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diff --git a/uni_notes/algebra.html b/uni_notes/algebra.html new file mode 100755 index 0000000..27c8d76 --- /dev/null +++ b/uni_notes/algebra.html @@ -0,0 +1,1295 @@ +<?xml version="1.0" encoding="utf-8"?> +<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" +"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> +<html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"> +<head> +<!-- 2023-10-11 Wed 19:04 --> +<meta http-equiv="Content-Type" content="text/html;charset=utf-8" /> +<meta name="viewport" content="width=device-width, initial-scale=1" /> +<title>Algebra 1</title> +<meta name="author" content="Crystal" /> +<meta name="generator" content="Org Mode" /> +<link rel="stylesheet" type="text/css" href="../src/css/colors.css"/> +<link rel="stylesheet" type="text/css" href="../src/css/style.css"/> +<script> + window.MathJax = { + tex: { + ams: { + multlineWidth: '85%' + }, + tags: 'ams', + tagSide: 'right', + tagIndent: '.8em' + }, + chtml: { + scale: 1.0, + displayAlign: 'center', + displayIndent: '0em' + }, + svg: { + scale: 1.0, + displayAlign: 'center', + displayIndent: '0em' + }, + output: { + font: 'mathjax-modern', + displayOverflow: 'overflow' + } + }; +</script> + +<script + id="MathJax-script" + async + src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"> +</script> +</head> +<body> +<div id="content" class="content"> +<h1 class="title">Algebra 1</h1> +<div id="outline-container-orgcc5f4d8" class="outline-2"> +<h2 id="orgcc5f4d8">Contenu de la Matiére</h2> +<div class="outline-text-2" id="text-orgcc5f4d8"> +</div> +<div id="outline-container-orgf4040f2" class="outline-3"> +<h3 id="orgf4040f2">Rappels et compléments (11H)</h3> +<div class="outline-text-3" id="text-orgf4040f2"> +<ul class="org-ul"> +<li>Logique mathématique et méthodes du raisonnement mathématique</li> +<li>Ensembles et Relations</li> +<li>Applications</li> +</ul> +</div> +</div> +<div id="outline-container-orge74cdc9" class="outline-3"> +<h3 id="orge74cdc9">Structures Algébriques (11H)</h3> +<div class="outline-text-3" id="text-orge74cdc9"> +<ul class="org-ul"> +<li>Groupes et morphisme de groupes</li> +<li>Anneaux et morphisme d’anneaux</li> +<li>Les corps</li> +</ul> +</div> +</div> +<div id="outline-container-org0b54650" class="outline-3"> +<h3 id="org0b54650">Polynômes et fractions rationnelles</h3> +<div class="outline-text-3" id="text-org0b54650"> +<ul class="org-ul"> +<li>Notion du polynôme à une indéterminée á coefficients dans un anneau</li> +<li>Opérations Algébriques sur les polynômes</li> +<li>Arithmétique dans l’anneau des polynômes</li> +<li>Polynôme dérivé et formule de Taylor</li> +<li>Notion de racine d’un polynôme</li> +<li>Notion de Fraction rationelle á une indéterminée</li> +<li>Décomposition des fractions rationelles en éléments simples</li> +</ul> +</div> +</div> +</div> +<div id="outline-container-org9dbc8bb" class="outline-2"> +<h2 id="org9dbc8bb">Premier cours : Logique mathématique et méthodes du raisonnement mathématique <i>Sep 25</i> :</h2> +<div class="outline-text-2" id="text-org9dbc8bb"> +<p> +Let <b>P</b> <b>Q</b> and <b>R</b> be propositions which can either be <b>True</b> or <b>False</b>. And let’s also give the value <b>1</b> to each <b>True</b> proposition and <b>0</b> to each false one. +</p> + +<p> +<i>Ex:</i> +</p> +<ul class="org-ul"> +<li><b>5 ≥ 2</b> is a proposition, a correct one !!!</li> +<li><b>The webmaster is a girl</b> is also a proposition, which is also correct.</li> +<li><b>x is always bigger than 5</b> is <b>not</b> a proposition, because we CAN’T determine if it’s correct or not as <b>x</b> changes.