Sometimes tested Paper 2 (Mathematical Reasoning)

Mathematical Proof: Direct Proof, Cases, Contradiction and Counterexample

Proof is the backbone of TMUA Paper 2. You must construct and follow direct proofs, split into cases, argue by contradiction, and disprove false claims with a single counterexample, all by hand and at speed. Match your method to what the statement is really asking.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • A proof is a watertight chain of steps from stated assumptions to a conclusion; each step must hold with certainty, not merely for the cases you happened to try.
  • Direct proof: assume the hypothesis and deduce the conclusion algebraically (write an even number as 2k, an odd number as 2k + 1).
  • Proof by cases: break every possibility into a finite list and prove each one.
    e.g. any integer is even, n = 2k, or odd, n = 2k + 1
  • Proof by contradiction: assume the statement is false, derive something impossible, and conclude the original assumption must have been wrong.
  • Disproof by counterexample: one case where the claim fails destroys a 'for all' statement.
    e.g. 23 + 1 = 9 is not prime
  • Conjecture then justify: use small cases to spot a pattern, then prove it in general - checking finitely many values of n is evidence, never proof.
  • An implication and its converse are different: 'P implies Q' can be true while 'Q implies P' is false, so never quietly assume the reverse direction.
  • Ordering a proof: start from the assumptions, let each line use only facts already established, and place the conclusion last.

Formulae

n even: n = 2k; n odd: n = 2k + 1

Parametrise integers by parity - the opening move for most direct and case proofs (k an integer).

n = d × k (n divisible by d)

Turn a 'divisible by d' hypothesis into an equation you can substitute and manipulate.

n2 - m2 = (n - m)(n + m)

Difference of two squares - a frequent tool for divisibility and factorisation proofs.

Definitions

Conjecture
A statement that looks true from evidence such as small cases, but has not yet been proved.
Counterexample
A single specific case in which a general claim fails; it disproves a 'for all' statement but can never prove one true.
Converse
The statement obtained by swapping hypothesis and conclusion: the converse of 'if P then Q' is 'if Q then P'. Its truth is independent of the original.
Contradiction
A conclusion that cannot possibly hold, such as an integer being both even and odd; reaching one completes a proof by contradiction.

Worked examples

1

Prove that for any integer n, if n2 is even then n is even.

  1. 1

    Argue by contradiction.

    Assume n2 is even but n is odd, so n = 2k + 1 for some integer k.

  2. 2

    Square both sides:

    n2 = 4k2 + 4k + 1
  3. 3

    Take out a factor of 2:

    n2 = 2(2k2 + 2k) + 1
  4. 4

    The right-hand side has the form 2*(integer) + 1, so n2 is odd.

    This contradicts the assumption that n2 is even.

  5. 5

    Hence the supposition was wrong, and n must be even.

Answer: n is even. (Equivalently, we have proved the contrapositive: n odd implies n2 odd.)

2

A student conjectures that 2n + 1 is prime for every positive integer n. Determine whether the conjecture is true.

  1. 1

    Test small cases.

    For n = 1:
    21 + 1 = 3, which is prime
  2. 2
    For n = 2:
    22 + 1 = 5, which is prime
  3. 3
    For n = 3:
    23 + 1 = 9, and 9 = 3 × 3
  4. 4

    Since 9 is not prime, the pattern seen at n = 1 and n = 2 fails; one counterexample is enough to disprove a 'for every n' claim.

Answer: The conjecture is false; n = 3 is a counterexample, since 23 + 1 = 9 = 3 × 3 is not prime.

Common mistakes

  • ×Treating checked cases as a proof: verifying n = 1, 2, 3 says nothing about all n - you still owe a general argument.
  • ×Proving the converse by mistake: showing 'Q implies P' when the question asked for 'P implies Q'. They are separate claims.
  • ×In proof by contradiction, never actually reaching an impossibility, or contradicting a fact you were not given.
  • ×Trying to prove a statement with an example: a counterexample can only disprove, and one example can never establish a 'for all' result.
  • ×Circular reasoning: quietly assuming the very thing you are meant to prove, so the conclusion appears among your assumptions.

No-calculator tips

  • ✓Parametrise at once: replace 'even' with 2k and 'odd' with 2k + 1 - most number proofs then fall straight out of routine algebra.
  • ✓To break a 'for all' claim, test small or awkward inputs first (0, 1, 2, a negative, or a fraction) before committing to a full proof.
  • ✓For 'put the steps in order' questions, find the line that uses only the given assumptions to start, chain each later line to facts already stated, and finish with the conclusion.
  • ✓Match method to quantifier: 'there exists' needs just one worked example, whereas 'for all' demands a general argument.
  • ✓Mind the direction of the arrow: if a question leans on a converse, try a quick example to see whether the converse can fail.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.A student conjectures: "For every real number x, x2 ≥ x." Which single value of x DISPROVES this conjecture (is a counterexample)?

  • A. x = -3
  • B. x = 0
  • C. x = 1/2
  • D. x = 1
  • E. x = 5
Show answer

Answer: C — x = 1/2

At x = 1/2, x2 = 1/4 which is less than 1/2, so the inequality fails. The common trap is to test negatives (e.g. x = -3 gives 9 ≥ -3, true), but the statement actually breaks only for values strictly between 0 and 1.

Q2.You want to prove by contradiction the statement: "For all integers n, if n2 is even then n is even." Which assumption should you begin with?

  • A. Assume n2 is odd and n is even.
  • B. Assume n2 is even and n is even.
  • C. Assume for all integers n, n2 is even and n is odd.
  • D. Assume there exists an integer n such that n2 is even and n is odd.
  • E. Assume n is even and deduce that n2 is even.
Show answer

Answer: D — Assume there exists an integer n such that n2 is even and n is odd.

The negation of "for all n, P(n) ⇒ Q(n)" is "there exists n with P(n) true and Q(n) false", i.e. an integer n with n2 even but n odd. Option 2 keeps the wrong (universal) quantifier, and option 1 assumes what you actually want to prove.

Q3.Exactly one of the following statements is FALSE (can be disproved by a counterexample). Which one?

  • A. For every integer n, if n is divisible by 6 then n is divisible by 3.
  • B. For every integer n, if n is divisible by 3 then n is divisible by 6.
  • C. For every real number x, (x - 1)2 ≥ 0.
  • D. For every integer n, n2 + n is even.
Show answer

Answer: B — For every integer n, if n is divisible by 3 then n is divisible by 6.

n = 9 is divisible by 3 but not by 6, so option 1 (index 1) is false. It is the converse of the true statement in index 0; confusing a true implication with its converse is the classic error here.

Read this topic in the official UAT-UK TMUA content specification →

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