558 lines
17 KiB
TeX
558 lines
17 KiB
TeX
\documentclass[../../script.tex]{subfiles}
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% !TEX root = ../../script.tex
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\begin{document}
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\section{Numbers}
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\begin{defi}
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The real numbers are a set $\realn$ with the following structure
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\begin{enumerate}[(i)]
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\item Addition
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\begin{align*}
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+: \realn \times \realn \longrightarrow \realn
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\end{align*}
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\item Multiplication
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\begin{align*}
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\cdot: \realn \times \realn \longrightarrow \realn
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\end{align*}
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Instead of $+(x, y)$ and $\cdot(x, y)$ we write $x+y$ and $x \cdot y$.
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\item Order relations
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$\le$ is a relation on $\realn$, i.e. $x \le y$ is a statement.
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\end{enumerate}
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\end{defi}
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\begin{defi}[Axioms of Addition]\leavevmode
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\begin{enumerate}[{A}1:]
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\item Associativity
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\[
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\forall a, b, c \in \realn: \quad (a + b) + c = a + (b + c)
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\]
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\item Existence of a neutral element
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\[
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\exists 0 \in \realn ~\forall x \in \realn: \quad x + 0 = x
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\]
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\item Existence of an inverse element
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\[
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\forall x \in \realn ~\exists (-x) \in \realn: \quad x + (-x) = 0
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\]
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\item Commutativity
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\[
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\forall x, y \in \realn: \quad x + y = y + x
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\]
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\end{enumerate}
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\end{defi}
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\begin{thm}\label{thm:addition}
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Let $x, y \in \realn$. Then
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\begin{enumerate}[(i)]
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\item The neutral element is unique
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\item $\forall x \in \realn$ the inverse is unique
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\item $-(-x) = x$
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\item $-(x + y) = (-x) + (-y)$
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\end{enumerate}
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\end{thm}
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\begin{proof}\leavevmode
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\begin{enumerate}[(i)]
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\item Assume $a, b \in \realn$ are both neutral elements, i.e.
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\begin{equation}
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\forall x \in \realn: x + a = x = x + b
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\end{equation}
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This also implies that $a + b = a$ and $b + a = b$.
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\begin{equation}
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\implies b = b + a \stackrel{\text{A4}}{=} a + b = a
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\end{equation}
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Therefore $a = b$.
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\item Assume $c, d \in \realn$ are both inverse elements of $x \in \realn$, i.e.
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\begin{equation}
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x + c = 0 = x + d
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\end{equation}
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\begin{equation}
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c = 0 + c = x + d + c \stackrel{\text{A4}}{=} x + c + d = 0 + d = d
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\end{equation}
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Therefore $c = d$.
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\item By definition, we know
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\begin{align}
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x + (-x) &\eqlbl{\text{A4}} (-x) + x \eqlbl{\text{A3}} 0 \\
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(-x) + (-(-x)) &\eqlbl{\text{A4}} (-(-x)) + (-x) \eqlbl{\text{A3}} 0
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\end{align}
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Thus follows
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\begin{equation}
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x + (-x) = (-(-x)) + (-x) \implies x = (-(-x))
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\end{equation}
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\item
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\begin{equation}
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\begin{split}
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x + y + ((-x) + (-y)) &= x + y + (-x) + (-y) \\
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&\eqlbl{A4} x + (-x) + y + (-y) = 0
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\end{split}
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\end{equation}
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Therefore $(-x) + (-y)$ is the inverse element of $(x+y)$, i.e. $-(x + y) = (-x) + (-y)$.
