Unit 6 Exam Practice
Mathematical Proofs
These questions are written in the style of IB examination questions, and they are harder than the review exercises. The marks for each part are shown in square brackets on the right, for example [3]. Paper 1 means no calculator; Paper 2 means a graphic display calculator (GDC) is allowed. Try each question under exam conditions first, then open the markscheme.
| Mark | Meaning |
|---|---|
| M | Method mark: a correct method is used, even with a small slip |
| A | Accuracy mark: a correct answer or line of working (usually needs the M mark before it) |
| R | Reasoning mark: a clear reason or justification is given |
| N | Mark for a correct final answer with no working shown |
| AG | “Answer given” in the question, so no extra mark is awarded for writing it |
The number after the letter is the number of marks, so A2 is worth 2 marks. In the exam, give answers exactly or correct to 3 significant figures (s.f.) unless the question says otherwise.
SL questions
1 Paper 1
Consider three consecutive integers \(n - 1\), \(n\) and \(n + 1\), where \(n \in \mathbb{Z}\).
(a) Show that the sum of their squares is \(3n^{2} + 2\). [2]
(b) Hence prove that the sum of the squares of three consecutive integers is never a multiple of \(3\). [2]
(c) Prove that, when the middle integer \(n\) is odd, the sum of the squares leaves a remainder of \(5\) when it is divided by \(24\). [4]
Total 8 marks
(a)
\((n - 1)^{2} + n^{2} + (n + 1)^{2} = \left(n^{2} - 2n + 1\right) + n^{2} + \left(n^{2} + 2n + 1\right)\) M1
\(= 3n^{2} + 2\) A1 AG
[2 marks]
(b)
\(3n^{2} + 2 = 3\left(n^{2}\right) + 2\), where \(n^{2} \in \mathbb{Z}\) M1
so the sum leaves a remainder of \(2\) when it is divided by \(3\), and therefore it is not a multiple of \(3\) R1
[2 marks]
(c)
\(n\) is odd, so \(n = 2m + 1\) for some \(m \in \mathbb{Z}\) M1
\(3(2m + 1)^{2} + 2 = 12m^{2} + 12m + 5\) A1
\(= 12m(m + 1) + 5\) A1
\(m\) and \(m + 1\) are consecutive integers, so one of them is even and \(m(m + 1) = 2j\), \(j \in \mathbb{Z}\); then the sum is \(24j + 5\), which leaves a remainder of \(5\) R1
Do not award the R1 if the reason why \(m(m + 1)\) is even is missing.
[4 marks]
See 6.2.
2 Paper 1
(a) Prove that \(\dfrac{1}{\sqrt{r} + \sqrt{r + 1}} \equiv \sqrt{r + 1} - \sqrt{r}\), for \(r \geq 0\). [2]
(b) Hence show that \(\displaystyle\sum_{r=1}^{99} \frac{1}{\sqrt{r} + \sqrt{r + 1}} = 9\). [3]
(c) Find the least value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} \frac{1}{\sqrt{r} + \sqrt{r + 1}} > 20\). [2]
Total 7 marks
(a)
\(\text{LHS} = \dfrac{\sqrt{r + 1} - \sqrt{r}}{\left(\sqrt{r + 1} + \sqrt{r}\right)\left(\sqrt{r + 1} - \sqrt{r}\right)}\) (multiplying by the conjugate) M1
\(= \dfrac{\sqrt{r + 1} - \sqrt{r}}{(r + 1) - r} = \sqrt{r + 1} - \sqrt{r} = \text{RHS}\) A1 AG
[2 marks]
(b)
\(\displaystyle\sum_{r=1}^{99} \frac{1}{\sqrt{r} + \sqrt{r + 1}} = \sum_{r=1}^{99} \left(\sqrt{r + 1} - \sqrt{r}\right)\) (by part (a)) M1
\(= \left(\sqrt{2} - \sqrt{1}\right) + \left(\sqrt{3} - \sqrt{2}\right) + \cdots + \left(\sqrt{100} - \sqrt{99}\right)\) M1
all the middle terms cancel, so the sum is \(\sqrt{100} - \sqrt{1} = 10 - 1 = 9\) A1 AG
[3 marks]
(c)
the sum is \(\sqrt{n + 1} - 1\) (in the same way as part (b)), so \(\sqrt{n + 1} > 21\) M1
\(n + 1 > 441\), so \(n = 441\) A1
[2 marks]
3 Paper 1
Let \(a\) and \(b\) be real numbers with \(a > 1\) and \(b > 1\).
