Exam2-Solutions

Exam2-Solutions - Math 326 - Exam 2 Solutions - 12 Oct 2010...

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Math 326 - Exam 2 Solutions - 12 Oct 2010 1. (a) (2 points) State precisely but concisely what it means when a | b . ( k Z )( b = ka ) (b) (3 points) Does 12 | 0? why or why not? Yes, because 0 = 12 · 0 (c) (2 points) State precisely but concisely what it means for a number to be rational. ( x Q ) ⇐⇒ ( a,b Z )( b 6 = 0)( x = b/a ) (d) (3 points) Is 4.613 rational? why or why not? Yes because 4 . 613 = 4613 1000 2. Prove by contradiction that for any integer n , it is impossible for n to equal both 3 q 1 + r 1 and 3 q 2 + r 2 , where q 1 ,q 2 ,r 1 ,r 2 Z and 0 r 1 ,r 2 < 3, and r 1 6 = r 2 . Proof 1. Let q 1 ,q 2 Z and suppose that for some r 1 and r 2 , 3 q 1 + r 1 = 3 q 2 + r 2 3( q 1 - q 2 ) = r 2 - r 1 If r 1 = 0 then r 2 = 1 or r 2 = 2. r 2 = 1 = r 2 - r 1 = 1 = q 1 - q 2 = 1 / 3 6∈ Z r 2 = 2 = r 2 - r 1 = 2 = q 1 - q 2 = 2 / 3 6∈ Z If r 1 = 1 then r 2 = 0 or r 2 = 2. r 2 = 0 = r 2 - r 1 = 0 - 1 = - 1 = q 1 - q 2 = - 1 / 3 6∈ Z r 2 = 2 = r 2 - r 1 = 2 - 1 = 1 = q 1 - q 2 = 1 / 3 6∈ Z If r 1 = 2 then r 2 = 0 or r 2 = 1. r 2 = 0 = r 2 - r 1 = 0 - 2 = - 2 = q 1 - q 2 = - 2 / 3 6∈ Z r 2 = 1 = r 2 - r 1 = 1 - 2 = - 1 = q 1 - q 2 = - 1 / 3 6∈ Z Hence all possible cases lead to contradiction so the conjecture is proven. Proof 2. Let q 1 ,q 2 Z and suppose that for some r 1 and r 2 , 3 q 1 + r 1 = 3 q 2 + r 2 3( q 1 - q 2 ) = r 2 - r 1 3 = r 2 - r 1 q 1 - q 2 1
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But 0 r 1 ,r 2 < 3, i.e., r 1 = 0 , 1 , or 2 and r 2 = 0 , 1 , or 2, where r 1 6 = r 2 ; hence | r 1 - r 2 | ≤ 2 and therefore 3 = r 2 - r 1 q 1 - q 2 2 q 1 - q 2 Since r 1 6 = r 2 , from the equation 3( q 1 - q 2 ) = r 2 - r 1 we conclude q 1 6 = q 2 . Hence | q 1 - q 2 | ≥ 1 = 1 | q 1 - q 2 | 1 Hence by substitution 3 2 which is a contradiction. . Proof 3.
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.

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Exam2-Solutions - Math 326 - Exam 2 Solutions - 12 Oct 2010...

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