Skip to main content

Proof methods and counterexamples: VCE Specialist Mathematics Unit 3

Syllabus dot point

“Methods of proof including direct proof, proof by contrapositive, proof by contradiction, and the use of a single counterexample to disprove a universal statement, together with the language of quantifiers and implication”

6 min read

Quick answer

A direct proof assumes the hypothesis and reasons forward to the conclusion. A proof by contrapositive proves "if PP then QQ" by instead proving the logically equivalent "if not QQ then not PP". A proof by contradiction assumes the statement is false, derives a logical impossibility, and concludes the statement must be true. To disprove a universal claim ("for all …\dots") a single counterexample suffices. An existential claim ("there exists …\dots") is proved by exhibiting one example and disproved only by ruling out every case. Knowing which method fits a statement is half the work.

Jump to a section
  1. What this dot point is asking
  2. Implication and quantifiers
  3. Direct proof
  4. Proof by contrapositive
  5. Proof by contradiction
  6. Disproof by counterexample
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

VCAA wants you to prove and disprove mathematical statements using the main methods: direct proof, proof by contrapositive, proof by contradiction, and disproof by a single counterexample. You also need the logical language of implication and quantifiers so you know exactly what a statement claims and what would establish or refute it.

Implication and quantifiers

Most statements to prove have the form "if PP then QQ", written P⇒QP \Rightarrow Q. The converse Q⇒PQ \Rightarrow P and the contrapositive "not Q⇒Q \Rightarrow not PP" are different statements: the contrapositive is logically equivalent to the original, but the converse is not.

A universal statement ("for all xx, ...") claims something about every case. An existential statement ("there exists xx such that ...") claims at least one case works. The way you prove or disprove each depends on this structure.

Direct proof

Assume the hypothesis PP and deduce QQ by valid steps. For example, to prove "the sum of two even integers is even", write the integers as 2m2m and 2n2n, add to get 2(m+n)2(m + n), and note this is even. Direct proof is the default when the path from PP to QQ is clear.

Proof by contrapositive

To prove P⇒QP \Rightarrow Q, prove the equivalent "not Q⇒Q \Rightarrow not PP". This helps when the negation of QQ is easier to work with. For instance, "if n2n^2 is even then nn is even" is awkward directly, but the contrapositive "if nn is odd then n2n^2 is odd" follows at once: n=2k+1n = 2k + 1 gives n2=4k2+4k+1=2(2k2+2k)+1n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, which is odd.

Proof by contradiction

Assume the statement is false, then derive something impossible. The classic example is proving 2\sqrt{2} is irrational: assume 2=pq\sqrt{2} = \frac{p}{q} in lowest terms, square to get p2=2q2p^2 = 2q^2, deduce pp is even, then qq is even, contradicting "lowest terms". Since the assumption leads to contradiction, the original statement holds.

Disproof by counterexample

A universal statement is false if even one case fails, so a single counterexample disproves it. To disprove "every prime is odd", give the example 22, which is prime and even. You do not need to explain why other primes are odd; one exception is enough.

Worked example

Choosing the right method

Prove that if n2n^2 is odd, then nn is odd, for any integer nn.

Step 1: Choose the proof method

A direct proof would require deducing the parity of nn from the parity of n2n^2, which is awkward because squaring obscures the factor of 22. The contrapositive of "n2n^2 odd ⇒\Rightarrow nn odd" is "nn even ⇒\Rightarrow n2n^2 even", which runs in the easier direction: starting from nn even, we can immediately write a formula for nn and square it.

Step 2: Prove the contrapositive

Suppose nn is even, so n=2kn = 2k for some integer kk. Then

n2=(2k)2=4k2=2(2k2).n^2 = (2k)^2 = 4k^2 = 2(2k^2).

Since 2k22k^2 is an integer, n2n^2 is twice an integer, hence even.

Step 3: Conclude

We have shown "nn even ⇒\Rightarrow n2n^2 even". Because the contrapositive is logically equivalent to the original implication, "if n2n^2 is odd then nn is odd" is proved for all integers nn.

As a check, n=3n = 3 gives n2=9n^2 = 9 (odd, nn odd) and n=4n = 4 gives n2=16n^2 = 16 (even, nn even), consistent with the result.

