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Sagot :
Let's analyze each of the given sets step-by-step to determine which one is an empty set.
1. Set 1: [tex]\(\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \}\)[/tex]
We are considering elements [tex]\( x \)[/tex] from the universal set [tex]\( U \)[/tex] such that [tex]\( \frac{1}{2} x \)[/tex] is a prime number. For [tex]\( \frac{1}{2} x \)[/tex] to be a prime number, [tex]\( x \)[/tex] must be an even number (since the half of an odd number isn't an integer). Hence, if [tex]\( x \)[/tex] is even, say [tex]\( x = 2k \)[/tex], then:
[tex]\[ \frac{1}{2} \cdot 2k = k \][/tex]
This means [tex]\( k \)[/tex] must be a prime number. We need to check the positive integers greater than 1 which after dividing by 2 give a prime number. We can find that the set is not empty as there exist such [tex]\( x \)[/tex] values that result in primes.
2. Set 2: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ is prime} \}\)[/tex]
Here, we look for elements [tex]\( x \)[/tex] from [tex]\( U \)[/tex] such that [tex]\( 2x \)[/tex] is a prime number. Since [tex]\( 2x \)[/tex] represents an even number greater than 2, and the only even prime number is 2 itself, there is a contradiction because for [tex]\( x \)[/tex] to be greater than 1, [tex]\( 2x \)[/tex] would be at least 4, which is not prime. Thus, this set is empty.
3. Set 3: [tex]\(\{ x \mid x \in U \text{ and } \frac{1}{2} x \text{ can be written as a fraction} \}\)[/tex]
For any positive integer [tex]\( x \)[/tex] in [tex]\( U \)[/tex], [tex]\( \frac{1}{2}x \)[/tex] is always a fraction. For example, if [tex]\( x = 4 \)[/tex], then [tex]\( \frac{1}{2} \cdot 4 = 2 \)[/tex], which is expressible as a fraction [tex]\( \frac{2}{1} \)[/tex]. Therefore, this set is not empty since every [tex]\( x \)[/tex] in [tex]\( U \)[/tex] satisfies this condition.
4. Set 4: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ can be written as a fraction} \}\)[/tex]
Any positive integer [tex]\( x \)[/tex] in [tex]\( U \)[/tex] when multiplied by 2 will result in an integer, which is inherently a fraction (e.g., 8 can be written as [tex]\( \frac{8}{1} \)[/tex]). Hence, this set is also not empty as all [tex]\( x \)[/tex] in [tex]\( U \)[/tex] will meet this condition.
Therefore, the only set that is empty is:
[tex]\[ \{ x \mid x \in U \text{ and } 2x \text{ is prime} \} \][/tex]
1. Set 1: [tex]\(\{ x \mid x \in U \text{ and } \frac{1}{2}x \text{ is prime} \}\)[/tex]
We are considering elements [tex]\( x \)[/tex] from the universal set [tex]\( U \)[/tex] such that [tex]\( \frac{1}{2} x \)[/tex] is a prime number. For [tex]\( \frac{1}{2} x \)[/tex] to be a prime number, [tex]\( x \)[/tex] must be an even number (since the half of an odd number isn't an integer). Hence, if [tex]\( x \)[/tex] is even, say [tex]\( x = 2k \)[/tex], then:
[tex]\[ \frac{1}{2} \cdot 2k = k \][/tex]
This means [tex]\( k \)[/tex] must be a prime number. We need to check the positive integers greater than 1 which after dividing by 2 give a prime number. We can find that the set is not empty as there exist such [tex]\( x \)[/tex] values that result in primes.
2. Set 2: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ is prime} \}\)[/tex]
Here, we look for elements [tex]\( x \)[/tex] from [tex]\( U \)[/tex] such that [tex]\( 2x \)[/tex] is a prime number. Since [tex]\( 2x \)[/tex] represents an even number greater than 2, and the only even prime number is 2 itself, there is a contradiction because for [tex]\( x \)[/tex] to be greater than 1, [tex]\( 2x \)[/tex] would be at least 4, which is not prime. Thus, this set is empty.
3. Set 3: [tex]\(\{ x \mid x \in U \text{ and } \frac{1}{2} x \text{ can be written as a fraction} \}\)[/tex]
For any positive integer [tex]\( x \)[/tex] in [tex]\( U \)[/tex], [tex]\( \frac{1}{2}x \)[/tex] is always a fraction. For example, if [tex]\( x = 4 \)[/tex], then [tex]\( \frac{1}{2} \cdot 4 = 2 \)[/tex], which is expressible as a fraction [tex]\( \frac{2}{1} \)[/tex]. Therefore, this set is not empty since every [tex]\( x \)[/tex] in [tex]\( U \)[/tex] satisfies this condition.
4. Set 4: [tex]\(\{ x \mid x \in U \text{ and } 2x \text{ can be written as a fraction} \}\)[/tex]
Any positive integer [tex]\( x \)[/tex] in [tex]\( U \)[/tex] when multiplied by 2 will result in an integer, which is inherently a fraction (e.g., 8 can be written as [tex]\( \frac{8}{1} \)[/tex]). Hence, this set is also not empty as all [tex]\( x \)[/tex] in [tex]\( U \)[/tex] will meet this condition.
Therefore, the only set that is empty is:
[tex]\[ \{ x \mid x \in U \text{ and } 2x \text{ is prime} \} \][/tex]
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