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CS1527/notes-2024-02-27.md
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CS1527/notes-2024-02-27.md
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\author{}
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\newcommand{\paren}[1]{\left(#1\right)}
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\newcommand{\powerset}[1]{\mathbb{P}\paren{#1}}
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\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
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\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
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\newcommand{\C}{\mathbb{C}}
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@@ -9,6 +9,7 @@
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\author{}
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\newcommand{\paren}[1]{\left(#1\right)}
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\newcommand{\powerset}[1]{\mathbb{P}\paren{#1}}
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\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
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\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
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\newcommand{\C}{\mathbb{C}}
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175
MA1511/sets.tex
175
MA1511/sets.tex
@@ -71,23 +71,23 @@ For sets $X$, $Y$:
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\end{description}
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For $f: X \to Y$, $A \subseteq A' \subseteq X$, $B \subseteq B' \subseteq Y$:
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\begin{align*}
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f(A) \subseteq~& f(A') \\
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f^{-1}(B) \subseteq ~& f^{-1}(B') \\
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f^{-1}(f(A)) \supseteq ~& A \\
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f(f^{-1}(B)) \subseteq ~& B
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f(A) \subseteq~ & f(A') \\
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f^{-1}(B) \subseteq ~ & f^{-1}(B') \\
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f^{-1}(f(A)) \supseteq ~ & A \\
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f(f^{-1}(B)) \subseteq ~ & B
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\end{align*}
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For set families $(A_i \subseteq X)_{i \in I}, (B_j \subseteq Y)_{j \in J}$:
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\begin{align*}
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f(\cup_{i \in I} A_i) = ~& \cup_{i \in I}f(A_i) \\
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f(\cap_{i \in I} A_i) \subseteq ~& \cap_{i \in I}f(A_i) \\
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f^{-1}(\cup_{j \in J} B_j) = ~& \cup_{j \in J} f^{-1}(B_j) \\
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f^{-1}(\cap_{j \in J} B_j) = ~& \cap_{j \in J} f^{-1}(B_j)
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f(\cup_{i \in I} A_i) = ~ & \cup_{i \in I}f(A_i) \\
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f(\cap_{i \in I} A_i) \subseteq ~ & \cap_{i \in I}f(A_i) \\
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f^{-1}(\cup_{j \in J} B_j) = ~ & \cup_{j \in J} f^{-1}(B_j) \\
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f^{-1}(\cap_{j \in J} B_j) = ~ & \cap_{j \in J} f^{-1}(B_j)
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\end{align*}
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\section*{Function Composition}
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For functions $f: X \to Y$, $g: Y \to Z$:
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\begin{align*}
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& (g \circ f): X \to Z \\
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& (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
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& (g \circ f): X \to Z \\
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& (g\circ f)(x) = g(f(x)) & \forall x \in X \\ \\
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\end{align*}
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\section*{Surjection and Injection}
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For $f: X \to Y$, $f$ is
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@@ -95,6 +95,159 @@ For $f: X \to Y$, $f$ is
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\item[surjective] iff $f(X) = Y$, i.e. $\forall y \in Y.~ \exists x \in X.~ f(x) = y$ \\
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Range is codomain, 'onto'
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\item[injective] iff $\forall x, x' \in X. ~ f(x) = f(x') \implies x = x'$
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\item[bijective] iff $f$ is injective and $f$ is surjective
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\item[bijective] iff $f$ is injective and $f$ is surjective \\
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one-to-one
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\end{description}
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Given $f: X \to Y$, $g: Y \to Z$: \\
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\quad If f and g are injective, so is $g\circ f$ \\
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\quad If f and g are surjective, so is $g \circ f$ \\
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\quad If $g \circ f$ is injective, so is $f$ \\
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\quad If $g \circ f$ is surjective, so is $g$
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\section*{Inverse Functions}
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The inverse function $f^{-1}$ of $f: X \to Y$ exists iff $f$ is bijective, and is defined by \[f^{-1}(y) = x \text{ where } \exists! x \in X. f(x) = y \qquad \forall y \in Y\]
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\begin{align*}
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(f \circ f^{-1}) = \operatorname{Id} \\
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(f^{-1} \circ f) = \operatorname{Id}
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\end{align*}
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\section*{Power Sets}
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Powerset of S: $\powerset{S}$ has $2^{\card{S}}$ elements is the set of all subsets of $S$ \\
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$Fun(X, \{0, 1\})$ is the set of functions $X \to \{0, 1\}$ \\
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$\Phi: Fun(X, \{0, 1\}) \to \powerset{X}$ \\
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$\Phi(f) = \{ x \in X | f(x) = 1 \}$ \\
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$\Phi$ is bijective.
