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Unformatted text preview: Homework 1 Solutions 1. Construct a truth table for the statement: [ p ⇒ ( q ∧ ∼ q )] ⇔∼ p. Solution: First note that q ∧ ∼ q is a contradiction, so replace it with the variable c which only takes on the truth value F. Then we get the truth table: p c p ⇒ c ∼ p [ p ⇒ c ] ⇒∼ p [ p ⇒ c ] ⇐∼ p [ p ⇒ c ] ⇔∼ p T F F F T T T F F T T T T T Because the truth values of the statement are always T, this says the statement is a tautology. In other words, whenever you want to prove that p ⇒ c where c is always F (i.e. c is a contradiction), you can instead prove ∼ p , and viceversa. 2. Consider the following statement: A function f is uniformly continuous on a set S if and only if for every > there is a δ > such that  f ( x ) f ( y )  < whenever x and y are in S and  x y  < δ . (a) Rewrite the statement using the logical symbols ⇒ , ⇔ , ∀ , ∃ , etc. Solution: [ f is uniformly continuous on a set S ] ⇔ [ ∀ > 0, ∃ δ > 0 so that ∀ x,y ∈ S and  x y  < δ ⇒  f ( x ) f ( y )  ≤ .] (b) Write the negation of the statement in words and in logical symbols....
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 Winter '11
 Wiley
 Differential Equations, Equations

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