### Solutions of Exercise A, D, and E

Added on - 25 Sep 2019

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Solutions of Exercise A, D, and E:Exercise A:LetP,Q, andRbe the propositions defined as follows:P: I am thirsty.Q: My glass is empty.R: It is three o’clock.Ifpis a statement variable, the negation ofpis “notp” or “It is not the case thatp” and is denoted¬p.The negation of above propositions is¬P: I am not thirsty.¬Q: My glass is not empty.¬R: It is not three o’clock.Let us write each of the following propositions as logical expression involvingP,Q, andR.1.I am thirsty and my glass is not empty.Ifpandqare statement variables, the conjunction ofpandqis “pandq,” denotedp∧q. It istrue when, and only when, bothpandqare true. If eitherporqis false, or if both are false,p∧qis false.Here, the proposition “I am thirsty” i.e.Pis true and the proposition “My glass is not empty.” i.e.¬Qis false. The logic word “and” between statement variablesPand¬Qimplies conjunction ofPand¬Q. That is, the logical expression of the statement isP∧¬Q.2.It is three o’clock and I am thirsty.Ifpandqare statement variables, the conjunction ofpandqis “pandq,” denotedp∧q. It istrue when, and only when, bothpandqare true.Here, the proposition “It is three o’clock” i.e.Ris true and the proposition “I am thirsty.” i.e.Pisalso true. The logic word “and” between statement variablesRandPimplies conjunction ofRandP. That is, the logical expression of the statement isR∧P.3.If it is three o’clock, then I am thirsty.Ifpandqare statement variables, the conditional ofqbypis “Ifpthenq” or “pimpliesq” and isdenotedp→q. It is false whenpis true andqis false; otherwise it is true;pis called thehypothesis of the conditional andqis called the conclusion.Here, the first proposition “It is three o’clock.” i.e.Ris true and the second proposition “I amthirsty.” i.e.Pis also true and the logic words “IfRthenP” implies conditional statement.Therefore, the logical expression of the statement isR→P.4.If I am thirsty, then my glass is empty.Ifpandqare statement variables, the conditional ofqbypis “Ifpthenq” or “pimpliesq” and isdenotedp→q. It is false whenpis true andqis false; otherwise it is true;pis called thehypothesis of the conditional andqis called the conclusion.Here, the first proposition “It is three o’clock.” i.e.Ris true and the second proposition “I amthirsty.” i.e.Pis also true and the logic words “IfRthenP” implies conditional statement.Therefore, the logical expression of the statement isR→P.5.If I am not thirsty, then my glass is not empty.

The contrapositive of a conditional statement of the form “Ifpthenq” is “If¬qthen ¬p”. Here,the first proposition is “I am not thirsty” which implies¬Pand the second proposition is “Myglass is not empty.” which implies ¬Q. It is contrapositive of conditional statementQ→Pthat isthe logical expression for this statement is¬Q→¬P.Exercise D:Given thatU={x∈N:x≤12}be a universal set then the elements of the set U are{1,2,3,4,5,6,7,8,9,10,11,12}.LetA={x:xisodd}then the elements of the setA={1,3,5,7,9,11},B={x:x>7}then theelements of the setB={8,9,10,11,12}, andC={x:xis divisible by 3}then the elements of setC={3,6,9,12}. The venn diagram of the sets is as follows:Let us write down following sets in enumerated form:i.A∩Bis intersection of setsAandBwhich contains all the elements that are common toboth setsAandB.We haveA={1,3,5,7,9,11},B={8,9,10,11,12}therefore,A∩B={1,3,5,7,9,11}∩{8,9,10,11,12}={9,11}i.e. the elements 9 and 11 arecommon to both setsAandB.ii.B∪Cis union of setsAandBwhich consist of all elements ofAand all the elements ofBsuch that no element is repeated.We haveB={8,9,10,11,12}andC={3,6,9,12}thenB∪C={8,9,10,11,12}∪{3,6,9,12}B∪C={3,6,8,9,10,11,12}iii.́Ais complement of setA, ifUis universal set andAis subset ofU, then the complement ofsetAi.e.́Ais the set of all the elements ofUwhich are not the elements ofA.We know thatU={1,2,3,4,5,6,7,8,9,10,11,12}andA={1,3,5,7,9,11}. Then,the complement ofAiśA={2,4,6,8,10,12}={x:xis even}.iv.(A∪́B)∩C