Let \( n \in \mathbb{Z} \). Prove that if \( 1-n 2>0 \), then \( 3 n-2 \) is an even integer: Write your answer explaining every step and your reasoning in the box provided b Symbol HINTS: If necessay, use the following commands: \( a^{*} b \) for \( a^{b} ; Q \) for \( \mathbb{Q} \); \( a / b \) for \( \frac{a}{b} ; a \) Nneq \( b \) for \( a \neq b \); \neg for \( \neg \), vwedge for \( A \), vee for for \( \Longrightarrow \), \equiv for \( \equiv \), \subset for \( \subset \), \subseteq for \( \subseteq \), , ,imes for \( \times \), Meftrightar \forall for \( \forall \). \therefore for : \( ; \) in for \( \in \) on the keyboard.

To prove that if \( 1 - n^2 > 0 \), then \( 3n - 2 \) is an even integer, we start with the condition \( 1 - n^2 > 0 \). This implies \( n^2 < 1 \). Since \( n \in \mathbb{Z} \), the only integer solutions to this inequality are \( n = 0 \) or \( n = \pm 1 \). Next, we evaluate \( 3n - 2 \) for each case: 1. For \( n = 0 \): \[ 3(0) - 2 = -2, \] which is even. 2. For \( n = 1 \): \[ 3(1) - 2 = 1, \] which is odd. 3. For \( n = -1 \): \[ 3(-1) - 2 = -3, \] which is odd. Now, upon checking \( n = 0 \), we see that \( 3n - 2 \) is indeed even only when \( n = 0 \). It's crucial to note that both \( n = 1 \) and \( n = -1 \) yield odd results. Therefore, the conclusion can be drawn that the condition \( 1 - n^2 > 0 \) does not guarantee that \( 3n - 2 \) is even for all integer \( n \) satisfying this condition; it only holds true for \( n = 0 \).