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diff --git a/src/org/uni_notes/algebra1.org b/src/org/uni_notes/algebra1.org index 21e41ef..865d5b1 100755 --- a/src/org/uni_notes/algebra1.org +++ b/src/org/uni_notes/algebra1.org @@ -1,13 +1,16 @@ #+title: Algebra 1 #+AUTHOR: Crystal #+OPTIONS: ^:{} +#+OPTIONS: \n:y #+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 - +#+OPTIONS: toc:4 +#+HTML_LINK_HOME: https://crystal.tilde.institute/ +#+HTML_LINK_UP: ../../../uni_notes/ +#+OPTIONS: tex:imagemagick * Contenu de la Matiére ** Rappels et compléments (11H) - Logique mathématique et méthodes du raisonnement mathématique @@ -505,7 +508,7 @@ Let E be a set. We define P(E) as the set of all parts of E : *P(E) = {X/X ⊂ 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}} +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: @@ -515,10 +518,10 @@ 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)} +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 +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=∅ @@ -552,3 +555,42 @@ Let E be a set and R be a relation defined in E. We say that R is a relation of ∀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 + +#+BEGIN_VERSE +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 +#+END_VERSE +** Exercice 4 : +*** DONE Question 1 : +#+BEGIN_VERSE +∀ 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 +#+END_VERSE |