Math / Algebra

QuestionExercice 1 .(04 Pts)
1. Show by induction : $\forall n \in \mathbb{N}: \sum_{k=0}^{n} 2^{k}=2^{n+1}-1 ;(02 \mathrm{Pts})$
2. Show by contrapositive : $n$ prime $\Rightarrow n=2$ or $n$ odd.(02 Pts)
Exercice 2 .(04 Pts) Let $\Re$ the relation in $\mathbb{R}$ defined by : $a \Re b \Leftrightarrow a^{3}-b^{3}=a-b$
1. Show that $\Re$ is an equivalence relation. ( 02 Pts )
2. Discuss according to $m$ the number of elements in the class of $m .(02 \mathrm{Pts})$
Exercice 3 .(12 Pts) Let $S \neq \emptyset, A, B \in \mathcal{P}(S)$ . We define a function $f$ as: $\begin{aligned} f: \mathcal{P}(S) & \rightarrow \mathcal{P}(A) \times \mathcal{P}(B) \\ X & \mapsto(X \cap A ; X \cap B) \end{aligned}$
1. Let $A=\{e,\{f, g\}\}, B=\{d\}$ . Calculate the power set of $A$ , and the Cartesian product of $\mathcal{P}(A)$ and $\mathcal{P}(B)$ . (03 Pts)
2. Show by contradiction that if $f$ is injective then $S \subset A \cup B$ .(02 Pts)
3. Show by contradiction that if $f$ is surjective then $A \cap B \subset 0 .(02 \mathrm{Pts})$
4. Show quickly that $A \cup B \subset S$ and $\emptyset \subset A \cap B .(01 \mathrm{Pts})$
5. Using the direct proof, show that if $A \cup B=S$ then $f$ is injective.( 01 Pts )
6. Using the direct proof, show that if $A \cap B=\emptyset$ then $f$ is surjective.(01 Pts)
7. Using the previous questions, give the necessary condition to $f$ to be a bijective function, then calculate $f^{-1} .(02 \mathrm{Pts})$

Studdy Solution

STEP 1

1. We are given three exercises with multiple parts.
2. We need to use mathematical induction, contrapositive proof, and properties of equivalence relations.
3. We need to explore properties of functions, specifically injectivity and surjectivity.

STEP 2

1. Solve Exercise 1 Part 1 using mathematical induction.
2. Solve Exercise 1 Part 2 using contrapositive proof.
3. Solve Exercise 2 Part 1 by showing the relation is an equivalence relation.
4. Solve Exercise 2 Part 2 by discussing the equivalence class.
5. Solve Exercise 3 Part 1 by calculating power sets and Cartesian products.
6. Solve Exercise 3 Part 2 by showing injectivity implies a condition.
7. Solve Exercise 3 Part 3 by showing surjectivity implies a condition.
8. Solve Exercise 3 Part 4 by showing set inclusions.
9. Solve Exercise 3 Part 5 by proving injectivity directly.
10. Solve Exercise 3 Part 6 by proving surjectivity directly.
11. Solve Exercise 3 Part 7 by determining bijectivity conditions and calculating the inverse function.

STEP 3

To prove $\sum_{k=0}^{n} 2^{k} = 2^{n+1} - 1$ by induction, we start with the base case $n = 0$ .
For $n = 0$ , the left-hand side is $\sum_{k=0}^{0} 2^{k} = 2^{0} = 1$ .
The right-hand side is $2^{0+1} - 1 = 2 - 1 = 1$ .
Thus, the base case holds.

STEP 4

Assume the statement is true for $n = m$ , i.e., $\sum_{k=0}^{m} 2^{k} = 2^{m+1} - 1$ .
We need to show it holds for $n = m + 1$ .
Consider $\sum_{k=0}^{m+1} 2^{k} = \sum_{k=0}^{m} 2^{k} + 2^{m+1}$ .
By the induction hypothesis, $\sum_{k=0}^{m} 2^{k} = 2^{m+1} - 1$ .
Thus, $\sum_{k=0}^{m+1} 2^{k} = (2^{m+1} - 1) + 2^{m+1} = 2 \cdot 2^{m+1} - 1 = 2^{m+2} - 1$ .
This completes the induction step.

