Practise 19 exam-style questions on Proof for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.
19 questions
2–8 marks each
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19 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
3 marks
For each of the following statements about an integer \(n\), state whether it is always true, never true or sometimes true. Give a reason for each answer.
Statement 1: \(n^2 - n\) is even. Statement 2: \(n^2 + 3\) is even. Statement 3: \(n^2 - n + 5\) is even.
Model answer
Statement 1: always true. \(n^2 - n = n(n - 1)\), the product of two consecutive integers. One of any two consecutive integers is even, so the product is even. Statement 2: sometimes true. If \(n\) is odd, \(n^2\) is odd and \(n^2 + 3\) is even (e.g. \(n = 1\) gives 4). If \(n\) is even, \(n^2\) is even and \(n^2 + 3\) is odd (e.g. \(n = 2\) gives 7). Statement 3: never true. \(n^2 - n + 5 = n(n - 1) + 5\), which is (even) + 5, so it is always odd.
Explanation
"Always" and "never" need a general reason, while "sometimes" needs one example where the statement is true and one where it is false. Factorising \(n^2 - n\) as \(n(n - 1)\), a product of consecutive integers, shows it is always even, and Statement 3 is that even number plus 5, so it is never even. Giving a verdict without a reason, or using examples to justify "always", earns no credit.
Mark scheme
1 mark
B1 Statement 1 is always true, with a reason, e.g. \(n^2 - n = n(n - 1)\) is the product of two consecutive integers, one of which is even
A verdict with no reason scores B0. Accept a correct odd/even case argument (\(n\) even: \(n(n - 1)\) has an even factor; \(n\) odd: \(n - 1\) is even). Numerical examples alone do not justify "always": B0.
1 mark
B1 Statement 2 is sometimes true, with a reason, e.g. it is true when \(n\) is odd (\(n = 1\): \(1 + 3 = 4\), even) and false when \(n\) is even (\(n = 2\): \(4 + 3 = 7\), odd)
Needs "sometimes" and evidence of both a true case and a false case (an example of each, or the general statement "even when \(n\) is odd, odd when \(n\) is even").
1 mark
B1 Statement 3 is never true, with a reason, e.g. \(n^2 - n + 5 = n(n - 1) + 5\), which is an even number plus 5, so it is always odd
The reason may refer back to Statement 1. Accept a correct odd/even case argument for \(n^2 - n + 5\). Examples alone do not justify "never": B0.
Question 2
4 marks
Prove that \(mn(m + n)\) is even for all integers \(m\) and \(n\).
Model answer
Case 1: at least one of \(m\) and \(n\) is even. If \(m = 2a\), where \(a\) is an integer, then \(mn(m + n) = 2[an(m + n)]\), which is even. If \(n\) is even the same argument applies. Case 2: \(m\) and \(n\) are both odd. Let \(m = 2a + 1\) and \(n = 2b + 1\), where \(a\) and \(b\) are integers. Then \(m + n = 2a + 2b + 2 = 2(a + b + 1)\), so \(mn(m + n) = 2[mn(a + b + 1)]\), which is even. These cases cover all integers \(m\) and \(n\), so \(mn(m + n)\) is even for all integers \(m\) and \(n\).
Explanation
Proof by exhaustion works well here: either at least one of \(m\) and \(n\) is even, which gives an even factor, or both are odd, in which case \(m + n\) is even. Each case needs a reason, and the final mark needs a conclusion that covers all integers. Common errors are leaving out a case, using the same letter for two different odd numbers and testing examples instead of proving.
Mark scheme
1 mark
M1 Splits into cases that cover every possibility, e.g. "at least one of \(m\) and \(n\) is even" and "\(m\) and \(n\) are both odd" (or all four odd/even cases). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof. Testing numerical values is not a proof and scores 0 marks. The question does not say "using algebra", so a clear argument in words for each case is acceptable, e.g. "if \(m\) is even then \(mn(m + n)\) has an even factor". Considering only one case (e.g. only both odd) scores M0.
