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<title>Algebra 1</title>
<meta name="author" content="Crystal" />
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<div id="content" class="content">
<h1 class="title">Algebra 1</h1>
<div id="outline-container-org76eaba1" class="outline-2">
<h2 id="org76eaba1">Contenu de la Matiére</h2>
<div class="outline-text-2" id="text-org76eaba1">
</div>
<div id="outline-container-org2a89be2" class="outline-3">
<h3 id="org2a89be2">Rappels et compléments (11H)</h3>
<div class="outline-text-3" id="text-org2a89be2">
<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-orgfcd7f3f" class="outline-3">
<h3 id="orgfcd7f3f">Structures Algébriques (11H)</h3>
<div class="outline-text-3" id="text-orgfcd7f3f">
<ul class="org-ul">
<li>Groupes et morphisme de groupes</li>
<li>Anneaux et morphisme d’anneaux</li>
<li>Les corps</li>
</ul>
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</div>
<div id="outline-container-org03cbf05" class="outline-3">
<h3 id="org03cbf05">Polynômes et fractions rationnelles</h3>
<div class="outline-text-3" id="text-org03cbf05">
<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-org7db21e0" class="outline-2">
<h2 id="org7db21e0">Premier cours : Logique mathématique et méthodes du raisonnement mathématique <i>Sep 25</i> :</h2>
<div class="outline-text-2" id="text-org7db21e0">
<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" />
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<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-org5604636" class="outline-3">
<h3 id="org5604636">Properties:</h3>
<div class="outline-text-3" id="text-org5604636">
</div>
<div id="outline-container-orgfffc23d" class="outline-4">
<h4 id="orgfffc23d"><b>Absorption</b>:</h4>
<div class="outline-text-4" id="text-orgfffc23d">
<p>
(P ∨ P) ⇔ P
</p>
<p>
(P ∧ P) ⇔ P
</p>
</div>
</div>
<div id="outline-container-orgd43aeb7" class="outline-4">
<h4 id="orgd43aeb7"><b>Commutativity</b>:</h4>
<div class="outline-text-4" id="text-orgd43aeb7">
<p>
(P ∧ Q) ⇔ (Q ∧ P)
</p>
<p>
(P ∨ Q) ⇔ (Q ∨ P)
</p>
</div>
</div>
<div id="outline-container-org9e5868e" class="outline-4">
<h4 id="org9e5868e"><b>Associativity</b>:</h4>
<div class="outline-text-4" id="text-org9e5868e">
<p>
P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R
</p>
<p>
P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R
</p>
</div>
</div>
<div id="outline-container-orga530d13" class="outline-4">
<h4 id="orga530d13"><b>Distributivity</b>:</h4>
<div class="outline-text-4" id="text-orga530d13">
<p>
P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)
</p>
<p>
P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R)
</p>
</div>
</div>
<div id="outline-container-org7d55048" class="outline-4">
<h4 id="org7d55048"><b>Neutral element</b>:</h4>
<div class="outline-text-4" id="text-org7d55048">
<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-org7422610" class="outline-4">
<h4 id="org7422610"><b>Negation of a conjunction & a disjunction</b>:</h4>
<div class="outline-text-4" id="text-org7422610">
<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-org4145760" class="outline-4">
<h4 id="org4145760"><b>Transitivity</b>:</h4>
<div class="outline-text-4" id="text-org4145760">
<p>
[(P ⇒ Q) AND (Q ⇒ R)] ⇔ P ⇒ R
</p>
</div>
</div>
<div id="outline-container-org245af1d" class="outline-4">
<h4 id="org245af1d"><b>Contraposition</b>:</h4>
<div class="outline-text-4" id="text-org245af1d">
<p>
(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)
