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2013 Paper 1 Q3
D: 1500.0 B: 1500.0
vectors binary operation associativity geometric proof sequences convex combination line segment division algebraic manipulation

For any two points \(X\) and \(Y\), with position vectors \(\bf x\) and \(\bf y\) respectively, \(X*Y\) is defined to be the point with position vector \(\lambda {\bf x}+ (1-\lambda){\bf y}\), where \(\lambda\) is a fixed number.

  1. If \(X\) and \(Y\) are distinct, show that \(X*Y\) and \(Y*X\) are distinct unless \(\lambda\) takes a certain value (which you should state).
  2. Under what conditions are \((X*Y)*Z\) and \(X*(Y*Z)\,\) distinct?
  3. Show that, for any points \(X\), \(Y\) and \(Z\), \[ (X*Y)*Z = (X*Z)*(Y*Z)\, \] and obtain the corresponding result for \(X*(Y*Z)\).
  4. The points \(P_1\), \(P_2\), \(\ldots\) are defined by \( P_1 = X*Y\) and, for \(n \ge2\), \(P_n= P_{n-1}*Y\,.\) Given that \(X\) and \(Y\) are distinct and that \(0<\lambda<1\), find the ratio in which \(P_n\) divides the line segment \(XY\).

Solution:

  1. Suppose \(X*Y = Y*X\), then \begin{align*} && X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\ && Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\ \Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y}) \end{align*} Therefore, either \(\mathbf{x} = \mathbf{y}\) or \(\lambda = \frac12\). Since we assumed \(X,Y\) were distinct, \(\mathbf{x} \neq \mathbf{y}\) and so \(X*Y\) and \(Y*X\) are distinct unless \(\lambda = \frac12\)
  2. Suppose \((X*Y)*Z = X*(Y*Z)\) \begin{align*} &&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\ &&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\ &&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\ &&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\ \Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\ &&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\ &&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x}) \end{align*} Therefore they are distinct unless \(\lambda = 1, 0\) or \(\mathbf{x} = \mathbf{z}\).
  3. Claim: \((X*Y)*Z = (X*Z)*(Y*Z)\) Proof: \begin{align*} && (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ && (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z} \end{align*} Claim: \(X*(Y*Z) = (X*Y)*(X*Z)\) Proof: \begin{align*} X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ (X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \end{align*}
  4. \(P_1 = X*Y\) divides the line segment into the ratio \(\lambda:(1-\lambda)\). \(P_n\) divides the line segment \(P_{n-1}Y\) into the ratio \(\lambda:(1-\lambda)\), therefore it divides the line segment \(XY\) in the ratio \(\lambda^n : 1- \lambda^n\) Alternatively, \begin{align*} P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\ P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\ &= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y} \end{align*} Suppose \(P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}\) then \begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\ &= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\ & = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}

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1999 Paper 3 Q7
D: 1680.5 B: 1516.0
groups binary operation associativity group axioms subgroup identity element inverse element

Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.

Solution: Claim: \(*\) is associative. Proof: Then \(x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz\) and \((x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz\) so \(x*(y*z) = (x*y)*z\) and we are done. Let \(G = \mathbb{R} \setminus \{-\frac1{a} \}\) In order to show that \((G, *)\) is a group we need to check:

  1. closure \(x*y = x + y + axy\) is a real number
  2. associativity we have already checked
  3. identity \(x*0 = x + 0 + 0 = x = 0*x\), so \(0\) is an identity
  4. inverses \(x*\left ( \frac{-x}{1+ax} \right ) = x - \frac{x}{1+ax} + ax \frac{-x}{1+ax} = \frac{x +ax^2 - x - ax^2}{1+ax} = 0\) so \(x\) has an identity, assuming \(1+ax \neq 0\) which is true for everything in our set
Consider the set \(\{0, \frac{-2}{a} \}\). Then \(\frac{-2}{a} * \frac{-2}{a} = \frac{-4}{a} + a \frac{4}{a^2} = 0\), so this is a group of order \(2\)

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1993 Paper 3 Q5
D: 1730.5 B: 1466.6
groups binary operation associativity commutativity complex numbers subgroup isomorphism order of elements identity element inverse matrix quaternion group