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\end{enumerate}
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\end{proof}
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\begin{defi}[Axioms of Multiplication]\leavevmode
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\begin{enumerate}[M1:]
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\item Associativity
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\[
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\forall x, y, z \in \realn: \quad (xy)z = x(yz)
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\]
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\item Existence of a neutral element
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\[
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\exists 1 \in \realn ~\forall x \in \realn: \quad x1 = x
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\]
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\item Existence of an inverse element
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\[
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\forall x \in \realn \setminus \set{0} ~\exists \inv{x} \in \realn: \quad x\inv{x} = 1
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\]
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\item Commutativity
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\[
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\forall x, y \in \realn: \quad xy = yx
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\]
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\end{enumerate}
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\end{defi}
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\begin{defi}[Compatibility of Addition and Multiplication]\leavevmode
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\begin{enumerate}[label=R1:]
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\item Distributivity
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\[
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\forall x, y, z \in \realn: \quad x\cdot(y + z) = (x \cdot y) + (x \cdot z)
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\]
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\item Uniqueness of the neutral elements
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\[
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0 \ne 1
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\]
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\end{enumerate}
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\end{defi}
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\begin{thm}
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$x, y \in \realn$
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\begin{enumerate}[(i)]
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\item $x \cdot 0 = 0$
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\item $-(x \cdot y) = x \cdot (-y) = (-x) \cdot y$
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\item $(-x) \cdot (-y) = x \cdot y$
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\item $\inv{(-x)} = -(\inv{x}) ~~(\text{only for } x \ne 0)$
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\item $xy = 0 \implies x = 0 \vee y = 0$
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\end{enumerate}
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\end{thm}
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\begin{proof}\leavevmode
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\begin{enumerate}[(i)]
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\item $x \in \realn$
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\begin{equation}
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x \cdot 0 \eqlbl{A2} x \cdot (0 + 0) \eqlbl{R1} x \cdot 0 + x \cdot 0
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\end{equation}
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\begin{equation}
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\stackrel{\text{A3}}{\implies} 0 = x \cdot 0
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\end{equation}
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\item $x,y \in \realn$
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\begin{equation}
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xy + (-(xy)) \eqlbl{A3} 0 \eqlbl{(i)} x \cdot 0 = x(y + (-y)) \eqlbl{R1} xy + x(-y)
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\end{equation}
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\begin{equation}
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\stackrel{\text{A3}}{\implies} -(xy) = x\cdot(-y)
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\end{equation}
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\item Consider the expression $xy + (-x)y + (-x)(-y)$. By using distributivity we can see
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\begin{align}
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xy + (-x)y + (-x)(-y) = xy + (-x)(y + (-y)) &\eqlbl{\text{A3}} xy \\
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xy + (-x)y + (-x)(-y) = (x + (-x))y + (-x)(-y) &\eqlbl{\text{A3}} (-x)(-y)
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\end{align}
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Thus follows the desired statement.
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\item $x \in \realn$
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\begin{equation}
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x \cdot (-\inv{(-x)}) \eqlbl{(ii)} -(x \cdot \inv{(-x)}) \eqlbl{(ii)} (-x) \cdot \inv{(-x)} \eqlbl{M3} 1 \eqlbl{M3} x \cdot \inv{x}
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\end{equation}
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\begin{equation}
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\stackrel{\text{M3}}{\implies} -\inv{(-x)} = \inv{x} \stackrel{\ref{thm:addition} (iii)}{\implies} \inv{(-x)} = -(\inv{x})
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\end{equation}
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\item $x, y \in \realn$ and $y \ne 0$. Then $\exists \inv{y} \in \realn$:
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\begin{equation}
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xy = 0 \implies xy\inv{y} \eqlbl{M3} x \cdot 1 \eqlbl{M2} x = 0 = 0 \cdot \inv{y}
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\end{equation}
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\end{enumerate}
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\end{proof}
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\begin{rem}
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A structure that fulfils all the previous axioms is called a field. We introduce the following notation for $x, y \in \realn, ~y \ne 0$
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\[
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\frac{x}{y} = x\inv{y}
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\]
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\end{rem}
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\begin{defi}[Order relations]\leavevmode
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\begin{enumerate}[label=O\arabic*:]
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\item Reflexivity
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\[
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\forall x \in \realn: ~~x \le x
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\]
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\item Transitivity
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\[
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\forall x, y, z \in \realn: ~~x \le y \wedge y \le z \implies x \le z
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\]
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\item Anti-Symmetry
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\[
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\forall x, y \in \realn: ~~x \le y \wedge y \le x \implies x = y
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\]
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\item Totality
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\[
