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2022 Paper 3 Q8
D: 1500.0 B: 1500.0
de moivre's theorem complex numbers binomial theorem trigonometric identities proof by contradiction irrationality number theory

  1. Use De Moivre's theorem to prove that for any positive integer \(k > 1\), \[ \sin(k\theta) = \sin\theta\cos^{k-1}\theta \left( k - \binom{k}{3}(\sec^2\theta - 1) + \binom{k}{5}(\sec^2\theta - 1)^2 - \cdots \right) \] and find a similar expression for \(\cos(k\theta)\).
  2. Let \(\theta = \cos^{-1}(\frac{1}{a})\), where \(\theta\) is measured in degrees, and \(a\) is an odd integer greater than \(1\). Suppose that there is a positive integer \(k\) such that \(\sin(k\theta) = 0\) and \(\sin(m\theta) \neq 0\) for all integers \(m\) with \(0 < m < k\). Show that it would be necessary to have \(k\) even and \(\cos(\frac{1}{2}k\theta) = 0\). Deduce that \(\theta\) is irrational.
  3. Show that if \(\phi = \cot^{-1}(\frac{1}{b})\), where \(\phi\) is measured in degrees, and \(b\) is an even integer greater than \(1\), then \(\phi\) is irrational.

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2021 Paper 2 Q8
D: 1500.0 B: 1500.0
integration by parts recurrence relation differentiation proof by induction irrationality sequences and series exponential function

  1. Show that, for \(n = 2, 3, 4, \ldots\), \[ \frac{d^2}{dt^2}\bigl[t^n(1-t)^n\bigr] = n\,t^{n-2}(1-t)^{n-2}\bigl[(n-1) - 2(2n-1)t(1-t)\bigr]. \]
  2. The sequence \(T_0, T_1, \ldots\) is defined by \[ T_n = \int_0^1 \frac{t^n(1-t)^n}{n!}\,e^t\,dt. \] Show that, for \(n \geqslant 2\), \[ T_n = T_{n-2} - 2(2n-1)T_{n-1}. \]
  3. Evaluate \(T_0\) and \(T_1\) and deduce that, for \(n \geqslant 0\), \(T_n\) can be written in the form \[ T_n = a_n + b_n e, \] where \(a_n\) and \(b_n\) are integers (which you should not attempt to evaluate).
  4. Show that \(0 < T_n < \dfrac{e}{n!}\) for \(n \geqslant 0\). Given that \(b_n\) is non-zero for all~\(n\), deduce that \(\dfrac{-a_n}{b_n}\) tends to \(e\) as \(n\) tends to infinity.

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2019 Paper 1 Q7
D: 1500.0 B: 1500.0
proof by contradiction irrationality divisibility surds number theory modular arithmetic coprime integers

Consider the following steps in a proof that \(\sqrt{2} + \sqrt{3}\) is irrational.

  1. If an integer \(a\) is not divisible by 3, then \(a = 3k \pm 1\), for some integer \(k\). In both cases, \(a^2\) is one more than a multiple of 3.
  2. Suppose that \(\sqrt{2} + \sqrt{3}\) is rational, and equal to \(\frac{a}{b}\), where \(a\) and \(b\) are positive integers with no common factor greater than one.
  3. Then \(a^4 + b^4 = 10a^2b^2\).
  4. So if \(a\) is divisible by 3, then \(b\) is divisible by 3.
  5. Hence \(\sqrt{2} + \sqrt{3}\) is irrational.
  1. Show clearly that steps 1, 3 and 4 are all valid and that the conclusion 5 follows from the previous steps of the argument.
  2. Prove, by means of a similar method but using divisibility by 5 instead of 3, that \(\sqrt{6} + \sqrt{7}\) is irrational. Why can divisibility by 3 not be used in this case?

Solution:

  1. Step 1: There are only three possibilities for the remainder of \(a\) when divided by \(3\), (\(0\), \(1\), \(2\)). \(a = 3m+r\). If \(r = 0\) we are done, if \(r = 1\) take \(k = m\), and \(r=2\) take \(k=(m+1)\) and we have \(a = 3k-1\) as required. Then \(a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1\) which is clearly \(1\) more than a square. Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\ \Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\ \Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\ \Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\ &&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\ \Rightarrow && -b^4 &= a^4-10a^2b^2 \\ \Rightarrow && a^4+b^4 &= 10a^2b^2 \end{align*} Step 4: If \(a\) is divisible by \(3\) then \(b^4 = 10a^2b^2-a^4\) is a multiple of \(3\), but if \(b\) was not a multiple of \(3\) then \(b^2\) would be \(1\) more than a multiple of \(3\) (by Step 3) and \(b^4\) would also be \(1\) more than a multiple of \(3\), and we would have a contradiction. Step 5: Follows since either \(a,b\) are both divisible by \(3\) (contradicting Step 2), or neither is, but then \(a^2\) and \(b^2\) are both one more than a multiple of \(3\) and the RHS is one more than a multiple of \(3\) but the LHS is \(2\) more than a multiple of \(3\) which is a contradiction.
  2. Step 1: If \(a\) is not divisible by \(5\) then \(a^2 \equiv \pm 1 \pmod{5}\) Step 2: Suppose \(\frac{a}{b} = \sqrt{6}+\sqrt{7}\) Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\ \Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\ \Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\ &&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\ \Rightarrow && a^4+b^4 &= 26a^2b^2 \end{align*} Step 4: If \(a\) is a multiple of \(5\) then so is \(b^4\) and hence so is \(b^2\) and \(b\). Step 5: But the left hand side is always \(2 \pmod{5}\) and the right hand side is never \(2 \pmod{5}\) contradiction. Divisibility by \(3\) doesn't work here since mod \(3\) we can have \(a = 1, b = 1\) and have a valid solution.