</li> +</ul> +<p> +…etc +</p> + +<p> +In order to avoid repetition, and rewriting the proposition over and over, we just assign a capital letter to them such as <b>P Q</b> or <b>R</b>. +</p> + +<p> +So now we could write : +<b>Let the proposition P be 5 ≥ 2, we notice that P is always True, therefor its validity is 1</b> +</p> + +<p> +We also have the opposite of <b>P</b>, which is <b>not(P)</b> but for simplicity we use <b>P̅</b> (A P with a bar on top, in case it doesn’t load for you), now let’s go back to the previous example: +</p> + +<p> +<b>Since we know that the proposition P is true, we can conclude that P̅ is false. As P and P̅ can NOT be true at the same time. It’s like saying 5 is greater and also lesser than 2…doesn’t make sense, does it ?</b> +</p> + +<p> +Now let’s say we have two propositions, and we want to test the validity of their disjunction….. Okay what is this “disjunction” ? <b>Great Question Billy !!!</b> A disjunction is true if either propositions are true +</p> + +<p> +Ex: +<b>Let proposition P be “The webmaster is asleep”, and Q be “The reader loves pufferfishes”. The disjunction of these two propositions can have 4 different values showed in this Table of truth (such a badass name):</b> +</p> + +<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides"> + + +<colgroup> +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> +</colgroup> +<thead> +<tr> +<th scope="col" class="org-right">P</th> +<th scope="col" class="org-right">Q</th> +<th scope="col" class="org-right">Disjunction</th> +</tr> +</thead> +<tbody> +<tr> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +</tr> +</tbody> +</table> + +<p> +<i>What the hell is this ?</i> +The first colomn is equivalent to saying : “The webmaster is asleep AND The reader loves pufferfishes” +The second one means : “The webmaster is asleep AND The reader DOESN’T love pufferfishes (if you are in this case, then <b>I HATE YOU</b>)” +The third one… <i>zzzzzzz</i> +</p> + +<p> +You got the idea !!! +And since we are talking about a disjunction here, <b>one of the propositions</b> need to be true in order for this disjunction to be true. +</p> + +<p> +You may be wondering…. Crystal, can’t we write a disjunction in magical math symbols ? And to this I respond with a big <b>YES</b>. A disjunction is symbolized by a <b>∨</b> . So the disjunction between proposition <b>P & Q</b> can be written this way : <b>P ∨ Q</b> +</p> + +<p> +What if, we want to test whether or not two propositions are true AT THE SAME TIME ? Long story short, we can, it’s called a conjunction, same concept, as before, only this time the symbol is <b>P ∧ Q</b>, and is only true if <b>P</b> and <b>Q</b> are true. So we get a Table like this : +</p> + +<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides"> + + +<colgroup> +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> +</colgroup> +<thead> +<tr> +<th scope="col" class="org-right">P</th> +<th scope="col" class="org-right">Q</th> +<th scope="col" class="org-right">P ∨ Q</th> +<th scope="col" class="org-right">P ∧ Q</th> +</tr> +</thead> +<tbody> +<tr> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +</tr> +</tbody> +</table> + +<p> +<b>Always remember: 1 means true and 0 means false</b> +</p> + +<p> +There are two more basics to cover here before going to some properties, the first one is implication symbolized by the double arrow <b>⇒</b> +</p> + +<p> +Implication is kinda hard for my little brain to explain, so I will just say what it means: +</p> + +<p> +<b>If P implies Q, this means that either Q, or the opposite of P are correct</b> +</p> + +<p> +or in math terms +</p> + +<p> +<b>P ⇒ Q translates to P̅ ∨ Q</b> +Let’s illustrate : +</p> + +<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides"> + + +<colgroup> +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> +</colgroup> +<thead> +<tr> +<th scope="col" class="org-right">P</th> +<th scope="col" class="org-right">Q</th> +<th scope="col" class="org-right">P̅</th> +<th scope="col" class="org-right">Q̅</th> +<th scope="col" class="org-right">P ∨ Q</th> +<th scope="col" class="org-right">P ∧ Q</th> +<th scope="col" class="org-right">P ⇒ Q (P̅ ∨ Q)</th> +</tr> +</thead> +<tbody> +<tr> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +</tr> +</tbody> +</table> + +<p> +<b>If you look clearly, there is only one case where an implication is false. therefor you just need to find it, and blindly say that the others are correct. A rule of thumb is that: “A correct never implies a false”, or “If a 1 tries to imply a 0, the implication is a 0”</b> +</p> + +<p> +Aight, a last one and we are done!!! Equivalence, which is fairly easy, symbolized by a <b>⇔</b> symbol. +</p> + +<p> +A proposition is equivalent to another only when both of them have <b>the same value of truth</b> AKA: both true or both false. a little table will help demonstrate what i mean. +</p> + +<table border="2" cellspacing="0" cellpadding="6" rules="groups" frame="hsides"> + + +<colgroup> +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> + +<col class="org-right" /> +</colgroup> +<thead> +<tr> +<th scope="col" class="org-right">P</th> +<th scope="col" class="org-right">Q</th> +<th scope="col" class="org-right">P̅</th> +<th scope="col" class="org-right">Q̅</th> +<th scope="col" class="org-right">P ∨ Q</th> +<th scope="col" class="org-right">P ∧ Q</th> +<th scope="col" class="org-right">P ⇒ Q (P̅ ∨ Q)</th> +<th scope="col" class="org-right">P ⇔ Q</th> +</tr> +</thead> +<tbody> +<tr> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> + +<tr> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +</tr> + +<tr> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +<td class="org-right">0</td> +<td class="org-right">0</td> +<td class="org-right">1</td> +<td class="org-right">1</td> +</tr> +</tbody> +</table> + +<p> +<i>Note: P implying Q is equivalent to P̅ implying Q̅, or: (P ⇒ Q) ⇔ (P̅ ⇒ Q̅)</i> +</p> +</div> +<div id="outline-container-org29099d5" class="outline-3"> +<h3 id="org29099d5">Properties:</h3> +<div class="outline-text-3" id="text-org29099d5"> +</div> +<div id="outline-container-orgfeb62c8" class="outline-4"> +<h4 id="orgfeb62c8"><b>Absorption</b>:</h4> +<div class="outline-text-4" id="text-orgfeb62c8"> +<p> +(P ∨ P) ⇔ P +</p> + +<p> +(P ∧ P) ⇔ P +</p> +</div> +</div> +<div id="outline-container-org2fa2b1f" class="outline-4"> +<h4 id="org2fa2b1f"><b>Commutativity</b>:</h4> +<div class="outline-text-4" id="text-org2fa2b1f"> +<p> +(P ∧ Q) ⇔ (Q ∧ P) +</p> + +<p> +(P ∨ Q) ⇔ (Q ∨ P) +</p> +</div> +</div> +<div id="outline-container-orge30be26" class="outline-4"> +<h4 id="orge30be26"><b>Associativity</b>:</h4> +<div class="outline-text-4" id="text-orge30be26"> +<p> +P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R +</p> + +<p> +P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R +</p> +</div> +</div> +<div id="outline-container-orgd1de8d5" class="outline-4"> +<h4 id="orgd1de8d5"><b>Distributivity</b>:</h4> +<div class="outline-text-4" id="text-orgd1de8d5"> +<p> +P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R) +</p> + +<p> +P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R) +</p> +</div> +</div> +<div id="outline-container-org8c5c9f6" class="outline-4"> +<h4 id="org8c5c9f6"><b>Neutral element</b>:</h4> +<div class="outline-text-4" id="text-org8c5c9f6"> +<p> +<i>We define proposition <b>T</b> to be always <b>true</b> and <b>F</b> to be always <b>false</b></i> +</p> + +<p> +P ∧ T ⇔ P +</p> + +<p> +P ∨ F ⇔ P +</p> +</div> +</div> +<div id="outline-container-orgf102dde" class="outline-4"> +<h4 id="orgf102dde"><b>Negation of a conjunction & a disjunction</b>:</h4> +<div class="outline-text-4" id="text-orgf102dde"> +<p> +Now we won’t use bars here because my lazy ass doesn’t know how, so instead I will use not()!!! +</p> + +<p> +not(<b>P ∧ Q</b>) ⇔ P̅ ∨ Q̅ +</p> + +<p> +not(<b>P ∨ Q</b>) ⇔ P̅ ∧ Q̅ +</p> + +<p> +<b>A rule I really like to use here is: Break