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\forall x, y \in \realn: ~~x \le y \vee y \le x
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\]
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\item
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\[
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\forall x, y, z \in \realn: ~~x \le y \implies x + z \le y + z
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\]
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\item
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\[
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\forall x, y \in \realn: ~~0 \le x \wedge 0 \le y \implies 0 \le x \cdot y
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\]
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\end{enumerate}
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We write $x < y$ for $x \le y \wedge x \ne y$
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\end{defi}
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\begin{thm}
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$x, y \in \realn$
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\begin{enumerate}[(i)]
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\item $x \le y \implies -y \le -x$
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\item $x \le 0 \wedge y \le 0 \implies 0 \le xy$
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\item $0 \le 1$
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\item $0 \le x \implies 0 \le \inv{x}$
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\item $0 < x \le y \implies \inv{y} \le \inv{x}$
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\end{enumerate}
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\end{thm}
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\begin{proof}\leavevmode
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\begin{enumerate}[(i)]
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\item
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\begin{equation}
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\begin{split}
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x \le y &\implbl{O5} x + (-x) + (-y) \le y + (-x) + (-y) \\
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&\iff -y \le -x
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\end{split}
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\end{equation}
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\item With $y \le 0 \implbl{(i)} 0 \le -y$ and $x \le 0 \implbl{(i)} 0 \le -x$ follows from O6:
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\begin{equation}
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0 \le (-x)(-y) = xy
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\end{equation}
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\item Assume $0 \le 1$ is not true. From O4 we know that
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\begin{equation}
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1 \le 0 \implbl{(ii)} 0 \le 1 \cdot 1 = 1
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\end{equation}
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\item Assume that $0 \le \inv{x}$ is not true. Then
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\begin{equation}
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\inv{x} \le 0 \implbl{(ii)} 0 \le \inv{x}\inv{x}
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\end{equation}
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We can then conclude
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\begin{equation}
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0 \le x \implbl{\text{O6}} 0 \le x\inv{x}\inv{x} \implbl{\text{M3}} 0 \le \inv{x} \implbl{\text{O3}} 0 = \inv{x}
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\end{equation}
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which would contradict M3.
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\item
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\begin{equation}
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0 \le \inv{x} \wedge 0 \le \inv{y} \implbl{O6} 0 \le \inv{x}\inv{y}
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\end{equation}
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From $x \le y$ follows $0 \le y - x$
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\begin{align}
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&\implbl{O6} 0 \le (y - x)\inv{x}\inv{y} \eqlbl{R1} y\inv{x}\inv{y} - x\inv{x}\inv{y} = \inv{x} - \inv{y} \\
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&\implbl{O5} \inv{y} \le \inv{x}
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\end{align}
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\end{enumerate}
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\end{proof}
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\begin{rem}
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A structure that fulfils all the previous axioms is called an ordered field.
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\end{rem}
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\begin{defi}
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Let $A \subset \mathbb{R}$, $x \in \realn$.
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\begin{enumerate}[(i)]
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\item $x$ is an upper bound of $A$ if $\forall y \in A: \quad y \le x$
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\item $x$ is a maximum of $A$ if $x$ is an upper bound of $A$ and $x \in A$
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\item $x$ is the supremum of $A$ if $x$ is an upper bound of $A$ and if for every other upper bound $y \in \realn$ the statement $x \le y$ holds. In other words, $x$ is the smallest upper bound of $A$.
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\end{enumerate}
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$A$ is called bounded above if it has an upper bound. Analogously, there exists a lower bound, a minimum and an infimum. We introduce the notation $\sup A$ for the supremum and $\inf A$ for the infimum.
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\end{defi}
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\begin{defi}
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Let $a, b \in \realn$, $a < b$. We define
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\begin{itemize}
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\item $(a, b) := \{x \in \realn \setvert a < x \wedge x < b\}$
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\item $[a, b] := \{x \in \realn \setvert a \le x \wedge x \le b\}$
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\item $(a, \infty) := \{x \in \realn \setvert a < x\}$
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\end{itemize}
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\end{defi}
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\begin{eg}
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$(-\infty, 1)$ is bounded above ($1$, $2$, $1000$, $\cdots$ are upper bounds), but has no maximum. $1$ is the supremum.
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\end{eg}
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\begin{defi}[Completeness of the real numbers]
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Every non-empty subset of $\realn$ with an upper bound has a supremum.
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\end{defi}
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\begin{defi}
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A set $A \subset \realn$ is called inductive if $1 \in A$ and
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\[
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x \in A \implies x + 1 \in A
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\]
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\end{defi}
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\begin{lem}
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Let $I$ be an index set, and let $A_i$ be inductive sets for every $i \in I$. Then $\bigcap_{i \in I} A_i$ is also inductive.
