1 Proof — revision question pack

1 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 1. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

1.1 · Understand and use the structure of mathematical proof, from assumptions through logical steps to a conclusion; use proof by deduction, exhaustion, disproof by counter example, and contradiction (irrationality of √2, infinity of primes).

Explanation

A proof begins with stated assumptions, uses valid implications at every step and ends with a conclusion that matches the claim.
Choose the proof form deliberately: deduction for a general algebraic chain, exhaustion for finitely many cases, a counterexample to disprove a universal claim, or contradiction by assuming the opposite.
In the classic contradiction for $\sqrt2$ , writing $\sqrt2=a/b$ in lowest terms leads to both $a$ and $b$ being even, contradicting the lowest-terms assumption.
Checking several examples is not a proof of a universal statement; in contradiction proofs, also make clear exactly why the derived result conflicts with the assumption.
To prove there are infinitely many primes, assume a finite list, form their product plus one, and note that its prime divisor is not on the list, giving a contradiction.

Worked example

Prove by exhaustion that $n^2+n+2$ is even whenever $n$ is an integer.

1.Every integer is even or odd.
2.If $n=2k$ , then $n^2+n+2=4k^2+2k+2=2(2k^2+k+1)$ .
3.If $n=2k+1$ , then $n^2+n+2=4k^2+6k+4=2(2k^2+3k+2)$ .
4.Both forms are divisible by $2$ , so the result holds for every integer $n$ .

Answer: The expression is even in both the even and odd cases for $n$ .

Common mistakes

Don't attempt proof by exhaustion over a pattern with infinitely many possible cases.
Don't assume the desired conclusion inside the argument instead of deriving it from stated assumptions.

Exam tip

For an exhaustion proof, identify the finite cases, show why each case is accepted or rejected, and finish with one conclusion. A cso final mark requires the complete proof to be correct, although earlier method marks depend on the question-specific scheme.

Tier 1 · Easy

1.
Disprove the claim that $n^2+n+41$ is prime for every non-negative integer $n$ .
(2)
(Total for Question 1 is 2 marks)
2.
Prove that the square of any odd integer is odd.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Prove by contradiction that $\sqrt3$ is irrational.
(5)
(Total for Question 1 is 5 marks)
2.
An integer $r$ satisfies $-2\leq r\leq3$ . Prove by exhaustion that $r^2-r\leq2$ holds exactly when $r\in\{-1,0,1,2\}$ .
(4)
(Total for Question 2 is 4 marks)
3.
The claim ‘if the product of two integers is even, then both integers are even’ is false. Give a counterexample, then prove that at least one of the two integers must be even.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Suppose someone lists all primes as $p_1,p_2,\ldots,p_k$ . Construct an integer from this list and use contradiction to prove that the list cannot be complete.
(5)
(Total for Question 1 is 5 marks)
2.
Prove by contradiction that $\sqrt2+\sqrt3$ is irrational. Your proof must also justify any claim you use about $\sqrt6$ .
(6)
(Total for Question 2 is 6 marks)
3.
Let $S=\{q\in\mathbb Q:0<q<1\}$ . Prove by contradiction that $S$ has neither a least element nor a greatest element.
(6)
(Total for Question 3 is 6 marks)
4.
Let $n$ be a positive odd integer. Prove that there is exactly one pair of consecutive non-negative integers whose squares have difference $n$ . Express this pair in terms of $n$ .
(5)
(Total for Question 4 is 5 marks)
5.
Prove that $a^2+b^2+c^2\geq ab+bc+ca$ for all real numbers $a$ , $b$ and $c$ . Determine precisely when equality holds.
(5)
(Total for Question 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

1.1 · Understand and use the structure of mathematical proof, from assumptions through logical steps to a conclusion; use proof by deduction, exhaustion, disproof by counter example, and contradiction (irrationality of √2, infinity of primes).

Tier 1 · Easy

Mark scheme for 1.1 Tier 1 · Easy
Question	Scheme	Marks
1	At $n=40$ , the expression equals $1681=41^2$ , which is not prime.	2
(2 marks)		2
Notes
A single counterexample defeats a universal claim. Taking $n=40$ gives $40^2+40+41=1681=41\times41$ , a composite number. Therefore the claim is false.
2	Writing the integer as $2k+1$ gives a square of the form $2m+1$ .	2
(2 marks)		2
Notes
Let the odd integer be $2k+1$ , where $k$ is an integer. Then $(2k+1)^2=4k^2+4k+1=2(2k^2+2k)+1$ . Since $2k^2+2k$ is an integer, the square is odd.

