Problems

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Solution:

1. Not that unless a side is the largest side, it is clearly shorter than the sum of the other two sides (since it's greater than or equal to one on its own). Note also that the distance from one vertex to the other (say \(c\)) is shorter than going via the other vertex \(a+b\), therefore \(c < a+b\).
2. Draw a line of the length of the largest number, say \(c\), then since \(c < a+b\) we must have circles radius \(a\) and \(b\) at the endpoints cross, and at their intersection we have a vertex of a \(c\)-\(a\)-\(b\) triangle.
   

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3. (A) always. Suppose \(c\) is the longest side, then \(c < a+b \Rightarrow c+1 < a + 1 + b+1\) so \((a+1,b+1,c+1)\) are still sides of a triangle. (B) sometimes, but not always. \((1,1,1) \to (1,1,1)\) is still a triangle, but \((10, 10, 1) \to (1, 10, \frac{1}{10})\) is not a triangle since \(10 > 1 + \frac{1}{10}\) (C) never, suppose \(a \leq b \leq c\) then the sides are \(b-a, c-b, c-a\) but \(c-a = (c-b)+(b-a)\) so the triangle inequality cannot be satisfied. (D) always - without loss of generality let \(c\) be the longest side, and since every term is homogeneous degree \(2\) we can divide through by \(c^2\) to see we have the sides \(a^2+b, b^2+a, 1+ab\) and note that \(1 + ab < a+b +ab < a+b+a^2+b^2\), also \(a^2+b < 1 + b < 1 + (a+b)b = 1 + b^2 + ab < (1+ab)+(b^2+a)\).
4. Suppose \(f\) is increasing and \(\f(x)/x\) is decreasing, and suppose \(a,b,c\) are side-lengths of a triangle. Wlog \(c\) is the longest side, then note \(f(c) > f(b), f(a)\), so it suffices to prove that \(f(c) < f(a)+f(b)\) \begin{align*} \frac{f(c)}{c} < \frac{f(a)}{a}: && f(a) &> \frac{a}{c} f(c) \\ \frac{f(c)}{c} < \frac{f(b)}{b}: && f(b) &> \frac{b}{c} f(c) \\ \Rightarrow && f(a)+f(b) &> f(c) \underbrace{\left ( \frac{a+b}{c} \right)}_{>1} \\ &&&> f(c) \end{align*} as required

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Solution:

1. \(S \cup T\) is the set of odd positive integers. \(S \cap T\) is the empty set.
2. Suppose we have two integers in \(S\), say \(4n+1\) and \(4m+1\), so \((4n+1)(4m+1) = 16nm + 4(n+m) + 1 = 4(4nm + n+m) + 1\) which clearly is in \(S\). The product of any two integers in \(T\) is in \(S\), since \((4n+3)(4m+3) = 16nm + 12(n+m) + 9 = 4(4nm+3(n+m) + 2) + 1\)
3. Suppose \(p \in T\) is not prime, then it must have prime factors. They must all be odd, so they are in \(T\) or \(S\). If they are all in \(S\) then \(p\) must also be in \(S\) (as the product of numbers in \(S\)) but this is a contraction. Therefore \(p\) must have a prime factor in \(T\).
4. • \(3, 7, 11, 19\) are all primes in \(T\). Notice therefore that any product of two of them must be \(S\)-prime. But then \((3 \times 7) \times (11 \times 19)\) and \((3 \times 11) \times (7 \times 19)\) are distinct factorisations into \(S\)-prime factors.

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Solution:

1. If \(m = 1000\), then \(n \geq m \Rightarrow n^2 \geq 1000n \Rightarrow (1000n) \leq (n^2)\)
2. This is false. Let \(s_i = 1,2,1,2,\cdots\) and \(t_i = 2,1,2,1,\cdots\).
3. Suppose that for \(n \geq m_1, s_n \le t_n\) and for \(n \geq m_2, s_t \le u_n\), then for \(n \geq m = \max(m_1, m_2), s_n \leq t_n \leq u_n \Rightarrow s_n \leq u_n \Rightarrow (s_n) \leq (u_n)\)
4. Let \(m = 6\), then if \(n \geq m, 2^n \geq 1 + n + \frac{n(n-1)}{2} + \frac{n(n-1)}{2} + n + 1 = n^2 + n + 2 \geq n^2\), so \((2^n) \geq (n^2)\)

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Solution:

