproved with one counterexample. The specification also covers making a conjecture from small cases and justifying it.

Paper 2 topic. Specification codes: Prf1, Prf3, Prf5. Last checked 12 November 2026.

Proof by cases (exhaustion)

Split the possibilities so every one falls in exactly one case, such as nn even or odd, or nn leaving remainder 0, 1 or 2 on division by 3. Check each case.

Disproof by counterexample

To show 'for all nn, P(n)P(n)' is false, give one value of nn where it fails. Checking that it works for several values proves nothing.

Conjecturing from small cases

A pattern from small cases is a guess until it is proved. A formula that fits the first few values can still fail later.

Worked examples

Example 1

Prove that n3−nn^3-n is divisible by 3 for every integer nn.

Show the worked solution

n3−n=(n−1)n(n+1)n^3-n=(n-1)n(n+1), a product of three consecutive integers. Cases: if n=3kn=3k, the middle factor is a multiple of 3; if n=3k+1n=3k+1, then n−1=3kn-1=3k; if n=3k+2n=3k+2, then n+1=3k+3n+1=3k+3. So 3 always divides it.

Answer: Proved by the three cases

Example 2

Find the smallest counterexample to 'if nn is prime then 2n+12n+1 is prime'.

Show the worked solution

Test primes: n=2n=2 gives 5, n=3n=3 gives 7, n=5n=5 gives 11, all prime. n=7n=7 gives 15=3×515=3\times5, which is not prime.

Answer: n=7n=7

Example 3

The sums 1⋅21\cdot2, 1⋅2+2⋅31\cdot2+2\cdot3, 1⋅2+2⋅3+3⋅41\cdot2+2\cdot3+3\cdot4 are 2, 8, 20. A student conjectures that the sum of the first nn terms is n(n+1)(n+2)3\tfrac{n(n+1)(n+2)}3. Check the next term, and say whether this proves the formula.

Show the worked solution

The next sum is 20+4⋅5=4020+4\cdot5=40, and the formula gives 4⋅5⋅63=40\tfrac{4\cdot5\cdot6}3=40. This supports the conjecture but does not prove it.

Answer: Fits (40), but not a proof

Common mistakes

  • Treating several examples that work as a proof.
  • Choosing cases that do not cover every possibility.
  • Giving a counterexample that does not meet the condition of the statement.
  • Proving a case but forgetting to state the others are similar or checked.

Test yourself on this topic with our free practice questions.

Quick answers

What is proof by exhaustion?

A proof that splits the possibilities into a finite list of cases and checks every one.

How do you disprove a statement with a counterexample?

Find one value that satisfies the conditions but fails the conclusion, then state it clearly.

Does a pattern in small cases prove a formula?

No. It suggests a conjecture, which still needs a proof.

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

Related topics

Paper 2TMUA Proof by ContradictionLearn proof by contradiction for TMUA Paper 2: how to start, what the contradiction looks like, with worked examples you can check by hand.Paper 1TMUA Number, Factors and DivisibilityRevise primes, prime factorisation, HCF, LCM, counting factors and divisibility for the TMUA, with worked no-calculator examples.Paper 2TMUA Spotting Errors in ProofsLearn to find the faulty step in a purported proof for TMUA Paper 2: illegal cancelling, sign errors and invalid inferences, with worked examples.

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