Final answer:
To prove that the product m × p is an even integer, given m × n and n × p are even, we can express m × n = 2k and n × p = 2l, then show that m and p are each half even integers, yielding their product is even.
Step-by-step explanation:
To prove that the product of m and p is an even integer given that the product of m and n and the product of n and p are even integers, let's use a direct proof with the definition of even integers.
An even integer is any integer that can be expressed as 2 times another integer. Given that m × n is even, we can say m × n = 2k for some integer k. Similarly, n × p is even, so n × p = 2l for some integer l.
Since n is an integer, there must be some integer q such that n = 2q. Now we can rewrite m × n = 2k as m × (2q) = 2k, which simplifies to m = k/q. Similarly, we rewrite n × p = 2l as (2q) × p = 2l, simplifying to p = l/q. Multiplying m and p together yields:
m × p = (k/q) × (l/q) = kl/q².
Since k and l are both products of integers and q, being an integer, when squared is still an integer, the product of kl/q² is an integer. We're not finished yet because we need to show that m × p is specifically even. Since k and l are both even (they were each set to 2 times some integer), their product, kl, will also be even. Thus, the entire fraction kl/q², which equals m × p, is even as it is 2 times some integer.
This completes the proof that m × p is an even integer.