1. Notice that \(u^2 = v+2wz\), so \(w,z\) are roots of the quadratic \(t^2 -ut+\frac{u^2-v}{2}\). Therefore they are both real if \(u^2 \geq 2(u^2-v) \Rightarrow 2v \geq u^2\).
2. \begin{align*} && w+z &= u \\ && w^2+z^2 &= u^2 - \tfrac23 \\ && w^3+z^3 &= \lambda(u-1) \\ \\ && wz &= \frac{u^2 - (u^2-\tfrac23)}{2} = \tfrac13\\ \\ && (w+z)(w^2+z^2) &= w^3+z^3+wz(w+z) \\ &&u(u^2-\tfrac23)&= \lambda(u-1)+\frac13u \\ \Rightarrow && u^3-u&= \lambda (u-1) \\ \Rightarrow && 0 &= (u-1)(u(u+1) - \lambda) \\ \Rightarrow && 0 &= (u-1)(u^2+u - \lambda) \end{align*} Therefore there will be at most 3 values for \(u\), unless \(1\) is a root of \(u^2+u-\lambda\), ie \(\lambda = 2\). Suppose \(u = 1\), then we have: \(w+z = 1, wz = 1/3 \Rightarrow w,z = \frac{-1 \pm \sqrt{-1/3}}{2}\) which are clearly complex.

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Solution:

1. If \(f(x) = \sin nx\) then \(f'(x) = n \cos n x\) and so \begin{align*} && LHS &= \int_0^\pi \sin^2 n x \d x \\ &&&= \left [ \frac{x+\frac1{2n}\sin 2n x}{2} \right ]_0^{\pi} \\ &&&= \frac{\pi}{2} \\ \\ && RHS &= \int_0^{\pi} n^2 \cos^2 n x \d x \\ &&&= n^2 \left [ \frac{\frac{1}{2n}\sin 2n x + x}{2} \right]_0^{\pi} \\ &&&= n^2\frac{\pi}{2} \geq LHS \end{align*} [\(f(0) = 0, f(\pi) \neq 0\)] Suppose \(f(x) = x\) then \(f'(x) = 1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\). [\(f(0) \neq 0, f(\pi) = 0\)] Suppose \(f(x) = \pi - x\) then \(f'(x) = -1\) and \(LHS = \frac{\pi^3}{3} > \pi = RHS\)
2. Suppose \(f(x) = x(\pi - x)\) then \(f'(x) = \pi - 2x\) and so \begin{align*} && \int_0^\pi x^2(\pi-x)^2 \d x &\leq \int_0^\pi (\pi-2x)^2 \d x \\ \Leftrightarrow && \left [\pi^2 \frac{x^3}{3} - 2\pi \frac{x^4}{4} + \frac{x^5}{5} \right]_0^{\pi} &\leq \left [ \pi^2x - 4\pi \frac{x^2}{2} + \frac{4x^3}{3} \right]_0^{\pi} \\ \Leftrightarrow && \pi^5 \left (\frac13 - \frac12+\frac15 \right) &\leq \pi^3 \left ( 1 - 2+\frac43 \right) \\ \Leftrightarrow && \pi^2 \frac{1}{30} &\leq \frac13 \\ \Leftrightarrow && \pi^2 &\leq 10 \end{align*} Suppose \(f(x) = p\sin \tfrac12 x + q \cos \tfrac12 x + r\), so \(f(0) = q + r\) and \(f(\pi) = p + r\), so say \(p = q = 1, r = -1\) \begin{align*} && LHS &= \int_0^{\pi} \left ( \sin \tfrac12 x + \cos \tfrac12 x-1\right)^2 \d x \\ &&&=\int_0^\pi \left ( \sin^2 \tfrac12 x + \cos^2 \tfrac12 x+1-2\sin \tfrac12 x - 2\cos \tfrac12 x+ \sin x \right)\\ &&&= \left [2x + 4\cos \tfrac12 x - 4\sin \tfrac12 x - \cos x \right]_0^{\pi} \\ &&&= \left ( 2\pi -4+1 \right) - \left ( 4-1 \right) \\ &&&= 2\pi -6\\ \\ && RHS&= \int_0^{\pi} \left ( \tfrac12 \cos \tfrac12 x -\tfrac12 \sin \tfrac12 x\right)^2 \d x \\ &&&= \int_0^{\pi} \left ( \tfrac14 \cos^2 \tfrac12 x +\tfrac14 \sin^2 \tfrac12 x-\tfrac14 \sin x\right) \d x \\ &&&= \frac{\pi}{4} - \frac12 \\ \Rightarrow && 2\pi -6 &\leq \frac{\pi}{4} - \frac12 \\ \Rightarrow && \frac{7\pi}{4} &\leq \frac{11}{2} \\ \Rightarrow && \pi &\leq \frac{22}{7} \end{align*} \(22^2/7^2 = 484/49 < 10\) therefore \(\pi \leq \frac{22}{7}\) is the better estimate.

