Example-1 :- Check whether “Or” used in the following compound statement is exclusive or inclusive? Write the component statements of the compound statements and use them to check whether the compound statement is true or not. Justify your answer. t: you are wet when it rains or you are in a river.Solution :-
“Or” used in the given statement is inclusive because it is possible that it rains and you are in the river. The component statements of the given statement are p : you are wet when it rains. q : You are wet when you are in a river. Here both the component statements are true and therefore, the compound statement is true.
Example-2 :- Write the negation of the following statements:
(i) p: For every real number x, x² > x.
(ii) q: There exists a rational number x such that x² = 2.
(iii) r: All birds have wings.
(iv) s: All students study mathematics at the elementary level.
(i) The negation of p is “It is false that p is” which means that the condition x² > x does not hold for all real numbers. This can be expressed as ∼p: There exists a real number x such that x² < x. (ii) Negation of q is “it is false that q”, Thus ∼q is the statement. ∼q: There does not exist a rational number x such that x² = 2. This statement can be rewritten as ∼q: For all real numbers x, x² ≠2 (iii) The negation of the statement is ∼r: There exists a bird which have no wings. (iv) The negation of the given statement is ∼s: There exists a student who does not study mathematics at the elementary level.
Example-3 :- Using the words “necessary and sufficient” rewrite the statement “The integer n is odd if and only if n² is odd”. Also check whether the statement is true.Solution :-
The necessary and sufficient condition that the integer n be odd is n² must be odd. Let p and q denote the statements p : the integer n is odd. q : n² is odd. To check the validity of “p if and only if q”, we have to check whether “if p then q” and “if q then p” is true. Case 1 : If p, then q If p, then q is the statement: If the integer n is odd, then n² is odd. We have to check whether this statement is true. Let us assume that n is odd. Then n = 2k + 1 when k is an integer. Thus n² = (2k + 1)² = 4k² + 4k + 1 Therefore, n² is one more than an even number and hence is odd. Case 2 : If q, then p If q, then p is the statement If n is an integer and n² is odd, then n is odd. We have to check whether this statement is true. We check this by contrapositive method. The contrapositive of the given statement is: If n is an even integer, then n² is an even integer n is even implies that n = 2k for some k. Then n² = 4k². Therefore, n² is even.
Example-4 :- For the given statements identify the necessary and sufficient conditions. t: If you drive over 80 km per hour, then you will get a fine.Solution :-
Let p and q denote the statements: p : you drive over 80 km per hour. q : you will get a fine. The implication if p, then q indicates that p is sufficient for q. That is driving over 80 km per hour is sufficient to get a fine. Here the sufficient condition is “driving over 80 km per hour”: Similarly, if p, then q also indicates that q is necessary for p. That is When you drive over 80 km per hour, you will necessarily get a fine. Here the necessary condition is “getting a fine”.