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algebra--gallian.tex

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+\section{Integers and Equivalence relations}
+
+\subsubsection{Positive integers less than n and relatively prime}
+
+5 =>  1, 2, 3, 4
+8 =>  1, 3, 5, 7
+12 => 1, 5, 7, 11
+20 => 1, 3, 7, 9, 11, 13, 17, 19
+25 => 1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24
+$\checkmark$
+
+\subsubsection{gcd, lcm}
+
+Let $p$ and $q$ be distinct primes.
+\begin{align*}
+  &\gcd(2, 10) = \gcd(2, 2 \cdot 5)   = 2
+  &\lcm(2, 10) = \lcm(2, 2 \cdot 5)  = 20\\
+  &\gcd(20, 8) = \gcd(2^2 \cdot 5, 2^3) = 2^2
+  &\lcm(20, 8) = \lcm(2^2 \cdot 5, 2^3) = 2^3 \cdot 5 = 40\\
+  &\gcd(12, 40) = \gcd(2^2 \cdot 3, 2^3 \cdot 5) = 2^2 = 4
+  &\lcm(12, 40) = \lcm(2^2 \cdot 3, 2^3 \cdot 5) = 2^3 \cdot 3 \cdot 5 = 120\\
+  &\gcd(p^2q^2, pq^3) = pq^2
+  &\lcm(p^2q^2, pq^3) = p^2q^3
+\end{align*}
+$\checkmark$
+
+\subsubsection{modular arithmetic}
+\begin{align*}
+  (47 + 68) \mod 11 = 5
+\end{align*}
+$\checkmark$

algebra--herstein.tex

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+\section{A few preliminary remarks}
+\section{Set theory}
+\begin{enumerate}
+\item[12.]
+  \begin{claim*}
+    $(A \isect B)' = A' \union B'$
+  \end{claim*}
+\end{enumerate}

