Typographical edits to probability

Signed-off-by: zeramorphic <[email protected]>

ia/de/01_differentiation.tex

@@ -172,7 +172,7 @@ \subsection{Order of magnitude}
 	then \(f(x) = O(g(x))\) as \(x \to \infty\).
 \end{definition}
 
-This is basically the same as the previous definition --- but obviously we can't pick a value slightly less than infinity to test, so we just provide a lower bound on \(x\) where the condition holds true.
+This is basically the same as the previous definition---but obviously we can't pick a value slightly less than infinity to test, so we just provide a lower bound on \(x\) where the condition holds true.
 
 For example, \(2x^3 + 4x + 12 = O(x^3)\) as \(x \to \infty\).
 This is because the function is a cubic, so can be bounded by a cubic as it shoots off to infinity.

ia/de/03_multivariate_functions.tex

@@ -133,7 +133,7 @@ \subsection{Implicit differentiation}
 	\eval{\frac{\partial f}{\partial z}}_{xy} \eval{\frac{\partial z}{\partial x}}_{y}
 \]
 The left hand side is zero because on the surface \(z(x, y)\), \(f\) is always equivalent to \(c\) so there is never any \(\delta f\).
-The \(\eval{\frac{\partial f}{\partial x}}_{yz}\) term, however, is not zero in general because we are not going across the \(z(x, y)\) surface --- just parallel to the \(x\) axis, because we fixed both \(y\) and \(z\).
+The \(\eval{\frac{\partial f}{\partial x}}_{yz}\) term, however, is not zero in general because we are not going across the \(z(x, y)\) surface---just parallel to the \(x\) axis, because we fixed both \(y\) and \(z\).
 Hence,
 \[
 	\eval{\frac{\partial z}{\partial x}}_y = \frac{-\eval{\frac{\partial f}{\partial x}}_{yz}}{\eval{\frac{\partial f}{\partial z}}_{xy}}

ia/de/06_isoclines_and_solution_curves.tex

@@ -133,7 +133,7 @@ \subsection{Isoclines}
 
 Two such solution curves are drawn on this graph; the one intersecting zero has \(A = 1\) in the solution for \(y\), and the one above it has \(A = -1\).
 Note how, as they intersect the isoclines in red, they have exactly the gradient defined by the isocline.
-Particularly, the lower solution curve intersects the same isocline twice, and therefore has this exact gradient at two distinct points --- we observe these points as the intersection points between the solution curve and the isocline.
+Particularly, the lower solution curve intersects the same isocline twice, and therefore has this exact gradient at two distinct points---we observe these points as the intersection points between the solution curve and the isocline.
 
 Note also that the solutions \(y = 1\) and \(y = -1\) lie on these isoclines for all \(t\).
 This is because the isoclines specify that the function has zero gradient on such a straight line, so it makes sense that the function and isocline coincide.

ia/de/07_phase_portraits.tex

@@ -76,7 +76,7 @@ \subsection{Phase portraits}
 \]
 This is an autonomous nonlinear first order ordinary differential equation.
 We can create a phase portrait by mapping out \(\frac{\dd{c}}{\dd{t}}\) as a function of \(c\), as shown in the first diagram here, which is known as a 2D phase portrait.
-The second diagram, known as a 1D phase portrait, shows similar information but helps us see the behaviour of fixed points --- essentially the arrows point in the direction of motion of \(c\); if \(\dot c\) is positive then the arrows point to the right, if \(\dot c\) is negative they point to the left.
+The second diagram, known as a 1D phase portrait, shows similar information but helps us see the behaviour of fixed points---essentially the arrows point in the direction of motion of \(c\); if \(\dot c\) is positive then the arrows point to the right, if \(\dot c\) is negative they point to the left.
 \newpage
 \begin{wrapfigure}{r}{0.5\textwidth}
 	\begin{tikzpicture}

ia/de/10_impulse_forcing.tex

@@ -61,7 +61,7 @@ \subsection{Delta function forcing}
 \end{equation}
 The key principle is that the highest order deriative `inherits' the level of discontinuity from the forcing term, since if any other derivative were to contain the discontinuous function, then the next higher derivative would only be more discontinuous.
 So, \(y''\) behaves somewhat like \(\delta\).
-Here, we will denote this \(y'' \sim \delta\) --- this is extremely non-standard notation, however.
+Here, we will denote this \(y'' \sim \delta\)---this is extremely non-standard notation, however.
 
