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+#+title: Algebra 1
+#+AUTHOR: Crystal
+#+OPTIONS: ^:{}
+#+OPTIONS: num:nil
+#+EXPORT_FILE_NAME: ../../../uni_notes/algebra.html
+#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="../src/css/colors.css"/>
+#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="../src/css/style.css"/>
+#+OPTIONS: html-style:nil
+#+OPTIONS: toc:nil
+
+* Contenu de la Matiére
+** Rappels et compléments (11H)
+- Logique mathématique et méthodes du raisonnement mathématique
+- Ensembles et Relations
+- Applications
+
+** Structures Algébriques (11H)
+- Groupes et morphisme de groupes
+- Anneaux et morphisme d'anneaux
+- Les corps
+
+** Polynômes et fractions rationnelles
+- Notion du polynôme à une indéterminée á coefficients dans un anneau
+- Opérations Algébriques sur les polynômes
+- Arithmétique dans l'anneau des polynômes
+- Polynôme dérivé et formule de Taylor
+- Notion de racine d'un polynôme
+- Notion de Fraction rationelle á une indéterminée
+- Décomposition des fractions rationelles en éléments simples
+
+* Premier cours : Logique mathématique et méthodes du raisonnement mathématique /Sep 25/ :
+
+Let *P* *Q* and *R* be propositions which can either be *True* or *False*. And let's also give the value *1* to each *True* proposition and *0* to each false one.
+
+/Ex:/
+- *5 ≥ 2* is a proposition, a correct one !!!
+- *The webmaster is a girl* is also a proposition, which is also correct.
+- *x is always bigger than 5* is *not* a proposition, because we CAN'T determine if it's correct or not as *x* changes.
+...etc
+
+In order to avoid repetition, and rewriting the proposition over and over, we just assign a capital letter to them such as *P Q* or *R*.
+
+So now we could write :
+*Let the proposition P be 5 ≥ 2, we notice that P is always True, therefor its validity is 1*
+
+We also have the opposite of *P*, which is *not(P)* but for simplicity we use *P̅* (A P with a bar on top, in case it doesn't load for you), now let's go back to the previous example:
+
+*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 ?*
+
+Now let's say we have two propositions, and we want to test the validity of their disjunction..... Okay what is this "disjunction" ? *Great Question Billy !!!* A disjunction is true if either propositions are true
+
+Ex:
+*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):*
+
+| P | Q | Disjunction |
+|---+---+-------------|
+| 1 | 1 |           1 |
+| 1 | 0 |           1 |
+| 0 | 1 |           1 |
+| 0 | 0 |           0 |
+
+/What the hell is this ?/
+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 *I HATE YOU*)"
+The third one... /zzzzzzz/
+
+You got the idea !!!
+And since we are talking about a disjunction here, *one of the propositions* need to be true in order for this disjunction to be true.
+
+You may be wondering.... Crystal, can't we write a disjunction in magical math symbols ? And to this I respond with a big *YES*. A disjunction is symbolized by a *∨* . So the disjunction between proposition *P & Q* can be written this way : *P ∨ Q*
+
+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 *P ∧ Q*, and is only true if *P* and *Q* are true. So we get a Table like this :
+
+| P | Q | P ∨ Q | P ∧ Q |
+|---+---+-------+-------|
+| 1 | 1 |     1 |     1 |
+| 1 | 0 |     1 |     0 |
+| 0 | 1 |     1 |     0 |
+| 0 | 0 |     0 |     0 |
+
+*Always remember: 1 means true and 0 means false*
+
+There are two more basics to cover here before going to some properties, the first one is implication symbolized by the double arrow *⇒*
+
+Implication is kinda hard for my little brain to explain, so I will just say what it means:
+
+*If P implies Q, this means that either Q, or the opposite of P are correct*
+
+or in math terms
+
+*P ⇒ Q translates to P̅ ∨ Q*
+Let's illustrate :
+
+| P | Q | P̅ | Q̅ | P ∨ Q | P ∧ Q | P ⇒ Q (P̅ ∨ Q) |
+|---+---+---+---+-------+-------+---------------|
+| 1 | 1 | 0 | 0 |     1 |     1 |             1 |
+| 1 | 0 | 0 | 1 |     1 |     0 |             0 |
+| 0 | 1 | 1 | 0 |     1 |     0 |             1 |
+| 0 | 0 | 1 | 1 |     0 |     0 |             1 |
+
+*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"*
+
+Aight, a last one and we are done!!! Equivalence, which is fairly easy, symbolized by a *⇔* symbol.
