Problems

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Solution:

1. If \(b > a\) and \(c > 1\) then \(c \geq 2 \underbrace{\Rightarrow}_{\times c} bc \geq 2c = c+c \underbrace{\geq}_{c \geq 2} 2 + c\)
2. If \(a < b\) and \(d < c\) then \(b \geq a+1\) and \(c \geq d+1\) so \begin{align*} bc - ad &\underbrace{\geq}_{b \geq a+1} (a+1)c - ad \\ &\underbrace{\geq}_{d \leq c-1} (a+1)c - a(c-1) \\ &= a+c \end{align*}
3. If \(\displaystyle \frac{a}{b} < p < \frac{c}{d}\) then \(a < pb\) and \(pd < c\) so by the previous part \((pb)c - a(pd) \geq a + c \Leftrightarrow (bc-ad)p \geq a+c\).
4. If \(\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}\) then \(\displaystyle \frac{qa}{b} < p < \frac{qc}{d}\) and so by the previous part we must have \((bc-ad)qp \geq q(a+c) \Rightarrow p \geq \frac{a+c}{bc-ad}\). Similarly we have \(\frac{d}{c} < \frac{q}{p} < \frac{b}{a}\) and so \(q \geq \frac{b+d}{bc-ad}\)

Suppose \(\frac{p}{q}\) is a fraction such that \(q \leq 20\) and \(\frac89 < \frac{p}q < \frac9{10}\) then: \begin{align*} q & \leq 20 \\ p & \geq \frac{17}{81-80} = 17 \\ q & \geq \frac{19}{81-80} = 19 \end{align*} Therefore the only fraction is \(\frac{17}{19}\) since \(\frac{18}{19} > \frac{18}{20} = \frac{9}{10}\)

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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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Solution: Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}

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Solution:

1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}

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Solution: Recall, \(\cosh^{-1} x = \ln (x + \sqrt{x^2-1})\) \begin{align*} \cosh(n \cosh^{-1} x) &= \frac12 \left ( \exp(n \cosh^{-1} x) + \exp(-n\cosh^{-1}x) \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x + \sqrt{x^2-1})^{-n} \right) \\ &= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ \end{align*} When \(n = 2k+1\) \begin{align*} \cosh(n \cosh^{-1} x)&= \frac12 \left ((x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^{n} \right) \\ &= \frac12 \left (\sum_{i=0}^{2k+1}\binom{2k+1}{i}x^{2k+1-i}\left ( (\sqrt{x^2-1}^{i} + (-\sqrt{x^2-1})^{i} \right) \right) \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2k+1-2i}(x^2-1)^i \\ &=\sum_{i=0}^{k} \binom{2k+1}{2i}x^{2(k-i)+1}(x^2-1)^i \\ \end{align*} Which is clearly a polynomial with only odd degree terms. \begin{align*} a_0 &= \frac{\d y}{\d x} \vert_{x=0} \\ &= \sum_{i=0}^k\binom{2k+1}{2i} \left ( (2(k-i)+1)x^{2(k-i)}(x^2-1)^i + 2i\cdot x^{2(k-i)+2}(x^2-1) \right) \\ &= \binom{2k+1}{2k} (-1)^{k} \\ &= (-1)^k(2k+1) \end{align*}

1. \begin{align*} a_1 &= \binom{2k+1}{2k}\binom{k}{1}(-1)^{k-1}+\binom{2k+1}{2(k-1)}(-1)^{k-1} \\ &=(-1)^{k-1}\cdot ( (2k+1)k + \frac{(2k+1)\cdot 2k \cdot (2k-1)}{3!}) \\ &= (-1)^{k-1}(2k+1)k\frac{3 + 2k-1}{3} \\ &= (-1)^{k-1}2(2k+1)k (k+1) \end{align*}
2. \begin{align*} a_2 &= \binom{2k+1}{2k} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{2(k-1)} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{2(k-2)} (-1)^{k-2} \\ &= \binom{2k+1}{1} \binom{k}{2}(-1)^{k-2} + \binom{2k+1}{3} \binom{k-1}{1} (-1)^{k-2}+\binom{2k+1}{5} (-1)^{k-2} \\ &= (-1)^{k} \left (\binom{2k+1}{1} \frac{k(k-1)}{2} + \binom{2k+1}{3}(k-1)+\binom{2k+1}{5} \right) \\ &= (-1)^{k} \left ( \frac{(2k+1)k(k-1)}{2} + \frac{(2k+1)k(2k-1)}{3} + \frac{(2k+1)k(2k-1)(k-1)(2k-3)}{5\cdot2\cdot3} \right) \\ &= (-1)^k (2k+1)k\frac{1}{30} \left ( 15(k-1) + 10(2k-1)+(2k-1)(k-1)(2k-3) \right) \end{align*}

\begin{align*} \sum_{r=0}^k a_k &= \frac12 \left ((1 + \sqrt{1^2-1})^n + (1 - \sqrt{1^2-1})^{n} \right) \\ &= 1 \end{align*}

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Solution:

1. There are \(m \cdot n\) line segments, since each of the \(m\) points on the \(y\)-axis correspond to \(n\) points of the \(x\)-axis.
2. Every pair of points on the \(x\)-axis and pair of points on the \(y\) axis will generate \(4\) lines, but only \(1\) of those points will meet in the middle, therefore there will be \(\binom{m}{2} \binom{n}{2}\) intersections.

