Question

consider u = {x|x is a positive integer greater than 0}. which is an empty set? {x|x ∈ u and 2x can be written as a fraction} {x|x ∈ u and 2x is prime} {x|x ∈ u and (1/2)x is prime} {x|x ∈ u and (1/2)x can be written as a fraction}

Explanation:

Step1: Analyze the first set

Any integer \(2x\) (where \(x\in U\), \(U\) is positive integers) can be written as a fraction (e.g., \(2x=\frac{2x}{1}\)), so this set is not empty.

Step2: Analyze the second set

A prime number is a positive integer greater than 1 with no divisors other than 1 and itself. If \(2x\) is prime, then \(2x>1\) (which is true as \(x\in U\)) and its only divisors are 1 and \(2x\). But \(2x\) has a divisor 2 (since \(x\) is a positive integer, \(x\geq1\), so \(2x\) is even and greater than 2 when \(x > 1\); if \(x = 1\), \(2x=2\) which is prime, but \(x\in U\) (assuming \(U\) is positive integers greater than, say, 0? Wait, the original \(U\) is "positive integer greater than..." (the text is cut, but usually \(U\) here is positive integers, maybe greater than 0 or 1). Wait, if \(x\in U\) is a positive integer, and \(2x\) is prime. The only even prime is 2. So \(2x = 2\) implies \(x = 1\). But if \(U\) is positive integers greater than 1 (maybe the original \(U\) is positive integers greater than 1), then \(x\geq2\), so \(2x\geq4\), which is even and greater than 2, so not prime. So if \(U\) is positive integers greater than 1, then \(2x\) (for \(x\in U\)) is at least 4, which is composite. So this set is empty. Let's check other sets to confirm.

Step3: Analyze the third set

If \(\frac{1}{2}x\) is prime, let \(x = 2p\) where \(p\) is prime. For example, \(x = 4\) (then \(\frac{1}{2}x = 2\), prime), so this set is not empty.

Step4: Analyze the fourth set

\(\frac{1}{2}x\) can be written as a fraction (e.g., \(\frac{x}{2}\)), so this set is not empty.

Answer:

\(\{x|x\in U\text{ and }2x\text{ is prime}\}\) (assuming \(U\) is positive integers greater than 1; if \(U\) includes 1, then \(x = 1\) would make \(2x = 2\) prime, but the original \(U\) is "positive integer greater than..." (the text is cut, but from the context of the problem, likely \(U\) is positive integers greater than 1, so this set is empty).