Fixed some notations mistakes
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@ -126,7 +126,7 @@ $\begin{Vmatrix} a & b \\ c & d \end{Vmatrix}$
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\langsection{Informatique}{Computer science}
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\langsection{Informatique}{Computer science}
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%TODO Complete section
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%TODO Complete section
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\subsection{LaTex}
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\subsection{LaTeX}
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\begin{verbatim}
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\begin{verbatim}
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\begin{verbatim}
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\begin{verbatim}
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@ -139,9 +139,9 @@ $\forall (p,q) \in \Q, \forall n \in \N^*, \frac{p}{q} \Leftrightarrow \frac{p \
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\langsubsection{Opérateurs}{Operators}
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\langsubsection{Opérateurs}{Operators}
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%TODO Complete subsection
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%TODO Complete subsection
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$\forall ((p,q), (a,b)) \in \Q^2, \frac{p}{q} + \frac{a}{b} = \frac{pb + aq}{qb}$
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$\forall ((p,q), (a,b)) \in \Q, \frac{p}{q} + \frac{a}{b} = \frac{pb + aq}{qb}$
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$\forall ((p,q), (a,b)) \in \Q^2, \frac{p}{q} \cdot \frac{a}{b} = \frac{pa}{qb}$
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$\forall ((p,q), (a,b)) \in \Q, \frac{p}{q} \cdot \frac{a}{b} = \frac{pa}{qb}$
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$\forall (p,q) \in \Q, \forall k \in \Z, (\frac{p}{q})^k = \frac{p^k}{q^k}$
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$\forall (p,q) \in \Q, \forall k \in \Z, (\frac{p}{q})^k = \frac{p^k}{q^k}$
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@ -183,7 +183,7 @@ $\functiondef{(p,q)}{P_1^{\frac{p}{|p|} + 1}P_2^pP_3^q}$
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\citeannexes{wikipedia_complex_numbers}
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\citeannexes{wikipedia_complex_numbers}
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$\C = (a,b) \in R^2, a + ib ~= \R^2 $
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$\C = (a,b) \in R, a + ib ~= \R $
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$i^2 = -1$
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$i^2 = -1$
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@ -204,7 +204,7 @@ $i^2 = -1$
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\langsubsection{Relations binaries}{Binary relations}
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\langsubsection{Relations binaries}{Binary relations}
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%TODO Complete subsection
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%TODO Complete subsection
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$\forall ((a,b), (c,d)) \in \C^2, a = c \land b = d \Leftrightarrow a + ib = c + id$
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$\forall ((a,b), (c,d)) \in \C, a = c \land b = d \Leftrightarrow a + ib = c + id$
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\langsubsection{Opérateurs}{Operators}
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\langsubsection{Opérateurs}{Operators}
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%TODO Complete subsection
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%TODO Complete subsection
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@ -213,7 +213,7 @@ Il est impossible d'avoir une relation d'ordre dans le corps des complexes mais
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\subsubsection{Ordre lexicographique}
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\subsubsection{Ordre lexicographique}
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$\forall((a,b),(c,d)) \in \C^2, a + ib \Rel_L c + id := \begin{cases}
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$\forall((a,b),(c,d)) \in \C, a + ib \Rel_L c + id := \begin{cases}
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a < c & \implies a + ib < c + id \\
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a < c & \implies a + ib < c + id \\
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\otherwise & \begin{cases}
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\otherwise & \begin{cases}
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b < d & \implies a + ib < c + id \\
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b < d & \implies a + ib < c + id \\
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@ -16,6 +16,10 @@ $\forall A\forall B(\forall X(X \in A \Leftrightarrow X \in B) \Rightarrow A = B
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\langsubsection{Spécification}{Specification}
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\langsubsection{Spécification}{Specification}
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%TODO Complete subsection
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%TODO Complete subsection
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\langsubsection{Ensemble vide}{Empty set}
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Il existe un ensemble vide notée $\emptyset$.
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\langsubsection{Paire}{Pairing}
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\langsubsection{Paire}{Pairing}
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%TODO Complete subsection
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%TODO Complete subsection
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@ -24,13 +28,13 @@ $\forall A\forall B(\forall X(X \in A \Leftrightarrow X \in B) \Rightarrow A = B
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Unite all elements of two given sets into one.
