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<h1 class="title">Algebra 1</h1>
<div id="outline-container-orgc134a5b" class="outline-2">
<h2 id="orgc134a5b">Contenu de la Matiére</h2>
<div class="outline-text-2" id="text-orgc134a5b">
</div>
<div id="outline-container-orgae00938" class="outline-3">
<h3 id="orgae00938">Rappels et compléments (11H)</h3>
<div class="outline-text-3" id="text-orgae00938">
<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-org0eb35c9" class="outline-3">
<h3 id="org0eb35c9">Structures Algébriques (11H)</h3>
<div class="outline-text-3" id="text-org0eb35c9">
<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-org4a088f6" class="outline-3">
<h3 id="org4a088f6">Polynômes et fractions rationnelles</h3>
<div class="outline-text-3" id="text-org4a088f6">
<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-org73264c6" class="outline-2">
<h2 id="org73264c6">Premier cours : Logique mathématique et méthodes du raisonnement mathématique <i>Sep 25</i> :</h2>
<div class="outline-text-2" id="text-org73264c6">
<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-org74d9557" class="outline-3">
<h3 id="org74d9557">Properties:</h3>
<div class="outline-text-3" id="text-org74d9557">
</div>
<div id="outline-container-orgc7f1d03" class="outline-4">
<h4 id="orgc7f1d03"><b>Absorption</b>:</h4>
<div class="outline-text-4" id="text-orgc7f1d03">
<p>
(P ∨ P) ⇔ P
</p>
<p>
(P ∧ P) ⇔ P
</p>
</div>
</div>
<div id="outline-container-orgcb729de" class="outline-4">
<h4 id="orgcb729de"><b>Commutativity</b>:</h4>
<div class="outline-text-4" id="text-orgcb729de">
<p>
(P ∧ Q) ⇔ (Q ∧ P)
</p>
<p>
(P ∨ Q) ⇔ (Q ∨ P)
</p>
</div>
</div>
<div id="outline-container-org4ae8933" class="outline-4">
<h4 id="org4ae8933"><b>Associativity</b>:</h4>
<div class="outline-text-4" id="text-org4ae8933">
<p>
P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R
</p>
<p>
P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R
</p>
</div>
</div>
<div id="outline-container-org095f4a6" class="outline-4">
<h4 id="org095f4a6"><b>Distributivity</b>:</h4>
<div class="outline-text-4" id="text-org095f4a6">
<p>
P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)
</p>
<p>
P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R)
</p>
</div>
</div>
<div id="outline-container-orgf7e29ee" class="outline-4">
<h4 id="orgf7e29ee"><b>Neutral element</b>:</h4>
<div class="outline-text-4" id="text-orgf7e29ee">
<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-org9bfee59" class="outline-4">
<h4 id="org9bfee59"><b>Negation of a conjunction & a disjunction</b>:</h4>
<div class="outline-text-4" id="text-org9bfee59">
<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-orgd144fb3" class="outline-4">
<h4 id="orgd144fb3"><b>Transitivity</b>:</h4>
<div class="outline-text-4" id="text-orgd144fb3">
<p>
[(P ⇒ Q) AND (Q ⇒ R)] ⇔ P ⇒ R
</p>
</div>
</div>
<div id="outline-container-orgbc26f01" class="outline-4">
<h4 id="orgbc26f01"><b>Contraposition</b>:</h4>
<div class="outline-text-4" id="text-orgbc26f01">
<p>
(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)
</p>
</div>
</div>
<div id="outline-container-org70fd37a" class="outline-4">
<h4 id="org70fd37a">God only knows what this property is called:</h4>
<div class="outline-text-4" id="text-org70fd37a">
<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-org6dd9c74" class="outline-3">
<h3 id="org6dd9c74">Some exercices I found online :</h3>
<div class="outline-text-3" id="text-org6dd9c74">
</div>
<div id="outline-container-org5a46794" class="outline-4">
<h4 id="org5a46794">USTHB 2022/2023 Section B :</h4>
<div class="outline-text-4" id="text-org5a46794">
</div>
<ul class="org-ul">
<li><a id="orgdcdfa08"></a>Exercice 1: Démontrer les équivalences suivantes:<br />
<div class="outline-text-5" id="text-orgdcdfa08">
<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="orgfc2dd28"></a>Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:<br />
<div class="outline-text-5" id="text-orgfc2dd28">