1 mark
A1 Case at least one of them even: e.g. \(m = 2a\) gives \(mn(m + n) = 2an(2a + n)\), which is 2 times an integer, so even (similarly if \(n\) is even)
Accept "a product with an even factor is even". The case \(n\) even may be dealt with by symmetry.
1 mark
A1 Case both odd: \(m = 2a + 1\), \(n = 2b + 1\) gives \(m + n = 2a + 2b + 2 = 2(a + b + 1)\), which is even, so \(mn(m + n)\) is even
Different letters must be used for \(m\) and \(n\); \(m = 2k + 1\), \(n = 2k + 1\) scores A0 for this case. Accept "odd + odd = even, so \(m + n\) is an even factor".
1 mark
A1 Conclusion: every case gives an even result, so \(mn(m + n)\) is even for all integers \(m\) and \(n\)
cso. All cases must have been correctly considered. A minimal conclusion such as "hence proven" or "so always even" following both cases is acceptable.
Question 3
4 marks
You may assume that \(\sqrt{7}\) is irrational.
Use proof by contradiction to prove that the equation \[x^2 - 4x - 3 = 0\] has no rational roots.
Model answer
Assume that the equation has a rational root \(r\). Then \(r^2 - 4r - 3 = 0\). \(r = \dfrac{4 \pm \sqrt{16 + 12}}{2} = \dfrac{4 \pm \sqrt{28}}{2} = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}\) So \(\sqrt{7} = r - 2\) or \(\sqrt{7} = 2 - r\). \(r\) is rational, so \(r - 2\) and \(2 - r\) are rational, as the difference of two rational numbers is rational. So \(\sqrt{7}\) would be rational. This contradicts the fact that \(\sqrt{7}\) is irrational. Hence the equation \(x^2 - 4x - 3 = 0\) has no rational roots.
Explanation
Assume a rational root \(r\). Solving gives \(r = 2 \pm \sqrt{7}\), so \(\sqrt{7} = \pm(r - 2)\), which would be rational because \(r\) is rational; that contradicts \(\sqrt{7}\) being irrational. The marks are for the assumption, solving the quadratic, making \(\sqrt{7}\) the subject and the contradiction with a conclusion. Common errors are working with decimals, not simplifying \(\sqrt{28} = 2\sqrt{7}\) and not saying why \(r - 2\) is rational.
Mark scheme
1 mark
B1 Sets up the contradiction: assume the equation has a rational root \(r\) (or \(r = \dfrac{p}{q}\), with \(p\) and \(q\) integers, \(q ≠ 0\)), so \(r^2 - 4r - 3 = 0\). An answer with no working scores 0 marks.
Max 2 if the proof is not by contradiction: solving to get \(2 \pm \sqrt{7}\) and stating that these are irrational, with a reason, scores M1 A1 only. An answer with no working scores 0 marks: this is a proof. Decimal values of the roots (\(-0.646\ldots\), \(4.646\ldots\)) do not show irrationality: 0 marks.
1 mark
M1 Solves \(r^2 - 4r - 3 = 0\) by the formula or completing the square: \(r = \dfrac{4 \pm \sqrt{28}}{2}\) or \((r - 2)^2 = 7\)
Usual rules for solving a three-term quadratic: the formula with the values in the correct places, or \((r \pm 2)^2 \pm 4 - 3 = 0\).
1 mark
A1 \(r = 2 \pm \sqrt{7}\), so \(\sqrt{7} = r - 2\) or \(\sqrt{7} = 2 - r\) (i.e. \(\sqrt{7} = \pm(r - 2)\))
Requires \(\sqrt{28} = 2\sqrt{7}\) (or \((r - 2)^2 = 7\)) and \(\sqrt{7}\) made the subject; both signs must be dealt with.