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\section*{Binary Operators}
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A binary operator is $X^2 \to X$ \\
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Union is $\powerset{X}^2 \to \powerset{X}$ \\
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$\square$ is the unknown or indeterminate binop sigil \\
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Unital: $\forall x \in X. \exists u \in X. u \square x = x \square u = x$
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\section*{Construction of the Natural Numbers}
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\begin{align*}
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x^{+} & = x \cup \{x\} & \text{successor of $x$}
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\end{align*}
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A set $X$ is \emph{inductive} if $\varnothing \in X \land (a \in X \implies a^+ \in X)$ \\
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Axiom: There exists an inductive set. \\
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Definition: A natural number is a set that is an element of all inductive sets. \\
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Theorem: There exists a set whose elements are the natural numbers. \\
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\begin{align*}
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& \text{Given an inductive set $A$} \\
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& \omega = \{ x \in A | x \text{ is a natural number}\} \\
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& \text{any natural number is in $A$ (since $A$ is inductive), and therefore in $\omega$} \\
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& \omega \text{ is the natural numbers}
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\end{align*}
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\\
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Define:
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\begin{align*}
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0 & = \varnothing \\
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1 & = 0^+ \\
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2 & = 1^+ \\
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\ldots
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\end{align*}
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$\in$ is an ordering over $\omega$, as is $\subseteq$ \\
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$\omega$ is an inductive set. Proof: \\
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$\varnothing$ is a natural number.
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$x \in \omega$ implies $x^+ \in \omega$, as $x$ is natural and is therefore in
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every inductive set, and so $x^+$ is in every inductive set and is therefore a
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natural number.
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\subsection*{Principle of Induction}
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If $A \subseteq \omega$ and $A$ is inductive, $A = \omega$.
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Since $A$ is inductive, it contains every natural number, so $\omega \subseteq A$. \\
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Since $A \subseteq \omega \land \omega \subseteq A$, $A = \omega$
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\section*{Recursion on $\omega$}
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\subsection*{Principle of Recursion}
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For a set $X$, $x_0 \in X$, $h: X \to X$, there exists a unique function $f: \omega \to X$ where $f(0) = x_0$, $f(n^+) = h(f(n))$.
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\section*{Relations}
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Total order: $(X, \prec)$, requires $\forall x, y, z \in X$: \\
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\begin{align*}
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& x \prec y \land y \prec z \implies x \prec z \\
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& x = y \oplus x \prec y \oplus y \prec x & \text{ (where $\oplus$ denotes XOR) }
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\end{align*}
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Lexicographic order $x <_L y$ on $\N\times\N = x_0 < y_0 \lor (x_0 = y_0 \land x_1 < y_1)$ \\
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For a non-empty subset $A$ of $X$ given a total order $(X, \prec)$, a minimum/least element $a_0 \in A$
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exists where $\forall a \in A.~a_0 \preceq a$.
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\section*{Ordering on $\omega$}
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$(\omega, \in)$ and $(\omega, \subset)$ are both total orderings on $\omega$, such that $\varnothing$ is the minimum and $\forall x \in \omega.~x < x^+$
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\section*{Strong Induction}
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Every non-empty subset of $\N$ has a minimum. \\ \\
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Let $\phi(x)$ be a predicate over $\N$ where $\forall n\in \N.~ (\forall m \in \N.~ m < n \implies \phi(m)) \implies \phi(n)$. \\
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Then $\phi(0)$ holds, as $\neg\exists n \in \N.~n<0$, $\phi(1)$ holds as $\phi(0)$ holds, etc.