</p>
</div>
</div>
<div id="outline-container-orga47b617" class="outline-4">
<h4 id="orga47b617">God only knows what this property is called:</h4>
<div class="outline-text-4" id="text-orga47b617">
<p>
<i>If</i>
</p>
<p>
(P ⇒ Q) is true
</p>
<p>
and
</p>
<p>
(P̅ ⇒ Q) is true
</p>
<p>
then
</p>
<p>
Q is always true
</p>
</div>
</div>
</div>
<div id="outline-container-org3cfbd88" class="outline-3">
<h3 id="org3cfbd88">Some exercices I found online :</h3>
<div class="outline-text-3" id="text-org3cfbd88">
</div>
<div id="outline-container-orge60008b" class="outline-4">
<h4 id="orge60008b">USTHB 2022/2023 Section B :</h4>
<div class="outline-text-4" id="text-orge60008b">
</div>
<ul class="org-ul">
<li><a id="orgd7d6ce9"></a>Exercice 1: Démontrer les équivalences suivantes:<br />
<div class="outline-text-5" id="text-orgd7d6ce9">
<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="orgd64e49a"></a>Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:<br />
<div class="outline-text-5" id="text-orgd64e49a">
<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-org980d3be" class="outline-2">
<h2 id="org980d3be">2éme cours <i>Oct 2</i></h2>
<div class="outline-text-2" id="text-org980d3be">
</div>
<div id="outline-container-org22b148b" class="outline-3">
<h3 id="org22b148b">Quantifiers</h3>
<div class="outline-text-3" id="text-org22b148b">
<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="org4afa1df"></a>Example<br />
<div class="outline-text-6" id="text-org4afa1df">
<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-org8b437f3" class="outline-4">
<h4 id="org8b437f3">Proprieties</h4>
<div class="outline-text-4" id="text-org8b437f3">
</div>
<ul class="org-ul">
<li><a id="org6d0c06f"></a>Propriety Number 1:<br />
<div class="outline-text-5" id="text-org6d0c06f">
<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="orgd242e81"></a>Example:<br />
<div class="outline-text-6" id="text-orgd242e81">
<p>
∀ x ≥ 1 x² > 5 ⇔ ∃ x ≥ 1 x² < 5
</p>
</div>
</li>
</ul>
</li>
<li><a id="orgd7c2c1d"></a>Propriety Number 2:<br />
<div class="outline-text-5" id="text-orgd7c2c1d">
<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="org5cb6921"></a>Example :<br />
<div class="outline-text-6" id="text-org5cb6921">
<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="orgd56cb14"></a>Propriety Number 3:<br />
<div class="outline-text-5" id="text-orgd56cb14">
<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="org52c3098"></a>Example of why it’s NOT an equivalence :<br />
<div class="outline-text-6" id="text-org52c3098">
<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="org9439534"></a>Propriety Number 4:<br />
<div class="outline-text-5" id="text-org9439534">
<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-orgcb2ff75" class="outline-3">
<h3 id="orgcb2ff75">Multi-parameter proprieties :</h3>
<div class="outline-text-3" id="text-orgcb2ff75">
<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="org309a152"></a>Example :<br />
<div class="outline-text-6" id="text-org309a152">
<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="orgfbf5cee"></a>WARNING :<br />
<div class="outline-text-6" id="text-orgfbf5cee">
<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="org21332e2"></a>Example :<br />
<div class="outline-text-7" id="text-org21332e2">
<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="org2fad1a6"></a>Proprieties :<br />
<div class="outline-text-5" id="text-org2fad1a6">
<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-org405d91a" class="outline-3">
