Proof by Contraposition
To prove the statement by contraposition, we need to show the equivalent statement: If n is odd, then 3n² is odd.
Let n be an odd integer. This means that n can be expressed in the form n = 2k + 1, where k is an integer. Now, we can compute 3n²:
3n² = 3(2k + 1)² = 3(4k² + 4k + 1) = 12k² + 12k + 3.
The expression 12k² + 12k is even because it is a multiple of 2, and adding 3 (which is odd) to an even number results in an odd number. Thus, we conclude that 3n² is odd when n is odd.
Since we have shown that if n is odd, then 3n² is odd, we have proven the original statement by contraposition: If 3n² is even, then n must be even.
Proof by Contradiction
To prove the statement by contradiction, we start by assuming the opposite: that n is odd, and then show that this assumption leads to a contradiction.
Assume n is odd. Then, similar to before, we can express n as n = 2k + 1 for some integer k. Next, we calculate 3n² again:
3n² = 3(2k + 1)² = 3(4k² + 4k + 1) = 12k² + 12k + 3.
As established, 12k² + 12k is even, and thus 3n² is odd. However, our initial assumption stated that 3n² is even. This is a clear contradiction since something cannot be both even and odd.
Since our assumption that n is odd leads to a contradiction, we conclude that n must be even. Therefore, if 3n² is even, it follows that n is even.