The problem asks us to prove the following statement using either proof by contraposition or proof by contradiction:

Suppose $a \in \mathbb{Z}$. If $a^2 - 3a + 5$ is even, then $a$ is odd.

Approach: Proof by Contraposition

We will proceed by proving the contrapositive statement:

• The contrapositive of the original statement is: "If $a$ is even, then $a^2 - 3a + 5$ is odd."

Step-by-Step Proof:

1. Assume $a$ is even:
   If $a$ is even, we can express $a$ as $a = 2k$ for some integer $k \in \mathbb{Z}$.

2. Substitute $a = 2k$ into $a^2 - 3a + 5$:
   $a^2 - 3a + 5 = (2k)^2 - 3(2k) + 5$ Simplifying this expression: $= 4k^2 - 6k + 5$

3. Analyze the parity of $4k^2 - 6k + 5$:

   • $4k^2$ is clearly even because it is a multiple of 4.
   • $6k$ is also even because it is a multiple of 2.
   • So, $4k^2 - 6k$ is the sum of two even numbers, which is even.
   • Finally, adding 5 (an odd number) to an even number results in an odd number.

   Therefore, $a^2 - 3a + 5 = 4k^2 - 6k + 5$ is odd.

4. Conclusion:
   We have shown that if $a$ is even, then $a^2 - 3a + 5$ is odd. This is the contrapositive of the original statement, so by the law of contrapositive, the original statement is true:

   • If $a^2 - 3a + 5$ is even, then $a$ must be odd.

Thus, the proof is complete.

Would you like more details on any part of this explanation, or have any questions?

Related Questions:

1. How does proof by contraposition differ from direct proof?
2. Can you apply proof by contradiction to this problem instead of contraposition?
3. Why does the parity of $a^2 - 3a + 5$ depend on whether $a$ is even or odd?
4. What are the properties of even and odd numbers that make such proofs possible?
5. How can you generalize this proof technique to other quadratic expressions?

Tip: Contrapositive proofs are often simpler when dealing with even and odd properties because they allow you to assume a more manageable hypothesis (such as "even" instead of "odd").