<h3 id="org405d91a">Methods of mathematical reasoning :</h3>
<div class="outline-text-3" id="text-org405d91a">
</div>
<div id="outline-container-org0e2120a" class="outline-4">
<h4 id="org0e2120a">Direct reasoning :</h4>
<div class="outline-text-4" id="text-org0e2120a">
<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="orge655791"></a>Example:<br />
<div class="outline-text-5" id="text-orge655791">
<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-org3318c18" class="outline-4">
<h4 id="org3318c18">Reasoning by the Absurd:</h4>
<div class="outline-text-4" id="text-org3318c18">
<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="org6217ba8"></a>Example:<br />
<div class="outline-text-5" id="text-org6217ba8">
<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-orgdca4b33" class="outline-4">
<h4 id="orgdca4b33">Reasoning by contraposition:</h4>
<div class="outline-text-4" id="text-orgdca4b33">
<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-org45373bc" class="outline-4">
<h4 id="org45373bc">Reasoning by counter example:</h4>
<div class="outline-text-4" id="text-org45373bc">
<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-org1ab01d8" class="outline-2">
<h2 id="org1ab01d8">3eme Cours : <i>Oct 9</i></h2>
<div class="outline-text-2" id="text-org1ab01d8">
</div>
<div id="outline-container-orgc3cdd55" class="outline-4">
<h4 id="orgc3cdd55">Reasoning by recurrence :</h4>
<div class="outline-text-4" id="text-orgc3cdd55">
<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="org74698a3"></a>Example:<br />
<div class="outline-text-5" id="text-org74698a3">
<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-org62bfe2a" class="outline-2">
<h2 id="org62bfe2a">4eme Cours : Chapitre 2 : Sets and Operations</h2>
<div class="outline-text-2" id="text-org62bfe2a">
</div>
<div id="outline-container-org5c29bea" class="outline-3">
<h3 id="org5c29bea">Definition of a set :</h3>
<div class="outline-text-3" id="text-org5c29bea">
<p>
A set is a collection of objects that share the sane propriety
</p>
</div>
</div>
<div id="outline-container-org7f4934f" class="outline-3">
<h3 id="org7f4934f">Belonging, inclusion, and equality :</h3>
<div class="outline-text-3" id="text-org7f4934f">
<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-orgd439312" class="outline-3">
<h3 id="orgd439312">Intersections and reunions :</h3>
<div class="outline-text-3" id="text-orgd439312">
</div>
<div id="outline-container-org2eaf0a6" class="outline-4">
<h4 id="org2eaf0a6">Intersection:</h4>
<div class="outline-text-4" id="text-org2eaf0a6">
<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-org8bfbedf" class="outline-4">
<h4 id="org8bfbedf">Union:</h4>
<div class="outline-text-4" id="text-org8bfbedf">
<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-orgf5d7c25" class="outline-4">
<h4 id="orgf5d7c25">Difference between two sets:</h4>
<div class="outline-text-4" id="text-orgf5d7c25">
<p>
E\F(Which is also written as : E - F) = {x / x ∈ E and x ∉ F}
</p>
</div>
</div>
<div id="outline-container-org16f26ee" class="outline-4">
<h4 id="org16f26ee">Complimentary set:</h4>
<div class="outline-text-4" id="text-org16f26ee">
<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-org67da9c0" class="outline-4">
<h4 id="org67da9c0">Symentrical difference</h4>
<div class="outline-text-4" id="text-org67da9c0">
<p>
E Δ F = (E - F) ∪ (F - E) ; = (E ∪ F) - (E ∩ F)
</p>
</div>
</div>
</div>
<div id="outline-container-org10858f6" class="outline-3">