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2015 Paper 3 Q5
D: 1700.0 B: 1516.0
proof by contradiction irrationality well-ordering principle sqrt(2) cube root set theory integer properties

  1. In the following argument to show that \(\sqrt2\) is irrational, give proofs appropriate for steps 3, 5 and 6.
    1. Assume that \(\sqrt2\) is rational.
    2. Define the set \(S\) to be the set of positive integers with the following property:
      \(n\) is in \(S\) if and only if \(n \sqrt2\) is an integer.
    3. Show that the set \(S\) contains at least one positive integer.
    4. Define the integer \(k\) to be the smallest positive integer in \(S\).
    5. Show that \((\sqrt2-1)k\) is in \(S\).
    6. Show that steps 4 and 5 are contradictory and hence that \(\sqrt2\) is irrational.
  2. Prove that \(2^{\frac13} \) is rational if and only if \(2^{\frac23}\) is rational. Use an argument similar to that of part (i) to prove that \(2^{\frac13}\) and \(2^{\frac23}\) are irrational.

Solution:

  1. For step 3, since we have assumed \(\sqrt{2}\) is rational we can write it in the form \(p/q\) with \(p, q\) coprime with \(q \geq 1\). Then \(q \in S\) since \(q\sqrt{2} = p\) which is an integer. For step 5, notice that \((\sqrt{2}-1)k\) is an integer (since \(\sqrt{2}k\) is an integer and so is \(-k\). It is also positive since \(\sqrt{2} > 1\). We must check that \((\sqrt{2}-1)k \cdot \sqrt{2} = 2k - \sqrt{2}k\) is also an integer, but clearly it is as both \(2k\) and \(-\sqrt{2}k\) are integers. Therefore \((\sqrt{2}-1)k \in S\). For step 6, notice that \((\sqrt{2}-1) < 1\) and therefore \((\sqrt{2}-1)k < k\), contradicting that \(k\) is the smallest element in our set. (And all non-empty sets of positive integers have a smallest element)
  2. Claim: \(2^{\frac13}\) is irrational \(\Leftrightarrow 2^{\frac23}\) is irrational. Proof: Since \(2^{\frac13} \cdot 2^{\frac23} = 2\) if one of them is rational, then the other one must also be rational. Which is the same as them both being irrational at the same time.
    1. Assume that \(\sqrt[3]{2}\) is rational, ie \(\sqrt[3]{2} = p/q\) for some integers.
    2. \(S := \{ n \in \mathbb{Z}_{>0} : n \sqrt[3]{2} \text{ and } n \sqrt[3]{4}\in \mathbb{Z}\}\)
    3. Suppose \(k\) is the smallest element in \(S\) (which must exist, consider \(q^2\)
    4. Consider \((\sqrt[3]{2}-1)k\) then clearly this is an integer, and \((\sqrt[3]{2}-1)\sqrt[3]{2}k = \sqrt[3]{4}k - \sqrt[3]{2}k \in \mathbb{Z}\) and \((\sqrt[3]{2}-1)\sqrt[3]{4}k = 2 k -\sqrt[3]{4}k \in \mathbb{Z}\).
    5. But this is a smaller element of \(S\), contradicting that \(k\) is the smallest element. Therefore, we have a contradiction.

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2013 Paper 3 Q5
D: 1700.0 B: 1487.0
proof number theory irrationality coprime divisibility proof by contradiction prime factorization rational and irrational numbers

In this question, you may assume that, if \(a\), \(b\) and \(c\) are positive integers such that \(a\) and \(b\) are coprime and \(a\) divides \(bc\), then \(a\) divides \(c\). (Two positive integers are said to be coprime if their highest common factor is 1.)