and Invert. Basically you break the bar into the three characters of the propositions, so you get not(P) not(∧ or ∨) <i>NOT AN ACTUAL MATH WRITING. DONT USE IT ANYWHERE ELSE OTHER THAN YOUR BRAIN</i> and not(Q)</b> +</p> +</div> +</div> +<div id="outline-container-org9534ece" class="outline-4"> +<h4 id="org9534ece"><b>Transitivity</b>:</h4> +<div class="outline-text-4" id="text-org9534ece"> +<p> +[(P ⇒ Q) (Q ⇒ R)] ⇔ P ⇒ R +</p> +</div> +</div> +<div id="outline-container-org90fa987" class="outline-4"> +<h4 id="org90fa987"><b>Contraposition</b>:</h4> +<div class="outline-text-4" id="text-org90fa987"> +<p> +(P ⇒ Q) ⇔ (Q̅ ⇒ P̅) +</p> +</div> +</div> +<div id="outline-container-orga2d0ece" class="outline-4"> +<h4 id="orga2d0ece">God only knows what this property is called:</h4> +<div class="outline-text-4" id="text-orga2d0ece"> +<p> +<i>If</i> +</p> + +<p> +(P ⇒ Q) is true +</p> + +<p> +and +</p> + +<p> +(Q̅ ⇒ Q) is true +</p> + +<p> +then +</p> + +<p> +Q is always true +</p> +</div> +</div> +</div> +<div id="outline-container-org35a43b7" class="outline-3"> +<h3 id="org35a43b7">Some exercices I found online :</h3> +<div class="outline-text-3" id="text-org35a43b7"> +</div> +<div id="outline-container-orgf619324" class="outline-4"> +<h4 id="orgf619324">USTHB 2022/2023 Section B :</h4> +<div class="outline-text-4" id="text-orgf619324"> +</div> +<ul class="org-ul"> +<li><a id="org1c47389"></a>Exercice 1: Démontrer les équivalences suivantes:<br /> +<div class="outline-text-5" id="text-org1c47389"> +<ol class="org-ol"> +<li><p> +(P ⇒ Q) ⇔ (Q̅ ⇒ P̅) +</p> + +<p> +Basically we are asked to prove contraposition, so here we have ( P ⇒ Q ) which is equivalent to P̅ ∨ Q <b>By definition : (P ⇒ Q) ⇔ (P̅ ∨ Q)</b> +</p></li> +</ol> + + +<p> +So we end up with : <b>(P̅ ∨ Q) ⇔ (Q̅ ⇒ P̅)</b>, now we just do the same with the second part of the contraposition. <b>(Q̅ ⇒ P̅) ⇔ (Q ∨ P̅)</b> therefor : +</p> + + +<p> +<b>(Q ∨ P̅) ⇔ (P̅ ∨ Q)</b>, which is true because of commutativity +</p> + +<ol class="org-ol"> +<li>not(P ⇒ Q) ⇔ P ∧ Q̅</li> +</ol> + + +<p> +Okaaaay so, let’s first get rid of the implication, because I don’t like it : <b>not(P̅ ∨ Q)</b> +</p> + + +<p> +Now that we got rid of it, we can negate the whole disjunction <b>not(P̅ ∨ Q) ⇔ (P ∧ Q̅)</b>. Which is the equivalence we needed to prove +</p> + +<ol class="org-ol"> +<li><p> +P ⇒ (Q ∧ R) ⇔ (P ⇒ Q) ∧ (P ⇒ R) +</p> + +<p> +One might be tempted to replace P with P̅ to get rid of the implication…sadly this isnt it. All we have to do here is resort to <b>Distributivity</b>, because yeah, we can distribute an implication across a {con/dis}junction +</p></li> + +<li><p> +P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R) +</p> + +<p> +Literally the same as above 🩷 +</p></li> +</ol> +</div> +</li> +<li><a id="org688fdcc"></a>Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:<br /> +<div class="outline-text-5" id="text-org688fdcc"> +<ol class="org-ol"> +<li><p> +∀x ∈ ℝ ,∃y ∈ ℝ*+, tels que e^x = y +</p> + +<p> +For each x from the set of Real numbers, there exists a number y from the set of non-zero positive Real numbers that satisfies the equation : e^x = y +</p></li> +</ol> + + +<p> +“The function f(x)=e^x is always positive and non-null”, the very definition of an exponential function !!!! +</p> + + +<p> +<b>So the proposition is true</b> +</p> + + +<ol class="org-ol"> +<li>∃x ∈ ℝ, tels que x^2 < x < x^3</li> +</ol> + + +<p> +We just need to find a value that satisifies this condition…thankfully its easy…. +</p> + +<p> +x² < x < x³ , we divide the three terms by x so we get : +</p> + + +<p> +x < 1 < x² , or : +</p> + + +<p> +<b>x < 1</b> ; <b>1 < x²</b> ⇔ <b>x < 1</b> ; <b>1 < x</b> <i>We square root both sides</i> +</p> + + +<p> +We end up with a contradiction, therefor its wrong +</p> + + +<ol class="org-ol"> +<li>∀x ∈ ℝ, ∃y ∈ ℝ tels que y = 3x - 