OCR A Level Mathematics A: Proof: deduction, exhaustion, counter-example, logical connectives and contradiction (H240) – Revision Notes

These revision notes condense section 1.01 Proof (1.01a to 1.01d) of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Stage 1 content (1.01a to 1.01c) is shared with AS. Proof by contradiction, 1.01d, is A Level only. The study guide explains each method at length. Test yourself with the practice questions.

Course links: the OCR A Level Mathematics hub, the printable checklist and the free diagnostics.

The four methods at a glance

Method	Ref	Idea	Use it when
Deduction	1.01a	Start from known facts, apply valid steps, reach the result	The statement is general and algebra can show a form
Exhaustion	1.01a	Split into a finite list of cases and prove each	The cases are few, or can be sorted by remainder
Counter-example	1.01c	One value that makes P true and Q false	You must show a statement is false
Contradiction	1.01d, A Level only	Assume the opposite and reach an impossibility	“No such number exists”, or “is irrational”

Structure of every proof

1. Define letters: “Let n be an integer.”
2. Work in steps, one idea per line.
3. Finish with a sentence that states what has been proved.

The specification describes a formal proof as having variables clearly defined, correct algebraic manipulation and a concise conclusion. Testing a few values is evidence, never proof.

Forms to know

Statement	Write it as
Even / odd integer	2n / 2n + 1
Three integers in a row	n, n + 1, n + 2
Two integers that may differ	m and n
Multiple of 5	5n
Remainder 3 on division by 5	5n + 3
Rational number	a/b, with a and b integers, b ≠ 0
Irrational number	real, but not rational

Deduction in steps

1. Write the objects in general form.
2. Expand and simplify.
3. Factorise, or complete the square, to show the required property.
4. Say why the property follows (for example, “2 times an integer is even”).

Reminder. Prove that the sum of an odd and an even integer is odd.

(2m + 1) + 2n = 2(m + n) + 1

Since m + n is an integer, the sum has the form 2k + 1 and is odd.

Exhaustion in steps

1. List every case and check that the list is complete.
2. Prove each case separately.
3. Say that the cases cover all possibilities.

Reminder. Prove that n⁴ leaves remainder 0 or 1 when divided by 5. Every integer is congruent to one of 0, 1, 2, 3 or 4 when divided by 5, so check each residue.

0⁴ = 0    1⁴ = 1    2⁴ = 16    3⁴ = 81    4⁴ = 256

These have remainders 0, 1, 1, 1 and 1 on division by 5. All five cases give 0 or 1, so the result holds for every integer.

Connectives and language (1.01b)

Symbol	Name	Read as
⇒	implies	if P then Q
⇐	is implied by	if Q then P
⇔	is equivalent to	P ⇒ Q and Q ⇒ P (if and only if, iff)
≡	identity	true for every value of the variable

• The converse of P ⇒ Q is Q ⇒ P. It may be false when P ⇒ Q is true.
• The contrapositive of P ⇒ Q is “not Q ⇒ not P”. The two are equivalent.
• To prove “P if and only if Q”, prove both directions and say so.
• Squaring both sides gives ⇒ only. Check answers in the original equation.
• An equation holds for some values. An identity (≡) holds for all.

Disproof by counter-example (1.01c)

Method:

1. Choose a value that satisfies P.
2. Show that Q fails for it.
3. State that the statement is false.

Reminder. Disprove: “If a > b then 1/a < 1/b.”

Take a = 2 and b = −1. Then 2 > −1, so P holds. But 1/a = 1/2 and 1/b = −1, and 1/2 < −1 is false. The statement is false.

Rules: the value must lie in the stated domain, and it must be checked, not just quoted.

Contradiction in steps (A Level only)

1. Assume the exact opposite of what you want to prove.
2. Deduce consequences with valid steps.
3. Reach a statement that is false, or contradicts the assumption.
4. Conclude that the assumption is false, so the original statement is true.

Standard results.