ou-entry.tex

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+https://mcs-notes2.open.ac.uk/files/m337dq.pdf
+
+\includegraphics[width=400pt]{img/ou-entry-2236.png}
+
+\begin{enumerate}
+\item
+\begin{align*}
+  3(2ac + 2bc + 4) - 6(ab + ac + 2)
+  &= 6ac + 6bc + 12 - 6ab -6ac -12 \\
+  &= 6bc - 6ac \\
+  &= 6c(b-a)
+\end{align*}
+\item (b)
+\item
+ \begin{align*}
+    \frac{1}{\sqrt{2} + 1}
+   &= \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} \\
+   &= \sqrt{2} - 1
+ \end{align*}
+\item (b)
+\item
+\begin{align*}
+  x(x^2 - 5x + 6) = x(x-2)(x-3)
+\end{align*}
+\item \begin{align*}
+        y &= x(x - 3) \\
+          &= x^2 - 3x
+      \end{align*}
+      Second derivative is positive.
+      $y > 0$ for $x \in (-\infty, 0) \cup (3, \infty)$.
+\item \begin{align*}
+        \frac{1}{x(x - 3)} &= \frac{A}{x} + \frac{B}{x-3} \\
+        1                  &= Ax - 3A + Bx \\
+        A &= -1/3 \\
+        B &= 1/3 \\
+        \frac{1}{x(x - 3)} &= \frac{-1}{3x} + \frac{1}{3(x - 3)}
+      \end{align*}
+      Check:
+      \begin{align*}
+        \frac{-1}{3x} + \frac{1}{3(x - 3)}
+        &= \frac{-3(x - 3) + 3x}{9x(x - 3)} \\
+        &= \frac{9}{9x(x - 3)} \checkmark
+      \end{align*}
+
+
+
+\includegraphics[width=400pt]{img/ou-entry-9439.png}
+
+\begin{enumerate}
+\item
+  \begin{table}[h!]
+    \centering
+    \begin{tabular}{c|ccccccccc}
+      $\theta$      & 0 & $\pi/6$      & $\pi/4$      & $\pi/3$     & $\pi/2$  & $2\pi/3$      & $3\pi/4$        & $5\pi/6$      & $\pi$ \\
+      \hline
+      $\sin \theta$ & 0 & 1/2          & $1/\sqrt{2}$ & $\sqrt{3}/2$ & 1        & $\sqrt{3}/2$ & $1/\sqrt{2}$    & 1/2           & 0\\
+      $\cos \theta$ & 1 & $\sqrt{3}/2$ & $1/\sqrt{2}$ & 1/2          & 0        & -1/2         & $-1/\sqrt{2}$   & $-\sqrt{3}/2$ & -1\\
+      $\tan \theta$ & 0 & $1/\sqrt{3}$ & 1            & $\sqrt{3}$   & $\infty$ & $-\sqrt{3}$  & -1              & $-1/\sqrt{3}$ & 0
+    \end{tabular}
+  \end{table}
+\end{enumerate}
+\item \begin{align*}
+    \sin^2(2\theta)
+    &= (2\sin\theta\cos\theta)^2 \\
+    &= 4\cos^2(1-\cos^2\theta)
+  \end{align*}
+\item $\cos(2\theta) = -1/2$ implies $2\theta = 4\pi/6$ or $2\theta = 8\pi/6$, i.e. $\theta \in \{\pi/3, 2\pi/3 \}$.
+\item
+  \begin{enumerate}[label=(\alph*)]
+  \item $e$
+  \item \begin{align*}
+          \frac{e^{2x}e^{3x+1}}{e^x} = e^xe^{3x+1}
+        \end{align*}
+  \item \begin{align*}
+      \frac{1}{3} \log_e 8 - \frac{1}{2} \log_e 4
+      &= 3 \cdot \frac{1}{3} \log_e 2 - 2 \cdot \frac{1}{2} \log_e 2 \\
+      &= 0
+    \end{align*}
+  \item \begin{align*}
+      e^{-\log_e 2} = 1/2
+    \end{align*}
+  \end{enumerate}
+\item \begin{align*}
+    e^{2x} + 2e^x - 3 = 0
+  \end{align*}
+  Let $y = e^x$. Then we have
+  \begin{align*}
+    y^2 + 2y - 3   &= 0 \\
+    (y + 3)(y - 1) &= 0
+  \end{align*}
+  hence $e^x = -3$ or $e^x = 1$,
+
+  hence the only real solution is $x = \ln 1$.
+\end{enumerate}
+
+\includegraphics[width=400pt]{img/ou-entry-cdc6.png}
+
+
+\includegraphics[width=400pt]{img/ou-entry-ce7c.png}
+\begin{enumerate}
+\item \begin{align*}
+    \sum_{n=0}^\infty \Big(\frac{1}{5}\Big)^n
+  \end{align*}
+  Geometric series formula...how to derive?
+\item
+  \begin{enumerate}[item=(\alph*)]
+  \item Ratio test:
+    \begin{align*}
+      \lim_{n\to\infty} \frac{1/(n+1)}{1/n}
+      &= \lim_{n\to\infty} \frac{n}{n+1} \\
+      &= \lim_{n\to\infty} \frac{1}{1 + 1/n} \\
+      &= 1
+    \end{align*}
+    Inconclusive by the ratio test.
+
+    This is the harmonic series and it diverges.
+  \item \begin{align*}
+      \frac{(-1)^{n+2}}{n+1} / \frac{(-1)^{n+1}}{n}
+      &= \frac{(-1)^{n+2}n}{(-1)^{n+1}(n+1)} \\
+      &= (1 + n)(-1) < 0
+    \end{align*}
+    Converges by Alternating Series test.
+  \end{enumerate}
+\item Ratio test:
+  \begin{align*}
+      \lim_{n\to\infty} \frac{1/(n+1)^2}{1/n^2}
+      &= \lim_{n\to\infty} \frac{n^2}{n^2 + 2n + 1} \\
+      &= \lim_{n\to\infty} \frac{1}{1 + 2/n + 1/n^2} \\
+      &= 1
+    \end{align*}
+    Inconclusive. This converges; every term is smaller than harmonic series.
+
+\end{enumerate}
+
+
+\includegraphics[width=400pt]{img/ou-entry-a82d.png}
+
+
+
+\includegraphics[width=400pt]{img/ou-entry-efb5.png}
+
+\begin{enumerate}
+\item False. E.g. the maximum of $f(x) = x$ occurs at $b$ and yet $f'(b) = 1$.
+\item False. This is only true if $f$ is continuous. Counter-example
+  \begin{align*}
+    f(x) =
+    \begin{cases}
+      -1 &~~~\text{if}~~~ x <= 0\\
+      1 &~~~\text{if}~~~ x > 0
+    \end{cases}
+  \end{align*}
+\item
+\end{enumerate}