To prove the statement "If an integer is odd and another integer is even, their product is even" using a direct proof, we proceed as follows:

Restating the Given Statement:

We are tasked with proving that:

• If $m$ is an odd integer and $n$ is an even integer, then the product $m \times n$ is even.

Definitions:

1. Odd integer: An integer $m$ is odd if it can be written as $m = 2k + 1$, where $k$ is some integer.
2. Even integer: An integer $n$ is even if it can be written as $n = 2l$, where $l$ is some integer.

Proof:

1. Assume $m$ is odd: Since $m$ is odd, by definition, we can write: $m = 2k + 1 \quad \text{for some integer } k.$

2. Assume $n$ is even: Since $n$ is even, by definition, we can write: $n = 2l \quad \text{for some integer } l.$

3. Multiply $m$ and $n$: We now find the product of $m$ and $n$: $m \times n = (2k + 1) \times (2l).$ Expanding this expression: $m \times n = 2k \times 2l + 1 \times 2l = 4kl + 2l.$

4. Factor the result: Notice that the product simplifies to: $m \times n = 2(2kl + l).$ Since $2kl + l$ is an integer, we can conclude that: $m \times n = 2 \times (\text{some integer}).$ Therefore, $m \times n$ is even.

Conclusion:

We have shown that the product of an odd integer and an even integer is always even. This concludes the direct proof.

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Would you like more details or have any questions?

Here are five related questions to extend this concept:

1. What happens if both integers are odd?
2. How would you prove that the product of two even numbers is always even?
3. Can an odd number ever result from multiplying two integers? Why or why not?
4. How does the distributive property apply to multiplying an odd and even number?
5. What are the properties of integers under multiplication involving even and odd numbers?

Tip: When proving statements about integers, using their algebraic definitions (like $2k$ for even and $2k + 1$ for odd) helps break down problems systematically.