(a) Prove that \(\log_a b + \log_b a \geq 2\). [5]
(b) State the condition on \(a\) and \(b\) for which \(\log_a b + \log_b a = 2\). [1]
Total 6 marks
(a)
let \(t = \log_a b\); then \(\log_b a = \dfrac{\log_a a}{\log_a b} = \dfrac{1}{t}\) (change of base) M1
\(t > 0\), because \(b > 1\) and \(a > 1\) (so \(a^{t} = b > 1\) needs \(t > 0\)) R1
\(t + \dfrac{1}{t} - 2 = \dfrac{t^{2} - 2t + 1}{t}\) M1
\(= \dfrac{(t - 1)^{2}}{t}\) A1
\((t - 1)^{2} \geq 0\) (a square) and \(t > 0\), so \(t + \dfrac{1}{t} - 2 \geq 0\), that is \(\log_a b + \log_b a \geq 2\) R1 AG
Accept a proof that starts from \((t - 1)^{2} \geq 0\) and divides by \(t\), if the reason \(t > 0\) is given.
[5 marks]
(b)
equality when \(t = 1\), that is when \(a = b\) A1
[1 mark]
4 Paper 1
Let \(m \in \mathbb{Z}\) and \(n \in \mathbb{Z}^{+}\). Let \(S\) be the sum of the \(n\) consecutive integers \(m,\ m + 1,\ m + 2,\ \ldots,\ m + n - 1\).
(a) Show that \(S = \dfrac{n(2m + n - 1)}{2}\). [2]
(b) Prove that the sum of any four consecutive integers is never a multiple of \(4\). [3]
(c) Prove that, if \(n\) is odd, then \(S\) is a multiple of \(n\). [3]
(d) Prove that, if \(n\) is even, then \(S\) is not a multiple of \(n\). [3]
(e) Write \(100\) as the sum of eight consecutive positive integers. [3]
Total 14 marks
(a)
arithmetic series with \(u_1 = m\) and \(d = 1\): \(S = \dfrac{n}{2}\big(2m + (n - 1) \times 1\big)\) M1
\(S = \dfrac{n(2m + n - 1)}{2}\) A1 AG
[2 marks]
(b)
\(n = 4\): \(S = \dfrac{4(2m + 3)}{2}\) M1
\(= 4m + 6\) A1
\(= 4(m + 1) + 2\), where \(m + 1 \in \mathbb{Z}\), so \(S\) leaves a remainder of \(2\) when divided by \(4\), and is not a multiple of \(4\) R1
[3 marks]
(c)
\(n\) is odd, so \(n = 2p + 1\) for some \(p \in \mathbb{Z}\), and \(2m + n - 1 = 2m + 2p\) M1
\(S = \dfrac{n(2m + 2p)}{2} = n(m + p)\) A1
\(m + p \in \mathbb{Z}\), so \(S\) is a multiple of \(n\) R1
[3 marks]
(d)
\(n\) is even, so \(n = 2p\) for some \(p \in \mathbb{Z}^{+}\) M1
\(\dfrac{S}{n} = \dfrac{2m + 2p - 1}{2} = m + p - \dfrac{1}{2}\) A1
\(m + p \in \mathbb{Z}\), so \(\dfrac{S}{n}\) is not an integer, and \(S\) is not a multiple of \(n\) R1
Accept the argument that \(2m + n - 1\) is odd, so \(S = p(2m + n - 1)\) is \(p\) times an odd number and cannot equal \(2p \times (\text{an integer})\).