The segment joining the smallest \(x\) to the smallest \(y\) and the segment joining the largest \(x\) to the largest \(y\) will not meet any other segments. The segment joining the smallest \(x\) to the largest \(y\) and the segment joining the largest \(x\) to the smallest \(y\) will have the largest number of intersections. They each intersect all the other segment coming out of other points (except the ones headed for the same point) ie \((m-1)(n-1)\) each. They also intersect each other, so \(2(m-1)(n-1)-1\) points. Therefore we are interested in: \begin{align*} \frac{\binom{n}{2}\binom{n}{2} - 2(n-1)^2+1}{n^2-4} &= \frac{\frac{n^2(n-1)^2}{4} - 2n^2+4n-1}{n^2-4} \\ &= \frac{n^4-2n^3+n^2-8n^2+16n-4}{4(n^2-4)} \\ &= \frac{(n - 2) (n^3 - 7 n + 2)}{4(n^2-4)} \\ &= \frac{n^3-7n+3}{4(n+2)} \end{align*}

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Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).

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Solution: After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days. Therefore the outstanding balance will be: \begin{align*} c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\ &= \frac{c[(1+k)^n-1]}{k} \end{align*} At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\). We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is: \begin{align*} (1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k} \end{align*} Therefore the net position, \(S_{T+m}\) will be: \begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\ &= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k} \end{align*} Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)

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Solution: \begin{questionparts} \item Show that, if \(o(g)=n\) and \(g^{N}=e,\) then \(n\) divides \(N.\) Using the division algorithm, write \(N = qn + r\) where \(0 \leq r < n\) to divide \(N\) by \(n\). Then we have \(e = g^N = g^{qn + r} = g^{qn}g^r = (g^{n})^qg^r = e^qg^r = g^r\) therefore \(r\) is a number smaller than \(n\) such that \(g^r = e\). Therefore either \(r = 0\) or \(o(g) = r\), but by definition \(o(g) = n\) therefore \(r = 0\) and \(n \mid N\). \item Let \(g\) and \(h\) be elements of \(G\). Prove that, for any integer \(m,\) \[ gh^{m}g^{-1}=(ghg^{-1})^{m}. \] \((ghg^{-1})^m = \underbrace{(ghg^{-1})(ghg^{-1})\cdots(ghg^{-1})}_{m \text{ times}} = gh(g^{-1}g)h(g^{-1}g)\cdots (g^{-1}g)hg^{-1} = gh^mg^{-1}\) \item Let \(g\) and \(h\) be elements of \(G\), such that \(g^{5}=e,h\neq e\) and \(ghg^{-1}=h^{2}.\) Prove that \(g^{2}hg^{-2}=h^{4}\) and find \(o(h).\) \(g^2hg^{-2} = g(ghg^{-1})g^{-1} = gh^2g^{-1} = (ghg^{-1})^2 = (h^2)^2 = h^4\). \(h = g^{5}hg^{-5} = g^4ghg^{-1}g^{-4} = g^4h^2g^{-4} = g^3(ghg^{-1})^2g^{-3} = g^3h^4g^{-3} = h^32\) Therefore \(e = h^{31}\). Therfore \(o(h) \mid 31 \Rightarrow \boxed{o(h) = 31}\) since \(31\) is prime and \(o(h) \neq 1\)

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Solution:

1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}

The general result is: If \(\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc\) then \((Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2\) Alternatively, if \(\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}\) then \(\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2\) which is easy to see is true.

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Solution:

1. If we can show that \(0 < a - \sqrt{a^2-b} < 1\) then we will be done, since raising a number in \([0,1)\) to a positive integer power will always remain in the same interval. Clearly \(\sqrt{a^2-b} < \sqrt{a^2} = a\) so we have \(a-\sqrt{a^2-b} > 0\) We also have that \(\sqrt{a^2-b} > \sqrt{a^2-(2a-1)} = (a-1)\). Therefore \(a - \sqrt{a^2-b} < a - (a-1) = 1\) as required.
2. If we can show that \(\l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n = N -r + s\) is an integer we will be done, since the only integer value \(-r+s\) can be is \(0\). This is easy to see, since \begin{align*} \l a + \sqrt{a^2-b} \r^n + \l a - \sqrt{a^2-b} \r^n &= \sum_{k=0}^n \binom{n}{k}a^{n-k}(\sqrt{a^2-b})^k +\sum_{k=0}^n \binom{n}{k}a^{n-k}(-\sqrt{a^2-b})^k \\ &= \sum_{k=0}^n \binom{n}{k}a^{n-k}\l (\sqrt{a^2-b})^k +(-\sqrt{a^2-b})^k \r \\ \end{align*} But every term where \(k\) is odd in this sum is \(0\) (since they cancel) and ever term where \(k\) is even in this sum is an integer. Therefore the sum is an integer and we're done.
3. \begin{align*} -r^2+rN &= -r(r-N) \\ &= s(r-N) \\ &=- \l a - \sqrt{a^2-b} \r^n \l a + \sqrt{a^2-b} \r^n \\ &= -\l \l a - \sqrt{a^2-b} \r \l a + \sqrt{a^2-b} \r\r^n \\ &= - \l a^2 - a^2+b\r^n \\ &= b^n \end{align*} Therefore \(r^2-rN + b^n = 0\)