<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-org8e69635" class="outline-2">
<h2 id="org8e69635">2éme cours <i>Oct 2</i></h2>
<div class="outline-text-2" id="text-org8e69635">
</div>
<div id="outline-container-org1d4ffa3" class="outline-3">
<h3 id="org1d4ffa3">Quantifiers</h3>
<div class="outline-text-3" id="text-org1d4ffa3">
<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="orgd0b7f53"></a>Example<br />
<div class="outline-text-6" id="text-orgd0b7f53">
<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-orga7cd185" class="outline-4">
<h4 id="orga7cd185">Proprieties</h4>
<div class="outline-text-4" id="text-orga7cd185">
</div>
<ul class="org-ul">
<li><a id="org7460082"></a>Propriety Number 1:<br />
<div class="outline-text-5" id="text-org7460082">
<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="org27b8375"></a>Example:<br />
<div class="outline-text-6" id="text-org27b8375">
<p>
∀ x ≥ 1 x² > 5 ⇔ ∃ x ≥ 1 x² < 5
</p>
</div>
</li>
</ul>
</li>
<li><a id="org21aa647"></a>Propriety Number 2:<br />
<div class="outline-text-5" id="text-org21aa647">
<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="orgb4e2845"></a>Example :<br />
<div class="outline-text-6" id="text-orgb4e2845">
<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="orgce6dd51"></a>Propriety Number 3:<br />
<div class="outline-text-5" id="text-orgce6dd51">
<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="org956d29b"></a>Example of why it’s NOT an equivalence :<br />
<div class="outline-text-6" id="text-org956d29b">
<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="orgce400d9"></a>Propriety Number 4:<br />
<div class="outline-text-5" id="text-orgce400d9">
<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-org48fa1b7" class="outline-3">
<h3 id="org48fa1b7">Multi-parameter proprieties :</h3>
<div class="outline-text-3" id="text-org48fa1b7">
<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="org985d3f3"></a>Example :<br />
<div class="outline-text-6" id="text-org985d3f3">
<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="orgd3167fe"></a>WARNING :<br />
<div class="outline-text-6" id="text-orgd3167fe">
<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="orge81043c"></a>Example :<br />
<div class="outline-text-7" id="text-orge81043c">
<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="org96c2514"></a>Proprieties :<br />
<div class="outline-text-5" id="text-org96c2514">
<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-org47ee190" class="outline-3">
<h3 id="org47ee190">Methods of mathematical reasoning :</h3>
<div class="outline-text-3" id="text-org47ee190">
</div>
<div id="outline-container-org24c7fa4" class="outline-4">
<h4 id="org24c7fa4">Direct reasoning :</h4>
<div class="outline-text-4" id="text-org24c7fa4">
<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="orgfa904f5"></a>Example:<br />
<div class="outline-text-5" id="text-orgfa904f5">
<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-orgb9d2c9e" class="outline-4">
<h4 id="orgb9d2c9e">Reasoning by the Absurd:</h4>
<div class="outline-text-4" id="text-orgb9d2c9e">
<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="orgfbfd0bb"></a>Example:<br />
<div class="outline-text-5" id="text-orgfbfd0bb">
<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-org8b209a6" class="outline-4">
<h4 id="org8b209a6">Reasoning by contraposition:</h4>
<div class="outline-text-4" id="text-org8b209a6">
<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-org94c9a28" class="outline-4">
<h4 id="org94c9a28">Reasoning by counter example:</h4>
<div class="outline-text-4" id="text-org94c9a28">
<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-org0eabee1" class="outline-2">
<h2 id="org0eabee1">3eme Cours : <i>Oct 9</i></h2>
<div class="outline-text-2" id="text-org0eabee1">
</div>
<div id="outline-container-orgdafe6b7" class="outline-4">
<h4 id="orgdafe6b7">Reasoning by recurrence :</h4>
