Answer
See explanation
Work Step by Step
### a. \( A = \{2, \{2\}, (\sqrt{2})^2\} \), \( B = \{2, \{2\}, \{\{2\}\}\} \)
- \( A \subseteq B \)? No, because \(\{\{2\}\} \in B\) is not in \( A \).
- \( B \subseteq A \)? No, because \(\{\{2\}\} \in B\) is not in \( A \).
- Neither \( A \) nor \( B \) is a proper subset of the other.
### b. \( A = \{3, \sqrt{5^2 - 4^2}, 24 \mod 7\} \), \( B = \{8 \mod 5\} \)
- Simplify the elements:
- \( \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \)
- \( 24 \mod 7 = 3 \)
- \( 8 \mod 5 = 3 \)
- So, \( A = \{3\} \), \( B = \{3\} \).
- \( A \subseteq B \)? Yes.
- \( B \subseteq A \)? Yes.
- \( A = B \), so neither is a proper subset.
### c. \( A = \{\{1, 2\}, \{2, 3\}\} \), \( B = \{1, 2, 3\} \)
- \( A \subseteq B \)? No, because the elements in \( A \) are sets, not individual numbers.
- \( B \subseteq A \)? No, the elements in \( B \) are numbers, not sets.
- Neither \( A \) nor \( B \) is a proper subset of the other.
### d. \( A = \{a, b, c\} \), \( B = \{\{a\}, \{b\}, \{c\}\} \)
- \( A \subseteq B \)? No, because the elements in \( A \) are not sets.
- \( B \subseteq A \)? No, because the elements in \( B \) are sets.
- Neither \( A \) nor \( B \) is a proper subset of the other.
### e. \( A = \{\sqrt{16}, \{4\}\} \), \( B = \{4\} \)
- Simplify the elements:
- \( \sqrt{16} = 4 \)
- So, \( A = \{4, \{4\}\} \).
- \( A \subseteq B \)? No, because \(\{4\}\in A\) is not in \( B \).
- \( B \subseteq A \)? Yes, because \( 4 \in A \).
- \( B \) is a proper subset of \( A \).
### f. \( A = \{x \in \mathbb{R} \mid \cos x \in \mathbb{Z}\} \), \( B = \{x \in \mathbb{R} \mid \sin x \in \mathbb{Z}\} \)
- The cosine function can only take integer values \( 1 \) and \( -1 \), and similarly for sine.
- \( A = \{2n\pi \mid n \in \mathbb{Z}\} \) and \( B = \{(2n + 1)\frac{\pi}{2} \mid n \in \mathbb{Z}\} \).
- These sets do not intersect.
- \( A \subseteq B \)? No.
- \( B \subseteq A \)? No.
- Neither \( A \) nor \( B \) is a proper subset of the other.
