13) Consider the set \( A=\{0,1,4,9,25,36,49,64,81,100\} \) Determine whether or not each or the following is a partition on \( A \). if it isn't give every reason why it isn't a partition. ( 12 points) a) \( \{\{100,81,64\},\{36,25,9\},\{4,49,0\} \) b) \( \{0,100,36\},\{1,81,49,4,25,64\},\{9\}\} \) c) \( \{\{0,9,25\},\{1,36,144,81\},\{49,4,9,64,100\}\} \)

To determine if each of the given subsets is a partition of \( A \), we need to check if they meet the criteria of covering the entire set \( A \) without overlaps. The set \( A = \{0, 1, 4, 9, 25, 36, 49, 64, 81, 100\} \) contains 10 distinct elements, and each subset in a partition must be disjoint (no elements should repeat across subsets) and collectively exhaustive (together they must cover all elements in \( A \)). a) The subset \( \{\{100,81,64\},\{36,25,9\},\{4,49,0\}\} \) contains: - First set: 100, 81, 64 - Second set: 36, 25, 9 - Third set: 4, 49, 0 Each of these subsets is disjoint and combined they cover all elements in \( A \) without leaving any behind. Therefore, this is a valid partition. b) The subset \( \{\{0,100,36\},\{1,81,49,4,25,64\},\{9\}\} \) contains: - First set: 0, 100, 36 - Second set: 1, 81, 49, 4, 25, 64 - Third set: 9 While the elements are partially covered, there is a problem here: the element 9 is only in the third set and is repeated in the second set. Hence, it fails the disjoint condition as it does not cover all ten elements uniquely. This is **not** a partition. c) The subset \( \{\{0,9,25\},\{1,36,144,81\},\{49,4,9,64,100\}\} \) contains: - First set: 0, 9, 25 - Second set: 1, 36, 144, 81 - Third set: 49, 4, 9, 64, 100 Here lies the issue: the number 9 appears in the first and third sets, thus violating the disjoint condition. Additionally, the set contains the number 144, which is not part of the original set A. Therefore, it's excluded from the set and cannot be considered a partition. This is also **not** a partition.