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\end{lem}
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\begin{proof}
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Since $A_i$ is inductive $\forall i \in I$, we know that $1 \in A_i$. Therefore
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\begin{equation}
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1 \in \bigcap_{i \in I} A_i
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\end{equation}
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Now let $x \in \bigcap_{i \in I} A_i$, this means that $x \in A_i ~~\forall i \in I$.
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\begin{equation}
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\implies x + 1 \in A_i ~~\forall i \in I \implies x + 1 \in \bigcap_{i \in I} A_i
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\end{equation}
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\end{proof}
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\begin{defi}
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The natural numbers are the smallest inductive subset of $\realn$. I.e.
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\[
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\bigcap_{A \text{ inductive}} A =: \natn
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\]
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\end{defi}
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\begin{thm}[The principle of induction]
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Let $\Phi(x)$ be a statement with a free variable $x$. If $\Phi(1)$ is true, and if $\Phi(x) \implies \Phi(x + 1)$, then $\Phi(x)$ holds for all $x \in \natn$.
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\end{thm}
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\begin{proof}
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Define $A = \{x \in \realn \setvert \Phi(x)\}$. According to the assumptions, $A$ is inductive and therefore $\natn \subset A$. This means that $\forall n \in \natn: ~~\Phi(n)$.
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\end{proof}
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\begin{cor}
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Let $m, n \in \natn$
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\begin{enumerate}[(i)]
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\item $m + n \in \natn$
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\item $mn \in \natn$
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\item $\forall n \in \natn: \quad 1 \le n$
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\end{enumerate}
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\end{cor}
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\begin{proof}\leavevmode
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\begin{enumerate}[(i)]
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\item
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Let $n \in \natn$. Define $A = \{m \in \natn \setvert m + n \in \natn\}$. Then $1 \in A$, since $\natn$ is inductive. Now let $m \in A$, therefore $n + m \in \natn$.
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\begin{align}
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\implies &n + m + 1 \in \natn \\
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\iff &m + 1 \in A
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\end{align}
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Hence $A$ is inductive, so $\natn \subset A$. From $A \subset \natn$ follows that $\natn = A$.
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\item
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Let $n \in \natn$. Define $A = \set[mn \in \natn]{m \in \natn}$. Then $1 \in A$ because of M2.
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Now let $m \in A$, therefore $nm \in \natn$
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\begin{align}
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\implies &(m + 1)n = mn + n \in \natn \\
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\iff &m + 1 \in A
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\end{align}
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Hence $A$ is inductive, so $\natn \subset A$. From $A \subset \natn$ follows that $\natn = A$.
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\item
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Define $A = \set[1 \le n]{n \in \natn}$. Then $1 \in A$ since $1 \le 1$. Now let $n \in A$.
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\begin{align}
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&0 \le 1 \implies n \le n + 1 \implies 1 \le n + 1
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\iff &n + 1 \in A
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\end{align}
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Hence $A$ is inductive, so $\natn \subset A$. But since $A \subset \natn$ it must follow that $\natn = A$.
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\end{enumerate}
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\end{proof}
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\begin{thm}
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Let $n \in \natn$. There are no natural numbers between $n$ and $n + 1$.
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\end{thm}
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\begin{proof}
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Let $x \in \natn \cap (1, 2)$, and define $A = \natn \setminus \set{x} = \set[n \ne x]{n \in \natn}$. Obviously, $1 \in A$ and $2 \in A$. Let $n \in A$, then
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\begin{align}
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&1 \le n \implies 2 \le n + 1 \\
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\iff &n + 1 \in A
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\end{align}
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Thus $A$ is inductive. Since $A \subset \natn$ we have $A = \natn$, so there are no natural numbers between $1$ and $2$.
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Now assume $\natn \cap (n, n+1) = \emptyset$, and consider $x \in \natn \cap (n + 1, n + 2)$. We will take another look at $A$ with this new $x$. Obviously $1 \in A$.
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Let $m \in A$, we want to show that $m + 1 \in A$. To do this, assume that $m + 1 \not\in A$, i.e.
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\begin{equation}
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m + 1 \in \natn \cap (n + 1, n + 2) \implies m \in \natn \cap (n, n + 1)
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\end{equation}
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However since $\natn \cap (n, n + 1) = \emptyset$ this $m$ can't exist, so $m + 1 \in A$.
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So $A$ is still inductive, and $A = \natn$ still holds.
