370hw1solnsFall2011

370hw1solnsFall2011 - Math 370 Solutions to Assignment 1 5...

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Math 370 Solutions to Assignment 1 5, 8.1, 8.2, 18, 22, 25, 28. 5. Let A = f n 2 N : n > 3 g : Then A has no maximal element. Proof. Suppose that A has a greatest element m: Then m 2 A so 3 < m < m +1 2 A: This contradicts the property that m is maximmal in A: Thus our supposition is false so A has no maximal element. 8.1 Let F be an ordered °eld and x 2 F: Then x > 0 i/ ° x < 0 : Proof. Suppose x > 0 : Then by O 2 , ° x = 0 + ( ° x ) < x + ( ° x ) = 0 : Conversely, suppose ° x < 0 : Then 0 = ° x + x < 0 + x = x: 8.2 Let e be the multiplicative identity in an ordered °eld F: Then e > 0 : Proof. We are given that e 6 = 0 : By the trichotomy law there remain only two mutually exclusive possiblities: 0 < e and e < 0 : Suppose that e < 0 : Adding ° e , we get 0 < ° e: Thus 0 < ( ° e ) ( ° e ) = e: This contradiction shows that 0 < e: 18 Let S be a subset of an ordered °eld F . The b = inf S i/ the following conditions hold ( i ) b ± x for all x 2 S and ( ii ) " > 0 ) b + " > x for some x 2 S: Proof. (= ) ) Suppose b = inf S: Then b is a lower bound for S so ( i ) holds. Suppose " > 0 : Then b < b + ": Since b is the greatest lower bound of S it follows that b + " cannot be a lower bound of
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