Certainly! Let's analyze each set step-by-step:
### Part (i)
[tex]\[ A = \{ x \mid x \in I, x^2 \text{ is not positive} \} \][/tex]
1. Identify the elements of the set: The set [tex]\( A \)[/tex] contains elements [tex]\( x \)[/tex] such that [tex]\( x \)[/tex] is an integer ([tex]\( x \in I \)[/tex]) and [tex]\( x^2 \)[/tex] is not positive.
2. Analyze [tex]\( x^2 \)[/tex] for integers:
- For any integer [tex]\( x \)[/tex], [tex]\( x^2 \)[/tex] represents [tex]\( x \)[/tex] squared.
- The square of any real number, including integers, is always non-negative. This means [tex]\( x^2 \)[/tex] is always either zero or positive.
3. Determine the possibility: Since [tex]\( x^2 \)[/tex] cannot be less than zero for any integer, there is no integer [tex]\( x \)[/tex] for which [tex]\( x^2 \)[/tex] is not positive.
4. Conclusion: As there are no integers satisfying the condition that [tex]\( x^2 \)[/tex] is not positive, the set [tex]\( A \)[/tex] is empty.
Therefore, the set [tex]\( A \)[/tex] is an empty set.
### Part (ii)
[tex]\[ B = \{ b \mid b \in N, 2b + 1 \text{ is even} \} \][/tex]
1. Identify the elements of the set: The set [tex]\( B \)[/tex] contains elements [tex]\( b \)[/tex] such that [tex]\( b \)[/tex] is a natural number ([tex]\( b \in N \)[/tex]) and [tex]\( 2b + 1 \)[/tex] is even.
2. Analyze [tex]\( 2b + 1 \)[/tex] for natural numbers:
- For any natural number [tex]\( b \)[/tex], [tex]\( 2b \)[/tex] is always an even number (since multiplying an integer by 2 results in an even number).
- Adding 1 to any even number results in an odd number. Therefore, [tex]\( 2b + 1 \)[/tex] will always be odd.
3. Determine the possibility: Since [tex]\( 2b + 1 \)[/tex] is always odd for any natural number [tex]\( b \)[/tex], it can never be even.
4. Conclusion: There are no natural numbers [tex]\( b \)[/tex] for which [tex]\( 2b + 1 \)[/tex] is even, hence the set [tex]\( B \)[/tex] is empty.
Therefore, the set [tex]\( B \)[/tex] is an empty set.
Both sets [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are empty sets.