[3 marks]
(e)
\(n = 8\): \(\dfrac{8(2m + 7)}{2} = 100\) M1
\(2m + 7 = 25\), so \(m = 9\) A1
\(100 = 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16\) A1
[3 marks]
5 Paper 1
Let \(n \in \mathbb{Z}^{+}\).
(a) Prove that \((1 - x)\left(1 + x + x^{2} + \cdots + x^{n-1}\right) \equiv 1 - x^{n}\). [2]
(b) Hence prove that \(7^{n} - 1\) is divisible by \(6\) for every \(n \in \mathbb{Z}^{+}\). [3]
(c) By replacing \(x\) with \(-a\) in part (a), show that, when \(n\) is odd, \[ a^{n} + 1 \equiv (a + 1)\left(1 - a + a^{2} - \cdots + a^{n-1}\right). \] [3]
(d) Hence prove that \(2^{n} + 1\) is divisible by \(3\) for every odd positive integer \(n\). [2]
(e) Show that \(2^{15} + 1\) is divisible by \(9\) and by \(11\). [4]
Total 14 marks
(a)
\(\text{LHS} = \left(1 + x + x^{2} + \cdots + x^{n-1}\right) - \left(x + x^{2} + \cdots + x^{n-1} + x^{n}\right)\) M1
all the terms from \(x\) to \(x^{n-1}\) cancel, so \(\text{LHS} = 1 - x^{n} = \text{RHS}\) A1 AG
[2 marks]
(b)
\(x = 7\): \((1 - 7)\left(1 + 7 + \cdots + 7^{n-1}\right) = 1 - 7^{n}\) M1
\(7^{n} - 1 = 6\left(1 + 7 + 7^{2} + \cdots + 7^{n-1}\right)\) A1
the bracket is a sum of integers, so it is an integer; hence \(7^{n} - 1\) is divisible by \(6\) R1
[3 marks]
(c)
\((1 + a)\left(1 - a + a^{2} - \cdots + (-a)^{n-1}\right) \equiv 1 - (-a)^{n}\) M1
\(n\) is odd, so \((-a)^{n} = -a^{n}\) and the right-hand side is \(1 + a^{n}\) A1
\(n - 1\) is even, so the last term \((-a)^{n-1} = a^{n-1}\), giving \(a^{n} + 1 \equiv (a + 1)\left(1 - a + a^{2} - \cdots + a^{n-1}\right)\) A1 AG
[3 marks]
(d)
\(a = 2\): \(2^{n} + 1 = 3\left(1 - 2 + 2^{2} - \cdots + 2^{n-1}\right)\) M1
the bracket is a sum of integers, so \(2^{n} + 1\) is divisible by \(3\) R1
[2 marks]
(e)
\(2^{15} + 1 = 8^{5} + 1\), and \(5\) is odd, so use \(a = 8\), \(n = 5\) M1
\(2^{15} + 1 = 9\left(1 - 8 + 8^{2} - 8^{3} + 8^{4}\right)\), so it is divisible by \(9\) A1
\(2^{15} + 1 = 32^{3} + 1\), and \(3\) is odd, so use \(a = 32\), \(n = 3\) M1
\(2^{15} + 1 = 33\left(1 - 32 + 32^{2}\right)\), which is divisible by \(33 = 3 \times 11\), so it is divisible by \(11\) A1
Accept a correct direct check, for example \(2^{15} + 1 = 32\,769 = 9 \times 3641 = 11 \times 2979\).
[4 marks]
HL questions
HL questions on this unit use the AHL methods of proof: mathematical induction, proof by contradiction and disproof by counterexample.
