Minor edits

Advanced/Continued Fractions/parts/02 part B.tex

@@ -115,7 +115,7 @@ Similarly derive the formula $p_nq_{n-2}-p_{n-2}q_n = (-1)^{n-2}a_n$.
 \problem{}<diff>
 Recall $C_n=p_n/q_n$.
 Show that $C_n-C_{n-1}=\frac{(-1)^{n-1}}{q_{n-1}q_n}$
-and $C_n-C_{n-2}=\frac{(-1)^{n-2}a_n}{q_{n-2}q_n}$.
+and $C_n-C_{n-2}=\frac{(-1)^{n-2}a_n}{q_{n-2}q_n}$. \par
 \hint{Use \ref{form1} and $p_nq_{n-2}-p_{n-2}q_n = (-1)^{n-2}a_n$ respectively}
 
 In \ref{sqrt5}, the value $\alpha-C_n$ alternated between negative and positive

Advanced/Cryptography/parts/4 DiffieHellman.tex

@@ -93,7 +93,7 @@ What is their shared secret?
 
 \problem{}
 Let $p = 11$, $g = 2$, $a = 9$, and $b = 4$. \par
-Run the algorithm. What is the resultingw shared secret?
+Run the algorithm. What is the resulting shared secret?
 
 \begin{solution}
 	$g^b = 5$\par

Advanced/Cryptography/parts/5 Elgamal.tex

@@ -117,9 +117,9 @@ Also, say Eve knows the value of $m_1 - m_2$. How can Eve find $m_1$ and $m_2$?\
 \note[Note]{If Bob doesn't change his key, Eve will also be able to decrypt future messages.}
 
 \begin{solution}
-	$c_2 - d_2 = (m_1 - m_2)A^k$. \par
-	So, $(c_2 - d_2)(m_1 - m_2)^{-1} = A^k$.\par
-	Now that we have $A^k$, we can compute $m_1 = c_2 \times A^{-k}$.
+	$c_2 - d_2 = (m_1 - m_2)A^k$ \par
+	So, $(c_2 - d_2)(m_1 - m_2)^{-1} = A^k$\par
+	Now that we have $A^k$, we can compute $m_1 = c_2 \times A^{-k}$
 \end{solution}
 
 \vfill