8</li> +</ol> + + +<p> +I dont really understand this one, so let me translate it “For any value of x from the set of Real numbers, 3x - 8 is a Real number”…. i mean….yeah, we are substracting a Real number from an other real number… +</p> + +<p> +<b>Since substraction is an Internal composition law in ℝ, therefor all results of a substraction between two Real numbers is…Real</b> +</p> + +<ol class="org-ol"> +<li><p> +∃x ∈ ℕ, ∀y ∈ ℕ, x > y ⇒ x + y < 8 +</p> + +<p> +“There exists a number x from the set of Natural numbers such as for all values of y from the set of Natural numbers, x > y implies x + y < 8” +</p></li> +</ol> + + +<p> +Let’s get rid of the implication : +</p> + +<p> +∃x ∈ ℕ, ∀y ∈ ℕ, (y > x) ∨ (x + y < 8) <i>There exists a number x from the set of Natural numbers such as for all values of y from the set of Natural numbers y > x OR x + y < 8</i> +</p> + +<p> +This proposition is true, because there exists a value of x that satisfies this condition, it’s <b>all numbers under 8</b> let’s take 3 as an example: +</p> + + +<p> +<b>x = 3 , if y > 3 then the first condition is true ; if y < 3 then the second one is true</b> +</p> + + +<p> +Meaning that the two propositions CAN NOT BE WRONG TOGETHER, either one is wrong, or the other +</p> + + +<p> +y > x +</p> + + +<p> +<b>y - x > 0</b> +</p> + + +<p> +y + x < 8 +</p> + + +<p> +<b>y < 8 - x</b> <i>This one is always true for all values of x below 8, since we are working in the set ℕ</i> +</p> + + +<ol class="org-ol"> +<li><p> +∀x ∈ ℝ, x² ≥ 1 ⇔ x ≥ 1 +</p> + +<p> +….This is getting stupid. of course it’s true it’s part of the definition of the power of 2 +</p></li> +</ol> +</div> +</li> +</ul> +</div> +</div> +</div> +<div id="outline-container-orgac834f2" class="outline-2"> +<h2 id="orgac834f2">2éme cours <i>Oct 2</i></h2> +<div class="outline-text-2" id="text-orgac834f2"> +</div> +<div id="outline-container-org42a8fad" class="outline-3"> +<h3 id="org42a8fad">Quantifiers</h3> +<div class="outline-text-3" id="text-org42a8fad"> +<p> +A propriety P can depend on a parameter x +</p> + + +<p> +∀ is the universal quantifier which stands for “For any value of…” +</p> + + +<p> +∃ is the existential quantifier which stands for “There exists at least one…” +</p> +</div> +<ul class="org-ul"> +<li><a id="org48c4773"></a>Example<br /> +<div class="outline-text-6" id="text-org48c4773"> +<p> +P(x) : x+1≥0 +</p> + +<p> +P(X) is True or False depending on the values of x +</p> +</div> +</li> +</ul> +<div id="outline-container-org923b5fe" class="outline-4"> +<h4 id="org923b5fe">Proprieties</h4> +<div class="outline-text-4" id="text-org923b5fe"> +</div> +<ul class="org-ul"> +<li><a id="orgaf93dd2"></a>Propriety Number 1:<br /> +<div class="outline-text-5" id="text-orgaf93dd2"> +<p> +The negation of the universal quantifier is the existential quantifier, and vice-versa : +</p> + +<ul class="org-ul"> +<li>not(∀x ∈ E , P(x)) ⇔ ∃ x ∈ E, not(P(x))</li> +<li>not(∃x ∈ E , P(x)) ⇔ ∀ x ∈ E, not(P(x))</li> +</ul> +</div> +<ul class="org-ul"> +<li><a id="orgf4f038f"></a>Example:<br /> +<div class="outline-text-6" id="text-orgf4f038f"> +<p> +∀ x ≥ 1 x² > 5 ⇔ ∃ x ≥ 1 x² < 5 +</p> +</div> +</li> +</ul> +</li> +<li><a id="org2c00dae"></a>Propriety Number 2:<br /> +<div class="outline-text-5" id="text-org2c00dae"> +<p> +<b>∀x ∈ E, [P(x) ∧ Q(x)] ⇔ [∀ x ∈ E, P(x)] ∧ [∀ x ∈ E, Q(x)]</b> +</p> + + +<p> +The propriety “For any value of x from a set E , P(x) and Q(x)” is equivalent to “For any value of x from a set E, P(x) AND for any value of x from a set E, Q(x)” +</p> +</div> +<ul class="org-ul"> +<li><a id="orga95ad05"></a>Example :<br /> +<div class="outline-text-6" id="text-orga95ad05"> +<p> +P(x) : sqrt(x) > 0 ; Q(x) : x ≥ 1 +</p> + + +<p> +∀x ∈ ℝ*+, [sqrt(x) > 0 , x ≥ 1] ⇔ [∀x ∈ R*+, sqrt(x) > 0] ∧ [∀x ∈ R*+, x ≥ 1] +</p> + + +<p> +<b>Which is true</b> +</p> +</div> +</li> +</ul> +</li> +<li><a