<h3 id="org10858f6">Proprieties :</h3>
<div class="outline-text-3" id="text-org10858f6">
<p>
Let E,F and G be 3 sets. We have :
</p>
</div>
<div id="outline-container-orgdeeff37" class="outline-4">
<h4 id="orgdeeff37">Commutativity:</h4>
<div class="outline-text-4" id="text-orgdeeff37">
<p>
E ∩ F = F ∩ E
E ∪ F = F ∪ E
</p>
</div>
</div>
<div id="outline-container-org6228f00" class="outline-4">
<h4 id="org6228f00">Associativity:</h4>
<div class="outline-text-4" id="text-org6228f00">
<p>
E ∩ (F ∩ G) = (E ∩ F) ∩ G
E ∪ (F ∪ G) = (E ∪ F) ∪ G
</p>
</div>
</div>
<div id="outline-container-org2523e0e" class="outline-4">
<h4 id="org2523e0e">Distributivity:</h4>
<div class="outline-text-4" id="text-org2523e0e">
<p>
E ∩ (F ∪ G) = (E ∩ F) ∪ (E ∩ G)
E ∪ (F ∩ G) = (E ∪ F) ∩ (E ∪ G)
</p>
</div>
</div>
<div id="outline-container-orgeb0c0a3" class="outline-4">
<h4 id="orgeb0c0a3">Lois de Morgan:</h4>
<div class="outline-text-4" id="text-orgeb0c0a3">
<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-orge638501" class="outline-4">
<h4 id="orge638501">An other one:</h4>
<div class="outline-text-4" id="text-orge638501">
<p>
E - (F ∩ G) = (E-F) ∪ (E-G) ; E - (F ∪ G) = (E-F) ∩ (E-G)
</p>
</div>
</div>
<div id="outline-container-orgfe0b562" class="outline-4">
<h4 id="orgfe0b562">An other one:</h4>
<div class="outline-text-4" id="text-orgfe0b562">
<p>
E ∩ ∅ = ∅ ; E ∪ ∅ = E
</p>
</div>
</div>
<div id="outline-container-org48afea2" class="outline-4">
<h4 id="org48afea2">And an other one:</h4>
<div class="outline-text-4" id="text-org48afea2">
<p>
E ∩ (F Δ G) = (E ∩ F) Δ (E ∩ G)
</p>
</div>
</div>
<div id="outline-container-org1138be8" class="outline-4">
<h4 id="org1138be8">And the last one:</h4>
<div class="outline-text-4" id="text-org1138be8">
<p>
E Δ ∅ = E ; E Δ E = ∅
</p>
</div>
</div>
</div>
</div>
<div id="outline-container-orgf188863" class="outline-2">
<h2 id="orgf188863">5eme cours: L’ensemble des parties d’un ensemble <i>Oct 16</i></h2>
<div class="outline-text-2" id="text-orgf188863">
<p>
Let E be a set. We define P(E) as the set of all parts of E : <b>P(E) = {X/X ⊂ E}</b>
</p>
</div>
<div id="outline-container-org6cfe0d7" class="outline-4">
<h4 id="org6cfe0d7">Notes :</h4>
<div class="outline-text-4" id="text-org6cfe0d7">
<p>
∅ ∈ P(E) ; E ∈ P(E)
</p>
<p>
cardinal E = n <i>The number of terms in E</i> , cardinal P(E) = 2^n <i>The number of all parts of E</i>
</p>
</div>
</div>
<div id="outline-container-orgd0b341d" class="outline-4">
<h4 id="orgd0b341d">Examples :</h4>
<div class="outline-text-4" id="text-orgd0b341d">
<p>
E = {a,b,c} // P(E)={∅, {a}, {b}, {c}, {a,b}, {b,c}, {a,c}, {a,b,c}}
</p>
</div>
</div>
<div id="outline-container-org7ec7b74" class="outline-3">
<h3 id="org7ec7b74">Partition of a set :</h3>
<div class="outline-text-3" id="text-org7ec7b74">
<p>
We say that <b>A</b> is a partition of E if:
</p>
<ol class="org-ol">
<li>∀ x ∈ A , x ≠ 0</li>
<li>All the elements of <b>A</b> are two by two disjoint. Or in other terms, there should not be two elements that intersects with each other.</li>
<li>The reunion of all elements of <b>A</b> is equal to E</li>
</ol>
</div>
</div>
<div id="outline-container-orgc0fd081" class="outline-3">
<h3 id="orgc0fd081">Cartesian products :</h3>
<div class="outline-text-3" id="text-orgc0fd081">
<p>
Let E and F be two sets, the set EXF = {(x,y)/ x ∈ E AND y ∈ F} is called the Cartesian product of E and F
</p>
</div>
<div id="outline-container-org4b0f328" class="outline-4">
<h4 id="org4b0f328">Example :</h4>