1 Paper 1
Prove by mathematical induction that \(5^{n+1} + 2 \times 3^{n} + 1\) is divisible by \(8\) for all \(n \in \mathbb{Z}^{+}\). [8]
Total 8 marks
\(n = 1\): \(5^{2} + 2 \times 3 + 1 = 32 = 8 \times 4\), so \(P(1)\) is true A1
assume \(P(k)\) is true for some \(k \in \mathbb{Z}^{+}\): \(5^{k+1} + 2 \times 3^{k} + 1 = 8M\), where \(M \in \mathbb{Z}\) M1
consider \(n = k + 1\): \(5^{k+2} + 2 \times 3^{k+1} + 1 = 5 \times 5^{k+1} + 6 \times 3^{k} + 1\) M1
\(= 5\left(8M - 2 \times 3^{k} - 1\right) + 6 \times 3^{k} + 1\) (by the assumption) A1
\(= 40M - 4 \times 3^{k} - 4 = 40M - 4\left(3^{k} + 1\right)\) A1
\(3^{k}\) is odd, so \(3^{k} + 1\) is even: \(3^{k} + 1 = 2q\), where \(q \in \mathbb{Z}\) R1
\(5^{k+2} + 2 \times 3^{k+1} + 1 = 40M - 8q = 8(5M - q)\), which is divisible by \(8\), so \(P(k + 1)\) is true A1
since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^{+}\) R1
Award the final R1 only if at least four of the previous marks have been awarded. Do not award the second M1 for writing \(P(k + 1)\) with no attempt to use \(P(k)\).
[8 marks]
See 6.4.
2 Paper 1
(a) Prove by contradiction that \(\log_4 12\) is irrational. [6]
(b) Hence, by considering \(x = \log_4 12\) and \(y = -\log_4 3\), show that the statement “if \(x\) and \(y\) are irrational, then \(x + y\) is irrational” is false. [2]
Total 8 marks
(a)
assume, for a contradiction, that \(\log_4 12 = \dfrac{p}{q}\), where \(p, q \in \mathbb{Z}^{+}\) M1
\(\log_4 12 > \log_4 4 = 1\), so \(p > q\) R1
\(4^{\frac{p}{q}} = 12\), so \(4^{p} = 12^{q}\) A1
\(2^{2p} = 2^{2q} \times 3^{q}\), so \(2^{2p - 2q} = 3^{q}\) A1
\(2p - 2q \geq 2\) (because \(p > q\)), so the left side is even; the right side is odd (a product of odd numbers) R1
an even number cannot equal an odd number, which is a contradiction, so \(\log_4 12\) is irrational R1
Do not award the final R1 if the assumption at the start is not stated clearly.
[6 marks]
(b)
\(y = 1 - \log_4 12\) (because \(\log_4 12 = 1 + \log_4 3\)); if \(y\) were rational, \(\log_4 12 = 1 - y\) would be rational, so \(y\) is irrational R1
\(x + y = \log_4 12 - \log_4 3 = \log_4 4 = 1\), which is rational, so \(x\) and \(y\) form a counterexample and the statement is false A1
[2 marks]
See 6.3.