• √2 is irrational: assume √2 = p/q in lowest terms, so p² = 2q². Then p is even, p = 2r, then q² = 2r², so q is even. Both have factor 2, a contradiction.
• There are infinitely many primes: assume a finite list p₁ to pₖ and form N = p₁ × … × pₖ + 1. No listed prime divides N, so N has a prime factor off the list.

Reminder. Prove that there are no positive integers m and n with m² − n² = 2.

Suppose there are such m and n. Then (m − n)(m + n) = 2. The two factors differ by 2n, which is even, so they are both even or both odd. If both are even, the product is a multiple of 4, not 2. If both are odd, the product is odd, not 2. Either way there is a contradiction, so no such integers exist.

Notation to use in your answers

Write ∴ for “therefore” only if you also state the reason in words. Use ℤ for the integers and ℚ for the rationals if you wish, but define any symbol you introduce. In a proof, “Let n be an integer” is a complete first line, and “QED” on its own is not a conclusion.

Must-know distinctions

• Proof or evidence. Many examples support a statement. Only an argument for all cases proves it.
• Disprove or not prove. One counter-example disproves. Failing to find a counter-example proves nothing.
• Contrapositive or converse. The contrapositive is equivalent to the original. The converse is a different statement.
• Rational or irrational. Rational means a ratio of integers with a non-zero denominator. Irrational means real but not rational.
• Identity or equation. Use ≡ only when the statement is true for all values.

Quick self-test

1. Insert ⇒, ⇐ or ⇔ for integers n: “n is a multiple of 3” … “n is a multiple of 6”.
2. Write the converse of “if x > 4 then x² > 16”. Is the converse true?
3. Find a value showing that “if a > b then a² > b²” is false.
4. What is the opposite of “√5 is irrational”, as the first line of a proof by contradiction?
5. Show that n² + n is even for every integer n.
6. Is (x + 3)² ≡ x² + 9? Justify with one value.
7. You want to show that every cube leaves remainder 0, 1 or 3 on division by 4, splitting by the remainder of n on division by 4. How many cases do you need?
8. In the proof that √2 is irrational, why does p² even force p to be even?
9. Disprove “all prime numbers are odd”.
10. A student proves “x² = 9 ⇒ x = 3 or x = −3” and concludes ⇔. Is ⇔ justified?

Answers

1. The statement “multiple of 6” implies “multiple of 3”, but not the other way round. So write “n is a multiple of 3” ⇐ “n is a multiple of 6”.
2. The converse is “if x² > 16 then x > 4”. It is false: x = −5 gives x² = 25 > 16, but −5 is not greater than 4.
3. Take a = 1 and b = −3. Then a > b, but a² = 1 and b² = 9, and 1 > 9 is false. a = 1, b = −3 works.
4. Assume that √5 is rational.
5. n² + n = n(n + 1). One of two consecutive integers is even, so the product is even.
6. No. For x = 1, (x + 3)² = 16 but x² + 9 = 10. Not an identity.
7. Four cases: n = 4k, 4k + 1, 4k + 2 and 4k + 3.
8. If p were odd, p² would be odd (odd × odd = odd). So an even p² means p is even.
9. The number 2 is prime and even, so 2 is a counter-example.
10. Yes, in this direction, since x = 3 or x = −3 gives x² = 9. Both directions hold, so ⇔ is justified provided you say why the reverse works.

Where marks are usually lost

• Writing “hence proved” after checking a few numbers.
• Not defining the letters, or using one letter for two different integers.
• Skipping the algebra between a correct start and the claimed form.
• Missing the concluding sentence in terms of the question.
• Treating ⇒ as ⇔, especially after squaring both sides.
• Proving only one direction of an “if and only if” statement.
• Giving a counter-example that fails P, or not checking that Q fails.
• In a contradiction proof, starting from the wrong opposite, or not saying where the contradiction is.
• Leaving a fraction a/b unreduced when the proof relies on lowest terms.

Official syllabus

OCR, A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, section 1.01 Proof (references 1.01a to 1.01d), published by OCR.