 Now, since \(\delta(x) = 0\) for all nonzero \(x\), then
 \[

ia/de/13_systems_of_odes.tex

@@ -225,7 +225,7 @@ \subsection{Nonlinear systems of ODEs}
 \[
 	\eqref{nonlinear1} \implies \dot \xi = f(x_0 + \xi, y_0 + \eta)
 \]
-We can expand this in a multivariate Taylor series, keeping the first three terms --- the constant term and the two linear terms.
+We can expand this in a multivariate Taylor series, keeping the first three terms---the constant term and the two linear terms.
 \begin{align*}
 	\dot \xi  & \approx f(x_0, y_0) + \xi f_x(x_0, y_0) + \eta f_y(x_0, y_0) \\
 	          & = \xi f_x(x_0, y_0) + \eta f_y(x_0, y_0)                     \\

ia/dr/01_basic_definitions_and_newton_s_laws.tex

@@ -98,7 +98,7 @@ \subsection{Galilean transformations}
 	\item at any constant velocity
 \end{itemize}
 Any set of equations which describe Newtonian physics must preserve this Galilean invariant.
-This shows that measurement of velocity cannot be absolute, it must be relative to a specific inertial frame of reference --- but conversely, measurement of acceleration \textit{is} absolute.
+This shows that measurement of velocity cannot be absolute, it must be relative to a specific inertial frame of reference---but conversely, measurement of acceleration \textit{is} absolute.
 
 \subsection{Newton's second law}
 For any particle subject to a force \(\vb F\), the momentum \(\vb p\) of the particle satisfies

ia/groups/01_axiomatic_definition.tex

@@ -65,7 +65,7 @@ \subsection{Definition}
 
 Here are a few examples of groups.
 \begin{enumerate}
-	\item \(G = \{ e \}\) --- this is the `trivial group'.
+	\item \(G = \{ e \}\)---this is the `trivial group'.
 	\item \(G = \{ \text{symmetries of the equilateral triangle} \} \); \(\ast\) is defined by: `\(g \ast h\) means doing \(h\) then \(g\)'.
 	\item \(G = (\mathbb Z, +)\).
 	      This is easy to prove by verifying the axioms.
@@ -151,7 +151,7 @@ \subsection{Subgroups}
 	A subset \(H \subseteq G\) is a subgroup of \(G\) if \((H, \ast)\) is a group.
 	We denote this \(H \leq G\).
 \end{definition}
-We must verify each group axiom on a subset to check if it is a subgroup --- with the notable exception of the associativity axiom, the property of associativity is inherited by subgroups.
+We must verify each group axiom on a subset to check if it is a subgroup---with the notable exception of the associativity axiom, the property of associativity is inherited by subgroups.
 Here are some examples of subgroups.
 
 \begin{enumerate}

ia/groups/06_cosets_and_lagrange_s_theorem.tex

@@ -129,7 +129,7 @@ \subsection{Lagrange's theorem}
 \end{proof}
 We can take Lagrange's theorem into the world of number theory, and specifically modular arithmetic, where we are dealing with finite groups.
 Clearly, \(\mathbb Z_n\) is a group under addition modulo \(n\), but what happens with multiplication modulo \(n\)?
-Clearly this is not a group --- for a start, 0 has no inverse.
+Clearly this is not a group---for a start, 0 has no inverse.
 By removing all elements of the group that have no inverse, we obtain \(\mathbb Z_n^*\).
 
 Note that for any \(x \in \mathbb Z_n\), \(x\) has a multiplicative inverse if and only if \(\HCF(x, n) = 1\), i.e.\ if \(x\) and \(n\) are coprime.

ia/groups/07_normal_subgroups_and_quotients.tex

@@ -67,7 +67,7 @@ \subsection{Normal subgroups}
 As another example, let \(k \in K\) and let \(g \in G \setminus K\).
 Then \(kg\) does not have this property, as \(kg \notin K\).
 