STEP 5

To prove by contrapositive that if $n$ is prime, then $n = 2$ or $n$ is odd, we consider the contrapositive: if $n \neq 2$ and $n$ is even, then $n$ is not prime.
If $n$ is even and $n \neq 2$ , then $n = 2k$ for some integer $k \geq 2$ .
Thus, $n$ has divisors other than 1 and itself, so $n$ is not prime.
This proves the contrapositive.

STEP 6

To show that $\Re$ is an equivalence relation, we need to check reflexivity, symmetry, and transitivity.
Reflexivity: For any $a \in \mathbb{R}$ , $a^3 - a^3 = a - a$ , so $a \Re a$ .
Symmetry: If $a \Re b$ , then $a^3 - b^3 = a - b$ . This implies $b^3 - a^3 = b - a$ , so $b \Re a$ .
Transitivity: If $a \Re b$ and $b \Re c$ , then $a^3 - b^3 = a - b$ and $b^3 - c^3 = b - c$ .
Adding these gives $a^3 - c^3 = a - c$ , so $a \Re c$ .
Thus, $\Re$ is an equivalence relation.

STEP 7

To discuss the equivalence class of $m$ , consider the equation $a^3 - m^3 = a - m$ .
This simplifies to $(a - m)(a^2 + am + m^2) = a - m$ .
If $a \neq m$ , then $a^2 + am + m^2 = 1$ .
The number of solutions depends on $m$ .
If $m = 0$ , then $a^2 = 1$ , so $a = \pm 1$ .
If $m \neq 0$ , solve $a^2 + am + m^2 = 1$ for $a$ .

STEP 8

Calculate $\mathcal{P}(A)$ where $A = \{e, \{f, g\}\}$ .
The power set $\mathcal{P}(A) = \{\emptyset, \{e\}, \{\{f, g\}\}, \{e, \{f, g\}\}\}$ .
Calculate $\mathcal{P}(B)$ where $B = \{d\}$ .
The power set $\mathcal{P}(B) = \{\emptyset, \{d\}\}$ .
The Cartesian product $\mathcal{P}(A) \times \mathcal{P}(B)$ consists of all ordered pairs of elements from $\mathcal{P}(A)$ and $\mathcal{P}(B)$ .

STEP 9

To show by contradiction that if $f$ is injective, then $S \subset A \cup B$ , assume $f$ is injective and $S \not\subset A \cup B$ .
Then there exists $x \in S$ such that $x \notin A \cup B$ .
Consider $X = \{x\}$ .
Then $f(X) = (X \cap A, X \cap B) = (\emptyset, \emptyset)$ .
Since $f$ is injective, no other subset of $S$ can map to $(\emptyset, \emptyset)$ , contradicting $x \notin A \cup B$ .
Thus, $S \subset A \cup B$ .

STEP 10

To show by contradiction that if $f$ is surjective, then $A \cap B \subset \emptyset$ , assume $f$ is surjective and $A \cap B \not\subset \emptyset$ .
Then there exists $x \in A \cap B$ .
Consider $(X \cap A, X \cap B) = (\{x\}, \{x\})$ .
Since $f$ is surjective, there exists $X \subset S$ such that $f(X) = (\{x\}, \{x\})$ .
This implies $x \in A \cap B$ , contradicting $A \cap B \subset \emptyset$ .
Thus, $A \cap B \subset \emptyset$ .

STEP 11

To show $A \cup B \subset S$ , note that $A, B \subset S$ implies $A \cup B \subset S$ .
To show $\emptyset \subset A \cap B$ , note that the empty set is a subset of any set.

