TEST1 Solutions

TEST1 Solutions - EXAM 1 SOLUTIONS Problem 1. If |A| = 4...

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EXAM 1 SOLUTIONS Problem 1. If | A | = 4 and | B | = 5 , then | A × B | = 9 . Proof. This is false. The product has size 4 · 5 = 20. It is straightforward to construct a counter-example if you want. ± Problem 2. Let x R . The sentence “If x > 3 , then x 2 < 2 .” is a statement. Proof. This is false. The sentence is an open sentence, but is not a statement. It would become a statement if it was quantified like “ x R , if x > 3 then x 2 < 2.” ± Problem 3. The converse of the statement “If x S then x T ” is the statement “If x / T then x / S .” Proof. This is false. That is the contrapositive. The converse would be “If x T then x S .” ± Problem 4. The negation of the statement “ x R , x 2 1 and x 3 < 2 ” is the statement x R such that x 2 < 1 and x 3 2 . Proof. This is false. We need to change the word “and” to the word “or” using DeMorgan’s law. ± Problem 5. If A is a set and P ( A ) is the power set of A , then A ⊆ P ( A ) . Proof. This is false in general. The set A = { 1 } is a counter-example. ± Problem 6. For i Z , define A i = { i - 2 ,i - 1 ,i + 2 ,i + 3 } . Then the set { A 3 ,A 5 ,A 11 ,A 13 } is a partition of { n Z : 1 n 16 } . Proof. This is true. (The collection of sets is pair-wise disjoint, their union is the set of integers between 1 and 16, and none of them is empty.) ± Problem 7. Let A , B , and C be sets. Then A - ( B C ) = ( A - B ) ( A - C ) . Proof. This is true, and is one of DeMorgan’s laws for sets. ± Problem 8. One way to read P = Q is “ P is necessary for Q .” Proof. This is false. It should read “ P is sufficient for Q ” or “ Q is necessary for P ”. ± Problem 9.
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TEST1 Solutions - EXAM 1 SOLUTIONS Problem 1. If |A| = 4...

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