Algebra 1

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:

(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
1. 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
1. 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
2. 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
1. ∃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
1. ∀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
1. ∃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
1. ∀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 :

not(∀x ∈ E ,∃y ∈ F P(x,y)) ⇔ ∃x ∈ E, ∀y ∈ F not(P(x,y))
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 :

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
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
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)
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:

∀ x ∈ A , x ≠ 0
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.
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:

ExF = ∅ ⇔ E=∅ OR F=∅
ExF = FxE ⇔ E=F OR E=∅ OR F=∅
E x (F∪G) = (ExF) ∪ (ExG)
(E∪F) x G = (ExG) ∪ (FxG)
(E∪F) ∩ (GxH) = (E ∩ G) x (F ∩ H)
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

We say that R is reflexive if ∀ x ∈ E, xRx (for any element x in E,x is related to itself)
We say that R is symmetrical if ∀ x,y ∈ E , xRy ⇒ yRx
We say that R is transitive if ∀ x,y,z ∈ E (xRy , yRz) ⇒ xRz
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.

The order R is called total if ∀ x,y ∈ E xRy OR yRx
The order R is called partial if ∃ x,y ∈ E xR̅y AND yR̅x

TODO Examples :

∀x,y ∈ ℝ , xRy ⇔ x²-y²=x-y

Prove that R is an equivalence relation
Let a ∈ ℝ, find ̅a