\documentclass[../../script.tex]{subfiles} % !TEX root = ../../script.tex \begin{document} \section{Numbers} \begin{defi} The real numbers are a set $\realn$ with the following structure \begin{enumerate}[(i)] \item Addition \begin{align*} +: \realn \times \realn \longrightarrow \realn \end{align*} \item Multiplication \begin{align*} \cdot: \realn \times \realn \longrightarrow \realn \end{align*} Instead of $+(x, y)$ and $\cdot(x, y)$ we write $x+y$ and $x \cdot y$. \item Order relations $\le$ is a relation on $\realn$, i.e. $x \le y$ is a statement. \end{enumerate} \end{defi} \begin{defi}[Axioms of Addition]\leavevmode \begin{enumerate}[label=A\arabic*:] \item Associativity \[ \forall a, b, c \in \realn: ~~(a + b) + c = a + (b + c) \] \item Existence of a neutral element \[ \exists 0 \in \realn ~\forall x \in \realn: ~~x + 0 = x \] \item Existence of an inverse element \[ \forall x \in \realn ~\exists (-x) \in \realn: ~~ x + (-x) = 0 \] \item Commutativity \[ \forall x, y \in \realn: ~~x + y = y + x \] \end{enumerate} \end{defi} \begin{thm}\label{thm:addition} $x, y \in \realn$ \begin{enumerate}[(i)] \item The neutral element is unique \item $\forall x \in \realn$ the inverse is unique \item $-(-x) = x$ \item $-(x + y) = (-x) + (-y)$ \end{enumerate} \end{thm} \begin{proof}\leavevmode \begin{enumerate}[(i)] \item Assume $a, b \in \realn$ are both neutral elements, i.e. \begin{equation} \forall x \in \realn: x + a = x = x + b \end{equation} This also implies that $a + b = a$ and $b + a = b$. \begin{equation} \implies b = b + a \stackrel{\text{A4}}{=} a + b = a \end{equation} Therefore $a = b$. \item Assume $c, d \in \realn$ are both inverse elements of $x \in \realn$, i.e. \begin{equation} x + c = 0 = x + d \end{equation} \begin{equation} c = 0 + c = x + d + c \stackrel{\text{A4}}{=} x + c + d = 0 + d = d \end{equation} Therefore $c = d$. \item \reader \item \begin{equation} \begin{split} x + y + ((-x) + (-y)) &= x + y + (-x) + (-y) \\ &\eqlbl{A4} x + (-x) + y + (-y) = 0 \end{split} \end{equation} Therefore $(-x) + (-y)$ is the inverse element of $(x+y)$, i.e. $-(x + y) = (-x) + (-y)$. \end{enumerate} \end{proof} \begin{defi}[Axioms of Multiplication]\leavevmode \begin{enumerate}[label=M\arabic*:] \item $\forall x, y, z \in \realn: ~~(xy)z = x(yz)$ \item $\exists 1 \in \realn ~\forall x \in \realn: ~~x1 = x$ \item $\forall x \in \realn \setminus \{0\} ~\exists \inv{x} \in \realn: ~~x\inv{x} = 1$ \item $\forall x, y \in \realn: ~~xy = yx$ \end{enumerate} \end{defi} \begin{defi}[Compatibility of Addition and Multiplication]\leavevmode \begin{enumerate}[label=R\arabic*:] \item Distributivity \[ \forall x, y, z \in \realn: ~~ x\cdot(y + z) = (x \cdot y) + (x \cdot z) \] \item $0 \ne 1$ \end{enumerate} \end{defi} \begin{thm} $x, y \in \realn$ \begin{enumerate}[(i)] \item $x \cdot 0 = 0$ \item $-(x \cdot y) = x \cdot (-y) = (-x) \cdot y$ \item $(-x) \cdot (-y) = x \cdot y$ \item $\inv{(-x)} = -(\inv{x}) ~~(\text{only for } x \ne 0)$ \item $xy = 0 \implies x = 0 \vee y = 0$ \end{enumerate} \end{thm} \begin{proof}\leavevmode \begin{enumerate}[(i)] \item $x \in \realn$ \begin{equation} x \cdot 0 \eqlbl{A2} x \cdot (0 + 0) \eqlbl{R1} x \cdot 0 + x \cdot 0 \end{equation} \begin{equation} \stackrel{\text{A3}}{\implies} 0 = x \cdot 0 \end{equation} \item $x,y \in \realn$ \begin{equation} xy + (-(xy)) \eqlbl{A3} 0 \eqlbl{(i)} x \cdot 0 = x(y + (-y)) \eqlbl{R1} xy + x(-y) \end{equation} \begin{equation} \stackrel{\text{A3}}{\implies} -(xy) = x\cdot(-y) \end{equation} \item \reader \item $x \in \realn$ \begin{equation} 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} \end{equation} \begin{equation} \stackrel{\text{M3}}{\implies} -\inv{(-x)} = \inv{x} \stackrel{\ref{thm:addition} (iii)}{\implies} \inv{(-x)} = -(\inv{x}) \end{equation} \item $x, y \in \realn$ and $y \ne 0$. Then $\exists \inv{y} \in \realn$: \begin{equation} xy = 0 \implies xy\inv{y} \eqlbl{M3} x \cdot 1 \eqlbl{M2} x = 0 = 0 \cdot \inv{y} \end{equation} \end{enumerate} \end{proof} \begin{rem} A structure that fulfils all the previous axioms is called a field. We introduce the following notation for $x, y \in \realn, ~y \ne 0$ \[ \frac{x}{y} = x\inv{y} \] \end{rem} \begin{defi}[Order relations]\leavevmode \begin{enumerate}[label=O\arabic*:] \item Reflexivity \[ \forall x \in \realn: ~~x \le x \] \item Transitivity \[ \forall x, y, z \in \realn: ~~x \le y \wedge y \le z \implies x \le z \] \item Anti-Symmetry \[ \forall x, y \in \realn: ~~x \le y \wedge y \le x \implies x = y \] \item Totality \[ \forall x, y \in \realn: ~~x \le y \vee y \le x \] \item \[ \forall x, y, z \in \realn: ~~x \le y \implies x + z \le y + z \] \item \[ \forall x, y \in \realn: ~~0 \le x \wedge 0 \le y \implies 0 \le x \cdot y \] \end{enumerate} We write $x < y$ for $x \le y \wedge x \ne y$ \end{defi} \begin{thm} $x, y \in \realn$ \begin{enumerate}[(i)] \item $x \le y \implies -y \le -x$ \item $x \le 0 \wedge y \le 0 \implies 0 \le xy$ \item $0 \le 1$ \item $0 \le x \implies 0 \le \inv{x}$ \item $0 < x \le y \implies \inv{y} \le \inv{x}$ \end{enumerate} \end{thm} \begin{proof}\leavevmode \begin{enumerate}[(i)] \item \begin{equation} \begin{split} x \le y &\implbl{O5} x + (-x) + (-y) \le y + (-x) + (-y) \\ &\iff -y \le -x \end{split} \end{equation} \item With $y \le 0 \implbl{(i)} 0 \le -y$ and $x \le 0 \implbl{(i)} 0 \le -x$ follows from O6: \begin{equation} 0 \le (-x)(-y) = xy \end{equation} \item Assume $0 \le 1$ is not true. From O4 we know that \begin{equation} 1 \le 0 \implbl{(ii)} 0 \le 1 \cdot 1 = 1 \end{equation} \item \reader \item \begin{equation} 0 \le \inv{x} \wedge 0 \le \inv{y} \implbl{O6} 0 \le \inv{x}\inv{y} \end{equation} From $x \le y$ follows $0 \le y - x$ \begin{align} &\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} \\ &\implbl{O5} \inv{y} \le \inv{x} \end{align} \end{enumerate} \end{proof} \begin{rem} A structure that fulfils all the previous axioms is called an ordered field. \end{rem} \begin{defi} Let $A \subset \mathbb{R}$, $x \in \realn$. \begin{enumerate}[(i)] \item $x$ is called an upper bound of $A$ if $\forall y \in A: ~y \le x$ \item $x$ is called a maximum of $A$ if $x$ is an upper bound of $A$ and $x \in A$ \item $x$ is called supremum of $A$ is $x$ 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 $A$. \end{enumerate} $A$ 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. \end{defi} \begin{defi} $a, b \in \realn$, $a < b$. We define \begin{itemize} \item $(a, b) := \{x \in \realn \setvert a < x \wedge x < b\}$ \item $[a, b] := \{x \in \realn \setvert a \le x \wedge x \le b\}$ \item $(a, \infty) := \{x \in \realn \setvert a < x\}$ \end{itemize} \end{defi} \begin{eg} $(-\infty, 1)$ is bounded above ($1$, $2$, $1000$, $\cdots$ are upper bounds), but has no maximum. $1$ is the supremum. \end{eg} \begin{defi}[Completeness of the real numbers] Every non-empty subset of $\realn$ with an upper bound has a supremum. \end{defi} \begin{defi} A set $A \subset \realn$ is called inductive if $1 \in A$ and \[ x \in A \implies x + 1 \in A \] \end{defi} \begin{lem} 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. \end{lem} \begin{proof} Since $A_i$ is inductive $\forall i \in I$, we know that $1 \in A_i$. Therefore \begin{equation} 1 \in \bigcap_{i \in I} A_i \end{equation} Now let $x \in \bigcap_{i \in I} A_i$, this means that $x \in A_i ~~\forall i \in I$. \begin{equation} \implies x + 1 \in A_i ~~\forall i \in I \implies x + 1 \in \bigcap_{i \in I} A_i \end{equation} \end{proof} \begin{defi} The natural numbers are the smallest inductive subset of $\realn$. I.e. \[ \bigcap_{A \text{ inductive}} A =: \natn \] \end{defi} \begin{thm}[The principle of induction] 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 $x \in \natn$. \end{thm} \begin{proof} Define $A = \{x \in \realn \setvert \Phi(x)\}$. According to the assumptions, $A$ is inductive and therefore $\natn \subset A$. This means that $\forall n \in \natn: ~~\Phi(n)$. \end{proof} \begin{cor} $m, n \in \natn$ \begin{enumerate}[(i)] \item $m + n \in \natn$ \item $mn \in \natn$ \item $1 \le n ~~\forall n \in \natn$ \end{enumerate} \end{cor} \begin{proof} We will only proof (i). (ii) and (iii) are left as an exercise for the reader. 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 $n + m \in \natn$. \begin{align} &\implies n + m + 1 \in \natn \\ &\iff m + 1 \in A \end{align} Hence $A$ is inductive, so $\natn \subset A$. From $A \subset N$ follows that $\natn = A$. \end{proof} \begin{thm} $n \in \natn$. There are no natural numbers between $n$ and $n + 1$. \end{thm} \begin{hproof} Show that $x \in \natn \cap (1, 2)$ implies that $\natn \setminus \{x\}$ is inductive. Now show that if $\natn \cap (n, n+1) = \varnothing$ and $x \in \natn \cap (n + 1, n + 2)$ then $\natn \setminus \{x\}$ is inductive. \end{hproof} \begin{thm}[Archimedian property] \[ \forall x \in \realn ~\exists n \in \natn: ~~x x$. \end{proof} \begin{cor}\label{cor:minimum} Every non-empty subset of $\natn$ has a minimum, and every non-empty subset of $\natn$ that is bounded above has a maximum. \end{cor} \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 $A = \varnothing$. \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 $0$ ($\natn_0 = \natn \cap \{0\}$). \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 \[ \exists r \in \ratn: ~~x < r < y \] \end{cor} \begin{proof} This proof relies on the Archimedian property. \begin{equation} \exists q \in \natn: ~~ \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 corollary \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} \reader \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. \end{proof} \end{document}