16 exam-style questions plus unlimited generated ones on this topic, at up to four difficulty levels, with full mark schemes. Notes and worksheets are open to everyone. Practice needs a free account.
Or take a quick quiz: 10 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
A proof is a chain of logical steps from facts you know to the statement you want. You need proof by deduction (general algebra that covers every case), proof by exhaustion (checking every possible case) and disproof by counter-example. Proof questions appear on both Pure papers, usually for 3 to 5 marks, and always need a clear concluding statement.
Key facts
Even: \(2n\); odd: \(2n + 1\); consecutive: \(n, n + 1, n + 2\) (\(n \in \mathbb{Z}\))
Multiple of \(k\): show the expression \(= k \times\) (integer)
\((x - a)^2 \ge 0\) for all real \(x\)
Rational number: \(\dfrac{a}{b}\) with \(a, b\) integers, \(b \ne 0\)
\(\Rightarrow\) implies; \(\Leftarrow\) is implied by; \(\Leftrightarrow\) if and only if
Every integer is \(3k\), \(3k + 1\) or \(3k + 2\)
\(n(n + 1)\) is always even; \(n(n + 1)(n + 2)\) is always a multiple of 6
Exhaustion: the cases must cover every possibility
Counter-example: ONE case where the statement fails
Notes
What a proof must do
Start from facts that are known to be true: definitions, given information and standard results.
Every step must follow from the one before. Only write \(\Rightarrow\) ('implies') when the next line really follows.
Checking a few examples is not a proof: examples only suggest a result, unless you check every possible case.
Finish with a conclusion that repeats the statement, e.g. 'so \(n^3 - n\) is a multiple of 6 for all integers \(n\)'.
Proof by deduction
diagram
Write general numbers algebraically, with \(n, m \in \mathbb{Z}\): even \(2n\), odd \(2n + 1\), consecutive integers \(n\),\(n + 1\), \(n + 2\)
Use different letters for numbers that are not related: two odd numbers are \(2n + 1\) and \(2m + 1\)
To show something is a multiple of \(k\), write it as \(k \times\) (an expression that is an integer).
To show an expression is always positive, complete the square: \(x^2 - 6x + 11 = (x - 3)^2 + 2 \ge 2 \gt 0\), because a square is never negative.
The lowest point is (3, 2), so the curve never reaches the x-axis: x² − 6x + 11 > 0 for all x.
For positive \(a\) and \(b\), start from a square: \((\sqrt{a} - \sqrt{b})^2 \ge 0\) gives \(a + b \ge 2\sqrt{ab}\) Grade A or above only
Proof by exhaustion
Split all possibilities into a finite number of cases that together cover everything, then prove the statement in each case.
Useful splits: \(n\) even or odd; \(n = 3k\), \(3k + 1\) or \(3k + 2\); or a short finite list, e.g. every integer from 1 to 6.
Say why your cases cover everything, and prove every case, not just some of them.
Disproof by counter-example
One example where the statement is false is enough to disprove it.
Show the calculation and then say clearly that it is a counter-example, so the statement is false.
Good values to try: 0, 1, 2 (the only even prime), negative numbers and fractions.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Disprove a statement by counter-example
1 to 2 marksGrade E
Try small values, including 1, 2, 0 and negative numbers, until the statement fails.
Show the calculation for that value.
State that this is a counter-example, so the statement is not true.
Example. Show that the statement '\(2^n + 1\) is prime for every positive integer \(n\)' is false.
Show the model answerHide the model answer
When \(n = 3\), \(2^3 + 1 = 9\)
(M1)
\(9 = 3 \times 3\) is not prime, so \(n = 3\) is a counter-example and the statement is false.
(A1)
Prove by deduction that a result holds for all integers
3 to 4 marksGrade C
Write the numbers in general form, e.g. \(2n + 1\) and \(2n + 3\), and say that \(n\) is an integer.
Expand and simplify.
Rearrange into the form needed, e.g. \(8 \times\) (integer) \(+ 2\)
Write a conclusion that repeats the statement.
Example. Prove that the sum of the squares of any two consecutive odd numbers is 2 more than a multiple of 8.
Show the model answerHide the model answer
Let the numbers be \(2n + 1\) and \(2n + 3\), where \(n\) is an integer.
(B1)
\((2n + 1)^2 + (2n + 3)^2\)
\(= 4n^2 + 4n + 1 + 4n^2 + 12n + 9\)
\(= 8n^2 + 16n + 10\)
(M1)
\(= 8(n^2 + 2n + 1) + 2\)
(A1)
\(n^2 + 2n + 1\) is an integer, so the sum is 2 more than a multiple of 8.
(A1)
Prove by exhaustion
3 to 4 marksGrade A
Choose cases that cover every possibility, e.g. \(n = 3k\),\(3k + 1\),\(3k + 2\), and say so.
Substitute each case and simplify.
Show the statement is true in every case.
Conclude for all integers.