Tier 2 · Standard

Mark scheme for 1.1 Tier 2 · Standard
Question	Scheme	Marks
1	Assuming $\sqrt3$ is rational forces both the numerator and denominator of a fraction in lowest terms to be divisible by $3$ , giving a contradiction.	5
(5 marks)		5
Notes
Assume, for contradiction, that $\sqrt3=a/b$ , where $a$ and $b$ are coprime positive integers. Squaring gives $a^2=3b^2$ , so $3$ divides $a^2$ . Because $3$ is prime, $3\mid a^2$ forces $3\mid a$ — this step needs primality and is where the proof does its work: $4\mid a^2$ does not give $4\mid a$ , as $a=2$ shows. Write $a=3k$ . Then $9k^2=3b^2$ , so $b^2=3k^2$ , and $3\mid b^2$ gives $3\mid b$ by the same primality argument. Thus $a$ and $b$ share a factor of $3$ , contradicting coprimality. Hence $\sqrt3$ is irrational.
2	The inequality holds exactly for $r=-1,0,1,2$ .	4
(4 marks)		4
Notes
The only cases are $r=-2,-1,0,1,2,3$ . The corresponding values of $r^2-r$ are $6,2,0,0,2,6$ . Therefore the value is at most $2$ precisely in the four cases $r=-1,0,1,2$ , which proves the statement by exhaustion.
3	For example, $2\times3=6$ is even although $3$ is odd. If both integers were odd, their product would be odd, so an even product requires at least one even factor.	4
(4 marks)		4
Notes
The integers $2$ and $3$ give a counterexample because their product is even but they are not both even. For the corrected statement, suppose instead that neither integer is even. They can then be written as $2r+1$ and $2s+1$ , where $r$ and $s$ are integers. Their product is $(2r+1)(2s+1)=2(2rs+r+s)+1$ , which is odd. This contradicts the product being even, so at least one of the two integers must be even.

Tier 3 · Hard

Mark scheme for 1.1 Tier 3 · Hard
Question	Scheme	Marks
1	$N=p_1p_2\cdots p_k+1$ leads to a prime not on the alleged complete list.	5
(5 marks)		5
Notes
Assume for contradiction that the finite list contains every prime and form $N=p_1p_2\cdots p_k+1$ . Dividing $N$ by any listed prime $p_i$ leaves remainder $1$ , so none of the listed primes divides $N$ . Yet $N>1$ , so $N$ is prime or has a prime factor. That prime factor is absent from the list, contradicting its completeness. Hence there are infinitely many primes.
2	A rational value of $\sqrt2+\sqrt3$ would make $\sqrt6$ rational, but a lowest-terms argument proves that $\sqrt6$ is irrational.	6
(6 marks)		6
Notes
First suppose $\sqrt6=a/b$ for coprime positive integers $a$ and $b$ . Then $a^2=6b^2$ , so $a$ is even; write $a=2c$ . This gives $2c^2=3b^2$ , so $b^2$ , and hence $b$ , is even. This contradicts the coprimality of $a$ and $b$ , proving that $\sqrt6$ is irrational. Now assume, for contradiction, that $q=\sqrt2+\sqrt3$ is rational. Squaring gives $q^2=5+2\sqrt6$ , so $\sqrt6=(q^2-5)/2$ would be rational. This contradiction proves that $\sqrt2+\sqrt3$ is irrational.
3	For any $q\in S$ , the rational numbers $q/2$ and $(q+1)/2$ lie in $S$ with $q/2<q<(q+1)/2$ ; these contradict leastness and greatestness respectively.	6
(6 marks)		6
Notes
Assume first that $q$ is the least element of $S$ . Since $q$ is rational, $q/2$ is rational, and $0<q/2<q<1$ . Thus $q/2\in S$ and is smaller than the alleged least element, a contradiction. Now assume that $r$ is the greatest element of $S$ . The number $(r+1)/2$ is rational and, because $0<r<1$ , it satisfies $r<(r+1)/2<1$ . Hence $(r+1)/2\in S$ and is greater than the alleged greatest element, another contradiction. Therefore $S$ has neither a least nor a greatest element.
4	The unique pair is $\dfrac{n-1}{2}$ and $\dfrac{n+1}{2}$ .	5
(5 marks)		5
Notes
Since $n$ is positive and odd, $a=(n-1)/2$ is a non-negative integer. The consecutive integer is $a+1=(n+1)/2$ , and $(a+1)^2-a^2=2a+1=n$ , proving existence. For uniqueness, if consecutive non-negative integers $b$ and $b+1$ have squares with difference $n$ , then $(b+1)^2-b^2=2b+1=n$ . Hence $b=(n-1)/2=a$ , so no other pair is possible.
5	$a^2+b^2+c^2\geq ab+bc+ca$ , with equality if and only if $a=b=c$ .	5
(5 marks)		5
Notes
The identity $2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2$ holds by expansion. The right-hand side is a sum of squares, so it is non-negative. Therefore $a^2+b^2+c^2-ab-bc-ca\geq0$ , which proves the inequality. Equality holds exactly when all three squares are zero, requiring $a=b$ , $b=c$ and $c=a$ ; equivalently, $a=b=c$ .