What is 1.1: Mathematical proof?

A proof begins with stated assumptions, uses valid implications at every step and ends with a conclusion that matches the claim.

How many marks are 1.1 questions worth?

1.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 1.1 appeared in?

We have not yet indexed a verified past-paper appearance for 1.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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).

Draft — not yet indexed

Mathematical proof

Worked answers and methods for 1.1 on Edexcel A-level Maths 9MA0.

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.

Worked practice

Tier 1 · Easy

Disprove the claim that

n^2+n+41

is prime for every non-negative integer

n

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	At $n=40$ , the expression equals $1681=41^2$ , which is not prime.	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 marks)

Tier 2 · Standard

Prove by contradiction that

\sqrt3

is irrational.

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	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
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.

(5 marks)

Tier 3 · Hard

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 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$N=p_1p_2\cdots p_k+1$ leads to a prime not on the alleged complete list.	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.

(5 marks)

Tier 1 · Easy

Prove that the square of any odd integer is odd.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Writing the integer as $2k+1$ gives a square of the form $2m+1$ .	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.

(2 marks)

Tier 2 · Standard

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 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	The inequality holds exactly for $r=-1,0,1,2$ .	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.

(4 marks)

Tier 3 · Hard

Prove by contradiction that

\sqrt2+\sqrt3

is irrational. Your proof must also justify any claim you use about

\sqrt6

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	A rational value of $\sqrt2+\sqrt3$ would make $\sqrt6$ rational, but a lowest-terms argument proves that $\sqrt6$ is irrational.	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.

(6 marks)

Tier 2 · Standard

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 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	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
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.

(4 marks)

Tier 3 · Hard

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 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	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
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.

(6 marks)

Tier 3 · Hard

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 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The unique pair is $\dfrac{n-1}{2}$ and $\dfrac{n+1}{2}$ .	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 marks)

Q10

Tier 3 · Hard

10.

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 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$a^2+b^2+c^2\geq ab+bc+ca$ , with equality if and only if $a=b=c$ .	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, $a=b=c$ .

(5 marks)

Verified exam appearances

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