Example. Prove by exhaustion that \(n^2 + 1\) is not a multiple of 3 for any integer \(n\)
Show the model answerHide the model answer
Every integer can be written as \(3k\), \(3k + 1\) or \(3k + 2\), where \(k\) is an integer.
In every case the remainder on division by 3 is 1 or 2, so \(n^2 + 1\) is never a multiple of 3.
(A1)
Prove that an expression is always positive
3 marksGrade B
Complete the square.
State that the squared bracket is never negative, for all real \(x\)
Conclude that the expression is at least a positive number, so it is always positive.
Example. Prove that \(x^2 - 8x + 19 \gt 0\) for all real values of \(x\)
Show the model answerHide the model answer
\(x^2 - 8x + 19 = (x - 4)^2 - 16 + 19\)
(M1)
\(= (x - 4)^2 + 3\)
(A1)
\((x - 4)^2 \ge 0\) for all real \(x\), so \((x - 4)^2 + 3 \ge 3 \gt 0\). Hence \(x^2 - 8x + 19 \gt 0\) for all real \(x\).
(A1)
Don’t lose marks
Using the same letter for two unrelated numbers, e.g. writing two odd numbers as \(2n + 1\) and \(2n + 1\), which only covers equal numbers.
Checking a few values and calling it a proof. Unless every case is checked, examples prove nothing.
Missing a case in exhaustion, e.g. forgetting zero or negative numbers when the statement says 'all integers'.
Leaving \(8n + 8\) instead of writing \(8(n + 1)\) and saying that \(n + 1\) is an integer.
No conclusion: the final mark usually needs a statement linked to the question.
More tips
Memory tricks
Stuck for a counter-example? Try 0, 1, 2, a negative number, a fraction and a large number.
For 'multiple of \(k\)' proofs, your whole aim is to take \(k\) out as a factor of the entire expression.
Product of two consecutive integers is always even: one of them must be even.
Sense check: substitute a number such as \(n = 2\) into your final factorised form and into the original expression.
Exam technique
Define your variables at the start, e.g. 'Let \(n\) be an integer, so \(2n + 1\) is odd'.
End every proof with a conclusion using the words of the question, e.g. '... so it is a multiple of 8 for all integers \(n\)'.
For 'disprove' questions, one clear counter-example with its working and a statement is enough.
When the result is printed in the question, every step must be shown: examiners mark the argument, not the answer.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Disprove a statement with a counter-exampleFind one value that makes the statement false, show the working and say that it disproves the statement.
Grade D
Write general even, odd and consecutive integersUse \(2n\),\(2n + 1\), and \(n\),\(n + 1\), where \(n\) is an integer, so the argument covers every case.
Grade C
Prove a statement by exhaustionSplit all possibilities into a finite set of cases (e.g. \(n\) even and \(n\) odd) and prove each one.
Grade C
Prove an algebraic result by deductionExpand and factorise to show, for example, that an expression equals \(8 \times\) an integer.
Grade B
Prove an expression is always positiveComplete the square to get \((x - a)^2 + k\) with \(k \gt 0\), then use \((x - a)^2 \ge 0\)
Grade A
Use remainders to cover every integerWrite \(n\) as \(3k\), \(3k + 1\) or \(3k + 2\) (or similar) and prove the result for each form.
Grade A*
Construct a full proof in an unfamiliar contextChoose a method, state every assumption, justify each step and finish with a precise conclusion.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) Ravi claims that \(n^2 + n + 11\) is a prime number for every positive integer \(n\). Show, by means of a counter-example, that Ravi’s claim is false.[2]
(b) Show that the following statement is false. “For all real numbers \(x\), if \(x^2 \gt 9\) then \(x \gt 3\).”[1]
(c) Correct the conclusion of the statement in part (b) so that it is true for all real numbers \(x\).[1]
Show the answer and mark scheme
(a)Answer: e.g. \(n = 10\): \(10^2 + 10 + 11 = 121 = 11 \times 11\), which is not prime.
B1 for choosing a value of \(n\) that gives a non-prime value, e.g. \(n = 10\) or \(n = 11\)
B1 for showing the value is not prime, e.g. \(121 = 11 \times 11\) (or \(143 = 11 \times 13\)), with a conclusion that the claim is false
Worked solution: When \(n = 10\), \(n^2 + n + 11 = 100 + 10 + 11 = 121 = 11 \times 11\). 121 is not prime, so the claim is false.
(b)Answer: e.g. \(x = -4\): \(x^2 = 16 \gt 9\) but \(-4 \lt 3\).
B1 for a correct counter-example with both facts shown, e.g. \(x = -4\): \((-4)^2 = 16 \gt 9\) but \(-4\) is not greater than 3
Worked solution: Take \(x = -4\). Then \(x^2 = 16 \gt 9\), but \(x = -4\) is not greater than 3. So the statement is false.
(c)Answer: If \(x^2 \gt 9\) then \(x \gt 3\) or \(x \lt -3\).
