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Unformatted text preview: Math 347G1. (Milin) Solutions to Exam 1 1 1. (20 points) For both claims below, rewrite them in logical notation, negate the logical expressions and then write out an English sentence which expresses the negated claim without using words of negation. (a) There exists a δ > , so that for all ε > and all real numbers x , whenever  x  < δ we have  f ( x )  < ε . Solution. In notation: ( ∃ δ > 0)( ∀ ε > 0)( ∀ x ∈ R )  x  < δ ⇒  f ( x )  < ε Negation: ¬ [( ∃ δ > 0)( ∀ ε > 0)( ∀ x ∈ R )  x  < δ ⇒  f ( x )  < ε ] ⇔ ( ∀ δ > 0)( ∃ ε > 0)( ∃ x ∈ R )  x  < δ ∧  f ( x )  ≥ ε The negation is: For all positive δ there exists a positive ε and a real number x so that  x  < δ and  f ( x )  ≥ ε . (b) For all real numbers a , if a 2 + 5 a + 6 > , then we must have a > 2 or a < 3 . Solution. In notation: ( ∀ a ∈ R ) a 2 + 5 a + 6 > ⇒ ( a > 2 ∨ a < 3) Negation: ¬ ( ∀ a ∈ R ) a 2 + 5 a + 6 > ⇒ ( a > 2 ∨ a < 3) ⇔ ( ∃ a ∈ R ) a 2 + 5 a + 6 > ∧ ¬ ( a > 2 ∨ a < 3) ⇔ ( ∃ a ∈ R ) a 2 + 5 a + 6 > ∧ a ≤  2 ∧ a ≥  3 The negation is: There exists a real number a so that a 2 + 5 a + 6 > 0, a ≤  2 and a ≥  3. 2. (20 points) Give brief proofs to the following two claims. (a) If f : R → R is a bounded function, then g ( x ) = f ( x ) + f ( x ) is bounded as well....
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 Fall '09
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 Math, Logic, Metric space, Rational number, Bounded function

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