3.4N_Proving_Conditional_Statements

3.4N_Proving_Conditional_Statements - Math 280 Strategies...

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Unformatted text preview: Math 280 Strategies of Proof Fall 2007 Class Notes 3.4 Proving Conditional Statements I: Direct Proofs As we’ve already seen, many definitions and theorems in mathematics have the logical form ∀ x [ p ( x ) → q ( x )] or, more generally, ∀ x 1 ∀ x 2 · · · ∀ x n [ p ( x 1 , x 2 , . . . , x n ) → q ( x 1 , x 2 , . . . , x n )] . To prove a statement of the form ∀ x [ p ( x ) → q ( x )] is true, we need to show that the conditional propositional function p ( x ) → q ( x ) is true for all x in its domain. For a given x , the conditional p ( x ) → q ( x ) is true except when p ( x ) is true and q ( x ) is false. To prove that p ( x ) → q ( x ) is true, we need to show that this situation cannot occur. That is, whenever the hypothesis p ( x ) is true, then the conclusion q ( x ) must also be true. This observation forms the basis of a direct proof of a conditional propositional function p ( x ) → q ( x ). We start by supposing the hypothesis p ( x ) is true. We then show that the conclusion q ( x ) must be true under this assumption. So to prove that a conditional quantified statement of the form ∀ x [ p ( x ) → q ( x )] is true, we combine a proof by arbitrary element with a direct proof of the conditional propositional function to obtain the following general procedure: Direct Proof of ∀ x [ p ( x ) → q ( x )] 1. Specify variables. Since the statement involves a universal quantifier, we start by supposing x is an arbitrary but fixed element of the domain. We then have to prove that the conditional propositional function p ( x ) → q ( x ) is true. 2. Suppose the hypothesis is true. Suppose that the hypothesis p ( x ) is true for the arbitrary element x . Note that since a conditional statement with a false hypothesis is always true, we do not need to be concerned with the case that the hypothesis is false. 3. Show that the conclusion is true. We need to show that the conclusion q ( x ) must then be true for this element x using definitions, axioms, previously established results, or the rules of logic. As usual with a proof by arbitrary element, we must be careful to treat x as an arbitrary element. We cannot assume it has any properties that are not shared by all elements in its domain. The procedure is generally shortened slightly by combining the first two steps into a single statement by writing at the start of the proof something similar to: Suppose p ( x ) is true for an arbitrary element x of the domain. Functions We start with some definitions about functions whose logical form involve conditional statements. Definition. A function f : A → B is injective , or one-to-one , provided that for all a 1 , a 2 ∈ A , if a 1 6 = a 2 , then f ( a 1 ) 6 = f ( a 2 ). This means that distinct elements a 1 and a 2 in the domain A are mapped into distinct elements in the codomain B . In symbolic form, f is injective provided that ( ∀ a 1 ∈ A ) ( ∀ a 2 ∈ A ) £ ( a 1 6 = a 2 ) → ° f ( a 1 ) 6 = f ( a 2 ) ¢§...
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This note was uploaded on 10/20/2009 for the course MATH 280 taught by Professor Solheid during the Spring '08 term at CSU Fullerton.

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3.4N_Proving_Conditional_Statements - Math 280 Strategies...

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