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Algebra 1

• Contenu de la Matiére
  • Rappels et compléments (11H)
  • Structures Algébriques (11H)
  • Polynômes et fractions rationnelles
• Premier cours : Logique mathématique et méthodes du raisonnement mathématique Sep 25 :
  • Properties:
    • Absorption:
    • Commutativity:
    • Associativity:
    • Distributivity:
    • Neutral element:
    • Negation of a conjunction & a disjunction:
    • Transitivity:
    • Contraposition:
    • God only knows what this property is called:
  • Some exercices I found online :
    • USTHB 2022/2023 Section B :
      • Exercice 1: Démontrer les équivalences suivantes:
      • Exercice 2: Dire si les propositions suivantes sont vraies ou fausses, et les nier:
• 2éme cours Oct 2
  • Quantifiers
    • Proprieties
      • Propriety Number 1:
      • Propriety Number 2:
      • Propriety Number 3:
      • Propriety Number 4:
  • Multi-parameter proprieties :
    • Proprieties :
  • Methods of mathematical reasoning :
    • Direct reasoning :
      • Example:
    • Reasoning by the Absurd:
      • Example:
    • Reasoning by contraposition:
    • Reasoning by counter example:
• 3eme Cours : Oct 9
  • Reasoning by recurrence :
    • Example:
• 4eme Cours : Chapitre 2 : Sets and Operations
  • Definition of a set :
  • Belonging, inclusion, and equality :
  • Intersections and reunions :
    • Intersection:
    • Union:
    • Difference between two sets:
    • Complimentary set:
    • Symmetrical difference
  • Proprieties :
    • Commutativity:
    • Associativity:
    • Distributivity:
    • Lois de Morgan:
    • An other one:
    • An other one:
    • And an other one:
    • And the last one:
• 5eme cours: L'ensemble des parties d'un ensemble Oct 16
  • Notes :
  • Examples :
  • Partition of a set :
  • Cartesian products :
    • Example :
    • Some proprieties:
• Binary relations in a set :
  • Definition :
  • Proprieties :
  • Equivalence relationship :
    • Equivalence class :
      • The quotient set :
  • Order relationship :
    • Examples :
• TP exercices Oct 20 :
  • Exercice 3 :
    • Question 3
  • Exercice 4 :
    • Question 1 :
• Chapter 3 : Applications
  • 3.1 Generalities about applications :
    • Definition :
      • Some examples :
    • Restriction and prolongation of an application :
      • Example
    • Composition of applications :
  • 3.2 Injection, surjection and bijection :
    • Proposition :
  • 3.3 Reciprocal applications :
    • Def :
    • Theorem :
    • Some proprieties :
  • 3.4 Direct Image and reciprocal Image :
    • Direct Image :
      • Example :
    • Reciprocal image :
      • Example :

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

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 :

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 :

1. 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
2. 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
3. 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)
4. 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

Symmetrical 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:

1. ∀ x ∈ A , x ≠ 0
2. 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.
3. 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

TP exercices Oct 20 :

Exercice 3 :

Question 3

Montrer par l'absurde que P : ∀x ∈ ℝ*, √(4+x³) ≠ 2 + x³/4 est vraies

On suppose que ∃ x ∈ ℝ* , √(4+x³) = 2 + x³/4 4+x³ = (2 + x³/4)² 4+x³ = 4 + x⁶/16 + 4*(x³/4) 4+x³ = 4 + x⁶/16 + x³ x⁶/16 = 0 x⁶ = 0 x = 0 . Or, x appartiens a ℝ\{0}, donc P̅ est fausse. Ce qui est equivalent a dire que P est vraie

Exercice 4 :

DONE Question 1 :

∀ n ∈ ℕ* , (n ,k=1)Σ1/k(k+1) = 1 - 1/1+n P(n) : (n ,k=1)Σ1/k(k+1) = 1 - 1/1+n

1. On vérifie P(n) pour n = 1

(1 ,k=1)Σ1/k(k+1) = 1/1(1+1) = 1/2 — (1) 1 - 1/1+1 = 1 - 1/2 = 1/2 — (2) De (1) et (2), P(0) est vraie —- (a)