-We can encapsulate this behaviour by making a homomorphism from the whole group \(G\) to some other group --- it \textit{doesn't matter where we end up}, just as long as anything with this particular property maps to the new group's identity element.
+We can encapsulate this behaviour by making a homomorphism from the whole group \(G\) to some other group---it \textit{doesn't matter where we end up}, just as long as anything with this particular property maps to the new group's identity element.
 Let \(\varphi: G \to H\), where \(H\) is some group that we don't really care about (apart from the identity).
 This means that any element of \(K\), i.e.\ any element with property \(P\), is mapped to \(e_H\).
 By the laws of homomorphisms, any product of \(k \in K\) with \(g \in G \setminus K\) does not give the identity element, so it does not have this property!
@@ -179,7 +179,7 @@ \subsection{Quotients}
 
 Now, given some element in one of the cosets (i.e.\ in \(G\)) we can do some transformation \(g\) to take us to another element.
 But because we made cosets out of a normal subgroup, multiplying by \(g\) is the same as swapping some of the rows, then maybe moving around the order of the elements in each row.
-It keeps the identity of each row consistent --- all elements in a given row are transformed to the same output row.
+It keeps the identity of each row consistent---all elements in a given row are transformed to the same output row.
 Remember that the word `row' basically means `coset'.
 
 This means that we can basically forget about the individual elements in these cosets, all that we really care about is how the rows are swapped with each other under a given transformation.
@@ -224,12 +224,12 @@ \subsection{Examples and properties}
 \begin{itemize}
 	\item We can check that certain properties are inherited into quotient groups from the original group, such as being abelian and being finite.
 	\item Quotients are not subgroups of the original group.
-	      They are associated with tha original group in a very different way to subgroups --- in general, a coset may not even be isomorphic to a subgroup in the group.
+	      They are associated with tha original group in a very different way to subgroups---in general, a coset may not even be isomorphic to a subgroup in the group.
 	      The example with direct products above was an example that is not true in general.
 	\item With normality, we need to specify in which group the subgroup is normal.
 	      For example, if \(K \leq N \leq G\), with \(K \trianglelefteq N\).
 	      This does not imply that \(K \trianglelefteq G\), this would require that \(g^{-1}Kg = K\) for all elements \(g\) in \(G\), but we only have that \(n^{-1}Kn = K\) for all elements \(n\) in \(N\), which is a weaker condition.
-	      Normality is not transitive --- for example, \(K \trianglelefteq N \trianglelefteq G\) does not imply \(K \trianglelefteq G\).
+	      Normality is not transitive---for example, \(K \trianglelefteq N \trianglelefteq G\) does not imply \(K \trianglelefteq G\).
 	\item However, if \(N \leq H \leq G\) and \(N \trianglelefteq G\), then the weaker condition \(N \trianglelefteq H\) is true.
 \end{itemize}
 

ia/groups/08_isomorphism_theorems.tex

@@ -23,7 +23,7 @@ \subsection{First isomorphism theorem}
 Now, multiplying together two rows, i.e.\ two elements from \(K\), we can apply the homomorphism \(\varphi\) to one of the coset representatives for each row to see how the entire row behaves under \(\varphi\).
 We know that all coset representatives give equal results, because each element in a given coset \(gN\) can be written as \(gn, n \in N\), so \(\varphi(gn) = \varphi(g)\).
 So all elements in the rows behave just like their coset representatives under the homomorphism.
-Further, all the cosets give different outputs under \(\varphi\) --- if they gave the same output they'd have to be part of the same coset.
+Further, all the cosets give different outputs under \(\varphi\)---if they gave the same output they'd have to be part of the same coset.
 So in some sense, each row represents a distinct output for \(\varphi\).
 So the quotient group must be isomorphic to the image of the homomorphism.
 

ia/groups/10_conjugation.tex

@@ -2,7 +2,7 @@ \subsection{Conjugation actions}
 \begin{definition}
 	Given \(g, h \in G\), the element \(hgh^{-1}\) is the conjugate of \(g\) by \(h\).
 \end{definition}
-We should think of conjugate elements as doing the same thing but from different `points of view' --- we change perspective by doing \(h^{-1}\), then do the action \(g\), then reset the perspective back to normal using \(h\).
+We should think of conjugate elements as doing the same thing but from different `points of view'---we change perspective by doing \(h^{-1}\), then do the action \(g\), then reset the perspective back to normal using \(h\).
 
