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Slide subtopic 4

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  • 1. UNIVERSITI PENDIDKAN SULTAN IDRIS PREPARED BY : MOHAMAD AL FAIZ BIN SELAMAT
  • 2. Tautology:A proposition that is always true for all possible value of its propositionalvariables.Example of a TautologyThe compound proposition p ˅¬p is a tautology because it is always true. P ¬p p ˅ ¬p T F T F T T
  • 3. P ¬p p ˅ ¬pT F FF T F
  • 4. A contingency table is a table of counts. A two-dimensional contingency table is formed by classifying subjects by two variables. One variable determines the row categories; the other variable defines the column categories. The combinations of row and column categories are called cells. Examples include classifying subjects by sex (male/female) and smoking status (current/former/never) or by "type of prenatal care" and "whether the birth required a neonatal ICU" (yes/no). For the1. mathematician, a two-dimensional contingency table with r rows and c columns is the set {xi j: i =1... r; j=1... c}.
  • 5. :, Definition: An argument is a sequence of propositions written The symbol ∴ is read “therefore.” The propositions, ,… are called the hypotheses (or premises) or the proposition q is called the conclusion. The argument is valid provide that if the proposition are all true, then q must also be true; otherwise, the argument is invalid (or a fallacy).
  • 6. p→qDetermine whether the argument p ∴qIs valid [First solution] We construct atruth table for all the propositionsinvolved. P q p→q p q T T T T T T F F T F F T T F T F F T F F
  • 7. Rule of inference Name p→q Modus ponens p ∴q p→q Modus tollens ⌐q ∴ ⌐p Addition p ∴p˅q Simplification p˅q ∴p p Conjunction q ∴p˅q p→q Hypothetical syllogism q→r ∴p→r p˅q Disjunctive syllogism ⌐p ∴q
  • 8. Represent the argument. The bug is either in module 17 or in module 81 The bug is a numerical error Module 81 has no numerical error ___________________________________________ ∴ the bug is in module 17.Given the beginning of this section symbolically and show that it is valid. If we let p : the bug is in module 17. q : the bug is in module 81. r : the bug is numerical error.
  • 9. The argument maybe written pVq r r → ⌐q ∴pFrom r → ⌐q and r, we may use modus ponens to conclude ⌐q. From r V q and ⌐q, we may use the disjunctive syllogism to conclude p. Thus the conclusion p follows from the hypotheses and the argument is valid.
  • 10. This method is based on Modus Ponens,[(p ⇒ q) ˄ p ] ⇒ qVirtually all mathematical theorems are composed of implication of the type, (The are called the hypothesis or premise, and q is called conclusion. To prove a theorem means to show the implication is a tautology. If all the are true, the q must be also true.
  • 11. Solution:Let p: x is odd, and q: x2 is odd. We want to prove p → q.Start: p: x is odd→ x = 2n + 1 for some integer n→ x2 = (2n + 1)2→ x2 = 4n2 + 4n + 1→ x2 = 2(2n2 + 2n) + 1→ x2 = 2m + 1, where m = (2n2 + 2n) is an integer→ x2 is odd→ q
  • 12. Definition:An indirect proof uses rules of inference on the negation ofthe conclusion and on some of the premises to derive thenegation of a premise. This result is called a contradiction.  Contradiction: to prove a conditional proposition p ⇒ q by contradiction, we first assume that the hypothesis p is true and the conclusion is false (p˄ ~ q). We then use the steps from the proof of ~q ⇒ ~p to show that ~p is true. This leads to a contradiction (p˄ ~ p), which complete the proof.
  • 13. Proof: Assume that x is even (negation of conclusion).Say x = 2n (definition of even).Then = (substitution)= 2n · 2n (definition of exponentiation)= 2 · 2n2 (commutatively of multiplication.)Which is an even number (definition of even)This contradicts the premise that is odd.

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