Yes, the product of two consecutive even integers is always divisible by 2.
Here's why:
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Understanding Even Integers: An even integer can be expressed in the form of 2q, where q is any integer.
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Consecutive Even Integers: Two consecutive even integers would then be 2q and 2q+2, where q is an integer.
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Their Product: The product of these two integers would be:
(2q) * (2q + 2) -
Simplifying the Product: Factoring out a 2 from the second term, we get:
(2q) * 2(q + 1) -
Final Product: Simplifying this, we have:
4q(q+1) -
Divisibility by 2 Since 4q(q+1) has a factor of 4, which is itself a multiple of 2, therefore the product must be divisible by 2.
Generalization based on the reference:
The provided reference states: "Hence the product of two consecutive integers is always divisible by 2."
This is not directly related to even integers specifically but reinforces the idea that at least one of two consecutive numbers is always divisible by 2, hence their product is always divisible by 2. This is because if n is an integer, it can be written in the form of 2q or 2q+1. So if we are looking at two consecutive integers n and n+1, one of these must be of the form 2q. Hence n(n+1) must be divisible by 2.
It is helpful to note that consecutive even integers are a subset of consecutive integers, thus if a general property exists for consecutive integers, then that property also applies to consecutive even integers.
Examples:
- 2 and 4: 2 * 4 = 8 (divisible by 2)
- 6 and 8: 6 * 8 = 48 (divisible by 2)
- 10 and 12: 10 * 12 = 120 (divisible by 2)
In fact, due to the existence of the factor of 4, the product of two consecutive even integers is divisible by 4 and 2.
