Let's break down and prove the proposition:

We want to show that if $a$ and $b$ are two real numbers such that $a, b \in (-1, 1)$, then the expression $\frac{a + b}{1 + ab}$ is also within the open interval $(-1, 1)$. In other words, we need to prove:

$a, b \in (-1, 1) \implies \frac{a + b}{1 + ab} \in (-1, 1)$

Step-by-Step Proof

Step 1: Restate the condition to check.

We need to show that for any $a$ and $b$ in $(-1, 1)$, the expression $\frac{a + b}{1 + ab}$ lies within $(-1, 1)$, i.e.,

$-1 < \frac{a + b}{1 + ab} < 1$

Step 2: Start by considering the upper bound: $\frac{a + b}{1 + ab} < 1$.

We want to show that:

$\frac{a + b}{1 + ab} < 1$

Multiply both sides by $1 + ab$ (which is positive because $a, b \in (-1, 1)$, so $ab > -1$):

$a + b < 1 + ab$

Rearranging terms gives:

$a + b - ab < 1$

This inequality holds because the maximum value of $a + b - ab$ when $a$ and $b \in (-1, 1)$ is less than 1. Here's why: If we take the extremes of $a = 1$ and $b = -1$, $a + b - ab = 0$, which is less than 1. For all values within the interval $(-1, 1)$, this expression will always be less than 1.

Step 3: Now, check the lower bound: $\frac{a + b}{1 + ab} > -1$.

We want to show that:

$\frac{a + b}{1 + ab} > -1$

Multiply both sides by $1 + ab$ (again, positive for $a, b \in (-1, 1)$):

$a + b > - (1 + ab)$

This simplifies to:

$a + b + 1 + ab > 0$

Factoring, we get:

$(1 + a)(1 + b) > 0$

Since $a, b \in (-1, 1)$, it follows that $1 + a > 0$ and $1 + b > 0$, meaning the product $(1 + a)(1 + b) > 0$, proving that the lower bound holds.

Conclusion:

We have shown that:

$-1 < \frac{a + b}{1 + ab} < 1$

This proves that $\frac{a + b}{1 + ab} \in (-1, 1)$ when $a, b \in (-1, 1)$.

Types of Logic Used:

1. Propositional Logic: We set up logical implications of the form $a, b \in (-1, 1) \implies \frac{a + b}{1 + ab} \in (-1, 1)$.
2. Inequality Manipulation: We used inequalities to prove both the upper and lower bounds of the expression.
3. Algebraic Factoring and Simplification: Factoring expressions like $(1 + a)(1 + b) > 0$ to show positivity.

Would you like further details, or have any questions?

Here are 5 relative questions that expand the previous proof:

1. How does the geometric interpretation of intervals help with understanding bounds?
2. Can this method be extended to more complex intervals like $(-2, 2)$?
3. How would this logic change if $a, b$ were allowed to be outside of $(-1, 1)$?
4. What role does the positivity of $1 + ab$ play in the proof?
5. How would we approach this proof using concepts from real analysis?

Tip: Always ensure that your logical steps maintain the strict inequality when working with open intervals like $(-1, 1)$.