+
+A proposition is equivalent to another only when both of them have *the same value of truth* AKA: both true or both false. a little table will help demonstrate what i mean.
+
+| P | Q | P̅ | Q̅ | P ∨ Q | P ∧ Q | P ⇒ Q (P̅ ∨ Q) | P ⇔ Q |
+|---+---+---+---+-------+-------+---------------+-------|
+| 1 | 1 | 0 | 0 |     1 |     1 |             1 |     1 |
+| 1 | 0 | 0 | 1 |     1 |     0 |             0 |     0 |
+| 0 | 1 | 1 | 0 |     1 |     0 |             1 |     0 |
+| 0 | 0 | 1 | 1 |     0 |     0 |             1 |     1 |
+
+/Note: P implying Q is equivalent to P̅ implying Q̅, or: (P ⇒ Q) ⇔ (P̅ ⇒ Q̅)/
+
+** Properties:
+*** *Absorption*:
+(P ∨ P) ⇔ P
+
+(P ∧ P) ⇔ P
+
+*** *Commutativity*:
+(P ∧ Q) ⇔ (Q ∧ P)
+
+(P ∨ Q) ⇔ (Q ∨ P)
+
+*** *Associativity*:
+P ∧ (Q ∧ R) ⇔ (P ∧ Q) ∧ R
+
+P ∨ (Q ∨ R) ⇔ (P ∨ Q) ∨ R
+
+*** *Distributivity*:
+P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)
+
+P ∨ (Q ∧ R) ⇔ (P ∨ Q) ∧ (P ∨ R)
+
+*** *Neutral element*:
+/We define proposition *T* to be always *true* and *F* to be always *false*/
+
+P ∧ T ⇔ P
+
+P ∨ F ⇔ P
+*** *Negation of a conjunction & a disjunction*:
+Now we won't use bars here because my lazy ass doesn't know how, so instead I will use not()!!!
+
+not(*P ∧ Q*) ⇔ P̅ ∨ Q̅
+
+not(*P ∨ Q*) ⇔ P̅ ∧ Q̅
+
+*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 ∨) /NOT AN ACTUAL MATH WRITING. DONT USE IT ANYWHERE ELSE OTHER THAN YOUR BRAIN/ and not(Q)*
+
+*** *Transitivity*:
+[(P ⇒ Q) (Q ⇒ R)] ⇔ P ⇒ R
+
+*** *Contraposition*:
+(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)
+
+*** God only knows what this property is called:
+/If/
+
+(P ⇒ Q) is true
+
+and
+
+(Q̅ ⇒ Q) is true
+
+then
+
+Q is always true
+
+** Some exercices I found online :
+
+*** USTHB 2022/2023 Section B :
+
+**** Exercice 1: Démontrer les équivalences suivantes:
+1. (P ⇒ Q) ⇔ (Q̅ ⇒ P̅)
+
+   Basically we are asked to prove contraposition, so here we have ( P ⇒ Q ) which is equivalent to P̅ ∨ Q *By definition : (P ⇒ Q) ⇔  (P̅ ∨ Q)*
+
+
+   So we end up with : *(P̅ ∨ Q) ⇔ (Q̅ ⇒ P̅)*, now we just do the same with the second part of the contraposition. *(Q̅ ⇒ P̅) ⇔ (Q ∨ P̅)* therefor :
+
+
+   *(Q ∨ P̅) ⇔ (P̅ ∨ Q)*, which is true because of commutativity
+
+2. not(P ⇒ Q) ⇔  P ∧ Q̅
+
+
+Okaaaay so, let's first get rid of the implication, because I don't like it : *not(P̅ ∨ Q)*
+
+
+Now that we got rid of it, we can negate the whole disjunction *not(P̅ ∨ Q) ⇔ (P ∧ Q̅)*. Which is the equivalence we needed to prove
+
+3. P ⇒ (Q ∧ R) ⇔ (P ⇒ Q) ∧ (P ⇒ R)
+
+   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 *Distributivity*, because yeah, we can distribute an implication across a {con/dis}junction
+
+4. P ∧ (Q ∨ R) ⇔ (P ∧ Q) ∨ (P ∧ R)
+
+   Literally the same as above 🩷
+
+
+**** Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:
+
+1. ∀x ∈ ℝ ,∃y ∈ ℝ*+, tels que e^x = y
+
+   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
+
+
+"The function f(x)=e^x is always positive and non-null", the very definition of an exponential function !!!!
