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Unformatted text preview: 3. Assume c > for contradiction 4. a < b Step 2, logic 5. ac > bc multiplying step 4 by c , Axiom 13 6. ac n > bc trichotomy, Step 2 7. Contradiction Steps 5 and 6 Theorem. If a < b and ac ≤ bc , then c ≤ 0. Proof. We will prove this theorem by contradiction. Thus suppose a , b , and c are real numbers such that a < b and ac ≤ bc , and assume for the sake of contradiction that c > 0. When we multiply both sides of the inequality a < b by the negative number c , the multiplicative law for inequalities yields ac > bc. But this contradicts our assumption that ac ≤ bc . Thus the assumption must have been false, and we must have c ≥ 0 as claimed....
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This note was uploaded on 12/08/2009 for the course MATH 300 taught by Professor Staff during the Spring '08 term at University of Washington.
 Spring '08
 Staff
 Math, Addition

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