3.6N_Proving_Conditional_Statements

# 3.6N_Proving_Conditional_Statements - Math 280 Strategies...

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Math 280 Strategies of Proof Fall 2007 Class Notes 3.6 Proving Conditional Statements II: Indirect Proofs In Section 3.4, we studied direct proofs of a conditional statement of the form x [ p ( x ) q ( x )]. We start by supposing x is an arbitrary element of the domain. We then need to show that the conditional predicate p ( x ) q ( x ) must be true for this arbitrary element x . From the deﬁnition, we know that a conditional statement is false only when the hypothesis is true and the conclusion is false. To prove that p ( x ) q ( x ) is true, we must then show that whenever the hypothesis p ( x ) is true, the conclusion q ( x ) must also be true. In a direct proof of p ( x ) q ( x ), we start with the hypothesis and then make one deduction after another until we can deduce that the conclusion must then be true. In an indirect proof, we show that a statement is true by showing that a logically equivalent statement must be true. If we can show the equivalent statement is true, then it follows from the equivalence that the original statement must also be true. As we’ll see, it is often easier to prove the equivalent statement and in many cases, we can only prove the equivalent statement and not the original statement. Proof by Contrapositive Since a conditional statement is logically equivalent to its contrapositive, then in order to prove a statement which has the form x [ p ( x ) q ( x )] , we may instead prove the contrapositive statement x £° q ( x ) ¢ ° p ( x ) ¢§ . This method of proof of a conditional statement is called a proof by contrapositive . Proof by Contrapositive of x [ p ( x ) q ( x )] Form the contrapositive of the given statement: x £° q ( x ) ¢ ° p ( x ) ¢§ . Then carry out the steps for a direct proof of the contrapositive statement: 1. Specify variables. Suppose x is an arbitrary but ﬁxed element of the domain. We then have to prove that the conditional propositional function ° q ( x ) ¢ ° p ( x ) ¢ is true. 2. Suppose the hypothesis is true. Suppose that the hypothesis q ( x ) of the contrapositive statement is true or, equivalently, the conclusion q ( x ) of the original statement is false for the arbitrary element x . 3. Show that the conclusion is true. We need to show that the conclusion p ( x ) of the contrapositive statement must then be true or, equivalently, the hypothesis p ( x ) of the original statement must be false for this element x using deﬁnitions, axioms, previously established results, or the rules of logic. We’ve already seen some examples of proofs by contrapositive in previous sections. In order to show that a function f : X Y is injective, we usually use the contrapositive version of the statement in the deﬁnition: ( x 1 X ) ( x 2 X ) [ x 1 6 = x 2 f ( x 1 ) 6 = f ( x 2 )] ( x 1 X ) ( x 2 X ) [ f ( x 1 ) = f ( x 2 ) x 1 = x 2 ] The reason for using the contrapositive statement is that it is often easier to prove an equality,

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3.6N_Proving_Conditional_Statements - Math 280 Strategies...

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