3.7N_Division_into_Cases

3.7N_Division_into_Cases - Math 280 Strategies of Proof...

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Unformatted text preview: Math 280 Strategies of Proof Fall 2007 Class Notes 3.7 Division into Cases In the previous section, we considered the proof of conditional statements with a disjunction, an “or,” in the conclusion. In this section, we want to look at conditional statements with a disjunction in the hypothesis. To prove statements of this form, we use the method of division into cases , which is based upon the equivalence ( p ∨ q ) → r ≡ ( p → r ) ∧ ( q → r ) contained in Theorem 1.2.2(p). According to this equivalence, in order to prove a conditional statement of the form ( p ∨ q ) → r, we can prove the two separate conditional statements p → r and q → r. We can think of p and q as representing two cases of the hypothesis. We then prove that the conclusion r follows from each of the two cases. Here is an example to illustrate the method. Example. Prove: For all integers a , b , and c , with a 6 = 0, if a | b or a | c , then a | bc . Solution. This has symbolic form ( ∀ a 6 = 0) ( ∀ b ) ( ∀ c ) [( a | b ) ∨ ( a | c ) → ( a | bc )] , where the domain for a , b , and c is the set of integers Z . According to the above equivalence, this is logically equivalent to ( ∀ a 6 = 0) ( ∀ b ) ( ∀ c ) £° Case 1 z }| { a | b → a | bc ¢ ∧ ° Case 2 z }| { a | c → a | bc ¢§ . Proof. Suppose a , b , and c are arbitrary integers with a 6 = 0. We need to show ° a | b → a | bc ¢ and ° a | c → a | bc ¢ . Case 1. Suppose a | b . We need to show a | bc. To show this, we need to show there exists an integer k such that bc = ka . Find a k . Since a | b , then there exists an integer j such that b = ja. Multiplying this equation by c , we obtain bc = ( ja ) c = ( jc ) a. Choose k = jc . Verify that k satisfies: (i) Domain: k ∈ Z (ii) Propositional Function: bc = ka Since j and c are integers, then k = jc is an integer. From the work on the left, bc = ( jc ) a = ka, as required. Therefore a | bc , which proves Case 1. Case 2. Suppose a | c . We again need to show a | bc. To show this, we need to show there exists an integer k such that bc = ka . 113 114 Chapter 3 Proof Techniques Example. (continued) Find a k . Since a | c , then there exists an integer l such that c = la. Multiplying this equation by b , we obtain bc = b ( la ) = ( bl ) a. Choose k = bl . Verify that k satisfies: (i) Domain: k ∈ Z (ii) Propositional Function: bc = ka Since b and l are integers, then k = bl is an integer. From the work on the left, bc = ( bl ) a = ka, as required. Therefore a | bc , which proves Case 2. Since we’ve proved both cases, then this proves the given statement. Note that in both cases of the proof in the preceding example, we need to prove the same conclusion, namely that a | bc . In this example, the proofs in the two cases are almost identical, with b replaced by c in the proof of the second case. When such a situation occurs in a textbook, the proof in the second case might be omitted with the comment that the proof is similar to that in...
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3.7N_Division_into_Cases - Math 280 Strategies of Proof...

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