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CS/MATH 1019 Discrete Math for Computer Science
Typology: Exams
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Name : YorkU email : Student Number : READ THE FOLLOWING INSTRUCTIONS.
- Do not open your exam until told to do so.
Extra Work Space. Multiple Choice. Circle the best answer. No work needed. No partial credit available.
Solution: Standard Response Questions. Show all work to receive credit.
Solution: ¬( p ∨ r ∨ (¬ p ∧ q )) ≡ ¬ p ∧ ¬ r ∧ ¬(¬ p ∧ q ) ≡ ¬ p ∧ ¬ r ∧ ( p ∨ ¬ q ) ≡ (¬ p ∧ ¬ r ∧ p ) ∨ (¬ p ∧ ¬ r ∧ ¬ q ) ≡ F ∨ (¬ p ∧ ¬ r ∧ ¬ q ) ≡ ¬ p ∧ ¬ r ∧ ¬ q ≡ ¬ p ∧ ¬ q ∧ ¬ r
Solution: One possible solution is S = {x, y, z, w} and T = {1, 2, 3, 4 }. The fact that |P(S)| = 16 means |S| = 4 and |S × T | = 16 means |S||T | = 16. In this case the members of S × T are:
Solution: We first see when x = 0, for any y ∈ Q we have that xy = 0 · y = 0 /= 1. Thus we can Now take any x ∈ Q with x = 0. This means x = conclude that ∃x, ∀y(xy /= 1) is true. We must now show the uniqueness. definition of a rational number b = 0, and a a b = 0 because x for a, b ∈ Z with both a /= 0 and b /= 0 (by the = 0). For such x we can then take y = b a which is also a rational number (hence in our domain) and see that xy = 1. Therefore we can conclude that ∃!x, ∀y(xy /= 1) is true as desired.
Extra Work Space.