Question

The product of an even integer and an odd integer is even.

"Proof: Suppose m is an even integer and n is an odd integer. If m • n is even, then by definition of even there exists an integer r such that m • n = 2r. Also since m is even, there exists an integer p such that m = 2p, and since n is odd there exists an integer q such that n = 2q + 1. Thus
mn = (2p)(2q + 1) = 2r,

Answer 1

A proof was provided demonstrating that the product of an even integer and an odd integer is even by expressing the product in the form 2 times some integer.

Step-by-step explanation:

The question asks for a proof that the product of an even integer and an odd integer is always even. An even integer can be represented as 2p, where p is an integer, and an odd integer can be represented as 2q + 1, where q is an integer. Multiplying an even integer m by an odd integer n, we have m • n = (2p) • (2q + 1). This product simplifies to 4pq + 2p, which can be rewritten as 2(2pq + p). Since 2pq + p is an integer, we have expressed the product mn as 2 times some integer, which by definition means mn is even. This confirms that the product of an even integer and an odd integer is even.