The set \(S\) consists of ordered pairs of complex numbers \((z_1,z_2)\) and a binary operation \(\circ\) on \(S\) is defined by $$ (z_1,z_2)\circ(w_1,w_2)= (z_1w_1-z_2w^*_2, \; z_1w_2+z_2w^*_1). $$ Show that the operation \(\circ\) is associative and determine whether it is commutative. Evaluate \((z,0)\circ(w,0)\), \((z,0)\circ(0,w)\), \((0,z)\circ(w,0)\) and \((0,z)\circ(0,w)\). The set \(S_1\) is the subset of \(S\) consisting of \(A\), \(B\), \(\ldots\,\), \(H\), where \(A=(1,0)\), \(B=(0,1)\), \(C=(i,0)\), \(D=(0,i)\), \(E=(-1,0)\), \(F=(0,-1)\), \(G=(-i,0)\) and \(H=(0,-i)\). Show that \(S_1\) is closed under \(\circ\) and that it has an identity element. Determine the inverse and order of each element of \(S_1\). Show that \(S_1\) is a group under \(\circ\). \hfil\break [You are not required to compute the multiplication table in full.] Show that \(\{A,B,E,F\}\) is a subgroup of \(S_1\) and determine whether it is isomorphic to the group generated by the \(2\times2\) matrix $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ under matrix multiplication.

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1990 Paper 3 Q3
D: 1667.9 B: 1490.9
groups group theory order of element commutativity conjugation proof inverse binary operation

The elements \(a,b,c,d\) belong to the group \(G\) with binary operation \(*.\) Show that

  1. if \(a,b\) and \(a*b\) are of order 2, then \(a\) and \(b\) commute;
  2. \(c*d\) and \(d*c\) have the same order;
  3. if \(c^{-1}*b*c=b^{r},\) then \(c^{-1}*b^{s}*c=b^{sr}\) and \(c^{-n}*b^{s}*c^{n}=b^{sr^{n}}.\)

Solution: \begin{questionparts} \item \((ab)^2 = abab = e\) (since \(ab\) has order \(2\)), but \(a^2 = e, b^2 = e \Rightarrow a^{-1} = a, b^{-1} = b\) (since \(a\) and \(b\) have order 2) so \(ba = ab\) by multiplication on the left by \(a\) and right by \(b\). \item Suppose \((cd)^n = e \Leftrightarrow d(cd)^nc = dc \Leftrightarrow (dc)^n(dc) = e \Leftrightarrow (dc)^n = e\) Therefore any number for which \((cd)^n = e\) has the property that \((dc)^n = e\) and vice-versa, in particular the smallest number for either \(cd\) or \(dc\) will also be the smallest number for the other. \item Given \(c^{-1}bc=b^r\), then \(b^{rs} = (b^r)^s = (c^{-1}bc)^s =\underbrace{(c^{-1}bc)(c^{-1}bc) \cdots (c^{-1}bc)}_{s \text{ times}} = c^{-1}\underbrace{bb\cdots b}_{s \text{ times}}c = c^{-1}b^sc\) We proceed by induction on \(n\). When \(n = 0\), we have \(b^s = b^{sr^0}\) so the base case is true. Suppose it is true for some \(n = k\), ie \(c^{-k}b^sc^k = b^{sr^k}\). Now consider \(c^{-{k+1}}b^sc^{k+1} = c^{-1}c^{-k}b^sc^kc = c^{-1}b^{sr^k}c = (b^{sr^k \cdot r}) = b^{sr^{k+1}}\) (where the second to last equality was by the previous part). Therefore if our statement is true for \(n=k\) it is true for \(n = k+1\). Therefore, since it is also true for \(n=0\), by the principle of mathematical induction it is true for all non-negative integers \(n\).

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1989 Paper 3 Q3
D: 1675.2 B: 1469.0
groups rotation matrix matrix geometric series proof by induction recurrence relation binary operation finite group

The matrix \(\mathbf{M}\) is given by \[ \mathbf{M}=\begin{pmatrix}\cos(2\pi/m) & -\sin(2\pi/m)\\ \sin(2\pi/m) & \cos(2\pi/m) \end{pmatrix}, \] where \(m\) is an integer greater than \(1.\) Prove that \[ \mathbf{M}^{m-1}+\mathbf{M}^{m-2}+\cdots+\mathbf{M}^{2}+\mathbf{M}+\mathbf{I}=\mathbf{O}, \] where $\mathbf{I}=\begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( and \)\mathbf{O}=\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}.$ The sequence \(\mathbf{X}_{0},\mathbf{X}_{1},\mathbf{X}_{2},\ldots\) is defined by \[ \mathbf{X}_{k+1}=\mathbf{PX}_{k}+\mathbf{Q}, \] where \(\mathbf{P,Q}\) and \(\mathbf{X}_{0}\) are given \(2\times2\) matrices. Suggest a suitable expression for \(\mathbf{X}_{k}\) in terms of \(\mathbf{P},\) \(\mathbf{Q}\) and \(\mathbf{X}_{0},\) and justify it by induction. The binary operation \(*\) is defined as follows: \[ \mathbf{X}_{i}*\mathbf{X}_{j}\mbox{ is the result of substituting \ensuremath{\mathbf{X}_{j}}for \ensuremath{\mathbf{X}_{0}}in the expression for \ensuremath{\mathbf{X}_{i}}. } \] Show that if \(\mathbf{P=M},\) the set \(\{\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3},\ldots\}\) forms a finite group under the operation \(*\).