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Unite all elements of two given sets into one.
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$n,m \in \N^+$
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$n,m \in \N$
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$A = \{a_1, \cdots, a_n\}$
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$A = \{a_0, \cdots, a_n\}$
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$B = \{b_1, \cdots, b_m\}$
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$B = \{b_0, \cdots, b_m\}$
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$A \cup B = \{a_1, \cdots, a_n, b_1, \cdots, b_m\}$
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$A \union B = \{a_0, \cdots, a_n, b_0, \cdots, b_m\}$
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\langsubsection{Scheme of replacement}{Scheme of replacement}
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\langsubsection{Scheme of replacement}{Scheme of replacement}
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%TODO Complete subsection
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%TODO Complete subsection
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@ -41,29 +45,43 @@ $A \cup B = \{a_1, \cdots, a_n, b_1, \cdots, b_m\}$
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\subsection{Power set}
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\subsection{Power set}
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%TODO Complete subsection
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%TODO Complete subsection
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For a set $S$ such that $|S| = n \Leftrightarrow \mathbf{P}(S) = 2^n$
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\langsubsection{Choix}{Choice}
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\langsubsection{Choix}{Choice}
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%TODO Complete subsection
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%TODO Complete subsection
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\section{Intersection}
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\section{Intersection}
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%TODO Complete subsection
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Unite all common elements of two given sets into one.
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$n,m,i \in \N$
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$A = \{a_0, \cdots, a_n, c_0, \cdots, c_n\}$
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$B = \{b_0, \cdots, b_m, c_0, \cdots, c_n\}$
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$A \cap B = \{c_0, \cdots, c_n\}$
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\langsection{Différence des sets}{Set difference}
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\langsection{Différence des sets}{Set difference}
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%TODO Complete section
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%TODO Complete section
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\langsection{Fonction}{Function}
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\langsection{Fonction}{Function}
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%TODO Complete section
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Une fonction $f$ est un opération qui permet de transformer un ou plusieurs éléments d'un set $A$ en d'autres éléments d'un set $B$.
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Source: \citeannexes{wikipedia_function_mathematics}
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Une fonction $f$ est un tuple d'un domaine \citeannexes{wikipedia_domain_function} $A$ et un codomaine \citeannexes{wikipedia_codomain} $B$.
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If the domain is the same as the codomain then the function is an endormorphsim \ref{definition:endomorphism} applied on set theory \ref{set_theory}.
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\subsection{Notation}
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\subsection{Notation}
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%TODO Complete subsection
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$A \longrightarrow B$
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$A \longrightarrow B$
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$ x \longrightarrow f(x)$
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$ x \longrightarrow f(x)$
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\langsubsection{Injectivité}{Injectivity}
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\langsubsection{Injectivité}{Injectivity} \label{definition:injective}
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%TODO Complete subsection
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Source: \citeannexes{wikipedia_injective_function}
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Une fonction $f$ de $E$ dans $F$ est dite \textbf{injective} si, et seulement si, $\forall (a,b) \in E, f(a) = f(b) \Rightarrow a = b$.
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Une fonction $f$ de $E$ dans $F$ est dite \textbf{injective} si, et seulement si, $\forall (a,b) \in E, f(a) = f(b) \Rightarrow a = b$.
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@ -130,7 +130,7 @@ Toute sous-suites (ou suites extraite) d'un suite convergente vers $l \in E$ con
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Montrer que l’ensemble $\{x_n, n \in \N\}$ est borné.
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Montrer que l’ensemble $\{x_n, n \in \N\}$ est borné.
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\\
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\\
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Sachant que $(x_n) \ in E$ converge vers $l \in E$ \&\& $\epsilon > 0$.
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Sachant que $(x_n) \in E$ converge vers $l \in E \land \epsilon > 0$.
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$\Leftrightarrow \exists y \in E$ tel que $\{\forall n \in \N, x_n, l\} \subset \bar{\mathbb{B}}(y, \epsilon) \subset E$.
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$\Leftrightarrow \exists y \in E$ tel que $\{\forall n \in \N, x_n, l\} \subset \bar{\mathbb{B}}(y, \epsilon) \subset E$.
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