Looking at \(\left(8+3\sqrt{7}\right)^{n}\) we have \(a = 8, b = 1\) (since \(8^2 - 1 = 9 \cdot 7\). So we can apply the result of the previous question to see that: \(\left(8+3\sqrt{7}\right)^{n}\) differs from an integer by \(\left(8-3\sqrt{7}\right)^{n}\). \begin{align*} 8-3\sqrt{7} &= \frac{1}{8+3\sqrt{7}} \\ &\approx \frac{1}{8 + 8} \\ &\approx 2^{-4} \end{align*} Therefore it differs by approximation \((2^{-4})^n = 2^{-4n}\)

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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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Solution: Claim: \(n \not \mid (n-1)!\) if and only if \(n\) is prime or \(4\) Proof: \((\Leftarrow)\)

1. \(4 \not \mid 3! = 6\).
2. If \(p\) is prime, then \(p \not \mid k\) for \(k < n\), therefore \(p \not \mid (n-1)!\)

\((\Rightarrow)\) If \(n = 1\) then \(1 \mid 0! = 1\) so \(1\) is not in our set. The numbers less than \(6\) are all accounted for (either primes, \(4\) or \(1\)), so let \(n\) be a composite number larger than \(6\), ie \(n = ab\). Suppose first \(a \neq b\) then \((n-1) = 1 \cdots a \cdots b \cdots (n-1)\) so \(n \mid (n-1)!\). Suppose instead that \(a = b\), then \(n = a^2\). Since we know \(a \geq 3\) we must have \(1 \cdots a \cdots (2a) \cdots (a^2-1)\) so \(a^2 \mid (n-1)!\) and we're done.

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Solution:

1. If \(z=x+iy\) then \(|z|^2 = x^2 + y^2 \leq x^2 + y^2 + 2|x||y| \leq (|x|+|y|)^2\). The LHS is Cauchy-Schwarz with the vectors \(\begin{pmatrix} |x| \\ |y| \end{pmatrix}\) and \(\begin{pmatrix} \cos \alpha \\ \sin \alpha \end{pmatrix}\), although that's not in the spirit of the question. Consider \(e^{i \alpha}z = (\cos \alpha x - \sin \alpha y) + i(\sin \alpha x + \cos \alpha y)\) then \(\left | \textrm{Re}(e^{i \alpha} z) \right | \leq |z|\) for all values of \(\alpha\) and in particular we can choose \(\alpha\) to match the signs of the \(x\) and \(y\) to prove the result in question.
2. Consider \((5-\mathrm{i})^{4}(1+\mathrm{i})\), then \begin{align*} \arg \l (5-\mathrm{i})^{4}(1+\mathrm{i}) \r &= \arg (5-i)^4 + \arg (1+i) \\ &= 4 \arg (5-i) + \arg (1+i) \\ &= -4 \tan^{-1} \frac{1}{5} + \tan^{-1} 1 \\ \\ &= \arg ( (24 - 10i)^2 (1+i)) \\ &= \arg (4 (12 - 5i)^2(1+i)) \\ &= \arg ((119 - 120i)(1+i)) \\ &= \arg (239 - i) \\ &= -\tan^{-1} \frac{1}{239} \end{align*} Therefore \(\displaystyle \frac{\pi}{4} =4 \tan^{-1} \frac{1}{5}- \tan^{-1} \frac{1}{239}\) Consider \((4-i)^3(1+i)(20-i)\) then \begin{align*} \arg \l(4-i)^3(1+i)(20-i) \r &= -3 \tan^{-1} \frac14 + \tan^{-1} 1 -\tan^{-1} \frac1{20} \\ \\ &= \arg \l(15-8i)(4-i)(1+i)(20-i) \r \\ &= \arg \l (52 - 47i)(1+i)(20-i) \r \\ &= \arg \l (99 + 5i)(20-i) \r \\ &= \arg (1985+i) \\ &= \tan^{-1} \frac1{1985} \end{align*} Therefore \(\displaystyle \frac{\pi}{4}=3\tan^{-1}\frac{1}{4}+\tan^{-1}\frac{1}{20}+\tan^{-1}\frac{1}{1985}\)