Answer :
the set E = {2, 4, 6, 8, ...} is countable.
Let's define a function f: J -> E as follows:
f(n) = 2n, for every positive integer n.
To show that f is one-to-one, we need to demonstrate that if f(m) = f(n), then m = n for any positive integers m and n.
Let's assume that f(m) = f(n), where m and n are positive integers. Then, we have:
2m = 2n.
Dividing both sides of the equation by 2, we get:
m = n.
Thus, we have shown that if f(m) = f(n), then m = n, proving that f is one-to-one.
Now, let's show that f is onto, which means that for every element y in E, there exists an element x in J such that f(x) = y.
Let y be an arbitrary element in E. Since E consists of even numbers, y must be an even number. Let's express y as y = 2k, where k is a positive integer.
Now, let's consider the positive integer x = k. Applying the function f to x, we get:
f(x) = f(k) = 2k = y.
Thus, for every element y in E, we have found an element x in J such that f(x) = y, proving that f is onto.
Since f is both one-to-one and onto, we have shown that there exists a function between J and E that satisfies the definition of countability. Therefore, the set E = {2, 4, 6, 8, ...} is countable.
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