1 mark
A1* Explains that \(r - 2\) and \(2 - r\) are rational (a difference of rational numbers is rational), so \(\sqrt{7}\) would be rational, which contradicts \(\sqrt{7}\) being irrational; concludes that the equation has no rational roots
cso. The reason why \(\pm(r - 2)\) is rational, the contradiction and the conclusion are all needed.
Question 4
7 marks
Prove that, for any integer \(n\), if \(n^2\) is even then \(n\) is even.
Hence prove by contradiction that \(\sqrt{2}\) is irrational.
Model answer
Suppose \(n\) is odd, so \(n = 2k + 1\) for some integer \(k\). \(n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\), which is odd. So if \(n\) is odd then \(n^2\) is odd. Therefore, if \(n^2\) is even, \(n\) cannot be odd, so \(n\) is even. Assume that \(\sqrt{2}\) is rational. Then \(\sqrt{2} = \dfrac{a}{b}\), where \(a\) and \(b\) are integers with no common factors. \(2 = \dfrac{a^2}{b^2}\), so \(a^2 = 2b^2\). So \(a^2\) is even, and by the first result \(a\) is even. Let \(a = 2c\). \((2c)^2 = 2b^2\), so \(4c^2 = 2b^2\) and \(b^2 = 2c^2\). So \(b^2\) is even, and so \(b\) is even. Then \(a\) and \(b\) are both even, so they have a common factor 2. This contradicts the assumption that \(a\) and \(b\) have no common factors. So the assumption is false, and \(\sqrt{2}\) is irrational.
Explanation
The first result is proved by showing that an odd number squared is odd, so an even square must come from an even number. In the proof by contradiction, assume \(\sqrt{2} = \dfrac{a}{b}\) in lowest terms; squaring and using the first result twice shows that \(a\) and \(b\) are both even, which contradicts "no common factors". Marks are lost for proving the converse, for leaving out "no common factors" and for not stating the contradiction and the conclusion.
Mark scheme
1 mark
M1 Considers \(n\) odd, \(n = 2k + 1\), and expands: \(n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\). An answer with no working scores 0 marks.
An answer with no working scores 0 marks: this is a proof. Testing values (e.g. \(4 = 2^2\), \(16 = 4^2\)) scores M0. Proving the converse ("if \(n\) is even then \(n^2\) is even", using \(n = 2k\)) scores M0 A0.
1 mark
A1 Concludes: if \(n\) is odd then \(n^2\) is odd, so if \(n^2\) is even then \(n\) cannot be odd, so \(n\) is even
Accept the argument set up as a contradiction ("assume \(n^2\) is even and \(n\) is odd ...") or as the contrapositive. The final link back to "if \(n^2\) is even then \(n\) is even" is needed.
1 mark
B1 Sets up the contradiction: assume \(\sqrt{2}\) is rational, so \(\sqrt{2} = \dfrac{a}{b}\), where \(a\) and \(b\) are integers, \(b ≠0\), with no common factors (the fraction is in its lowest terms)
"No common factors" (or "in its simplest form", "lowest terms", "coprime") is required for this mark. Accept "Let \(\sqrt{2} = \dfrac{p}{q}\) ..." with any letters.
1 mark
M1 Squares and rearranges to \(a^2 = 2b^2\)
1 mark
A1 States that \(a^2\) is even, so \(a\) is even (by the first result), and writes \(a = 2c\)
The reason must be given: "\(a^2 = 2b^2\) is even, so by the first result \(a\) is even".
1 mark
A1 \(4c^2 = 2b^2\), so \(b^2 = 2c^2\), so \(b^2\) is even and \(b\) is even
1 mark
A1* Contradiction: \(a\) and \(b\) are both even, so they have a common factor of 2, which contradicts the assumption that they have no common factors. Hence \(\sqrt{2}\) is irrational
cso. Requires the contradiction to be identified (with what it contradicts) and a final statement that \(\sqrt{2}\) is irrational.
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