 Here is an example using \(D_{10}\), where the vertices of the regular pentagon are \(v_1 \dots v_5\) clockwise.
 Consider the conjugates \(s\) and \(rsr^{-1}\), where \(s\) is a reflection through \(v_1\) and the centre, and \(r\) is a rotation by \(\frac{2\pi}{5}\) clockwise.

ia/ns/01_proofs.tex

@@ -88,7 +88,7 @@ \subsection{Proofs and non-proofs}
 \end{proof}
 \begin{itemize}
 	\item We prove things to show \textit{why} something is true.
-	      We can see why this claim was true here --- it's really a statement about the properties of odd numbers, not the properties of even numbers.
+	      We can see why this claim was true here---it's really a statement about the properties of odd numbers, not the properties of even numbers.
 	\item We started by saying that we need something tangible to work with: just stating that `\(n^2\) is even' is really hard to work with because square roots just get messy and don't yield any result.
 	      So we had to choose a clever first step.
 	\item The symbol \contradiction{} shows that we have a contradiction.

ia/ns/02_elementary_number_theory.tex

@@ -51,7 +51,7 @@ \subsection{Strong induction}
 This provides a very useful alternative way of looking at induction.
 Instead of just considering a process from \(n\) to \(n+1\), we can inject an inductive viewpoint into any proof.
 When proving something on the natural numbers, we can always assume that the hypothesis is true for smaller \(n\) than what we are currently using.
-This allows us to write very powerful proofs because in the general case we are allowed to refer back to other smaller cases --- but not just \(n-1\), any \(k\) less than \(n\).
+This allows us to write very powerful proofs because in the general case we are allowed to refer back to other smaller cases---but not just \(n-1\), any \(k\) less than \(n\).
 
 We may rewrite the principle of strong induction in the following ways:
 \begin{enumerate}
@@ -105,7 +105,7 @@ \subsection{Primes}
 
 We want to prove that prime factorisation is unique (up to the ordering).
 We need that \(p \mid ab \implies p \mid a \lor p \mid b\).
-However, this is hard to answer --- \(p\) is defined in terms of what divides it, not what it divides.
+However, this is hard to answer---\(p\) is defined in terms of what divides it, not what it divides.
 This is the reverse of its definition, so we need to prove it in a more round-about way.
 
