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Unformatted text preview: 347G1, Fall 2009Practice Exam 1 Solutions11.(15 points) For each of the claims below, rewrite them in logical notation, negate themand then write out an English sentence which expresses the negated claim (withoutusing words of negation).(a)For every natural numbernwe can find anx∈Rso thatx <1n3.Solution.In logical notation: (∀n∈N)(∃x∈R)x <1n3¬(∀n∈N)(∃x∈R)x <1n3⇔(∃n∈N)¬(∃x∈R)x <1n3⇔(∃n∈N)(∀x∈R)x≥1n3The negation is: There exists a natural numbernso that for all real numbersx,x≥1n3.(b)There existsM∈Rso that for all real numbersx,y≤M, ifx < ythenf(x)< f(y).Solution.In logical notation: (∃M∈R)(∀x,y≤M)x < y⇒f(x)< f(y)¬[(∃M∈R)(∀x,y≤M)x < y⇒f(x)< f(y)]⇔(∀M∈R)¬[(∀x,y≤M)x < y⇒f(x)< f(y)]⇔(∀M∈R)(∃x,y≤M)x < y∧f(x)≥f(y)The negation is: For all real numbersM, there arex,y≤Mso thatx < yandf(x)≥f(y).(c)For all real numbersx, ifx2<2xholds, then we must havex >2orx <.Solution.In logical notation: (∀x∈R)x2<2x⇒(x >2∨x <0)¬(∀x∈R)x2<2x⇒(x >2∨x <0)⇔(∃x∈R)¬x2<2x⇒(x >2∨x <0)⇔(∃x∈R)x2≥2x∧ ¬(x >2∨x <0)⇔(∃x∈R)x2≥2x∧x≤2∧x≥The negation is: There exists a real number...
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This note was uploaded on 01/27/2010 for the course MATH 347 taught by Professor ? during the Fall '09 term at University of Illinois at Urbana–Champaign.
 Fall '09
 ?
 Math, Logic

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