id="orgfe5ddf2"></a>Propriety Number 3:<br /> +<div class="outline-text-5" id="text-orgfe5ddf2"> +<p> +<b>∃ x ∈ E, [P(x) ∧ Q(x)] <i>⇒</i> [∃ x ∈ E, P(x)] ∧ [∃ x ∈ E, Q(x)]</b> +</p> + + +<p> +<i>Here its an implication and not an equivalence</i> +</p> +</div> +<ul class="org-ul"> +<li><a id="orgca7bea7"></a>Example of why it’s NOT an equivalence :<br /> +<div class="outline-text-6" id="text-orgca7bea7"> +<p> +P(x) : x > 5 ; Q(x) : x < 5 +</p> + + +<p> +Of course there is no value of x such as its inferior and superior to 5 at the same time, so obviously the proposition is false. However, the two propositions separated are correct on their own, because there is a value of x such as its superior to 5, and there is also a value of x such as its inferior to 5. This is why it’s an implication and NOT AN EQUIVALENCE!!! +</p> +</div> +</li> +</ul> +</li> +<li><a id="orgf4ecdc0"></a>Propriety Number 4:<br /> +<div class="outline-text-5" id="text-orgf4ecdc0"> +<p> +<b>[∀ x ∈ E, P(x)] ∨ [∀ x ∈ E, Q(x)] <i>⇒</i> ∀x ∈ E, [P(x) ∨ Q(x)]</b> +</p> + + +<p> +<i>Same here, implication and NOT en equivalence</i> +</p> +</div> +</li> +</ul> +</div> +</div> +<div id="outline-container-org6421557" class="outline-3"> +<h3 id="org6421557">Multi-parameter proprieties :</h3> +<div class="outline-text-3" id="text-org6421557"> +<p> +A propriety P can depend on two or more parameters, for convenience we call them x,y,z…etc +</p> +</div> +<ul class="org-ul"> +<li><a id="org314cff3"></a>Example :<br /> +<div class="outline-text-6" id="text-org314cff3"> +<p> +P(x,y): x+y > 0 +</p> + + +<p> +P(0,1) is a True proposition +</p> + + +<p> +P(-2,-1) is a False one +</p> +</div> +</li> +<li><a id="orga9b6089"></a>WARNING :<br /> +<div class="outline-text-6" id="text-orga9b6089"> +<p> +∀x ∈ E, ∃y ∈ F , P(x,y) +</p> + + +<p> +∃y ∈ F, ∀x ∈ E , P(x,y) +</p> + + +<p> +Are different because in the first one y depends on x, while in the second one, it doesn’t +</p> +</div> +<ul class="org-ul"> +<li><a id="org3c5e5a8"></a>Example :<br /> +<div class="outline-text-7" id="text-org3c5e5a8"> +<p> +∀ x ∈ ℕ , ∃ y ∈ ℕ y > x -–— True +</p> + + +<p> +∃ y ∈ ℕ , ∀ x ∈ ℕ y > x -–— False +</p> +</div> +</li> +</ul> +</li> +</ul> +<li><a id="org2f3208e"></a>Proprieties :<br /> +<div class="outline-text-5" id="text-org2f3208e"> +<ol class="org-ol"> +<li>not(∀x ∈ E ,∃y ∈ F P(x,y)) ⇔ ∃x ∈ E, ∀y ∈ F not(P(x,y))</li> +<li>not(∃x ∈ E ,∀y ∈ F P(x,y)) ⇔ ∀x ∈ E, ∃y ∈ F not(P(x,y))</li> +</ol> +</div> +</li> +</ul> +</div> +<div id="outline-container-orgc25dd7a" class="outline-3"> +<h3 id="orgc25dd7a">Methods of mathematical reasoning :</h3> +<div class="outline-text-3" id="text-orgc25dd7a"> +</div> +<div id="outline-container-orgf843851" class="outline-4"> +<h4 id="orgf843851">Direct reasoning :</h4> +<div class="outline-text-4" id="text-orgf843851"> +<p> +To show that an implication P ⇒ Q is true, we suppose that P is true and we show that Q is true +</p> +</div> +<ul class="org-ul"> +<li><a id="org6dd2136"></a>Example:<br /> +<div class="outline-text-5" id="text-org6dd2136"> +<p> +Let a,b be two Real numbers, we have to prove that <b>a² + b² = 1 ⇒ |a + b| ≤ 2</b> +</p> + + +<p> +We suppose that a²+b² = 1 and we prove that |a + b| ≤ 2 +</p> + + +<p> +a²+b²=1 ⇒ b² = 1 - a² ; a² = 1 - b² +</p> + + +<p> +a²+b²=1 ⇒ 1 - a² ≥ 0 ; 1 - b² ≥ 0 +</p> + + +<p> +a²+b²=1 ⇒ a² ≤ 1 ; b² ≤ 1 +</p> + + +<p> +a²+b²=1 ⇒ -1 ≤ a ≤ 1 ; -1 ≤ b ≤ 1 +</p> + + +<p> +a²+b²=1 ⇒ -2 ≤ a + b ≤ 2 +</p> + + +<p> +a²+b²=1 ⇒ |a + b| ≤ 2 <b>Which is what we wanted to prove, therefor the implication is correct</b> +</p> +</div> +</li> +</ul> +</div> +<div id="outline-container-orga59c0ad" class="outline-4"> +<h4 id="orga59c0ad">Reasoning by the Absurd:</h4> +<div class="outline-text-4" id="text-orga59c0ad"> +<p> +To prove that a proposition is True, we suppose that it’s False and we must come