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Solution:

• A: Suppose for sake of contradiction that neither \(p\) nor \(q\) is irrational, then \(pq\) is the product of two rational numbers, ie is also rational. Therefore \(pq\) is rational. Contradiction.
• B: Suppose for the sake of contradiction both \(p\) and \(q\) are rational, but then \(p+q\) is also rational, contradicting \(p+q\) is irrational.
• C: Note that \(\sqrt{2}\) and \(-\sqrt{2}\) are both irrational, but \(\sqrt{2}+(-\sqrt{2}) = 0\) which is rational.

Since \((\pi + \e) + (\pi - \e) = 2\pi\) is irrational, at most one of \(\pi+\e\) and \(\pi - \e\) can be rational. Since \((\pi+\e)(\pi-\e) = \pi^2 - \e^2\) is is the product of a (non-zero) rational and an irrational, \(\pi^2 - \e^2\) cannot be rational. Therefore for two of these numbers to be irrational, we need \(\pi^2 + \e^2\) to be rational. But then squaring whichever of \(\pi \pm \e\) is rational and subtracting \(\pi^2+\e^2\) we obtain \(\pm 2\pi \e\) which is irrational. But the product and sum of rationals is irrational. Therefore it cannot be the case that more than one of these numbers is rational.

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Solution:

1. \(f(12) = f(2^2 \cdot 3) = 12 \cdot (1-\frac12)\cdot(1-\frac13) = 12 \cdot \frac12 \cdot \frac 23 = 4\) \(f(180) = f(2^2 \cdot 3^2 \cdot 5) = 180 \cdot \frac12 \cdot \frac23 \cdot \frac 45 = 12 \cdot 4 = 48\) Suppose \(N\) has prime decomposition \(p_1^{a_1} \cdots p_k^{a_k}\), then \(f(N) = p_1^{a_1} \cdots p_k^{a_k} (1 - \frac{1}{p_1})\cdots ( 1- \frac{1}{p_k}) = p_1^{a_1-1} \cdots p_k^{a_k-1}(p_1-1) \cdots (p_k-1)\) which is clearly an integer.
2. \(f(2) = 1, f(4) = 2 \neq f(2) \cdot f(2)\) If \(p, q\) are distinct primes then \(f(p) = p \cdot \frac{p-1}{p} = p-1\) and \(f(q) = q-1\). \(f(pq) = pq \frac{p-1}{p} \cdot \frac{q-1}{q} = (p-1)(q-1) = f(p)f(q)\) \(f(12) = 4 = 2\cdot 2 = f(4) \cdot f(3)\)
3. \(f(p^m) = p^{m-1} (p-1)\) \(146410 = 10 \cdot 14641 = 10 \cdot 11^4\). Therefore \(p = 11, m = 5\)

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Solution: \begin{align*} \mathbb{P}(X = x_1) &= a + q - a = q \\ \mathbb{P}(X = x_2) &= 1 - q \\ \mathbb{P}(Y = y_1) & = a + p - a = p \\ \mathbb{P}(Y = y_2) & = 1 - p \end{align*} \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &= \l qx_1 + (1-q)x_2 \r \l p y_1 + (1-p)y_2\r \\ &= qpx_1y_1 + q(1-p)x_1y_2 + (1-q)px_2y_1 + (1-q)(1-p)x_2y_2 \\ \mathbb{E}(XY) &= ax_1y_1 + (q-a)x_1y_2 + (p-a)x_2y_1 + (1 + a - p - q)x_2y_2 &= \end{align*} Therefore \(\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)\) is a degree 2 polynomial in the \(x_i, y_i\). If \(x_1 = x_2\) then we have: \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &=x_1 \l p y_1 + (1-p)y_2\r \\ \mathbb{E}(XY) &= x_1(ay_1 + (q-a)y_2 + (p-a)y_1 + (1 + a - p - q)y_2) \\ &= x_1 (py_1 + (1-p)y_2) \end{align*} Therefore \(x_1 - x_2\) is a root and by symmetry \(y_1 - y_2\) is a root. Therefore it remains to check the coefficient of \(x_1y_1\) which is \(a - pq\) to complete the factorisation. For any two random variables taking two distinct values, we can find \(a, q, p\) satisfying the relations above. We also note that \(X\) and \(Y\) are independent if \(\mathbb{P}(X = x_i, Y = y_i) = \mathbb{P}(X = x_i)\mathbb{P}(Y = y_i)\). Since \(x_1 \neq x_2\) and \(y_1 \neq y_2\) and \(\E(A B) = \E( A)\E (B) \Rightarrow a = pq\). But if \(a = pq\), we have \(\mathbb{P}(X = x_1, Y = y_1) = \mathbb{P}(X = x_1)\mathbb{P}(Y = y_1)\) and all the other relations drop out similarly. Consider \begin{align*} \mathbb{P}(A = -1, B = 1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = 0, B = 0) &= \frac{1}{3} \\ \mathbb{P}(A = 1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \end{align*}

