Predicate: 1 ≤ x2 ≤ 9, domain: R Predicate: 1 ≤ x2 ≤ 9, domain: Z
The Correct Answer and Explanation is :
To analyze the predicate ( 1 \leq x^2 \leq 9 ) under the domains of the real numbers ( \mathbb{R} ) and the integers ( \mathbb{Z} ), we need to consider the implications of the inequality for each domain.
Domain: ( \mathbb{R} )
In the real number domain, the inequality ( 1 \leq x^2 \leq 9 ) can be separated into two parts:
- From ( 1 \leq x^2 ): This implies that ( x^2 ) is at least 1. Therefore, ( x ) must be either greater than or equal to 1 or less than or equal to -1. In mathematical terms:
- ( x \geq 1 ) or ( x \leq -1 ).
- From ( x^2 \leq 9 ): This implies that ( x^2 ) is at most 9. Hence, ( x ) must lie between -3 and 3. Formally, this can be expressed as:
- ( -3 \leq x \leq 3 ).
Combining the Results
Now, combining the results from the two inequalities:
- For ( 1 \leq x^2 ): ( x \leq -1 ) or ( x \geq 1 ).
- For ( x^2 \leq 9 ): ( -3 \leq x \leq 3 ).
Thus, the valid intervals for ( x ) are:
- ( -3 \leq x \leq -1 )
- ( 1 \leq x \leq 3 )
This gives us the solution set ( x \in [-3, -1] \cup [1, 3] ).
Domain: ( \mathbb{Z} )
When considering the predicate in the integer domain ( \mathbb{Z} ), we can evaluate the integers that satisfy the combined conditions:
- From ( -3 \leq x \leq -1 ), the integer solutions are ( -3, -2, -1 ).
- From ( 1 \leq x \leq 3 ), the integer solutions are ( 1, 2, 3 ).
Conclusion
In ( \mathbb{R} ), the solution is the interval ( [-3, -1] \cup [1, 3] ). In ( \mathbb{Z} ), the valid integer solutions are ( {-3, -2, -1, 1, 2, 3} ).