3 Paper 1
(a) Prove by mathematical induction that \[ \sum_{r=1}^{n} \frac{1}{\sqrt{r}} < 2\sqrt{n}, \qquad n \in \mathbb{Z}^{+}. \] [7]
(b) Hence show that \(1 + \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}} + \cdots + \dfrac{1}{\sqrt{400}} < 40\). [1]
Total 8 marks
(a)
\(n = 1\): \(\text{LHS} = 1\) and \(\text{RHS} = 2\), and \(1 < 2\), so \(P(1)\) is true A1
assume \(P(k)\) is true for some \(k \in \mathbb{Z}^{+}\): \(\displaystyle\sum_{r=1}^{k} \frac{1}{\sqrt{r}} < 2\sqrt{k}\) M1
\(\displaystyle\sum_{r=1}^{k+1} \frac{1}{\sqrt{r}} = \sum_{r=1}^{k} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{k + 1}} < 2\sqrt{k} + \frac{1}{\sqrt{k + 1}}\) (by the assumption) M1
\(4k^{2} + 4k < 4k^{2} + 4k + 1\), so \(2\sqrt{k^{2} + k} < 2k + 1\) (both sides are positive) M1 A1
\(2\sqrt{k}\sqrt{k + 1} + 1 < 2(k + 1)\); dividing by \(\sqrt{k + 1} > 0\): \(2\sqrt{k} + \dfrac{1}{\sqrt{k + 1}} < 2\sqrt{k + 1}\), so \(P(k + 1)\) is true R1
since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^{+}\) R1
Accept working backwards from the target \(2\sqrt{k} + \dfrac{1}{\sqrt{k + 1}} < 2\sqrt{k + 1}\) only if every step is clearly reversible (written with \(\Leftrightarrow\)).
[7 marks]
(b)
\(n = 400\): the sum is less than \(2\sqrt{400} = 40\) A1
[1 mark]
See 6.4.
4 Paper 1
(a) Prove by mathematical induction that \[ \sum_{r=1}^{n} (2r - 1)\,3^{r-1} = (n - 1)\,3^{n} + 1, \qquad n \in \mathbb{Z}^{+}. \] [6]
(b) Hence show that \(\displaystyle\sum_{r=1}^{n} r \times 3^{r-1} = \frac{(2n - 1)\,3^{n} + 1}{4}\). [5]
(c) Hence prove that \((2n - 1)\,3^{n} + 1\) is divisible by \(4\) for all \(n \in \mathbb{Z}^{+}\). [2]
(d) Show that the statement “\((2n - 1)\,3^{n} + 1\) is divisible by \(8\) for all \(n \in \mathbb{Z}^{+}\)” is false. [1]
Total 14 marks
(a)
\(n = 1\): \(\text{LHS} = 1 \times 3^{0} = 1\) and \(\text{RHS} = 0 \times 3 + 1 = 1\), so \(P(1)\) is true A1
assume \(P(k)\) is true for some \(k \in \mathbb{Z}^{+}\): \(\displaystyle\sum_{r=1}^{k} (2r - 1)\,3^{r-1} = (k - 1)\,3^{k} + 1\) M1
\(\displaystyle\sum_{r=1}^{k+1} (2r - 1)\,3^{r-1} = \sum_{r=1}^{k} (2r - 1)\,3^{r-1} + (2k + 1)\,3^{k}\) M1
\(= (k - 1)\,3^{k} + 1 + (2k + 1)\,3^{k}\) (by the assumption) A1
\(= 3k \times 3^{k} + 1 = k \times 3^{k+1} + 1\), which is \(P(k + 1)\) A1
since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^{+}\) R1
[6 marks]
(b)
\(r \times 3^{r-1} = \dfrac{1}{2}(2r - 1)\,3^{r-1} + \dfrac{1}{2} \times 3^{r-1}\) M1
\(\displaystyle\sum_{r=1}^{n} 3^{r-1} = \dfrac{3^{n} - 1}{3 - 1} = \dfrac{3^{n} - 1}{2}\) (geometric series) A1
\(\displaystyle\sum_{r=1}^{n} r \times 3^{r-1} = \dfrac{1}{2}\big((n - 1)\,3^{n} + 1\big) + \dfrac{1}{4}\left(3^{n} - 1\right)\) (using part (a)) M1
\(= \dfrac{2(n - 1)\,3^{n} + 2 + 3^{n} - 1}{4}\) A1
\(= \dfrac{(2n - 1)\,3^{n} + 1}{4}\) A1 AG
[5 marks]
(c)
\((2n - 1)\,3^{n} + 1 = 4\displaystyle\sum_{r=1}^{n} r \times 3^{r-1}\) M1
the sum is a sum of integers, so it is an integer; hence \((2n - 1)\,3^{n} + 1\) is divisible by \(4\) R1
[2 marks]
(d)
\(n = 1\): \(1 \times 3 + 1 = 4\), which is not divisible by \(8\), so \(n = 1\) is a counterexample A1
Accept any correct counterexample, for example \(n = 2\) gives \(28\).
