To prove that if $x$ is odd and $y$ is even, then $xy + 3$ is odd, we begin by using the definitions of odd and even integers.

Definitions:

• An odd integer can be expressed as $x = 2k + 1$, where $k$ is some integer.
• An even integer can be expressed as $y = 2m$, where $m$ is some integer.

Proof:

1. Since $x$ is odd, we have: $x = 2k + 1$ for some integer $k$.

2. Since $y$ is even, we have: $y = 2m$ for some integer $m$.

3. Now, consider the product $xy$: $xy = (2k + 1)(2m) = 2k \cdot 2m + 1 \cdot 2m = 4km + 2m$ Notice that $4km + 2m$ is clearly an even number, since both terms are divisible by 2.

4. Now, we calculate $xy + 3$: $xy + 3 = (4km + 2m) + 3$ Since $4km + 2m$ is even, adding 3 (which is odd) to an even number results in an odd number. Therefore: $xy + 3 \text{ is odd.}$

Conclusion:

Thus, we have directly proved that if $x$ is odd and $y$ is even, then $xy + 3$ is odd.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the proof change if both $x$ and $y$ were odd?
2. What happens if both $x$ and $y$ are even in this context?
3. Can you generalize this proof for any odd and even products?
4. How can you use modular arithmetic to simplify the reasoning?
5. What other properties of odd and even numbers can be derived from their products and sums?

Tip: When proving statements involving integers, using modular arithmetic (e.g., modulo 2) can often simplify the logic and reasoning.