hey i have a math question; use logic to proof it; and say what kinds of logic you used; prove that for every a and b in the closed interval 1 and -1; that a+b/1+ab is also in the intervel 1 and -1

To prove that for every \( a \) and \( b \) in the closed interval \([-1, 1]\), the expression

\[
\frac{a + b}{1 + ab}
\]

is also within the interval \([-1, 1]\), we can use a combination of algebraic manipulation, logical reasoning, and properties of inequalities.

### Steps to prove:

#### 1. **Algebraic Setup and Assumptions:**
We are given that \( a \in [-1, 1] \) and \( b \in [-1, 1] \). This means:
\[
-1 \leq a \leq 1 \quad \text{and} \quad -1 \leq b \leq 1
\]
We need to show that:
\[
-1 \leq \frac{a + b}{1 + ab} \leq 1
\]
under these conditions.

#### 2. **Apply Bounds:**
Let \( x = \frac{a + b}{1 + ab} \). We aim to check if \( -1 \leq x \leq 1 \) holds. We will first check both ends of the inequality separately.

#### 3. **Upper Bound Check \( x \leq 1 \):**
We want to prove:
\[
\frac{a + b}{1 + ab} \leq 1
\]
Multiply both sides by \( 1 + ab \) (which is non-negative since \( a, b \in [-1, 1] \) ensures \( 1 + ab \geq 0 \)) to eliminate the denominator:
\[
a + b \leq 1 + ab
\]
Rearranging terms:
\[
a + b - ab \leq 1
\]
This can be factored as:
\[
(a - 1)(b - 1) \leq 0
\]
Since \( a \in [-1, 1] \) and \( b \in [-1, 1] \), it is clear that \( (a - 1)(b - 1) \leq 0 \) holds because both terms are negative or zero. Thus, the upper bound \( x \leq 1 \) is satisfied.

#### 4. **Lower Bound Check \( x \geq -1 \):**
Now, we want to prove:
\[
\frac{a + b}{1 + ab} \geq -1
\]
Again, multiply both sides by \( 1 + ab \) (which is non-negative):
\[
a + b \geq - (1 + ab)
\]
Simplify:
\[
a + b \geq -1 - ab
\]
Rearranging terms:
\[
a + b + ab \geq -1
\]
This can be factored as:
\[
(1 + a)(1 + b) \geq 0
\]
Since \( a \in [-1, 1] \) and \( b \in [-1, 1] \), it follows that \( (1 + a)(1 + b) \geq 0 \), as both factors are non-negative. Therefore, the lower bound \( x \geq -1 \) is satisfied.

### Conclusion:
Both bounds \( -1 \leq \frac{a + b}{1 + ab} \leq 1 \) are satisfied, and thus we have proven that the expression \( \frac{a + b}{1 + ab} \) remains in the interval \([-1, 1]\) for any \( a, b \in [-1, 1] \).

### Types of Logic Used:
1. **Inequality Manipulation:** We used logical steps involving algebraic manipulation of inequalities.
2. **Algebraic Reasoning:** Rearranging and factoring inequalities.
3. **Interval Arithmetic:** Leveraging the bounds of \( a \) and \( b \) in the interval \([-1, 1]\).
4. **Case Checking:** Ensuring that the expression holds by analyzing the upper and lower bounds separately.

Do you want more details or explanations on any step?

Here are some related questions you might want to explore:
1. What are the general rules for manipulating inequalities?
2. Can we generalize this result to any interval other than \([-1, 1]\)?
3. How would the proof change if we had open intervals \((-1, 1)\)?
4. What happens when one or both of \( a \) and \( b \) are exactly 1 or -1?
5. How can we extend this result to more than two variables?

**Tip:** When proving inequalities, always carefully analyze each boundary condition and whether equality holds at the extremes.

hey i have a math question; use logic to proof it; and say what kinds of logic you used; prove that for every a and b in the closed interval 1 and -1; that a+b/1+ab is also in the interval 1 and -1

Learn how to prove that for any a and b in the interval [-1, 1], the expression (a + b) / (1 + ab) remains in the same interval. This solution covers algebraic manipulation, inequality reasoning, and interval arithmetic. Suitable for high school students in Grades 10-12.

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