Solution: \(\mathbf{M}^m = \mathbf{I}\), we also have \(\mathrm{det}(\mathbf{M - I}) = \cos^2(2\pi/m) - 2\cos(2\pi/m) + 1 + \sin^2(2\pi/m) = 2(1-\cos(2\pi/m))\) therefore \(\mathbf{M-I}\) is invertible. Therefore since \(\mathbf{(M-I)(M^{m-1} + M^{m-1} + \cdots + M^2 + M + I)= M^m-I = 0}\) we can cancel the \(\mathbf{M-I}\) to obtain the desired result. \(\mathbf{X_0 = X_0}\) \(\mathbf{X_1 = PX_0+Q}\) \(\mathbf{X_2 = P(PX_0+Q)+Q = P^2X_0 + PQ + Q}\) Claim: \(\mathbf{X_k = P^k X_0 + (P^{k-1} + P^{k-2} + \cdots + I)Q}\) Proof: (By induction on \(k\)). Base case \(k = 0\) is true. Assume it's true for some \(k = l\), then consider \(k = l+1\) \(\mathbf{X_{l+1} = PX_l + Q = P( P^l X_0 + (P^{l-1} + P^{l-2} + \cdots + I)Q) + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P)Q + Q = P^{l+1}X_0 + (P^l + P^{l-1} + \cdots + P + I)Q}\) Suppose \(\mathbf{P} = \mathbf{M}\), then consider the set \(\{\mathbf{X_1, X_2}, \ldots\}\) with the operation \(*\) as defined. \(\mathbf{X_i * X_j} = M^{i}(X_j) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i}(M^jX_0 + (M^{j-1} + M^{j-2} + \cdots + M + I)Q) + (M^{i-1} + M^{i-2} + \cdots + M + I)Q = M^{i+j}X_0 + (M^{i+j-1}+\cdots + M + I)Q = X_{i+j}\) Since \(X_m = X_0\) we can check all the requirements of the group, but this is going to be isomorphic to the cyclic group with \(m\) elements.

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1987 Paper 2 Q10
D: 1500.0 B: 1500.0
groups group axioms binary operation closure associativity identity element inverse proof abstract algebra

The set \(S\) consists of \(N(>2)\) elements \(a_{1},a_{2},\ldots,a_{N}.\) \(S\) is acted upon by a binary operation \(\circ,\) defined by \[ a_{j}\circ a_{k}=a_{m}, \] where \(m\) is equal to the greater of \(j\) and \(k\). Determine, giving reasons, which of the four group axioms hold for \(S\) under \(\circ,\) and which do not. Determine also, giving reasons, which of the group axioms hold for \(S\) under \(*\), where \(*\) is defined by \[ a_{j}*a_{k}=a_{n}, \] where \(n=\left|j-k\right|+1\).

Solution:

  1. (Closure) This operation is clearly closed by construction
  2. (Associative) \(a_j \circ (a_k \circ a_l) = a_j \circ a_{\max(k,l)} = a_{\max(j,k,l)} = a_{\max(j,k)} \circ a_l = (a_j \circ a_k) \circ a_l\), so it is associative
  3. (Identity) \(a_1 \circ a_k = a_{\max(1,k)} = a_k = a_{\max(k,1)} = a_k \circ a_1\) so \(a_1\) is an identity.
  4. (Inverses) There is no inverse, since \(a_N \circ a_k = a_N\) for all \(k\), and hence \(a_N\) can have no inverse.
  1. (Closure) \(n = |j-k|+1 \geq 1\) so we need to show that \(n \leq N\) to ensure closure. This is true since the largest \(j-k\) can be is if \(j = N\) and \(k = 1\), and this also satisfies \(|j-k| + 1 \leq N\). Hence the operation is closed.
  2. (Associative) \(a_j * (a_k * a_l) = a_j * (a_{|k-l|+1}) = a_{|j-|k-l|-1|+1}\). \((a_j * a_k) * a_l = a_{|j-k|+1}*a_l = a_{|l-|j-k|-1|+1}\). \(a_2 * (a_2 * a_3) = a_2 * a_2 = a_1\). \((a_2 *a_2)*a_3 = a_1 * a_3 = a_3 \neq a_2\) therefore this isn't associative for any \(N > 2\)
  3. (Identity) \(a_1\) is an identity, since \(a_1 * a_k = a_{|k-1|+1} = a_{k-1+1} = a_k\).
  4. (Inverse) Every element is self-inverse since \(a_k * a_k = a_{|k-k|+1} = a_1\)

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