Question

Use the method of direct proof to prove the following statements. Suppose \(x, y \in \mathbb{Z} .\) If \(x\) and \(y\) are odd, then \(x y\) is odd.

Short answer

The product of two odd integers is indeed odd.

Step-by-step solution

Understand the properties of odd integers

An odd integer can be represented in the general form \(2n+1\), where \(n\) is any integer. The '+1' distinguishes odd numbers from even numbers, which are represented as \(2m\). To start the proof, let's represent \(x\) and \(y\) in this format. So, let \(x = 2p+1\) and \(y = 2q+1\) where \(p\) and \(q\) are any integers.

Calculate the product \(xy\)

The next step is to calculate the product of \(x\) and \(y\). Substitute \(2p+1\) for \(x\) and \(2q+1\) for \(y\) to get \(xy = (2p+1)(2q+1)\).

Simplify the product

Then simplify the equation to: \(xy = 4pq+2p+2q+1 = 2(2pq+p+q) +1\). As \(2pq, p, q\) are all integers, the expression \(2pq+p+q\) is also integer. Let's denote it as \(r\), where \(r = 2pq+p+q\). So, \(xy = 2r + 1\).

Conclusion

Since \(r\) is an integer, \(xy\) fits the structure of an odd number, \(2n+1\). Therefore, we have proved that the product of two odd integers \(x\) and \(y\) is itself an odd integer. Our initial assumption about \(x\) and \(y\) being odd integers has led to \(xy\) also being an odd integer.