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## Here's the Solution to this Question

1a)

By definition, we have

$A×(B∩C)={(x,y):x∈A; y∈B∩C}\\ ={(x,y):x∈A and (y∈B and y∈C)}.\\$

Similarly,

$(A×B)∩(A×C)={(x,y):(x,y)∈A×B and (x,y)∈A×C}\\ ={(x,y):(x∈A and y∈B) and (x∈A and y∈C)}.$

With that established, what we need to show is that the two statements

$x∈A; (y∈B ; y∈C),(x∈A ; y∈B) ; (x∈A ; y∈C)$

are logically equivalent.

1b)

The bit string that represents the set of odd integers in U, namely, {1, 3, 5, 7, 9}, has one bit in the first, third, fifth, seventh, and ninth positions, and zero elsewhere. It is

10 1010 1010.

(We have split this bit string of length 10 into two blocks of length five for easy reading since long bitstrings are difficult to read.) Similarly, we represent the subset of all even integers in U, namely, {2, 4, 6, 8,10}, by the string

01 0101 0101.

The set of all integers in U that do not exceed 5, namely, {1, 2, 3, 4, 5}, is represented by the string

11 1110 0000.

Using bit strings to represent sets, it is easy to find complements of sets and unions, intersections, and differences of sets. To find the bit string for the complement of a set from the bit string for that set, we simply change each 1 to a 0 and each 0 to 1, since x E A if and only if x A. Note that this operation corresponds to taking the negation of each bit when we associate a bit with a truth value—with 1 representing true and 0, false.

3a)

As the OP said in his comment the answer is the following list of partitions of S={1,2,3}:

1. {{1,2,3}} (one set, just itself )
2. {{1,2},{3}} (two sets, one of two elements , one singleton)
3. {{1,3},{2}}
4. {{2,3},{1}}
5. {{1},{2},{3}} (all singletons)

and these are all.