HW2_Sol - COT 3100 Discrete Mathematics HW#2 Solution...

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Unformatted text preview: COT 3100 Discrete Mathematics HW #2 Solution Section 1.3 8. a. If an animal is a rabbit, then that animal hops. (Or every rabbit hops) b. Every animal is a rabbit and hops. c. There exists an animal such that if it is a rabbit, then it hops. (This is trivially true, satisfied for example, by lions, so it is not the sort of thing one would say) d. There exists an animal that is a rabbit and hops. (Or some rabbits hop Or some hopping animals are rabbits) Note that part (b) and (c) are not the sorts of things one would normally say. 9. a. We assume that this sentence is asserting that the same person has both talents. Therefore we can write ∃? ( ¡ ?¢ ∧ ? ?¢ ) b. Since `but‘ really means the same things as `and‘ logically, this is ∃? ( ¡ ?¢ ∧ ¬ ? ?¢ ) c. This time we make the universal statement: ∀? ( ¡ ?¢ ∨ ? ?¢ ) d. This sentence is asserting the nonexistence of anyone with either talent, so we could write it as ¬ ∃? ( ¡ ?¢ ∨ ? ?¢ ) . Alternatively, we can think of this as asserting that everyone fails to have either of these talents, and we obtain the logically equivalent answer ∀? ¬( ¡ ?¢ ∨ ? ?¢ ) . Failing to have either talent is equivalent to having neither talent (by De Morgan’s law), so we can also write this as ∀? ((¬ ¡ ?¢ ) ∧ (¬ ? ?¢ )) . Note that it would not be correct to write ∀? ((¬ ¡ ?¢ ) ∨ (¬ ? ?¢ )) nor to write ∀? ¬( ¡ ?¢ ∧ ? ?¢ ) . 16. a. true ( ? = £ 2 ) b. false ( £− 1 is not a real number) c. true (the left-hand side is always at least 2) d. false (not true for ? = 1 or ? = 0 ) 17. a. ? ¡ ∨ ? 1 ¡ ∨ ? 2 ¡ ∨ ? 3 ¡ ∨ ? 4 ¡ b. ? ¡ ∧ ? 1 ¡ ∧ ? 2 ¡ ∧ ? 3 ¡ ∧ ? 4 ¡ c. ¬ ? ¡ ∨ ¬ ? 1 ¡ ∨ ¬ ? 2 ¡ ∨ ¬ ? 3 ¡ ∨ ¬ ? 4 ¡ d. ¬ ? ¡ ∧ ¬ ? 1 ¡ ∧ ¬ ? 2 ¡ ∧ ¬ ? 3 ¡ ∧ ¬ ? 4 ¡ e. ¬( ? ¡ ∨ ? 1 ¡ ∨ ? 2 ¡ ∨ ? 3 ¡ ∨ ? 4 ¡ ) f. ¬( ? ¡ ∧ ? 1 ¡ ∧ ? 2 ¡ ∧ ? 3 ¡ ∧ ? 4 ¡ ) 20. c. ¬ ? − 5 ¡ ∧ ¬ ? ( − 3) ∧ ¬ ? ( − 1) ∧ ¬ ? (3) ∧ ¬ ? (5) d. ? (1) ∨ ? (3) ∨ ? (5) e. ¢ ¬ ? 1 ¡ ∨ ¬ ? 3 ¡ ∨ ¬ ? 5 ¡£ ∧ ? − 1 ¡ ∧ ? − 3 ¡ ∧ ? − 5 ¡ 44. NO, they aren’t equivalent. For example, if ? ¤¡ = ` ¤ is even‘, ? ¤¡ = ` ¤ is odd‘ and the domain of ¤ is 1 and 2. Then, ∀ ( ? ¡ ↔ ¥ ¡ ) is false (e.g. with ¤ = 1 ); while both ∀¤? ¤¡ and ∀¤? ¤¡ are false (since P(1) is false and Q(2) is false) that implies ∀? ¡ ↔ ∀¥ ¡ is true ....
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HW2_Sol - COT 3100 Discrete Mathematics HW#2 Solution...

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