<div class="outline-text-4" id="text-org4b0f328">
<p>
A = {4,5} ; B= {4,5,6} // AxB = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6)}
</p>
<p>
BxA = {(4,4), (4,5), (5,4), (5,5), (6,4), (6,5)} // Therefore AxB ≠ BxA
</p>
</div>
</div>
<div id="outline-container-orgc924520" class="outline-4">
<h4 id="orgc924520">Some proprieties:</h4>
<div class="outline-text-4" id="text-orgc924520">
<ol class="org-ol">
<li>ExF = ∅ ⇔ E=∅ OR F=∅</li>
<li>ExF = FxE ⇔ E=F OR E=∅ OR F=∅</li>
<li>E x (F∪G) = (ExF) ∪ (ExG)</li>
<li>(E∪F) x G = (ExG) ∪ (FxG)</li>
<li>(E∪F) ∩ (GxH) = (E ∩ G) x (F ∩ H)</li>
<li>Generally speaking : (ExF) ∪ (GxH) ≠ (E∪G) x (F∪H)</li>
</ol>
</div>
</div>
</div>
</div>
<div id="outline-container-org8f809af" class="outline-2">
<h2 id="org8f809af">Binary relations in a set :</h2>
<div class="outline-text-2" id="text-org8f809af">
</div>
<div id="outline-container-orgeb6cba6" class="outline-3">
<h3 id="orgeb6cba6">Definition :</h3>
<div class="outline-text-3" id="text-orgeb6cba6">
<p>
Let E be a set and x,y ∈ E. If there exists a link between x and y, we say that they are tied by a relation <b>R</b> and we write <b>xRy</b>
</p>
</div>
</div>
<div id="outline-container-org1696189" class="outline-3">
<h3 id="org1696189">Proprieties :</h3>
<div class="outline-text-3" id="text-org1696189">
<p>
Let E be a set and R a relation defined in E
</p>
<ol class="org-ol">
<li>We say that R is reflexive if ∀ x ∈ E, xRx (for any element x in E,x is related to itself)</li>
<li>We say that R is symmetrical if ∀ x,y ∈ E , xRy ⇒ yRx</li>
<li>We say that R is transitive if ∀ x,y,z ∈ E (xRy , yRz) ⇒ xRz</li>
<li>We say that R is anti-symmetrical if ∀ x,y ∈ E xRy AND yRx ⇒ x = y</li>
</ol>
</div>
</div>
<div id="outline-container-org38c5183" class="outline-3">
<h3 id="org38c5183">Equivalence relationship :</h3>
<div class="outline-text-3" id="text-org38c5183">
<p>
We say that R is a relation of equivalence in E if its reflexive, symetrical and transitive
</p>
</div>
<div id="outline-container-org110e6fa" class="outline-4">
<h4 id="org110e6fa">Equivalence class :</h4>
<div class="outline-text-4" id="text-org110e6fa">
<p>
Let R be a relation of equivalence in E and a ∈ E, we call equivalence class of <b>a</b>, and we write ̅a or ȧ, or cl a the following set :
</p>
<p>
<b>a̅ = {y ∈ E/ y R a}</b>
</p>
</div>
<ul class="org-ul">
<li><a id="org20e3b3b"></a>The quotient set :<br />
<div class="outline-text-5" id="text-org20e3b3b">
<p>
E/R = {̅a , a ∈ E}
</p>
</div>
</li>
</ul>
</div>
</div>
<div id="outline-container-org25fec1b" class="outline-3">
<h3 id="org25fec1b">Order relationship :</h3>
<div class="outline-text-3" id="text-org25fec1b">
<p>
Let E be a set and R be a relation defined in E. We say that R is a relation of order if its reflexive, anti-symetrical and transitive.
</p>
<ol class="org-ol">
<li>The order R is called total if ∀ x,y ∈ E xRy OR yRx</li>
<li>The order R is called partial if ∃ x,y ∈ E xR̅y AND yR̅x</li>
</ol>
</div>
<div id="outline-container-orgc094acc" class="outline-4">
<h4 id="orgc094acc"><span class="todo TODO">TODO</span> Examples :</h4>
<div class="outline-text-4" id="text-orgc094acc">
<p>
∀x,y ∈ ℝ , xRy ⇔ x²-y²=x-y
</p>
<ol class="org-ol">
<li>Prove that R is an equivalence relation</li>
<li>Let a ∈ ℝ, find ̅a</li>
</ol>
</div>
</div>
</div>
</div>
</div>
<div id="postamble" class="status">
<p class="author">Author: Crystal</p>
<p class="date">Created: 2023-10-17 Tue 22:32</p>
</div>
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