+
+
+*So the proposition is true*
+
+
+2. ∃x ∈ ℝ, tels que x^2 < x < x^3
+
+
+We just need to find a value that satisifies this condition...thankfully its easy....
+
+x² < x < x³ , we divide the three terms by x so we get :
+
+
+x < 1 < x² , or :
+
+
+*x < 1* ; *1 < x²* ⇔  *x < 1* ; *1 < x* /We square root both sides/
+
+
+We end up with a contradiction, therefor its wrong
+
+
+3. ∀x ∈ ℝ, ∃y ∈ ℝ tels que y = 3x - 8
+
+
+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...
+
+*Since substraction is an  Internal composition law in ℝ, therefor all results of a substraction between two Real numbers is...Real*
+
+4. ∃x ∈ ℕ, ∀y ∈ ℕ, x > y ⇒ x + y < 8
+
+   "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"
+
+
+Let's get rid of the implication :
+
+∃x ∈ ℕ, ∀y ∈ ℕ, (y > x) ∨ (x + y < 8) /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/
+
+This proposition is true, because there exists a value of x that satisfies this condition, it's *all numbers under 8* let's take 3 as an example:
+
+
+*x = 3 , if y > 3 then the first condition is true ; if y < 3 then the second one is true*
+
+
+Meaning that the two propositions CAN NOT BE WRONG TOGETHER, either one is wrong, or the other
+
+
+y > x
+
+
+*y - x > 0*
+
+
+y + x < 8
+
+
+*y < 8 - x* /This one is always true for all values of x below 8, since we are working in the set ℕ/
+
+
+5. ∀x ∈ ℝ, x² ≥ 1 ⇔  x ≥ 1
+
+   ....This is getting stupid. of course it's true it's part of the definition of the power of 2
+
+
+* 2éme cours /Oct 2/
+
+** Quantifiers
+
+A propriety P can depend on a parameter x
+
+
+∀ is the universal quantifier which stands for "For any value of..."
+
+
+∃ is the existential quantifier which stands for "There exists at least one..."
+
+
+***** Example
+P(x) : x+1≥0
+
+P(X) is True or False depending on the values of x
+
+
+*** Proprieties
+**** Propriety Number 1:
+The negation of the universal quantifier is the existential quantifier, and vice-versa :
+
+- not(∀x ∈ E , P(x)) ⇔ ∃ x ∈ E, not(P(x))
+- not(∃x ∈ E , P(x)) ⇔ ∀ x ∈ E, not(P(x))
+
+***** Example:
+∀ x ≥ 1  x² > 5 ⇔ ∃ x ≥ 1 x² < 5
+**** Propriety Number 2:
+
+*∀x ∈ E, [P(x) ∧ Q(x)] ⇔ [∀ x ∈ E, P(x)] ∧ [∀ x ∈ E, Q(x)]*
+
+
+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)"
+***** Example :
+P(x) : sqrt(x) > 0 ;  Q(x) : x ≥ 1
+
+
+∀x ∈ ℝ*+, [sqrt(x) > 0 , x ≥ 1] ⇔ [∀x ∈ R*+, sqrt(x) > 0] ∧ [∀x ∈ R*+, x ≥ 1]
+
+
+*Which is true*
+**** Propriety Number 3:
+
+*∃ x ∈ E, [P(x) ∧ Q(x)] /⇒/ [∃ x ∈ E, P(x)] ∧ [∃ x ∈ E, Q(x)]*
+
+
+/Here its an implication and not an equivalence/
+
+***** Example of why it's NOT an equivalence :
+P(x) : x > 5  ;  Q(x) : x < 5
+
+
+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!!!