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\end{proof}
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\begin{thm}[Archimedian property]
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\[
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\forall x \in \realn ~\exists n \in \natn: ~~x<n
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\]
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\end{thm}
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\begin{proof}
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If $x < 1$ there is nothing to prove, so let $x \ge 1$. Define the set
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\begin{equation}
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A = \{n \in \natn \setvert n \le x\}
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\end{equation}
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$A$ is bounded above by definition. There exists the supremum $s = \sup A$. By definition, $s-1$ is not an upper bound of $A$, i.e. $\exists m \in A: ~~s-1 < m$. Therefore $s \le m + 1$.
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\begin{equation}
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m \in A \subset \natn \implies m + 1 \in \natn
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\end{equation}
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Since $s$ is an upper bound of $A$, this implies that $m+1 \not\subset A$, so therefore $m + 1 > x$.
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\end{proof}
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\begin{cor}\label{cor:minimum}
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Every non-empty subset of $\natn$ has a minimum, and every non-empty subset of $\natn$ that is bounded above has a maximum.
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\end{cor}
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|
\begin{proof}
|
|
Let $A \subset \natn$. Propose that $A$ has no minimum. Define the set
|
|
\begin{equation}
|
|
\tilde{A} := \{n \in \natn \setvert \forall m \in A: ~n < m\}
|
|
\end{equation}
|
|
$1$ is a lower bound of $A$, but according to the proposition $A$ has no minimum, so therefore $1 \notin A$. This implies that $1 \in \tilde{A}$.
|
|
\begin{equation}
|
|
n \in \tilde{A} \implies n < m ~\forall m \in A
|
|
\end{equation}
|
|
But since there exists no natural number between $n$ and $n+1$, this means that $n+1$ is also a lower bound of $A$, and therefore
|
|
\begin{equation}
|
|
n+1 \le m ~\forall m \in A \implies n+1 \in \tilde{A}
|
|
\end{equation}
|
|
So $\tilde{A}$ is an inductive set, hence $\tilde{A} = \natn$. Therefore $A = \emptyset$.
|
|
The proof that a bounded above $A$ has a maximum works in the same way.
|
|
\end{proof}
|
|
|
|
\begin{defi}
|
|
We define the following new sets:
|
|
\begin{align*}
|
|
&\intn := \{x \in \realn \setvert x \in \natn_0 \vee (-x) \in \natn_0\}\\
|
|
&\ratn := \left\{\frac{p}{q} \setvert p, q \in \intn \wedge q \ne 0\right\}
|
|
\end{align*}
|
|
$\intn$ are called integers, and $\ratn$ are called the rational numbers. $\natn_0$ are the natural numbers with the $0$ ($\natn_0 = \natn \cup \{0\}$).
|
|
\end{defi}
|
|
|
|
\begin{rem}
|
|
\begin{align*}
|
|
x, y \in \intn &\implies x+y, x\cdot y, (-x) \in \intn \\
|
|
x, y \in \ratn &\implies x+y, x\cdot y, (-x) \in \ratn \text{ and } \inv{x} \in \ratn \text{ if } x \ne 0
|
|
\end{align*}
|
|
The second statement implies that $\ratn$ is a field.
|
|
\end{rem}
|
|
|
|
\begin{cor}[Density of the rationals]\label{cor:densityrats}
|
|
$x, y \in \realn, ~x < y$. Then
|
|
\[
|
|
\exists r \in \ratn:\quad ~x < r < y
|
|
\]
|
|
\end{cor}
|
|
\begin{proof}
|
|
This proof relies on the Archimedian property.
