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Even integers are numbers that are divisible by 2 without any remainder. This means that any integer can be expressed in the form of \(2k\), where \(k\) is also an integer. For instance, numbers like 2, 4, 6, and even negative numbers like -8 are all even integers.
Understanding even integers is crucial because they have specific properties that help in many mathematical applications, such as simplifying calculations and proving statements.
In the context of our given exercise, when we talk about \(a\) and \(b\) being even, it means we can represent them specifically as \(a = 2x\) and \(b = 2y\). By representing them this way, it becomes much easier to handle the problem using their form of \(2\) times some integer.

Even integers come with some interesting properties:

• Any operation involving two even numbers will always yield another even number—for example, adding, subtracting, and multiplying them.
• If you divide any even integer by 2, the result is always an integer. This makes halving even numbers a straightforward task.
• Every even integer contains at least one factor of 2, which is why they fit into the general form \(2k\).

For our proof, these properties help us understand that both \(a\) and \(b\) definitely share a factor of 2, facilitating simpler computation of the greatest common divisor (GCD).
This shared factor is a key aspect because their GCD will also necessarily include that same factor of 2, which we can mathematically confirm.

Proof by divisibility involves demonstrating that a particular number divides another with no remainder. In our problem, we need to show that the GCD of \(a = 2x\) and \(b = 2y\) involves the factor of 2 in an expected way.
Let's recall the given statement \((a, b) = 2(a / 2, b / 2)\). Here, using divisibility, we verify if dividing \(a\) and \(b\) by their common factor 2 correctly leads to an equivalent expression for the GCD.
By rewriting \(a\) and \(b\) as \(2x\) and \(2y\), and applying the rule of divisibility, we determine:

• The original GCD, \((a, b) = 2k\), where \(k\) is the GCD of \(x\) and \(y\).
• The equation \((x, y) = k\) holds true because when we divide both sides of \((2x, 2y) = 2k\) by 2, we find that the factor 2 consistently aligns.

Thus, these steps confirm both intuitively and mathematically that the shared factor 2 remains present as expected, ensuring that the GCD is effectively computed as 2 times the GCD of the halved values \(x\) and \(y\). This verification through divisibility satisfies the proof required by the problem.