1. On suppose que P(n) est vraie pour n ≥ n1 puis on vérifie pour n+1

(n ,k=1)Σ1/k(k+1) = 1 - 1/1+n (n ,k=1)Σ1/k(k+1) + 1/(n+1)(n+2) = 1 - (1/(1+n)) + 1/(n+1)(n+2) (n+1 ,k=1)Σ1/k(k+1) = 1 - 1/(n+1) + 1/[(n+1)(n+2)] (n+1 ,k=1)Σ1/k(k+1) = 1 + 1/[(n+1)(n+2)] - (n+2)/[(n+1)(n+2)] (n+1 ,k=1)Σ1/k(k+1) = 1 + [1-(n+2)]/[(n+1)(n+2)] (n+1 ,k=1)Σ1/k(k+1) = 1 + [-n-1]/[(n+1)(n+2)] (n+1 ,k=1)Σ1/k(k+1) = 1 - [n+1]/[(n+1)(n+2)] (n+1 ,k=1)Σ1/k(k+1) = 1 - 1/(n+1+1) CQFD

Donc P(n+1) est vraie. —- (b) De (a) et (b) on conclus que la proposition de départ est vraie

Chapter 3 : Applications

3.1 Generalities about applications :

Definition :

Let E and F be two sets.

1. We call a function of the set E to the set F any relation from E to F such as for any element of E, we can find at most one element of F that corresponds to it.
2. We call an application of the set E to the set F a relation from E to F such as for any element of E, we can find one and only one element of F that corresponds to it.

3. f: E₁ —> F₁ ; g: E₂ —> F₂ ; f ≡ g ⇔ [E₁ = E₂ ; F₁ = F₂ ; f(x) = g(x) ∀x ∈ E₁

   Generally speaking, we schematize a function or an application by this writing :

   f : E —> F x —> f(x)=y Γ = {(x , f(x))/ x ∈ E ; f(x) ∈ F} is the graph of f

Some examples :

Ex1:

f : ℝ —> ℝ x —> f(x) = (x-1)/x is a function, because 0 does NOT have a corresponding element using that relation.

Ex2:

f : ℝ^* —> ℝ x —> f(x)= (x-1)/x is, however, an application

Restriction and prolongation of an application :

Let f : E -> F an application and E₁ ⊂ E therefore :

g : E₁ -> F g(x) = f(x) ∀x ∈ E₁

g is called the restriction of f to E₁. And f is called the prolongation of g to E.

Example

f : ℝ —> ℝ x —> f(x) = x²

g : [0 , +∞[ —> ℝ x —> g(x) = x²

g is called the restriction of f to ℝ⁺. And f is called the prolongation of g to ℝ.

Composition of applications :

Let E,F, and G be three sets, f: E -> F and g: F -> G are two applications. We define their composition, symbolized by g_of as follow :

g_of : E -> G . ∀x ∈ E (g_of)_(x)= g(f(x))

3.2 Injection, surjection and bijection :

Let f: E -> F be an application :

1. We say that f is injective if : ∀x,x' ∈ E : f(x) = f(x') ⇒ x = x'
2. We say that f is surjective if : ∀ y ∈ F , ∃ x ∈ E : y = f(x)
3. We say that if is bijective if it's both injective and surjective at the same time.

Proposition :

Let f : E -> F be an application. Therefore:

1. f is injective ⇔ y = f(x) has at most one solution.
2. f is surjective ⇔ y = f(x) has at least one solution.
3. f is bijective ⇔ y = f(x) has a single and unique solution.

3.3 Reciprocal applications :

Def :

Let f : E -> F a bijective application. So there exists an application named f^-1 : F -> E such as : y = f(x) ⇔ x = f^-1(y)

Theorem :

Let f : E -> F be a bijective application. Therefore its reciprocal f^-1 verifies : f^-1_of=Id_E ; f_of^-1=Id_F Or :

Id_E : E -> E ; x -> Id_E(x) = x

Some proprieties :

1. (f^-1)^-1 = f
2. (g_of)⁻¹ = f⁻¹_og⁻¹
3. The graphs of f and f⁻¹ are symmetrical to each other by the first bis-sectrice of the equation y = x

3.4 Direct Image and reciprocal Image :

Direct Image :

Let f: E-> F be an application and A ⊂ E. We call a direct image of A by f, and we symbolize as f(A) the subset of F defined by :

f(A) = {f(x)/ x ∈ A} ; = { y ∈ F ∃ x ∈ A y=f(x)}

Example :

f: ℝ -> ℝ x -> f(x) = x² A = {0,4} f(A) = {f(0), f(4)} = {0, 16}

Reciprocal image :

Let f: E -> F be an application and B ⊂ F. We call the reciprocal image of E by F the subset f^-1(B) :

f^-1(B) = {x ∈ E/f(x) ∈ B} ; x ∈ f^-1(B) ⇔ f(x) ∈ B

Example :

f: ℝ -> ℝ x -> f(x) = x² B = {1,9,4} f^-1(B) = {1,-1,2,-2,3,-3} = {x ∈ ℝ/x² ∈ {1,4,9}}