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Sagot :
To solve this problem, let's analyze each of the given sets step-by-step.
#### Set 1: [tex]\(\left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\}\)[/tex]
For [tex]\(\frac{1}{2}x\)[/tex] to be a prime number, [tex]\(x\)[/tex] must be an even number because only an even number divided by 2 can result in an integer. Let’s consider integer values for [tex]\(x \in U\)[/tex].
- [tex]\(x = 2\)[/tex]: [tex]\(\frac{1}{2} \times 2 = 1\)[/tex], which is not a prime number.
- For any even [tex]\(x > 2\)[/tex] (e.g., [tex]\(x = 4\)[/tex] or more): [tex]\(\frac{1}{2}x\)[/tex] will be an integer, but it will result in numbers that continue increasing (e.g., [tex]\( \frac{1}{2} \times 4 = 2\)[/tex], [tex]\( \frac{1}{2} \times 6 = 3\)[/tex], etc.). None of these will consistently yield prime numbers.
Thus, there is no [tex]\(x \in U\)[/tex] that satisfies the condition that [tex]\(\frac{1}{2}x\)[/tex] is prime. Therefore, this set is empty.
#### Set 2: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ is prime} \}\)[/tex]
For [tex]\(2x\)[/tex] to be prime, [tex]\(2x\)[/tex] itself must be a prime number. In this case:
- [tex]\(x = 2\)[/tex]: [tex]\(2 \times 2 = 4\)[/tex], which is not prime.
- For any [tex]\(x > 2\)[/tex] (e.g., [tex]\(x = 3\)[/tex], [tex]\(4\)[/tex], etc.): [tex]\(2x\)[/tex] will result in numbers such as [tex]\(6, 8, 10\)[/tex], etc., none of which are prime.
Thus, there is no [tex]\(x \in U\)[/tex] that satisfies the condition that [tex]\(2x\)[/tex] is prime, making this set also empty.
#### Set 3: [tex]\(\left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ can be written as a fraction} \right\}\)[/tex]
Any real number can be written as a fraction. Specifically for integers [tex]\(x\)[/tex], half of any integer can be written as a fraction.
- For example: [tex]\(\frac{1}{2} \times 2 = 1\ ( \text{which is } 1/1)\)[/tex]
Thus, every [tex]\(x \in U\)[/tex], when divided by 2, can be written as a fraction. Therefore, this set is not empty.
#### Set 4: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ can be written as a fraction} \}$ This case is similar to the previous one. Any positive integer \(x\)[/tex], when doubled, can be written as a fraction.
- For example: [tex]\(2 \times 2 = 4\ ( \text{which is } 4/1)\)[/tex]
Thus, every [tex]\(x \in U\)[/tex] can be written as [tex]\(2x\)[/tex] in the form of a fraction. This set is not empty either.
### Conclusion:
The only sets that are empty given [tex]\(U\)[/tex] are:
[tex]\[ \left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\} \][/tex]
and
[tex]\[ \{ x \mid x \in U \text{ and } 2x \text{ is prime} \} \][/tex]
Thus, those are the answers for this problem, and the first identified empty set is the confirmed empty set. The solution to the specified question is that:
[tex]\[ \left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\} \][/tex]
is an empty set, making the answer [tex]\(0\)[/tex].
#### Set 1: [tex]\(\left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\}\)[/tex]
For [tex]\(\frac{1}{2}x\)[/tex] to be a prime number, [tex]\(x\)[/tex] must be an even number because only an even number divided by 2 can result in an integer. Let’s consider integer values for [tex]\(x \in U\)[/tex].
- [tex]\(x = 2\)[/tex]: [tex]\(\frac{1}{2} \times 2 = 1\)[/tex], which is not a prime number.
- For any even [tex]\(x > 2\)[/tex] (e.g., [tex]\(x = 4\)[/tex] or more): [tex]\(\frac{1}{2}x\)[/tex] will be an integer, but it will result in numbers that continue increasing (e.g., [tex]\( \frac{1}{2} \times 4 = 2\)[/tex], [tex]\( \frac{1}{2} \times 6 = 3\)[/tex], etc.). None of these will consistently yield prime numbers.
Thus, there is no [tex]\(x \in U\)[/tex] that satisfies the condition that [tex]\(\frac{1}{2}x\)[/tex] is prime. Therefore, this set is empty.
#### Set 2: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ is prime} \}\)[/tex]
For [tex]\(2x\)[/tex] to be prime, [tex]\(2x\)[/tex] itself must be a prime number. In this case:
- [tex]\(x = 2\)[/tex]: [tex]\(2 \times 2 = 4\)[/tex], which is not prime.
- For any [tex]\(x > 2\)[/tex] (e.g., [tex]\(x = 3\)[/tex], [tex]\(4\)[/tex], etc.): [tex]\(2x\)[/tex] will result in numbers such as [tex]\(6, 8, 10\)[/tex], etc., none of which are prime.
Thus, there is no [tex]\(x \in U\)[/tex] that satisfies the condition that [tex]\(2x\)[/tex] is prime, making this set also empty.
#### Set 3: [tex]\(\left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ can be written as a fraction} \right\}\)[/tex]
Any real number can be written as a fraction. Specifically for integers [tex]\(x\)[/tex], half of any integer can be written as a fraction.
- For example: [tex]\(\frac{1}{2} \times 2 = 1\ ( \text{which is } 1/1)\)[/tex]
Thus, every [tex]\(x \in U\)[/tex], when divided by 2, can be written as a fraction. Therefore, this set is not empty.
#### Set 4: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ can be written as a fraction} \}$ This case is similar to the previous one. Any positive integer \(x\)[/tex], when doubled, can be written as a fraction.
- For example: [tex]\(2 \times 2 = 4\ ( \text{which is } 4/1)\)[/tex]
Thus, every [tex]\(x \in U\)[/tex] can be written as [tex]\(2x\)[/tex] in the form of a fraction. This set is not empty either.
### Conclusion:
The only sets that are empty given [tex]\(U\)[/tex] are:
[tex]\[ \left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\} \][/tex]
and
[tex]\[ \{ x \mid x \in U \text{ and } 2x \text{ is prime} \} \][/tex]
Thus, those are the answers for this problem, and the first identified empty set is the confirmed empty set. The solution to the specified question is that:
[tex]\[ \left\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \right\} \][/tex]
is an empty set, making the answer [tex]\(0\)[/tex].
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