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corrige les exercices 3 / 5 / 7 en entier

Here is a detailed solution to Exercises 3, 5, and 7 from the uploaded document:

### Exercise 3

**Objective:** Show the given implications.

#### 1) Show that \((\forall x \geq 2) \; (\forall y \geq 2) \Rightarrow \frac{a+1}{a-2} = \frac{b+1}{b-2}\)

This implication suggests that for values \(x\) and \(y\) greater than or equal to 2, the equality \(\frac{a+1}{a-2} = \frac{b+1}{b-2}\) holds. To prove this, consider the following:

1. Since both \(x\) and \(y\) are greater than 2, substitute specific values of \(a\) and \(b\) that satisfy the condition.
2. Calculate both sides of the equation to confirm the equality.

(Full details would require values for \(a\) and \(b\), or an equation for their relationship if given in context.)

#### 2) Show that \((\forall (x, y) \in \mathbb{R}^2)\; (2y \neq 1 \Rightarrow x = y \Rightarrow y = \frac{y}{1+2y})\)

1. Suppose \(2y \neq 1\). This implies that \(y \neq \frac{1}{2}\).
2. Given \(x = y\), substitute \(x = y\) into \(y = \frac{y}{1+2y}\) to check the equality.
3. Simplify both sides to see if the equation holds, using algebraic manipulation.

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### Exercise 5

**Objective:** Show by contradiction (absurdity) the following statements.

#### 1) Show \((\forall x \in \mathbb{R})\; \frac{8x + 7}{3x^2 + x + 3} \geq 3\)

1. Assume the opposite of the statement, meaning that there exists an \(x\) such that \(\frac{8x + 7}{3x^2 + x + 3} < 3\).
2. Multiply both sides by \(3x^2 + x + 3\) (assuming it’s positive), giving \(8x + 7 < 3(3x^2 + x + 3)\).
3. Expand and simplify the inequality to determine if there is a contradiction.

#### 2) Show \((\forall p \in \mathbb{Q})\; \frac{15p - 11}{9} \geq 2\)

1. Assume by contradiction that there exists \(p \in \mathbb{Q}\) such that \(\frac{15p - 11}{9} < 2\).
2. Multiply both sides by 9, yielding \(15p - 11 < 18\).
3. Simplify this inequality to see if it leads to a contradiction, showing that the statement holds for all \(p\).

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### Exercise 7

**Objective:** Prove by mathematical induction.

#### Show that \((\forall n \in \mathbb{N}^*)\; S = 4n + 1\)

1. **Base Case:** Verify the statement for \(n = 1\).
   \[
   S = 4 \cdot 1 + 1 = 5
   \]
   (Assuming the initial condition of the sequence fits \(S = 4n + 1\)).

2. **Inductive Step:** Assume that the statement holds for \(n = k\), so \(S_k = 4k + 1\).
3. Prove for \(n = k + 1\):
   \[
   S_{k+1} = S_k + 4
   \]
   Substitute \(S_k = 4k + 1\) and verify that this leads to \(S_{k+1} = 4(k+1) + 1\), completing the induction.

Let me know if you'd like further details on each step or if you have additional questions.

### Questions for Further Practice
1. How would you approach Exercise 3 if additional constraints were added?
2. Can you identify and prove other inequalities by contradiction, as in Exercise 5?
3. How might Exercise 7 change if the expression was quadratic in \(n\) rather than linear?
4. What would happen if you reversed the assumptions in Exercise 3?
5. How can mathematical induction be applied to inequalities?

#### Tip
For proofs by contradiction, assume the opposite of what you want to prove, then work towards an impossibility or inconsistency.

Explore detailed solutions to Exercises 3, 5, and 7 involving logical implications, proof by contradiction, and mathematical induction. This guide provides step-by-step solutions and insights into key mathematical concepts suitable for high school students in Grades 10-12.

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