[1 mark]
5 Paper 1
For \(n \in \mathbb{N}\), let \(u_n = 2^{2^{n}} + 1\). So \(u_0 = 3\), \(u_1 = 5\), \(u_2 = 17\), \(u_3 = 257\), \(\ldots\)
(a) Show that \(u_{n+1} = (u_n - 1)^{2} + 1\). [2]
(b) Prove by mathematical induction that \[ u_0\, u_1\, u_2 \cdots u_{n-1} = u_n - 2, \qquad n \in \mathbb{Z}^{+}. \] [6]
(c) Prove by contradiction that, if \(m\) and \(n\) are natural numbers with \(m < n\), then \(u_m\) and \(u_n\) have no common factor greater than \(1\). [4]
(d) Hence prove that there are infinitely many prime numbers. [2]
(e) You are given that \(2^{32} = 4\,294\,967\,296\) and that \(641 \times 6\,700\,417 = 4\,294\,967\,297\). Show that the statement “\(u_n\) is prime for every \(n \in \mathbb{N}\)” is false. [2]
Total 16 marks
(a)
\((u_n - 1)^{2} = \left(2^{2^{n}}\right)^{2} = 2^{2 \times 2^{n}}\) M1
\(= 2^{2^{n+1}}\), so \((u_n - 1)^{2} + 1 = 2^{2^{n+1}} + 1 = u_{n+1}\) A1 AG
[2 marks]
(b)
\(n = 1\): \(\text{LHS} = u_0 = 3\) and \(\text{RHS} = u_1 - 2 = 5 - 2 = 3\), so \(P(1)\) is true A1
assume \(P(k)\) is true for some \(k \in \mathbb{Z}^{+}\): \(u_0\, u_1 \cdots u_{k-1} = u_k - 2\) M1
\(u_0\, u_1 \cdots u_{k-1}\, u_k = (u_k - 2)\, u_k\) (by the assumption) M1
\(= u_k^{2} - 2u_k = (u_k - 1)^{2} - 1\) A1
\(= \left((u_k - 1)^{2} + 1\right) - 2 = u_{k+1} - 2\) (by part (a)), so \(P(k + 1)\) is true A1
since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^{+}\) R1
[6 marks]
(c)
assume, for a contradiction, that \(u_m\) and \(u_n\) (\(m < n\)) have a common factor \(d > 1\) M1
\(u_m\) is one of the factors in \(u_0\, u_1 \cdots u_{n-1}\), so \(d\) divides \(u_0\, u_1 \cdots u_{n-1} = u_n - 2\) (by part (b)) R1
\(d\) divides both \(u_n\) and \(u_n - 2\), so \(d\) divides their difference \(2\), and \(d = 2\) R1
but \(2^{2^{n}}\) is even, so \(u_n\) is odd and \(2\) does not divide \(u_n\); this is a contradiction, so \(u_m\) and \(u_n\) have no common factor greater than \(1\) R1
[4 marks]
(d)
each \(u_n > 1\), so each \(u_n\) has at least one prime factor \(p_n\) R1
by part (c) the primes \(p_0, p_1, p_2, \ldots\) are all different, and there are infinitely many terms \(u_n\), so there are infinitely many primes R1
[2 marks]
(e)
\(u_5 = 2^{2^{5}} + 1 = 2^{32} + 1 = 4\,294\,967\,297\) M1
\(= 641 \times 6\,700\,417\), so \(u_5\) is not prime, and \(n = 5\) is a counterexample A1
[2 marks]