To prove that if n is an integer and 3n² is even, then n must be even, we can use a direct proof by contradiction.
1. **Understanding Even and Odd Numbers:**
An integer is considered even if it can be expressed as 2k, where k is an integer. An integer is odd if it can be expressed as 2k + 1.
2. **Assuming the Opposite:**
Let’s assume for the sake of contradiction that n is odd. Then, based on our definition of odd integers, we can express n as:
n = 2m + 1, where m is an integer.
3. **Substituting n into 3n²:**
We substitute this expression for n into 3n²:
3n² = 3(2m + 1)²
4. **Calculating n²:**
Now, let’s calculate (2m + 1)²:
(2m + 1)² = 4m² + 4m + 1
5. **Plugging this back in:**
Now, replace (2m + 1)² in our equation for 3n²:
3n² = 3(4m² + 4m + 1) = 12m² + 12m + 3
6. **Analyzing the Result:**
Now, notice that 12m² and 12m are both even since they are multiples of 2. However, the number 3 is odd.
7. **Conclusion about 3n²:**
Therefore, 3n² = 12m² + 12m + 3 can be rewritten in the form of even + even + odd = odd, showing that 3n² must be odd.
8. **Contradiction:**
We assumed that n is odd, which led us to conclude that 3n² is odd. However, this contradicts our initial premise that 3n² is even.
Thus, our initial assumption that n is odd must be false, which means that n must be even. We have proven that if 3n² is even, then n must also be even.