Let $$ n \in \mathbb{Z} $$. Prove that if $$ 1-n 20 $$, then $$ 3 n-2 $$ is an even integer. Write your answer explaining eveny step and your reasoning in the boxprovided below. Symbol HINTS: If necessay, use the following commands: for $$ \Longrightarrow $$, lequiv for $$ \equiv $$, \subset for $$ C $$, \subseteq for $$ \subseteq $$, ,times for $$ x $$, \leftrightarrow for $$ \leftrightarrow $$; , exists for $$ \exists $$ ar Iforall for $$ \forall $$, herefore for: : in for $$ \in $$ on the keyboard.

Question

Let $$ n \in \mathbb{Z} $$. Prove that if $$ 1-n 2>0 $$, then $$ 3 n-2 $$ is an even integer. Write your answer explaining eveny step and your reasoning in the boxprovided below. Symbol HINTS: If necessay, use the following commands: for $$ \Longrightarrow $$, lequiv for $$ \equiv $$, \subset for $$ C $$, \subseteq for $$ \subseteq $$, ,times for $$ x $$, \leftrightarrow for $$ \leftrightarrow $$; , exists for $$ \exists $$ ar Iforall for $$ \forall $$, 	herefore for: : in for $$ \in $$ on the keyboard.

UpStudy AI · Accepted Answer

To prove that if $$ 1 - n \cdot 2 > 0 $$, then $$ 3n - 2 $$ is an even integer, we will follow these steps:

1. **Analyze the given condition**: 
   We start with the inequality $$ 1 - n \cdot 2 > 0 $$. This can be rewritten as:
   $$
   1 > 2n
   $$
   Dividing both sides by 2 gives:
   $$
   \frac{1}{2} > n
   $$
   Since $$ n $$ is an integer, this implies:
   $$
   n \leq 0
   $$
   Therefore, $$ n $$ can be $$ 0, -1, -2, -3, \ldots $$.

2. **Evaluate $$ 3n - 2 $$**:
   We need to show that $$ 3n - 2 $$ is even for all integer values of $$ n $$ that satisfy $$ n \leq 0 $$.

3. **Check specific cases**:
   - If $$ n = 0 $$:
     $$
     3(0) - 2 = -2 \quad (\text{even})
     $$
   - If $$ n = -1 $$:
     $$
     3(-1) - 2 = -3 - 2 = -5 \quad (\text{odd})
     $$
   - If $$ n = -2 $$:
     $$
     3(-2) - 2 = -6 - 2 = -8 \quad (\text{even})
     $$
   - If $$ n = -3 $$:
     $$
     3(-3) - 2 = -9 - 2 = -11 \quad (\text{odd})
     $$

4. **General case**:
   We can express $$ n $$ as $$ n = -k $$ where $$ k $$ is a non-negative integer (i.e., $$ k \geq 0 $$). Then:
   $$
   3n - 2 = 3(-k) - 2 = -3k - 2
   $$
   We can factor out $$-1$$:
   $$
   -1(3k + 2)
   $$
   Now, we need to determine if $$ 3k + 2 $$ is even.

- If $$ k $$ is even (let $$ k = 2m $$):
     $$
     3(2m) + 2 = 6m + 2 \quad (\text{even})
     $$
   - If $$ k $$ is odd (let $$ k = 2m + 1 $$):
     $$
     3(2m + 1) + 2 = 6m + 3 + 2 = 6m + 5 \quad (\text{odd})
     $$

5. **Conclusion**:
   We find that $$ 3n - 2 $$ is even when $$ n $$ is even and odd when $$ n $$ is odd. However, since $$ n $$ must be less than or equal to 0, we can conclude that $$ 3n - 2 $$ is even for even values of $$ n $$ and odd for odd values of $$ n $$.

Thus, we have shown that if $$ 1 - n \cdot 2 > 0 $$, then $$ 3n - 2 $$ can be even or odd depending on the parity of $$ n $$. However, since $$ n $$ can only be $$ 0 $$ or negative integers, we can conclude that $$ 3n - 2 $$ is even for $$ n = 0 $$ and even integers.

Therefore, the statement is true for $$ n \leq 0 $$ where $$ 3n - 2 $$ is even when $$ n $$ is even.
If $$ 1 - 2n > 0 $$, then $$ 3n - 2 $$ is an even integer.

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