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\begin{align*}
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& \neg\exists x \in \N.~\neg\phi(x) \\
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& \text{Let } A \text{ be a subset of }N\text{ such that }\phi(n)\text{ is false }\forall n \in A \\
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& \text{If not, $\exists a_0 \in A$} \\
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& \forall n \in N.~n < a_0 \implies n \not\in A \implies \phi(n) \\
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& \text{Then $\phi(n)$ holds $\forall n < a_0$, so $\phi(a_0)$, then $a_0 \not\in A$, which is a contradiction, so $A = \varnothing$}
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\end{align*}
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\section*{Fibonacci}
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Fibonacci Sequence: $F_n$ \\
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Roots of $x^2 - x - 1$ are $\phi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$ \\
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$F_n = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}}$ \\
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For $n=0$:
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$F_0 = 1 = \frac{\frac{1 + \sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}} = \frac{2\sqrt{5}}{2} = 1$ \\
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For $n=1$: \\
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$F_1 = 1 = \frac{\paren{\frac{1 + \sqrt{5}}{2}}^2 - \paren{\frac{1 - \sqrt{5}}{2}}^2}{\sqrt{5}} = \frac{\phi + 1 - \psi - 1}{\sqrt{5}}$ \\
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For $n\geq 2$:
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\begin{align*}
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& F_n = F_{n-1} + F_{n-2} = \frac{\phi^n - \psi^n}{\sqrt{5}} + \frac{\phi^{n-1} - \psi^{n-1}}{\sqrt{5}} \\
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& = \frac{\phi^{n-1}\paren{\phi + 1} - \psi^{n-1}\paren{\psi + 1}}{\sqrt{5}} \\
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& = \frac{\phi^{n-1}\phi^2 - \psi^{n-1}\psi^2}{\sqrt{5}} \\
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& = \frac{\phi^{n+1} - \psi^{n+1}}{\sqrt{5}} & \text{as required}
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\end{align*}
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\subsection*{Zeckendorff's Theorem}
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Every natural number can be written as a sum of non-adjacent Fibonacci numbers in a unique way (excluding $F_0$). \\
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A finite subset $I$ of $\N^+$ is \emph{Zeckendorff} if it contains no adjacent elements ($\forall x \in I.~x^+ \not\in I$) \\
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Define $\mathcal{Z}$ as the set of all Zeckendorff sets, and $\sigma: \mathcal{Z} \to \N^+$ by $\sigma(I) = \sum_{i \in I}F_i$.
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The theorem claims $\sigma$ is bijective. \\ \\
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For nonempty $I \in \mathcal{Z}$ with largest element $k$:
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\begin{align*}
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&\text{Let } J = I \setminus \{k\} \\
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&\sigma(I) = F_k + \sigma(J) \geq F_k \\
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&\text{If } J = \varnothing\text{, } \sigma(I) = F_k \leq F_{k + 1} \\
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&\text{Otherwise, we must show } \sigma(I) < F_{k + 1} \\
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&\equiv F_k + \sigma(J) < F_{k + 1} \\
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&\equiv \sigma(J) < F_{k + 1} - F_k = F_{k-1} \\
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&\text{But if $k' = \operatorname{max}(J)$, } \\
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&\sigma(J) < F_{k' + 1} \land k' < k - 2 \text{ (since $I$ is \emph{Zeckendorff})} \\
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& \sigma(J) < F_{k - 1} \text{ as required.}
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\end{align*}
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Proof of theorem: $\forall n \in N.~ \sigma$ is bijective.
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\begin{align*}
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n = 0: \quad& I = \varnothing \\
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n > 0: \quad& \text{Let } F_k \leq n < F_{k + 1},~ m = n - F_k \\
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& m = \sigma(J) \text{ for some } J \\
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& \text{If } J = \varnothing, I = \{k\}. \\
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& \text{Otherwise, } k' = \operatorname{max}(J). \\
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& \text{If } k' \leq k - 2, I = J \cup \{k\} \\
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\end{align*}
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\section*{Equivalence Relations}
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Reflexive, Symmetric, Transitive \\
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Equivalence relations are usually called $\sim$ \\
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In a set $X$,
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\[ [x] = \{ y \in X | x \sim y \} \]
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\[ [x] = [y] \equiv x \sim y \]
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Any two equivalence classes of $X$ are either disjoint or equal. \\
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\subsection*{Quotient Sets}
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$X/\sim~ = \{[x] | x \in X \} \subset \powerset{X}$ \\
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A complete set of representatives is a subset $A$ of $X$ where $\forall x \in X. \exists! a \in A. a \in [x].$ \\
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I.E. a complete set of representatives contains exactly one element from each element of $X/\sim$ \\
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$f: A \to X/\sim$ defined by $f(a) = [a]$
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\\\\\\
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A function $f: X \to Y$ is \emph{compatible} iff $x \sim y \implies f(x) = f(y)~\forall x, y \in X$ \\
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For a \emph{compatible} function, $\Bar{f}: X/\sim~\to Y$ exists and is defined by $\Bar{f}([x]) = f(x)$
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\section*{Integers modulo $k$}
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Fix some $k \in \N^+$ \\
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Define $\sim$ on $\Z$ by $n \sim m \iff n - m$ is a multiple of $k$ \\
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$\sim$ is an equivalence relation \\
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$[0, k)\cap\N$ is a complete set of representatives for $\sim$ \\
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$Z/k$ is the set of integers modulo $k$, $n \equiv m~(\operatorname{mod} k) \iff n \sim m$ \\
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$+$ and $\times$ on $\Z/k$:
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\begin{align*}
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[n] + [m] = [n + m] \\
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[n][m] = [nm]
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\end{align*}
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\end{document}
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