|
|
\begin{equation}
|
|
\exists q \in \natn: \quad \frac{1}{y-x} < q \left( \iff \frac{1}{q} < y - x \right)
|
|
\end{equation}
|
|
Let $p \in \intn$ be the greatest integer that is smaller than $y \cdot q$. The existence of $p$ is ensured by \Cref{cor:minimum}. Then $\frac{p}{q} < y$ and
|
|
\begin{equation}
|
|
p + 1 \ge y \cdot q \implies y \le \frac{p}{q} + \frac{1}{q} < \frac{p}{q} + (y - x)
|
|
\end{equation}
|
|
\begin{equation}
|
|
\implies x < \frac{p}{q} < y
|
|
\end{equation}
|
|
\end{proof}
|
|
|
|
\begin{defi}[Absolute values]
|
|
We define the following function
|
|
\begin{align*}
|
|
|\cdot|: \realn &\longrightarrow [0, \infty) \\
|
|
x &\longmapsto
|
|
\begin{cases}
|
|
x &, x \ge 0 \\
|
|
-x &, x < 0
|
|
\end{cases}
|
|
\end{align*}
|
|
\end{defi}
|
|
|
|
\begin{thm}
|
|
\[
|
|
x, y \in \realn \implies |xy| = |x||y|
|
|
\]
|
|
\end{thm}
|
|
\begin{proof}
|
|
Let $x, y \in \realn$. If $x$ and $y$ are both positive or both negative, their product is positive and the statement is trivial:
|
|
\begin{equation}
|
|
\abs{xy} = xy = \abs{x} \abs{y}
|
|
\end{equation}
|
|
So let w.l.o.g. $0 \le x$ and $y < 0$. Then
|
|
\begin{equation}
|
|
\abs{x}\abs{y} = x(-y) = -(xy)
|
|
\end{equation}
|
|
Since $xy$ is negative, the absolute value is
|
|
\begin{equation}
|
|
\abs{xy} = -(xy) = \abs{x}\abs{y}
|
|
\end{equation}
|
|
\end{proof}
|
|
|
|
\begin{defi}[Complex numbers]
|
|
Complex numbers are defined as the set $\cmpln = \realn^2$. Addition and multiplication are defined as mappings $\cmpln \times \cmpln \rightarrow \cmpln$. Let $(x, y), (\tilde{x}, \tilde{y}) \in \cmpln$.
|
|
\begin{align*}
|
|
(x, y) + (\tilde{x}, \tilde{y}) &:= (x + \tilde{x}, y + \tilde{y}) \\
|
|
(x, y) \cdot (\tilde{x}, \tilde{y}) &:= (x\tilde{x} - y\tilde{y}, x\tilde{y} + \tilde{x}y)
|
|
\end{align*}
|
|
$\cmpln$ is a field. Let $z = (x, y) \in \cmpln$. We define
|
|
\begin{align*}
|
|
\real(z) = \Re(z) = x& ~~\text{ the real part} \\
|
|
\imag(z) = \Im(z) = y& ~~\text{ the imaginary part}
|
|
\end{align*}
|
|
\end{defi}
|
|
|
|
\begin{rem}\leavevmode
|
|
\begin{enumerate}[(i)]
|
|
\item We will not prove that $\cmpln$ fulfils the field axioms here, this can be left as an exercise to the reader. However, we will note the following statements
|
|
\begin{itemize}
|
|
\item Additive neutral element: $(0, 0)$
|
|
\item Additive inverse of $(x, y)$: $(-x, -y)$
|
|
\item Multiplicative neutral element: $(1, 0)$
|
|
\item Multiplicative inverse of $(x, y) \ne (0, 0)$: $\left( \frac{x}{x^2 + y^2}, -\frac{y}{x^2 + y^2} \right)$
|
|
\end{itemize}
|
|
|
|
\item Numbers with $y = 0$ are called real.
|
|
|
|
\item The imaginary unit is defined as $i = (0, 1)$
|
|
\[
|
|
(0, 1) \cdot (x, y) = (-y, x)
|
|
\]
|
|
Especially
|
|
\[
|
|
i^2 = (0, 1)^2 = (-1, 0) = -(1, 0) = -1
|
|
\]
|
|
\end{enumerate}
|
|
We also introduce the following notation
|
|
\[
|
|
(x, y) = (x, 0) + i\cdot(y, 0) = x + iy
|
|
\]
|
|
\end{rem}
|
|
|
|
\begin{thm}[Fundamental theorem of algebra]
|
|
Every non-constant, complex polynomial has a complex root. I.e. for $n \in \natn$, $\alpha_0, \cdots, \alpha_n \in \cmpln$, $\alpha_n \ne 0$ there is some $x \in \cmpln$ such that
|
|
\[
|
|
\sum_{i = 0}^n \alpha_i x^i = \alpha_0 + \alpha_1 x + \alpha_2 x^2 + \cdots + \alpha_n x^n = 0
|
|
\]
|
|
\end{thm}
|
|
\begin{proof}
|
|
Not here. It is proven in \Cref{thm:fundamental}.
|
|
\end{proof}
|
|
\end{document} |