Common exam traps
Proving the converse
Proving Q⇒PQ \Rightarrow P does not establish P⇒QP \Rightarrow Q. The converse is a different statement.
Generalising from examples
Checking a universal statement on a few cases is not a proof. One counterexample disproves, but no finite list of examples proves a "for all" claim.
Forgetting to reach a contradiction
In proof by contradiction you must actually derive an impossibility, then explicitly conclude the assumption was false.
Negating quantifiers wrongly
The negation of "for all xx, P(x)P(x)" is "there exists xx with not P(x)P(x)", not "for all xx, not P(x)P(x)".

Examples in context

Example 1. To disprove "n2+n+41n^2 + n + 41 is prime for all positive integers nn", use n=41n = 41, which gives a multiple of 4141.

Example 2. "The product of two odd integers is odd" is a direct proof: (2a+1)(2b+1)=2(2ab+a+b)+1(2a+1)(2b+1) = 2(2ab + a + b) + 1.

Try this

Q1. State the contrapositive of "if it rains then the ground is wet". [1 mark]

  • Cue. If the ground is not wet then it did not rain.

Q2. Disprove "every multiple of 33 is odd". [1 mark]

  • Cue. 66 is a multiple of 33 and is even.

Q3. Outline a proof by contradiction that there is no largest integer. [2 marks]

  • Cue. Assume NN is largest; then N+1N + 1 is larger, a contradiction.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style1 mark
Consider the following statement. 'If my football team plays badly, then they are not training enough.' Which one of the following statements is the contrapositive of the statement above? A. If they are not training enough, then my football team plays badly. B. If my football team plays badly, then they need more training. C. If they are training enough, then my football team does not play badly. D. If my football team doesn't play badly, then they are training enough. E. If they are training enough, then my football team will most likely win.
Show worked answer →

The answer is C.

The contrapositive of "if P then Q" is "if not Q then not P". It is logically equivalent to the original statement.

Identify the parts: P is "my football team plays badly" and Q is "they are not training enough".

Negate each: not Q is "they are training enough", and not P is "my football team does not play badly".

Assemble "if not Q then not P": "If they are training enough, then my football team does not play badly." That is option C.

Common trap: option A is the converse and option D is the inverse, neither of which is logically equivalent to the original.

2025 VCAA-style1 mark
A tiger is a type of cat. Consider the following statement. 'If I have a tiger, then I have a cat.' The contrapositive of this statement is A. if I do not have a tiger, then I do not have a cat. B. if I have a cat, then I have a tiger. C. if I do not have a cat, then I do not have a tiger. D. if I do not have a tiger, then I have a different type of cat.
Show worked answer →

The answer is C.

Write the statement as "if P then Q" with P = "I have a tiger" and Q = "I have a cat". The contrapositive is "if not Q then not P".

Negate each part: not Q is "I do not have a cat" and not P is "I do not have a tiger".

So the contrapositive is "if I do not have a cat, then I do not have a tiger", which is option C, and it is true.

Option A is the inverse and option B is the converse; neither is guaranteed true (a cat need not be a tiger).

2025 VCAA-style1 mark
Consider the following statement. 'If f''(0) = 0, then the graph of f necessarily has a point of inflection at x = 0.' A counter-example that disproves this statement is when A. f(x) = arcsin(x). B. f(x) = x^2 / 12. C. f(x) = x^(1/3). D. f(x) = x^4.
Show worked answer →

The answer is D.

A counterexample must satisfy the hypothesis f''(0) = 0 yet fail the conclusion, that is, have no point of inflection at x = 0.

Test D, f(x) = x^4. Then f'(x) = 4x^3 and f''(x) = 12x^2, so f''(0) = 0, satisfying the hypothesis. But f''(x) = 12x^2 is greater than or equal to 0 for all x and does not change sign at x = 0, so the concavity does not change: x = 0 is a minimum, not a point of inflection. This disproves the statement.

Why the others fail: for A, arcsin(x) has f''(0) = 0 but the concavity does change there, so it is a genuine inflection (not a counterexample). For C, f''(0) does not exist. For B, f''(x) = 1/6, so f''(0) is not 0.

Practise this

Sources & how we know this

ExamExplained