to a contradiction +</p> + + +<p> +And to prove that an implication P ⇒ Q is true using the reasoning by the absurd, we suppose that P ∧ not(Q) is true, and then we come to a contradiction as well +</p> +</div> +<ul class="org-ul"> +<li><a id="orgf8b9f83"></a>Example:<br /> +<div class="outline-text-5" id="text-orgf8b9f83"> +<p> +Prove that this proposition is correct using the reasoning by the absurd : ∀x ∈ ℝ* , sqrt(1+x²) ≠ 1 + x²/2 +</p> + + +<p> +We assume that ∃ x ℝ* , sqrt(1+x²) = 1 + x²/2 +</p> + + +<p> +sqrt(1+x²) = 1 + x²/2 ; 1 + x² = (1+x²/2)² ; 1 + x² = 1 + x^4/4 + x² ; x^(4)/4 = 0 … Which contradicts with our proposition, since x = 4 and we are working on the ℝ* set +</p> +</div> +</li> +</ul> +</div> +<div id="outline-container-orgcc285c2" class="outline-4"> +<h4 id="orgcc285c2">Reasoning by contraposition:</h4> +<div class="outline-text-4" id="text-orgcc285c2"> +<p> +If an implication P ⇒ Q is too hard to prove, we just have to prove not(Q) ⇒ not(P) is true !!! or in other words that both not(P) and not(Q) are true +</p> +</div> +</div> +<div id="outline-container-org2e67808" class="outline-4"> +<h4 id="org2e67808">Reasoning by counter example:</h4> +<div class="outline-text-4" id="text-org2e67808"> +<p> +To prove that a proposition ∀x ∈ E, P(x) is false, all we have to do is find a single value of x from E such as not(P(x)) is true +</p> +</div> +</div> +</div> +</div> +<div id="outline-container-org7440601" class="outline-2"> +<h2 id="org7440601">3eme Cours : <i>Oct 9</i></h2> +<div class="outline-text-2" id="text-org7440601"> +</div> +<div id="outline-container-org70aa2db" class="outline-4"> +<h4 id="org70aa2db">Reasoning by recurrence :</h4> +<div class="outline-text-4" id="text-org70aa2db"> +<p> +P is a propriety dependent of <b>n ∈ ℕ</b>. If for n0 ∈ ℕ P(n0) is true, and if for n ≥ n0 (P(n) ⇒ P(n+1)) is true. Then P(n) is true for n ≥ n0 +</p> +</div> +<ul class="org-ul"> +<li><a id="org52e5298"></a>Example:<br /> +<div class="outline-text-5" id="text-org52e5298"> +<p> +Let’s prove that ∀ n ≥ 1 , (n,k=1)Σk = [n(n+1)]/2 +</p> + + +<p> +P(n) : (n,k=1)Σk = [n(n+1)]/2 +</p> + + + +<p> +<b>Pour n = 1:</b> (1,k=1)Σk = 1 ; [n(n+1)]/2 = 1 . <b>So P(1) is true</b> +</p> + + + +<p> +For n ≥ 1. We assume that P(n) is true, OR : <b>(n, k=1)Σk = n(n+1)/2</b>. We now have to prove that P(n+1) is true, Or : <b>(n+1, k=1)Σk = (n+1)(n+2)/2</b> +</p> + + +<p> +(n+1, k=1)Σk = 1 + 2 + …. + n + (n+1) ; (n+1, k=1)Σk = (n, k=1)Σk + (n+1) ; = n(n+1)/2 + (n+1) ; = [n(n+1) + 2(n+1)]/2 ; = <b>[(n+2)(n+1)]/2</b> <i>WHICH IS WHAT WE NEEDED TO FIND</i> +</p> + + +<p> +<b>Conclusion: ∀n ≥ 1 , (n,k=1)Σk = n(n+1)/2</b> +</p> +</div> +</li> +</ul> +</div> +</div> +<div id="outline-container-orga6a518d" class="outline-2"> +<h2 id="orga6a518d">4eme Cours : Chapitre 2 : Sets and Operations</h2> +<div class="outline-text-2" id="text-orga6a518d"> +</div> +<div id="outline-container-org31e3615" class="outline-3"> +<h3 id="org31e3615">Definition of a set :</h3> +<div class="outline-text-3" id="text-org31e3615"> +<p> +A set is a collection of objects that share the sane propriety +</p> +</div> +</div> +<div id="outline-container-orgfa9bfd1" class="outline-3"> +<h3 id="orgfa9bfd1">Belonging, inclusion, and equality :</h3> +<div class="outline-text-3" id="text-orgfa9bfd1"> +<ol class="org-ol"> +<li>Let E be a set. If x is an element of E, we say that x belongs to E we write <b>x ∈ E</b>, and if it doesn’t, we write <b>x ∉ E</b></li> +<li>A set E is included in a set F if all elements of E are elements of F and we write <b>E ⊂ F ⇔ (∀x , x ∈ E ⇒ x ∈ F)</b>. We say that E is a subset of F, or a part of F. The negation of this propriety is : <b>E ⊄ F ⇔ ∃x , x ∈ E and x ⊄ F</b></li> +<li>E and F are equal if E is included in F and F is included in E, and we write <b>E = F ⇔ (E ⊂ F) et (F ⊂ E)</b></li> +<li>The empty set (symbolized