#+title: Algebra 1
#+AUTHOR: Crystal
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* 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) AND (Q ⇒ R)] ⇔ P ⇒ R
*** *Contraposition*:
(P ⇒ Q) ⇔ (Q̅ ⇒ P̅)
*** God only knows what this property is called:
/If/
(P ⇒ Q) is true
and
(P̅ ⇒ 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 = ∅
* 5eme cours: L'ensemble des parties d'un ensemble /Oct 16/
Let E be a set. We define P(E) as the set of all parts of E : *P(E) = {X/X ⊂ E}*
*** Notes :
∅ ∈ P(E) ; E ∈ P(E)
cardinal E = n /The number of terms in E/ , cardinal P(E) = 2^n /The number of all parts of E/
*** Examples :
E = {a,b,c} // P(E)={∅, {a}, {b}, {c}, {a,b}, {b,c}, {a,c}, {a,b,c}}
** Partition of a set :
We say that *A* is a partition of E if:
a. ∀ x ∈ A , x ≠ 0
b. All the elements of *A* are two by two disjoint. Or in other terms, there should not be two elements that intersects with each other.
c. The reunion of all elements of *A* is equal to E
** Cartesian products :
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
*** Example :
A = {4,5} ; B= {4,5,6} // AxB = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6)}
BxA = {(4,4), (4,5), (5,4), (5,5), (6,4), (6,5)} // Therefore AxB ≠ BxA
*** Some proprieties:
1. ExF = ∅ ⇔ E=∅ OR F=∅
2. ExF = FxE ⇔ E=F OR E=∅ OR F=∅
3. E x (F∪G) = (ExF) ∪ (ExG)
4. (E∪F) x G = (ExG) ∪ (FxG)
5. (E∪F) ∩ (GxH) = (E ∩ G) x (F ∩ H)
6. Generally speaking : (ExF) ∪ (GxH) ≠ (E∪G) x (F∪H)
* Binary relations in a set :
** Definition :
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 *R* and we write *xRy*
** Proprieties :
Let E be a set and R a relation defined in E
1. We say that R is reflexive if ∀ x ∈ E, xRx (for any element x in E,x is related to itself)
2. We say that R is symmetrical if ∀ x,y ∈ E , xRy ⇒ yRx
3. We say that R is transitive if ∀ x,y,z ∈ E (xRy , yRz) ⇒ xRz
4. We say that R is anti-symmetrical if ∀ x,y ∈ E xRy AND yRx ⇒ x = y
** Equivalence relationship :
We say that R is a relation of equivalence in E if its reflexive, symetrical and transitive
*** Equivalence class :
Let R be a relation of equivalence in E and a ∈ E, we call equivalence class of *a*, and we write ̅a or ȧ, or cl a the following set :
*a̅ = {y ∈ E/ y R a}*
**** The quotient set :
E/R = {̅a , a ∈ E}
** Order relationship :
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.
1. The order R is called total if ∀ x,y ∈ E xRy OR yRx
2. The order R is called partial if ∃ x,y ∈ E xR̅y AND yR̅x
*** TODO Examples :
∀x,y ∈ ℝ , xRy ⇔ x²-y²=x-y
1. Prove that R is an equivalence relation
2. Let a ∈ ℝ, find ̅a