+**** Propriety Number 4:
+
+*[∀ x ∈ E, P(x)] ∨ [∀ x ∈ E, Q(x)] /⇒/ ∀x ∈ E, [P(x) ∨ Q(x)]*
+
+
+/Same here, implication and NOT en equivalence/
+
+
+** Multi-parameter proprieties :
+
+A propriety P can depend on two or more parameters, for convenience we call them x,y,z...etc
+
+***** Example :
+P(x,y): x+y > 0
+
+
+P(0,1) is a True proposition
+
+
+P(-2,-1) is a False one
+
+***** WARNING :
+
+∀x ∈ E, ∃y ∈ F , P(x,y)
+
+
+∃y ∈ F, ∀x ∈ E , P(x,y)
+
+
+Are different because in the first one y depends on x, while in the second one, it doesn't
+****** Example :
+∀ x ∈ ℕ , ∃ y ∈ ℕ y > x ------ True
+
+
+∃ y ∈ ℕ , ∀ x ∈ ℕ y > x ------ False
+
+**** Proprieties :
+1. not(∀x ∈ E ,∃y ∈ F P(x,y)) ⇔ ∃x ∈ E, ∀y ∈ F not(P(x,y))
+2. not(∃x ∈ E ,∀y ∈ F P(x,y)) ⇔ ∀x ∈ E, ∃y ∈ F not(P(x,y))
+
+** Methods of mathematical reasoning :
+*** Direct reasoning :
+
+To show that an implication P ⇒ Q is true, we suppose that P is true and we show that Q is true
+
+**** Example:
+Let a,b be two Real numbers, we have to prove that *a² + b² = 1 ⇒ |a + b| ≤ 2*
+
+
+We suppose that a²+b² = 1 and we prove that |a + b| ≤ 2
+
+
+a²+b²=1 ⇒  b² = 1 - a² ; a² = 1 - b²
+
+
+a²+b²=1 ⇒  1 - a² ≥ 0 ; 1 - b² ≥ 0
+
+
+a²+b²=1 ⇒  a² ≤ 1 ; b² ≤ 1
+
+
+a²+b²=1 ⇒ -1 ≤ a ≤ 1 ; -1 ≤ b ≤ 1
+
+
+a²+b²=1 ⇒ -2 ≤ a + b ≤ 2
+
+
+a²+b²=1 ⇒ |a + b| ≤ 2 *Which is what we wanted to prove, therefor the implication is correct*
+
+*** Reasoning by the Absurd:
+To prove that a proposition is True, we suppose that it's False and we must come to a contradiction
+
+
+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
+**** Example:
+Prove that this proposition is correct using the reasoning by the absurd : ∀x ∈ ℝ* , sqrt(1+x²) ≠ 1 + x²/2
+
+
+We assume that ∃ x ℝ* , sqrt(1+x²) = 1 + x²/2
+
+
+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
+
+
+*** Reasoning by contraposition:
+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
+
+
+*** Reasoning by counter example:
+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
+* 3eme Cours : /Oct 9/
+*** Reasoning by recurrence :
+P is a propriety dependent of *n ∈ ℕ*. 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
+
+**** Example:
+Let's prove that ∀ n ≥ 1 , (n,k=1)Σk = [n(n+1)]/2
+
+
+P(n) : (n,k=1)Σk = [n(n+1)]/2
+
+
+