by ∅) is a set without elements, and is included in all sets (by convention) : <b>∅ ⊂ E</b></li> +</ol> +</div> +</div> +<div id="outline-container-org2a19707" class="outline-3"> +<h3 id="org2a19707">Intersections and reunions :</h3> +<div class="outline-text-3" id="text-org2a19707"> +</div> +<div id="outline-container-org6a5f566" class="outline-4"> +<h4 id="org6a5f566">Intersection:</h4> +<div class="outline-text-4" id="text-org6a5f566"> +<p> +E ∩ F = {x / x ∈ E AND x ∈ F} ; x ∈ E ∩ F ⇔ x ∈ F AND x ∈ F +</p> + + +<p> +x ∉ E ∩ F ⇔ x ∉ E OR x ∉ F +</p> +</div> +</div> +<div id="outline-container-org9bc9aeb" class="outline-4"> +<h4 id="org9bc9aeb">Union:</h4> +<div class="outline-text-4" id="text-org9bc9aeb"> +<p> +E ∪ F = {x / x ∈ E OR x ∈ F} ; x ∈ E ∪ F ⇔ x ∈ F OR x ∈ F +</p> + + +<p> +x ∉ E ∪ F ⇔ x ∉ E AND x ∉ F +</p> +</div> +</div> +<div id="outline-container-org9a7f719" class="outline-4"> +<h4 id="org9a7f719">Difference between two sets:</h4> +<div class="outline-text-4" id="text-org9a7f719"> +<p> +E\F(Which is also written as : E - F) = {x / x ∈ E and x ∉ F} +</p> +</div> +</div> +<div id="outline-container-org5f5c721" class="outline-4"> +<h4 id="org5f5c721">Complimentary set:</h4> +<div class="outline-text-4" id="text-org5f5c721"> +<p> +If F ⊂ E. E - F is the complimentary of F in E. +</p> + + +<p> +FCE = {x /x ∈ E AND x ∉ F} <b>ONLY WHEN F IS A SUBSET OF E</b> +</p> +</div> +</div> +<div id="outline-container-orga285d1d" class="outline-4"> +<h4 id="orga285d1d">Symentrical difference</h4> +<div class="outline-text-4" id="text-orga285d1d"> +<p> +E Δ F = (E - F) ∪ (F - E) ; = (E ∪ F) - (E ∩ F) +</p> +</div> +</div> +</div> +<div id="outline-container-orgc12e73b" class="outline-3"> +<h3 id="orgc12e73b">Proprieties :</h3> +<div class="outline-text-3" id="text-orgc12e73b"> +<p> +Let E,F and G be 3 sets. We have : +</p> +</div> +<div id="outline-container-org31d8697" class="outline-4"> +<h4 id="org31d8697">Commutativity:</h4> +<div class="outline-text-4" id="text-org31d8697"> +<p> +E ∩ F = F ∩ E +E ∪ F = F ∪ E +</p> +</div> +</div> +<div id="outline-container-org7080d99" class="outline-4"> +<h4 id="org7080d99">Associativity:</h4> +<div class="outline-text-4" id="text-org7080d99"> +<p> +E ∩ (F ∩ G) = (E ∩ F) ∩ G +E ∪ (F ∪ G) = (E ∪ F) ∪ G +</p> +</div> +</div> +<div id="outline-container-org13da04d" class="outline-4"> +<h4 id="org13da04d">Distributivity:</h4> +<div class="outline-text-4" id="text-org13da04d"> +<p> +E ∩ (F ∪ G) = (E ∩ F) ∪ (E ∩ G) +E ∪ (F ∩ G) = (E ∪ F) ∩ (E ∪ G) +</p> +</div> +</div> +<div id="outline-container-orgaa33b71" class="outline-4"> +<h4 id="orgaa33b71">Lois de Morgan:</h4> +<div class="outline-text-4" id="text-orgaa33b71"> +<p> +If E ⊂ G and F ⊂ G ; +</p> + +<p> +(E ∩ F)CG = ECG ∪ FCG ; (E ∪ F)CG = ECG ∩ FCG +</p> +</div> +</div> +<div id="outline-container-org7e7db42" class="outline-4"> +<h4 id="org7e7db42">An other one:</h4> +<div class="outline-text-4" id="text-org7e7db42"> +<p> +E - (F ∩ G) = (E-F) ∪ (E-G) ; E - (F ∪ G) = (E-F) ∩ (E-G) +</p> +</div> +</div> +<div id="outline-container-orgd02bd7f" class="outline-4"> +<h4 id="orgd02bd7f">An other one:</h4> +<div class="outline-text-4" id="text-orgd02bd7f"> +<p> +E ∩ ∅ = ∅ ; E ∪ ∅ = E +</p> +</div> +</div> +<div id="outline-container-org99eb39a" class="outline-4"> +<h4 id="org99eb39a">And an other one:</h4> +<div class="outline-text-4" id="text-org99eb39a"> +<p> +E ∩ (F Δ G) = (E ∩ F) Δ (E ∩ G) +</p> +</div> +</div> +<div id="outline-container-org3e9b2ef" class="outline-4"> +<h4 id="org3e9b2ef">And the last one:</h4> +<div class="outline-text-4" id="text-org3e9b2ef"> +<p> +E Δ ∅ = E ; E Δ E = ∅ +</p> +</div> +</div> +</div> +</div> +</div> +<div id="postamble" class="status"> +<p class="author">Author: Crystal</p> +<p class="date">Created: 2023-10-11 Wed 19:04</p> +</div> +</body> +</html> \ No newline at end of file |