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.