B1 for “\(x \gt 3\) or \(x \lt -3\)” (accept \(|x| \gt 3\))
Worked solution: \(x^2 \gt 9 \iff |x| \gt 3\), so a true statement is: if \(x^2 \gt 9\) then \(x \gt 3\) or \(x \lt -3\).
Question 2Medium6 marks
(a) Prove that the square of any integer can be written in the form \(3k\) or \(3k + 1\), where \(k\) is an integer.[4]
(b) Hence prove that, if neither of the integers \(a\) and \(b\) is a multiple of 3, then \(a^2 + b^2\) is not a multiple of 3.[2]
Show the answer and mark scheme
(a)Answer: Consider \(n = 3m\), \(3m + 1\), \(3m + 2\): the squares are \(3(3m^2)\), \(3(3m^2 + 2m) + 1\) and \(3(3m^2 + 4m + 1) + 1\).
M1 for considering the cases \(n = 3m\), \(n = 3m + 1\) and \(n = 3m + 2\) (or \(3m - 1\)), where \(m\) is an integer
A1* for a conclusion covering all cases: every integer is of one of these forms, so every square is of the form \(3k\) or \(3k + 1\)
Worked solution: Every integer \(n\) can be written as \(3m\), \(3m + 1\) or \(3m + 2\) for some integer \(m\). \((3m)^2 = 9m^2 = 3(3m^2)\), of the form \(3k\). \((3m + 1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1\), of the form \(3k + 1\). \((3m + 2)^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1\), of the form \(3k + 1\). So the square of any integer is of the form \(3k\) or \(3k + 1\).
M1 for using part (a): as \(a\) and \(b\) are of the form \(3m + 1\) or \(3m + 2\), \(a^2 = 3k_1 + 1\) and \(b^2 = 3k_2 + 1\)
A1* for \(a^2 + b^2 = 3(k_1 + k_2) + 2\), which is not a multiple of 3, with a conclusion
Worked solution: Since \(a\) is not a multiple of 3, it is of the form \(3m + 1\) or \(3m + 2\), so from part (a) \(a^2 = 3k_1 + 1\). Similarly \(b^2 = 3k_2 + 1\). Then \(a^2 + b^2 = 3(k_1 + k_2) + 2\), which leaves remainder 2 on division by 3, so it is not a multiple of 3.
Question 3Hard6 marks
(a) Explain why every prime number greater than 3 can be written in the form \(6k + 1\) or \(6k - 1\), where \(k\) is a positive integer.[2]
(b) Hence prove that, for every prime number \(p\) greater than 3, \(p^2 - 1\) is a multiple of 24.[4]
Show the answer and mark scheme
(a)Answer: Integers of the forms \(6k\), \(6k \pm 2\) are even and \(6k + 3\) is a multiple of 3, so primes greater than 3 must be \(6k \pm 1\).
M1 for considering every integer as one of \(6k\), \(6k \pm 1\), \(6k \pm 2\), \(6k + 3\) (or \(6k, 6k + 1, \ldots, 6k + 5\))
A1 for explaining that \(6k\) and \(6k \pm 2\) are divisible by 2 and \(6k + 3\) is divisible by 3, so a prime greater than 3 must be \(6k \pm 1\)
Worked solution: Every integer is of one of the forms \(6k, 6k \pm 1, 6k \pm 2, 6k + 3\). \(6k = 2(3k)\) and \(6k \pm 2 = 2(3k \pm 1)\) are even, and \(6k + 3 = 3(2k + 1)\) is a multiple of 3, so none of these is a prime greater than 3. Hence any prime greater than 3 is of the form \(6k + 1\) or \(6k - 1\), and since the prime is at least 5, \(k \ge 1\).
(b)Answer: \(p^2 - 1 = (6k \pm 1)^2 - 1 = 12k(3k \pm 1)\), and \(k(3k \pm 1)\) is even.
M1 for expanding \((6k \pm 1)^2 - 1\) to obtain \(36k^2 \pm 12k\)
A1 for \(12k(3k \pm 1)\)
M1 for arguing that \(k(3k \pm 1)\) is even: if \(k\) is even it is even; if \(k\) is odd then \(3k\) is odd and \(3k \pm 1\) is even
A1* for \(p^2 - 1 = 24 \times \frac{k(3k \pm 1)}{2}\), a multiple of 24, with a conclusion
Worked solution: \(p = 6k \pm 1\), so \(p^2 - 1 = 36k^2 \pm 12k + 1 - 1 = 12k(3k \pm 1)\). If \(k\) is even, \(k(3k \pm 1)\) is even. If \(k\) is odd, \(3k\) is odd so \(3k \pm 1\) is even. Either way \(k(3k \pm 1) = 2j\) for an integer \(j\). So \(p^2 - 1 = 24j\), a multiple of 24.
Stuck? Get 1-to-1 help.
Chhetri Academy tutors GCSE and A level Maths and Science online, with a free 30-minute trial lesson.