## Try It Now

Work these exercises to see how well you understand this material.

### Solutions

- Answer: {{a}, {b}, {c}}, {{a, b}, {c}}, {{a, c}, {b}}, {{a}, {b, c}}, {{a, b, c}}
- Answer: No. By this definition, it is possible that an element of A might belong to two of the subsets.
- Answer: The first subset is all the even integers and the second is all the odd integers. These two sets do not intersect and they cover the integers completely.
- Answer: Since 17 participated in both activities, 30 of the tennis players only played tennis and 25 of the swimmers only swam. Therefore, 17 + 30 + 25 = 72 of those who were surveyed participated in an activity and so 18 did not.
- Solution: We assume that |A
_{1}∪ A_{2}| = |A_{1}| + |A_{2}| − |A_{1}∩ A_{2}|.

|A_{1}∪ A_{2}∪ A_{3}| = |(A_{1}∪ A_{2}) ∪ A_{3}|*Why?*

= |A_{1}∪ A_{2}| + |A_{3}| − |(A_{1}∪ A_{2}) ∩ A_{3}|*Why?*

= A_{1}∪ A_{2}| + |A_{3}| − |(A_{1}∩ A_{3}) ∪ (A_{2}∩ A_{3})|*Why?*The law for four sets is:

= |A_{1}| + |A_{2}| − |A_{1}∩ A_{2}| + |A_{3}| − (|A_{1}∩ A_{3}| + |A_{2}∩ A_{3}| − |(A_{1}∩ A_{3}) ∩ (A_{2}∩ A_{3})|*Why**?*

= |A_{1}| + |A_{2}| + |A_{3}| − |A_{1}∩ A_{2}| − |A_{1}∩ A_{3}| − |A_{2}∩ A_{3}| + |A_{1}∩ A_{2}∩ A_{3}|*Why**?*

|A_{1}∪ A_{2}∪ A_{3}∪ A_{4}| = |A_{1}| + |A_{2}| + |A_{3}| + |A_{4}| − |A_{1}∩ A_{2}| − |A_{1}∩ A_{3}| − |A_{1}∩ A_{4}| − |A_{2}∩ A_{3}| − |A_{2}∩ A_{4}| − |A_{3}∩ A_{4}| + |A_{1}∩ A_{2}∩ A_{3}| + |A_{1}∩ A_{2 }∩ A_{4}| + |A_{1}∩ A_{3}∩ A_{4}| + |A_{2}∩ A_{3}∩ A_{4}| − |A_{1}∩ A_{2}∩ A_{3}∩ A_{4}|

Derivation:

|A_{1}∪ A_{2}∪ A_{3}∪ A_{4}| = |(A_{1}∪ A_{2}∪ A_{3}) ∪ A_{4}|

= |(A_{1}∪ A_{2}∪ A_{3}| + |A_{4}| − |(A_{1}∪ A_{2}∪ A_{3}) ∩ A_{4 }= |(A_{1}∪ A_{2}∪ A_{3}| + |A_{4}| − |(A_{1}∩ A_{4}) ∪ (A_{2}∩ A_{4}) ∪ (A_{3}∩ A_{4})|

= |A_{1}| + |A_{2}| + |A_{3}| − |A_{1}∩ A_{2}| − |A_{1}∩ A_{3}| − |A_{2}∩ A_{3}| + |A_{1}∩ A_{2}∩ A_{3}| + |A_{4}| − |A_{1}∩ A_{4}| + |A_{2}∩ A_{4}| + |A_{3}∩ A_{4}| − |(A_{1}∩ A_{4}) ∩ (A_{2}∩ A_{4})| − |(A_{1}∩ A_{4}) ∩ (A_{3}∩ A_{4})| − |(A_{2}∩ A_{4}) ∩ (A_{3}∩ A_{4})| + |(A_{1}∩ A_{4}) ∩ (A_{2}∩ A_{4}) ∩ (A_{3}∩ A_{4})|

= |A_{1}| + |A_{2}| + |A_{3}| + |A_{4}| − |A_{1}∩ A_{2}| − |A_{1}∩ A_{3}| − |A_{2}∩ A_{3}| − |A_{1}∩ A_{4}| − |A_{2}∩ A_{4}| − |A_{3}∩ A_{4}| + |A_{1}∩ A_{2}∩ A_{3}| + |A_{1}∩ A_{2}∩ A_{4}| + |A_{1}∩ A_{3}∩ A_{4}| + |A_{2}∩ A_{3}∩ A_{4}| − |A_{1}∩ A_{2}∩ A_{3}∩ A_{4}| - Answer: Partition the set of fractions into blocks, where each block contains fractions that are numerically equivalent. Describe how you would determine whether two fractions belong to the same block. Redefine the rational numbers to be this partition. Each rational number is a set of fractions.