# Solutions of Exercise A, D, and E

Added on 2019-09-25

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Solutions of Exercise A, D, and E:Exercise A:Let P, Q, and R be the propositions defined as follows: P: I am thirsty. Q: My glass is empty. R: It is three o’clock.If p is a statement variable, the negation of p is “not p” or “It is not the case that p” 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 involving P, Q, and R.1.I am thirsty and my glass is not empty.If p and q are statement variables, the conjunction of p and q is “p and q,” denoted p∧q. It is true when, and only when, both p and q are true. If either p or q is false, or if both are false, p∧q is false.Here, the proposition “I am thirsty” i.e. P is true and the proposition “My glass is not empty.” i.e.¬Q is false. The logic word “and” between statement variables P and ¬Q implies conjunction of Pand ¬Q. That is, the logical expression of the statement is P∧¬Q.2.It is three o’clock and I am thirsty.If p and q are statement variables, the conjunction of p and q is “p and q,” denoted p∧q. It is true when, and only when, both p and q are true.Here, the proposition “It is three o’clock” i.e. R is true and the proposition “I am thirsty.” i.e. P is also true. The logic word “and” between statement variables R and P implies conjunction of Rand P. That is, the logical expression of the statement is R∧P.3.If it is three o’clock, then I am thirsty.If p and q are statement variables, the conditional of q by p is “If p then q” or “p implies q” and isdenoted p →q. It is false when p is true and q is false; otherwise it is true; p is called the hypothesis of the conditional and q is called the conclusion.Here, the first proposition “It is three o’clock.” i.e. R is true and the second proposition “I am thirsty.” i.e. P is also true and the logic words “If R then P” implies conditional statement. Therefore, the logical expression of the statement is R→ P.4.If I am thirsty, then my glass is empty.If p and q are statement variables, the conditional of q by p is “If p then q” or “p implies q” and isdenoted p →q. It is false when p is true and q is false; otherwise it is true; p is called the hypothesis of the conditional and q is called the conclusion.Here, the first proposition “It is three o’clock.” i.e. R is true and the second proposition “I am thirsty.” i.e. P is also true and the logic words “If R then P” implies conditional statement. Therefore, the logical expression of the statement is R→ P.5.If I am not thirsty, then my glass is not empty.

The contrapositive of a conditional statement of the form “If p then q” is “If ¬q then ¬p”. Here, the first proposition is “I am not thirsty” which implies ¬P and the second proposition is “My glass is not empty.” which implies ¬Q. It is contrapositive of conditional statement Q→ P that is the logical expression for this statement is ¬Q→ ¬P.Exercise D:Given that U={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}.Let A={x:xisodd} then the elements of the set A={1,3,5,7,9,11}, B={x:x>7} then the elements of the set B={8,9,10,11,12}, and C={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∩B is intersection of sets A and B which contains all the elements that are common to both sets A and B.We have A={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 are common to both sets A and B.ii.B∪C is union of sets A and B which consist of all elements of A and all the elements of Bsuch that no element is repeated.We have B={8,9,10,11,12} and C={3,6,9,12} then B∪C={8,9,10,11,12}∪{3,6,9,12}B∪C={3,6,8,9,10,11,12}iii. ́A is complement of set A, if U is universal set and A is subset of U, then the complement ofset A i.e. ́A is the set of all the elements of U which are not the elements of A.We know that U={1,2,3,4,5,6,7,8,9,10,11,12} and A={1,3,5,7,9,11}. Then, the complement of A is ́A={2,4,6,8,10,12}={x:xis even}.iv.(A∪ ́B)∩C

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