Algebra 1
Table of Contents
- Contenu de la Matiére
- Premier cours : Logique mathématique et méthodes du raisonnement mathématique Sep 25 :
- 2éme cours Oct 2
- 3eme Cours : Oct 9
- 4eme Cours : Chapitre 2 : Sets and Operations
- 5eme cours: L’ensemble des parties d’un ensemble Oct 16
- Binary relations in a set :
- TP exercices Oct 20 :
- Chapter 3 : Applications
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:
(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
- 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
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
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:
∀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
- ∃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
- ∀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
∃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
∀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…”
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))
- not(∀x ∈ E , P(x)) ⇔ ∃ x ∈ E, not(P(x))
- 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)”
- 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!!!
- Example of why it’s 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
- 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(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:
- ∀ 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
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
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)
2. 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.
- 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.
- 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.
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
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.
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 gof as follow :
gof : E -> G . ∀x ∈ E (gof)(x)= g(f(x))
3.2 Injection, surjection and bijection :
Let f: E -> F be an application :
- We say that f is injective if : ∀x,x’ ∈ E : f(x) = f(x’) ⇒ x = x’
- We say that f is surjective if : ∀ y ∈ F , ∃ x ∈ E : y = f(x)
- 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:
- f is injective ⇔ y = f(x) has at most one solution.
- f is surjective ⇔ y = f(x) has at least one solution.
- 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-1of=IdE ; fof-1=IdF Or :
IdE : E -> E ; x -> IdE(x) = x
Some proprieties :
- (f-1)-1 = f
- (gof)⁻¹ = f⁻¹og⁻¹
- 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)}