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# Chapter 3

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### Chapter 3

1. 1. SUBTOPIC 3 : QUANTIFIERS The statement P: n is odd integer. A proposition is a statement that is either true or false. The statement p is not propositionbecause whether p is true or false depends on the value of n. For example, p is true if n = 104 andfalse if n = 8. Since, most of the statements in mathematics and computer a science use variable,we must extend the system of logic to include such statements. 1.Quantifiers Definition: Let P (x) be a statement involving the variable x and let D be a set. We call P a proportional function or predicate (with respect to D ) , if for each x ∈ D , P (x) is a proposition. We call D the domain of discourse of P.Example 1: Let P(n) be the statement n is an odd integer Then P is a propositional function with the domain of discourse since for each n ∈ , P(n) is a proposition [for each n ∈ , P(n) is true or false but not both]. For example, if n =1, we obtain the proposition. P (1): 1 is an odd integer(Which is true) If n = 2, we obtain the proposition/ P (2): 2 is an odd integer(Which is false) 14
2. 2. A propositional function P, by itself, is neither true nor false. However, for each x isdomain of discourse, P (x) is a proposition and is, therefore, either true or false. We can think ofpropositional function as defining a class of propositions, one for each element in the domain ofdiscourse. For example, if P is a propositional function with domain of discourse , we obtainthe class of propositions. P (1), P (2), …..Each of P (1), P (2), …. Is either true or false. 2. Universal Quantification Definition: Let P be a propositional function with the domain of discourse D. The universal quantification of P (x) is the statement. “For all values of x, P is true.” ∀x, P (x)Similar expressions: - For each… - For every… - For any… 3. Counterexample A counterexample is an example chosen to show that a universal statement is FALSE. To verify: - ∀x, P (x) is true - ∀x, P (x) is false 15
3. 3. Example 1:Consider the universally quantified statement. ∀x ( ≥ 0)The domain of discourse is R. The statement is true because for every real number x, it is truethat the square of x is positive or zero. According the definition, the universally quantified statement. ∀x, P (x)Is false for at least one x in the domain of discourse that makes P (x) is false. A value x in thedomain of discourse that makes P (x) false is called a counterexample to the statement.Example 2:Consider the universally quantified statement. ∀x ( -1 ≥ 0)The domain of discourse is R. The statement is false since, if x = 1, the proposition -1 > 0Is false. The value1 is counterexample of the statement. ∀x ( -1 ≥ 0)Although there are values of x that make the propositional function true, the counterexampleprovide show that the universally quantified statement is false. 16
4. 4. 4. Existential Quantification Let P be a proportional function with the domain of discourse D. The existential quantification of P (x) is the statement. “There exists a value of x for which P (x) is true. ∃x, P(x) Similar expressions: - There is some… - There exist… There is at least…Example 1:Consider the existentially quantified statement.∃x (The domain of discourse is R. the statement is true because it is possible to find at least one realnumber x for which the propositionIs true. For example, if x = 2, we obtain the true proposition. 17
5. 5. Is not the case that every value of x results in a true proposition. For example, if x = 1, thepropositionIs false. According to definition, the existentially quantified statement ∃x, P(x)Is false for every x in the domain of discourse, the proposition P (x) is false. 18
6. 6. 5. De Morgan’s Law for Logic Theorem: (∀x, P (x)) ≡ (∃x, (P(x)) (∃x, (P(x)) ≡ (∀x, P (x)) The statement “The sum of any two positive real numbers is positive”. ∀x > 0∀y > 0,Example 1: Let P(x) be the statementWe show that ∃x, P(x)Is false by verifying that ∀x, ⌐ P (x)Is true. The technique can be justified by appealing to theorem. After we prove that propositionis true, we may negate and conclude that is false. By theorem, ∃x, ⌐⌐P(x)Or equivalently ∃x, P(x)Is also false. 19
7. 7. 1. EXERCISE. a) let P (x) be the propositional function “x ≥ .” Tell whether each proposition is true or false. The domain of discourse is R i. P (1) ii. ⌐∃x P(x) b) Suppose that the domain of discourse of the propositional function P is {1, 2, 3, 4}. Rewrite each propositional function using only negation, disjunction and conjunction. i. ∀x P (x) c) Determine the truth value, the domain discourse R x R, justify the answer. i. ∀x ∀y ( < y + 1) d) Assume that ∀x ∃y P(x, y) is false and that the domain of discourse is nonempty. It must also be false? Prove the answer. i. ∀x ∀y P (x,y) 20