Solns1 - Math 55 Discrete Mathematics U.C Berkeley Dept of Mathematics Summer 2007 Instructor Jared Weinstein 1075 Evans Hall

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Instructor Jared Weinstein 1075 Evans Hall www.math.berkeley.edu/ jared [email protected] Math 55 Discrete Mathematics U.C. Berkeley Dept. of Mathematics Summer 2007 Homework #1 Solutions 1.1.20 Write each of the following sentences in “if . . . then” form, in English. Answer : (a) “If you do not send me an e-mail message, then I will not remember to send you the address.” Or: “If I remember to send you the address, then you will have sent me an e-mail message.” Or “If I’m going to remember to send you the address, then you must send me an e-mail message.” (b) “If you were born in the United States then you are a citizen of this country.” (f) “If there is a storm then the beach erodes.” (But notice how the original form “The beach erodes whenever there is a storm” conveys the meaning with greater clarity. My answer seems to leave open the possibility that there are two storms and the beach erodes once, while the origianl formulation makes it clear that each storm results in erosion and more directly suggests cause and effect. That’s why we would say it that way.) (g) “If someone logs onto the server then they must have a valid password.” 1.1.32(a) Construct a truth table for the compound proposition ( p q ) r . Solution : p q r p q ( p q ) r T T T T T T T F T T T F T T T T F F T T F T T T T F T F T T F F T F T F F F F F ± 1.1.44 A barber shaves those who do not shave themselves, but no one else. Is this possible? Answer : No. Let b be the barber. Let S be the set of all people who do not shave themselves. Then b shaves p if and only if p S . Now either b shaves b , or not. Case (1): If b shaves b then by assumption b S . But this means that b does not shave himself. Thus b doesn’t shave b , a contradiction. Case (2): If b doesn’t shave b then b doesn’t shave himself, so b S , so b shaves b by hypothesis. This is again a contradiction. Since one of the two cases must occur, and both cases lead to contradictions, this situation is impossible. (Note that this is essentially the same paradox as the set of all sets that are not elements of themselves.) ± 1.1.46a A group of people come in two types: those who always lie, and those who always tell the truth. By asking one of them exactly one question, we must determine which of the two types he/she is. (a) If we ask “Are you a liar?” then the truth-tellers will answer “No”, and liars will answer “No” as well, so we won’t gain any information from their responses. (b) First solution. If we ask “Is 1 + 1 = 2?” then truth-tellers must answer “Yes”, while liars must answer “No.” So we can tell them apart. Alternative solution. We ask “If you are a truth-teller, do you always tell the truth?”. A truth-teller would answer “Yes”. If a liar is answering, then the correct answer is “Yes”, because this if-then statement is automatically true, since its premise is false. The liar must therefore answer lie and answer
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This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.

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Solns1 - Math 55 Discrete Mathematics U.C Berkeley Dept of Mathematics Summer 2007 Instructor Jared Weinstein 1075 Evans Hall

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