STEP 12

To show that if $A \cup B = S$ , then $f$ is injective, assume $A \cup B = S$ .
For $f$ to be injective, $f(X_1) = f(X_2)$ implies $X_1 = X_2$ .
If $f(X_1) = f(X_2)$ , then $(X_1 \cap A, X_1 \cap B) = (X_2 \cap A, X_2 \cap B)$ .
Since $A \cup B = S$ , $X_1 = X_2$ .

STEP 13

0 To show that if $A \cap B = \emptyset$ , then $f$ is surjective, assume $A \cap B = \emptyset$ .
For $f$ to be surjective, for each pair $(Y, Z) \in \mathcal{P}(A) \times \mathcal{P}(B)$ , there exists $X \subset S$ such that $f(X) = (Y, Z)$ .
Since $A \cap B = \emptyset$ , $X = Y \cup Z$ satisfies $f(X) = (Y, Z)$ .

STEP 14

1 The necessary condition for $f$ to be bijective is $A \cup B = S$ and $A \cap B = \emptyset$ .
Calculate $f^{-1}$ by solving $f(X) = (Y, Z)$ for $X$ .
Given $(Y, Z)$ , $X = Y \cup Z$ because $A \cap B = \emptyset$ .
The necessary condition for $f$ to be bijective is $A \cup B = S$ and $A \cap B = \emptyset$ .
The inverse function $f^{-1}$ is given by $f^{-1}(Y, Z) = Y \cup Z$ .

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Studdy Solution

STEP 1

1. We are given three exercises with multiple parts.
2. We need to use mathematical induction, contrapositive proof, and properties of equivalence relations.
3. We need to explore properties of functions, specifically injectivity and surjectivity.

STEP 2

STEP 3

To prove $\sum_{k=0}^{n} 2^{k} = 2^{n+1} - 1$ by induction, we start with the base case $n = 0$ .
For $n = 0$ , the left-hand side is $\sum_{k=0}^{0} 2^{k} = 2^{0} = 1$ .
The right-hand side is $2^{0+1} - 1 = 2 - 1 = 1$ .
Thus, the base case holds.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

STEP 11

To show $A \cup B \subset S$ , note that $A, B \subset S$ implies $A \cup B \subset S$ .
To show $\emptyset \subset A \cap B$ , note that the empty set is a subset of any set.

STEP 12

To show that if $A \cup B = S$ , then $f$ is injective, assume $A \cup B = S$ .
For $f$ to be injective, $f(X_1) = f(X_2)$ implies $X_1 = X_2$ .
If $f(X_1) = f(X_2)$ , then $(X_1 \cap A, X_1 \cap B) = (X_2 \cap A, X_2 \cap B)$ .
Since $A \cup B = S$ , $X_1 = X_2$ .

STEP 13

STEP 14

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Studdy Solution

STEP 1

1. We are given three exercises with multiple parts.2. We need to use mathematical induction, contrapositive proof, and properties of equivalence relations.3. We need to explore properties of functions, specifically injectivity and surjectivity.

STEP 2

STEP 3

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

STEP 11

To show A∪B⊂S A \cup B \subset S A∪B⊂S, note that A,B⊂S A, B \subset S A,B⊂S implies A∪B⊂S A \cup B \subset S A∪B⊂S.To show ∅⊂A∩B \emptyset \subset A \cap B ∅⊂A∩B, note that the empty set is a subset of any set.

STEP 12

STEP 13

STEP 14

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are given three exercises with multiple parts.
2. We need to use mathematical induction, contrapositive proof, and properties of equivalence relations.
3. We need to explore properties of functions, specifically injectivity and surjectivity.

To show $A \cup B \subset S$ , note that $A, B \subset S$ implies $A \cup B \subset S$ .
To show $\emptyset \subset A \cap B$ , note that the empty set is a subset of any set.