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hw4c - 3 Assume c> for contradiction 4 a< b Step 2...

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Math 300A&B Introduction to Mathematical Reasoning Fall 2009 Assignment #4: Due Wednesday, 10/28/09 Part III: 3. In addition to the two-column proofs you were asked to do in Part I of this assignment, write paragraph-style proofs of the following two theorems, making sure to follow the guidelines in the handout Conventions for Writing Mathematical Proofs . Theorem 7(c). Theorem 9(b). As a sample, here’s my two-column proof of one of the theorems I proved in class today, together with a paragraph-style version of the same proof. Of course, this is just one way to write it up; there are many, many others. Theorem. If a < b and ac bc , then c 0. Statement Reason 1. Let a, b, c R hypothesis 2. Assume a < b and ac bc hypothesis
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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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