  1. Suppose that there are positive integers \(p\), \(q\), \(n\) and \(N\) such that \(p\) and \(q\) are coprime and \(q^nN=p^n\). Show that \(N=kp^n\) for some positive integer \(k\) and deduce the value of \(q\). Hence prove that, for any positive integers \(n\) and \(N\), \(\sqrt[n]N\) is either a positive integer or irrational.
  2. Suppose that there are positive integers \(a\), \(b\), \(c\) and \(d\) such that \(a\) and \(b\) are coprime and \(c\) and \(d\) are coprime, and \(a^ad^b = b^a c^b \,\). Prove that \(d^b = b^a\) and deduce that, if \(p\) is a prime factor of \(d\), then \(p\) is also a prime factor of \(b\). If \(p^m\) and \(p^n\) are the highest powers of the prime number \(p\) that divide \(d\) and \(b\), respectively, express \(b\) in terms of \(a\), \(m\) and \(n\) and hence show that \(p^n\le n\). Deduce the value of \(b\). (You may assume that if \(x > 0\) and \(y\ge2\) then \(y^x > x\).) Hence prove that, if \(r\) is a positive rational number such that \(r^r\) is rational, then \(r\) is a positive integer.

Solution:

  1. Suppose \(q^nN = p^n\) then since \((p,q) =1\) we must have \(p \mid N\), and then by dividing both \(p^n\) and \(N\) by \(p\) we can repeat this process \(n\) times to find that \(N = kp^n\) and in particular \(q = 1\). Suppose \(\sqrt[n]{N} = \frac{p}q\) for \(p,q\) coprime positive integers (ie it is not irrational), then \(q^nN = p^n\) and so \(q = 1\) and in fact \(\sqrt[n]{N}\) is an integer so \(N\).
  2. Suppose \((a,b) = 1, (c,d) = 1\) and \(a^ad^b = b^ac^b\), then since \((a,b) = 1\) we must have \((b^a, a) = 1\) so \(b^a \mid d^b\). Similarly since \((c,d) = 1\) we must have \((d^b, c) = 1\) so \(d^b \mid b^a\). Therefore \(d^b = b^a\). Suppose \(p \mid d\) then \(p \mid d^b = b^a \Rightarrow p \mid b\). Suppose \(\nu_p(d) = m, \nu_p(b) = n\) we must have \(bm = \nu_p(d^b) = \nu_p(b^a) = an\), ie \(b = \frac{an}{m}\). Note that \(p^n \mid b \Rightarrow p^n \mid n \frac{a}{m} \Rightarrow p^n \mid n \Rightarrow p^n \leq n\). Since \((p,a) = 1\).. But since \(p^n > n\) if \(p \geq 2\) we must have that \(b = 1\). Therefore suppose \(r = \frac{a}{b}\) with \((a,b) = 1\) an \(r^r = \frac{c}{d}\) we must ahve \(a^ac^b = b^ad^b\) and so \(b = 1\) implying \(r\) is an integer.

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2011 Paper 3 Q2
D: 1700.0 B: 1516.0
polynomials rational root theorem proof by contradiction irrationality integer roots divisibility monic polynomial

The polynomial \(\f(x)\) is defined by \[ \f(x) = x^n + a_{{n-1}}x^{n-1} + \cdots + a_{2} x^2+ a_{1} x + a_{0}\,, \] where \(n\ge2\) and the coefficients \(a_{0}\), \(\ldots,\) \(a_{{n-1}}\) are integers, with \(a_0\ne0\). Suppose that the equation \(\f(x)=0\) has a rational root \(p/q\), where \(p\) and \(q\) are integers with no common factor greater than \(1\), and \(q>0\). By considering \(q^{n-1}\f(p/q)\), find the value of \(q\) and deduce that any rational root of the equation \(\f(x)=0\) must be an integer.

  1. Show that the \(n\)th root of \(2\) is irrational for \(n\ge2\).
  2. Show that the cubic equation \[ x^3- x +1 =0 \] has no rational roots.
  3. Show that the polynomial equation \[ x^n- 5x +7 =0 \] has no rational roots for \(n\ge2\).

Solution: Let \(\f(x) = x^n + a_{{n-1}}x^{n-1}+ \cdots + a_{2} x^2+ a_{1} x + a_{0}\), and suppose \(f(p/q) = 0\) with \((p,q) = 1\), the consider \begin{align*} && 0 &= q^{n-1}f(p/q) \\ &&&= \frac{p^n}{q} + \underbrace{a_{n-1}p^{n-1} + a_{n-2}p^{n-2}q + \cdots + a_0q^{n-1}}_{\in \mathbb{Z}} \\ \end{align*} But \(p^n/q \not \in \mathbb{Z}\) unless \(q = 1\) therefore \(p/q\) must be an integer, ie all rational roots are integers.

  1. Note that \(\sqrt[n]2\) is a root of \(x^n - 2 =0\), but this has no integer solutions. (We can try all factors of \(2\)). Therefore all its roots must be irrational, ie \(\sqrt[n]2\) is irrational for \(n \geq 2\)
  2. If \(n\) is a root of \(x^3 - x+1\) then it must be \(1\) or \(-1\) by the rational root theorem, ie \(1-1+1 \neq 0\) and \(-1 + 1 +1 \neq 0\), therefore no integer roots, therefore no rational roots.
  3. Suppose \(m\) is an integer root of \(x^n - 5x + 7 = 0\) then by considering parity we must have \(m^n - 5m + 7 \equiv 1 \pmod{2}\) therefore we cannot have any rational roots.

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