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Solution:

1. This is false, for example let \(\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), then \begin{align*} \mathbf{AB} &= \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ &= \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \\ \mathbf{BA} &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ \end{align*}
2. This is also false, using the same matrices from part (i), we find: \begin{align*} (\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\ &= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ &\neq \mathbf{A}^2-\mathbf{B}^2 \end{align*}
3. This is true. Claim: The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\) Proof: \((\Rightarrow)\) Suppose \(\det\mathbf{A} \neq 0\) then \(\mathbf{A}\) has an inverse, and so we must have \(\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}\). \((\Leftarrow)\) Suppose \(\det \mathbf{A} = 0\) then \(ad-bc=0\), so consider the matrix \(\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}\) (or if this is zero, \(\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}\))
4. This is false. Consider \(\mathbf{A} = \mathbf{B} = \mathbf{0}\), then \(\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}\) has the property that \(\mathbf{X}^2 = \mathbf{0}\) for all \(x\), so at least more than 2 values

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Solution: Since \(0 = \det \mathbf{0} = \det \mathbf{AB} = \det \mathbf{A} \det\mathbf{B}\) at least one of \(\det \mathbf{A}\) or \(\det \mathbf{B}\) is zero. If \(\det \mathbf{B} \neq 0\) then \(\mathbf{B}\) is invertible, and multiplying on the right by \(\mathbf{B}^{-1}\) gives us \(\mathbf{A} = \mathbf{0}\). If \(\mathbf{A} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix}\), then \(\det \mathbf{A} = \det \mathbf{B} = 0\), but \(\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \mathbf{0}\) Since \(\mathbf{M}\) commutes with itself and the identity matrix, this is equivalent to factorising the polynomial over the reals. Therefore $$\mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}-2\mathbf{I})(\mathbf{M}+\mathbf{I})(\mathbf{M}+3\mathbf{I}),$$ Since we now know at least one of \(\det (\mathbf{M}-2\mathbf{I})\), \(\det (\mathbf{M}+\mathbf{I})\), \(\det (\mathbf{M}+3\mathbf{I})\), we should look at cases: Since at least one of those must be non-zero, we must have the following cases: \((a,b) = (2,-1), (-1,2), (-1,-3), (-3,-1), (2,-3), (-3,2)\) In each of those cases, we will have: \(\begin{pmatrix} 0 & c \\ 0 & b+k \end{pmatrix}\begin{pmatrix} a+l & c \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\) and so all of those solutions are valid. So \(c\) can be anything as long as \((a,b)\) are in that set of solutions

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Solution: Suppose \(|G_1|, |G_2| < |G|\) for sake of contraction. Then by Lagrange's theorem \(|G_1| \mid |G|\) and \(|G_2| \mid |G|\), so \(|G_1|, |G_2| \leq \frac{|G|}{2}\). But \(|G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|\). \(|G_1 \cup G_2| = |G|\) by assumption, and \(e \in G_1 \cap G_2\), so \(|G_1 \cap G_2| \geq 1\). Therefore \(|G| = |G_1| + |G_2| - |G_1 \cap G_2| \leq \frac{|G|}{2} + \frac{|G|}{2} - 1 = |G| - 1 < |G|\), contradiction! Let \(H = K_4 = \{e, a, b, c\}\) with \(a^2 = b^2 = c^2 = e\). then \(H = \{e, a\} \cup \{e, b\} \cup \{e, c\}\) with all subgroups distinct