 \subsection{Highest common factors}

ia/ns/04_the_reals.tex

@@ -103,7 +103,7 @@ \subsection{Examples of sets and least upper bounds}
 \begin{remark}
 	If \(S\) has a greatest element, then this element is the supremum of the set: \(\sup S \in S\).
 	But if \(S\) does not have a greatest element, then \(\sup S \notin S\).
-	Also, we do not need any kind of `greatest lower bound' axiom --- if \(S\) is a non-empty, bounded below set of reals, then the set \(\{ -x: x \in S \}\) is non-empty and bounded above, and so has a least upper bound, so \(S\) has a greatest lower bound equivalent to its additive inverse.
+	Also, we do not need any kind of `greatest lower bound' axiom---if \(S\) is a non-empty, bounded below set of reals, then the set \(\{ -x: x \in S \}\) is non-empty and bounded above, and so has a least upper bound, so \(S\) has a greatest lower bound equivalent to its additive inverse.
 	This is commonly called the `infimum', or \(\inf S\).
 \end{remark}
 \begin{theorem}
@@ -117,10 +117,10 @@ \subsection{Examples of sets and least upper bounds}
 	\begin{itemize}
 		\item (\(c^2 < 2\)) We want to prove that \((c+t)^2 < 2\) for some small \(t\).
 		      For \(0<t<1\), we have \((c+t)^2 = c^2 + 2ct + t^2 \leq c^2 + 5t\), since \(c\) is at most 2, and \(t^2\) is at most \(t\).
-		      So this value is less than 2 for some suitably small \(t\), contradicting the least upper bound --- we have just shown that \((c+t) \in S\).
+		      So this value is less than 2 for some suitably small \(t\), contradicting the least upper bound---we have just shown that \((c+t) \in S\).
 		\item (\(c^2 > 2\)) We want to prove that \((c-t)^2 > 2\) for some small \(t\).
 		      For \(0<t<1\), we have \((c-t)^2 = c^2 - 2ct + t^2 \geq c^2 - 4t\), since \(c\) is at most 2, and \(t^2\) is at least zero.
-		      So this value is greater than 2 for some suitably small \(t\), contradicting the least upper bound --- we have just created a lower upper bound.
+		      So this value is greater than 2 for some suitably small \(t\), contradicting the least upper bound---we have just created a lower upper bound.
 	\end{itemize}
 	So \(c^2 = 2\).
 \end{proof}
@@ -165,7 +165,7 @@ \subsection{Sequences and limits}
 	\item Consider now \(x_n = (-1)^n\), i.e.\ \(-1, 1, -1, 1, \dots\).
 	      We want to show that this does not tend to a limit.
 	      Suppose \(x_n \to c\) as \(n \to \infty\).
-	      We may choose some \(\varepsilon\) that acts as a counterexample --- for example, \(\varepsilon = 1\).
+	      We may choose some \(\varepsilon\) that acts as a counterexample---for example, \(\varepsilon = 1\).
 	      So \(\exists N \in \mathbb N\) such that \(\forall n \geq n\) we have \(\abs{x_n - c} < 1\).
 	      In particular, \(\abs{1 - c} < 1\) and \(\abs{-1 - c} < 1\) so \(\abs{1 - (-1)} < 2\), by the triangle inequality.
 	      This is a contradiction.
@@ -214,7 +214,7 @@ \subsection{Series}
 	So for every \(n \geq \max(M, N)\), by the triangle inequality, \(\abs{(x_n + y_n) - (c + d)} < 2\zeta = \varepsilon\) as required.
 \end{proof}
 This is commonly known as an \(\varepsilon/2\) argument.
-Also, if we had instead not taken any \(\zeta\) value and just stuck with \(\varepsilon\), it would still be a good proof because we could just have divided \(\varepsilon\) at the beginning --- it's not expected that you completely rewrite the proof to add in this division.
+Also, if we had instead not taken any \(\zeta\) value and just stuck with \(\varepsilon\), it would still be a good proof because we could just have divided \(\varepsilon\) at the beginning---it's not expected that you completely rewrite the proof to add in this division.
 
 \subsection{Testing convergence of a sequence}
 A sequence \(x_1, x_2, \dots\) is called `increasing' if \(x_{n+1} \geq x_n\) for all \(n\).

ia/ns/05_sets.tex

@@ -128,7 +128,7 @@ \subsection{Computing binomial coefficients}
 \end{proposition}
 \begin{proof}
 	The number of ways to name a \(k\)-set is \(n(n-1)(n-2)\cdots(n-k+1)\) because there are \(n\) ways to choose a first element, \(n-1\) ways to choose a second element, and so on.
-	We have overcounted the \(k\)-sets, though --- there are \(k(k-1)(k-2)\cdots(1)\) ways to name a given \(k\)-set because you have \(k\) choices for the first element, \(k-1\) choices for the second element, and so on.
+	We have overcounted the \(k\)-sets, though---there are \(k(k-1)(k-2)\cdots(1)\) ways to name a given \(k\)-set because you have \(k\) choices for the first element, \(k-1\) choices for the second element, and so on.
 	Hence the number of \(k\)-sets in \(\{ 1, 2, \dots, n \}\) is the required result.
 \end{proof}
 Note that we can also write

ia/ns/06_functions.tex

@@ -186,7 +186,7 @@ \subsection{Equivalence classes as partitions}
 	So \([x] = [y]\).
 \end{proof}
 As an example, does there exist an equivalence relation on \(\mathbb N\) with three equivalence classes, two of which are infinite, and one of which is finite?
-Yes --- we can break up \(\mathbb N\) into three parts, for example positive numbers, negative numbers and zero.
+Yes---we can break up \(\mathbb N\) into three parts, for example positive numbers, negative numbers and zero.
 This defines an equivalence relation.
 
 \subsection{Quotients}