<div class="outline-text-4" id="text-orgdafe6b7">
<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="org9970baf"></a>Example:<br />
<div class="outline-text-5" id="text-org9970baf">
<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-org4d1906f" class="outline-2">
<h2 id="org4d1906f">4eme Cours : Chapitre 2 : Sets and Operations</h2>
<div class="outline-text-2" id="text-org4d1906f">
</div>
<div id="outline-container-orgd19c38e" class="outline-3">
<h3 id="orgd19c38e">Definition of a set :</h3>
<div class="outline-text-3" id="text-orgd19c38e">
<p>
A set is a collection of objects that share the sane propriety
</p>
</div>
</div>
<div id="outline-container-orgcf58c48" class="outline-3">
<h3 id="orgcf58c48">Belonging, inclusion, and equality :</h3>
<div class="outline-text-3" id="text-orgcf58c48">
<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-org939fd93" class="outline-3">
<h3 id="org939fd93">Intersections and reunions :</h3>
<div class="outline-text-3" id="text-org939fd93">
</div>
<div id="outline-container-orge8ae0b6" class="outline-4">
<h4 id="orge8ae0b6">Intersection:</h4>
<div class="outline-text-4" id="text-orge8ae0b6">
<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-org07c050a" class="outline-4">
<h4 id="org07c050a">Union:</h4>
<div class="outline-text-4" id="text-org07c050a">
<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-org7ecf856" class="outline-4">
<h4 id="org7ecf856">Difference between two sets:</h4>
<div class="outline-text-4" id="text-org7ecf856">
<p>
E\F(Which is also written as : E - F) = {x / x ∈ E and x ∉ F}
</p>
</div>
</div>
<div id="outline-container-orgad5f4da" class="outline-4">
<h4 id="orgad5f4da">Complimentary set:</h4>
<div class="outline-text-4" id="text-orgad5f4da">
<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-org3e8e3b3" class="outline-4">
<h4 id="org3e8e3b3">Symentrical difference</h4>
<div class="outline-text-4" id="text-org3e8e3b3">
<p>
E Δ F = (E - F) ∪ (F - E) ; = (E ∪ F) - (E ∩ F)
</p>
</div>
</div>
</div>
<div id="outline-container-org8920c77" class="outline-3">
<h3 id="org8920c77">Proprieties :</h3>
<div class="outline-text-3" id="text-org8920c77">
<p>
Let E,F and G be 3 sets. We have :
</p>
</div>
<div id="outline-container-orgcb406ce" class="outline-4">
<h4 id="orgcb406ce">Commutativity:</h4>
<div class="outline-text-4" id="text-orgcb406ce">
<p>
E ∩ F = F ∩ E
E ∪ F = F ∪ E
</p>
</div>
</div>
<div id="outline-container-orgfcaf63a" class="outline-4">
<h4 id="orgfcaf63a">Associativity:</h4>
<div class="outline-text-4" id="text-orgfcaf63a">
<p>
E ∩ (F ∩ G) = (E ∩ F) ∩ G
E ∪ (F ∪ G) = (E ∪ F) ∪ G
</p>
</div>
</div>
<div id="outline-container-org6ad9182" class="outline-4">
<h4 id="org6ad9182">Distributivity:</h4>
<div class="outline-text-4" id="text-org6ad9182">
<p>
E ∩ (F ∪ G) = (E ∩ F) ∪ (E ∩ G)
E ∪ (F ∩ G) = (E ∪ F) ∩ (E ∪ G)
</p>
</div>
</div>
<div id="outline-container-org7a0450e" class="outline-4">
<h4 id="org7a0450e">Lois de Morgan:</h4>
<div class="outline-text-4" id="text-org7a0450e">
<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-org44fd147" class="outline-4">
<h4 id="org44fd147">An other one:</h4>
<div class="outline-text-4" id="text-org44fd147">
<p>
E - (F ∩ G) = (E-F) ∪ (E-G) ; E - (F ∪ G) = (E-F) ∩ (E-G)
</p>
</div>
</div>
<div id="outline-container-orgca3a4c6" class="outline-4">
<h4 id="orgca3a4c6">An other one:</h4>
<div class="outline-text-4" id="text-orgca3a4c6">
<p>
E ∩ ∅ = ∅ ; E ∪ ∅ = E
</p>
</div>
</div>
<div id="outline-container-org6cd18a3" class="outline-4">
<h4 id="org6cd18a3">And an other one:</h4>
<div class="outline-text-4" id="text-org6cd18a3">
<p>
E ∩ (F Δ G) = (E ∩ F) Δ (E ∩ G)
</p>
</div>
</div>
<div id="outline-container-org0889163" class="outline-4">
<h4 id="org0889163">And the last one:</h4>
<div class="outline-text-4" id="text-org0889163">
<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: 2023-10-13 Fri 16:58</p>
</div>
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