( forall a in]-1;1])( forall b in]-1;1[), show that (a + b)/(1 + ab) in]-1;1[

We need to prove that for all \( a \in (-1, 1) \) and \( b \in (-1, 1) \), the expression 

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

is also in the interval \( (-1, 1) \). Here's how we can approach the proof step by step:

### Step 1: Define the boundaries of \( a \) and \( b \)
We are given that \( a \in (-1, 1) \) and \( b \in (-1, 1) \), meaning:

\[
-1 < a < 1 \quad \text{and} \quad -1 < b < 1
\]

### Step 2: Analyze the expression

The expression we need to examine is:

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

Our goal is to prove that \( x \in (-1, 1) \), i.e.,

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

### Step 3: Start by checking the lower bound \( x > -1 \)
To prove that \( x > -1 \), we need:

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

Multiply both sides by \( 1 + ab \) (which is positive because \( -1 < ab < 1 \)):

\[
a + b > -(1 + ab)
\]

This simplifies to:

\[
a + b > -1 - ab
\]

Rearranging terms, we get:

\[
a + b + ab > -1
\]

This inequality holds because both \( a \) and \( b \) are in \( (-1, 1) \), and their sum and product will always result in a value greater than \(-1\). Hence, the lower bound is satisfied:

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

### Step 4: Check the upper bound \( x < 1 \)
Now, we need to prove that:

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

Multiply both sides by \( 1 + ab \) (again, positive):

\[
a + b < 1 + ab
\]

Rearranging terms:

\[
a + b - ab < 1
\]

This inequality also holds because \( a \) and \( b \) are both in \( (-1, 1) \), so their sum and product will always satisfy the inequality. Therefore, the upper bound is satisfied:

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

### Conclusion
Since both the lower and upper bounds are satisfied, we have:

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

Thus, the expression \( \frac{a + b}{1 + ab} \) lies in the interval \( (-1, 1) \) for all \( a \in (-1, 1) \) and \( b \in (-1, 1) \).

Would you like a more detailed explanation on any step, or do you have questions?

Here are 5 related questions:
1. How does the product \( ab \) affect the result of the fraction?
2. What happens to the expression if \( a \) or \( b \) equals 0?
3. Can the method used in this proof be applied to other similar expressions?
4. How would the result change if \( a \) and \( b \) were allowed to be outside the interval \( (-1, 1) \)?
5. What geometric interpretation can we give to the expression \( \frac{a + b}{1 + ab} \)?

**Tip:** When working with inequalities, it's helpful to analyze both upper and lower bounds independently, as we did here.

forall a in]-1;1])( forall b in]-1;1[), show that (a + b)/(1 + ab) in]-1;1[

This detailed proof demonstrates that for any real numbers a and b in the interval (-1, 1), the expression (a + b) / (1 + ab) lies within the interval (-1, 1). The solution uses algebraic manipulation and inequalities, making it suitable for students in Grades 10-12.

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