hey i have a math question; use propositional logic and negatiion to proof it; and say what kinds of logic you used; use another tzchnic than topology and algeber but simple prove that for every a and b in the open interval 1 and -1; that a+b/1+ab is also in the intervel 1 and -1

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 \( (-1, 1) \).

Explore a detailed proof using propositional logic and inequality manipulation to show that for any a and b in the interval (-1, 1), the expression (a + b)/(1 + ab) also lies in the interval (-1, 1). This proof uses basic algebra and interval properties, suitable for college-level mathematics.

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