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packages/macros.sty : Added convinences macros

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saundersp
2024-11-07 05:29:13 +01:00
parent 94cdd07e84
commit aafcec6a3e
5 changed files with 46 additions and 36 deletions
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contents/number_theory.tex
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@ -94,7 +94,7 @@ Il existe toujours un élément minimum pour n'importe quel sous-ensemble de $\N
\langsection{Construction des entiers relatifs $(\Z)$}{Construction of relative numbers}
%TODO Complete section
$\Z := \{\dots,-3,-2,-1,0,1,2,3,\dots\}$
$\Z := \{\dots,-3,-2,-1,0,1,2,3,\dots\} = \Union_{n \in \N} n \union \Union_{n \in \N^*} -n$
\langsubsection{Relations binaries}{Binary relations}
%TODO Complete subsection
@ -191,7 +191,7 @@ $i^2 = -1$
\begin{tabular}{|c||c|c|}
\hline
& 1 & i \\
$\cartesianProduct$ & 1 & i \\
\hline
\hline
1 & 1 & i \\
@ -229,7 +229,7 @@ Source: \citeannexes{wikipedia_quaternion}
\begin{tabular}{|c||c|c|c|c|}
\hline
& 1 & i & j & k \\
$\cartesianProduct$ & 1 & i & j & k \\
\hline
\hline
1 & 1 & i & j & k \\
@ -251,7 +251,7 @@ Source: \citeannexes{wikipedia_octonion}
\begin{tabular}{|c||c|c|c|c|c|c|c|c|}
\hline
$e_i/e_j $ & $e_0$ & $e_1$ & $e_2$ & $e_3$ & $e_4$ & $e_5$ & $e_6$ & $e_7$ \\
$\cartesianProduct$ & $e_0$ & $e_1$ & $e_2$ & $e_3$ & $e_4$ & $e_5$ & $e_6$ & $e_7$ \\
\hline
\hline
$e_0$ & $e_0$ & $e_1$ & $e_2$ & $e_3$ & $e_4$ & $e_5$ & $e_6$ & $e_7$ \\
@ -289,7 +289,7 @@ Source: \citeannexes{wikipedia_sedenion}
\begin{tabular}{|c|c|c|c|}
\hline
& i & j & k \\
$\cartesianProduct$ & i & j & k \\
\hline
i & -1 & k & -j \\
\hline
@ -324,15 +324,13 @@ $\Pn = \{p | p \in \N^* \land p \text{ est premier}\} = p_0, p_1, \dots p_{n-1},
$\omega = (\prod_{p\in \Pn} p) + 1$
$\forall p \in \Pn, \lnot(\omega \div p)$
$\implies \forall p \in \Pn$, $\lnot(p \divides \omega)$
$\omega \notin \Pn \land \omega \in \Pn$
$\implies (\omega \notin \Pn \land \omega \in \Pn) \implies \bot$
$\rightarrow\leftarrow$
$\implies \card{P} = \infty$
$\implies |P| = \infty$
Il existe une infinité de nombre premiers.
\end{proof}
\langsubsection{Irrationnalité}{Irrationality}
@ -351,22 +349,22 @@ By contradiction let's assume $\sqrt{p} \in \Q$
$a \in \Z, b \in \N^*, \text{PGCD}(a,b) = 1, \sqrt{p} = \frac{a}{b}$
$\Rightarrow p = (\frac{a}{b})^2 = \frac{a^2}{b^2}$
$\implies p = (\frac{a}{b})^2 = \frac{a^2}{b^2}$
$\Rightarrow b^2p = a^2$
$\implies b^2p = a^2$
$\Rightarrow p|a$
$\implies p \divides a$
Let $c \in \N^*$, $a = pc$
$\Rightarrow b^2 p = (pc)^2=p^2c^2$
$\implies b^2 p = (pc)^2=p^2c^2$
$\Rightarrow b^2 = pc^2$
$\implies b^2 = pc^2$
$\Rightarrow p|b$
$\implies p \divides b$
$\Rightarrow (p|b \land p|a \land \text{PGCD}(a,b)=1) \Rightarrow \bot$
$\implies (p \divides b \land p \divides a \land \text{PGCD}(a,b)=1) \implies \bot$
$\Rightarrow \sqrt{p} \notin \Q$
$\implies \sqrt{p} \notin \Q$
\end{proof}