+*Pour n = 1:* (1,k=1)Σk = 1 ; [n(n+1)]/2 = 1 . *So P(1) is true*
+
+
+
+For n ≥ 1. We assume that P(n) is true, OR : *(n, k=1)Σk = n(n+1)/2*. We now have to prove that P(n+1) is true, Or : *(n+1, k=1)Σk = (n+1)(n+2)/2*
+
+
+(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 ; = *[(n+2)(n+1)]/2* /WHICH IS WHAT WE NEEDED TO FIND/
+
+
+*Conclusion: ∀n ≥ 1 , (n,k=1)Σk = n(n+1)/2*
+
+* 4eme Cours : Chapitre 2 : Sets and Operations
+** Definition of a set :
+A set is a collection of objects that share the sane propriety
+
+** Belonging, inclusion, and equality :
+a. Let E be a set. If x is an element of E, we say that x belongs to E we write *x ∈ E*, and if it doesn't, we write *x ∉ E*
+b. A set E is included in a set F if all elements of E are elements of F and we write *E ⊂ F ⇔ (∀x , x ∈ E ⇒ x ∈ F)*. We say that E is a subset of F, or a part of F. The negation of this propriety is : *E ⊄ F ⇔ ∃x , x ∈ E and x ⊄ F*
+c. E and F are equal if E is included in F and F is included in E, and we write *E = F ⇔ (E ⊂ F) et (F ⊂ E)*
+d. The empty set (symbolized by ∅) is a set without elements, and is included in all sets (by convention) : *∅ ⊂ E*
+
+** Intersections and reunions :
+*** Intersection:
+E ∩ F = {x / x ∈ E AND x ∈ F} ; x ∈ E ∩ F ⇔ x ∈ F AND x ∈ F
+
+
+x ∉ E ∩ F ⇔ x ∉ E OR x ∉ F
+
+*** Union:
+E ∪ F = {x / x ∈ E OR x ∈ F} ;  x ∈ E ∪ F ⇔ x ∈ F OR x ∈ F
+
+
+x ∉ E ∪ F ⇔ x ∉ E AND x ∉ F
+*** Difference between two sets:
+E\F(Which is also written as : E - F) = {x / x ∈ E and x ∉ F}
+*** Complimentary set:
+If F ⊂ E. E - F is the complimentary of F in E.
+
+
+FCE = {x /x ∈ E AND x ∉ F} *ONLY WHEN F IS A SUBSET OF E*
+*** Symentrical difference
+E Δ F = (E - F) ∪ (F - E) ; = (E ∪ F) - (E ∩ F)
+** Proprieties :
+Let E,F and G be 3 sets. We have :
+*** Commutativity:
+E ∩ F = F ∩ E
+E ∪ F = F ∪ E
+*** Associativity:
+E ∩ (F ∩ G) = (E ∩ F) ∩ G
+E ∪ (F ∪ G) = (E ∪ F) ∪ G
+*** Distributivity:
+E ∩ (F ∪ G) = (E ∩ F) ∪ (E ∩ G)
+E ∪ (F ∩ G) = (E ∪ F) ∩ (E ∪ G)
+*** Lois de Morgan:
+If E ⊂ G and F ⊂ G ;
+
+(E ∩ F)CG = ECG ∪ FCG ; (E ∪ F)CG = ECG ∩ FCG
+*** An other one:
+E - (F ∩ G) = (E-F) ∪ (E-G) ;  E - (F ∪ G) = (E-F) ∩ (E-G)
+*** An other one:
+E ∩ ∅ = ∅ ; E ∪ ∅ = E
+*** And an other one:
+E ∩ (F Δ G) = (E ∩ F) Δ (E ∩ G)
+*** And the last one:
+E Δ ∅ = E ; E Δ E = ∅