solutionshw5-554-2010

solutionshw5-554-2010 - k This implies k 2 ≥ m 2 2 m 1>...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions homework 5. Page 25 Problem 3. : a. Take ( a n ,b n ) = (0 , 1 n ). Then n ( a n ,b n ) = , as there is no x > 0 which also satisfies x < 1 n for all n . b. Let [ a,b ] = [ a n ,b n ]. We claim A = [ a,b ]. As ( a n ,b n ) [ a n ,b n ], it is obvious that A [ a,b ]. Let now a x b . Then observe first that from ( a n ) strictly increasing it follows that a n < a for all n . Similarly b < b n for all n . Hence a n < a x b < b n for all n , i.e., x ( a n ,b n ) for all n and it follows that x A . Hence A = [ a,b ]. Page 29 Problem 3. Assume a b . Then by Theorem 1.2.8 (ii) and (iii) we conclude that a = ( a ) 2 ( b ) 2 = b , which is a contradiction. Page 29 Problem 4. Let x = 2 + 3. Then x = 2 + 3 2 - 3 · ( 2 - 3) = 1 3 - 2 . If x rational then this shows that both 2 - 3 and 2 + 3 are rational. This implies (by adding these two numbers and dividing by 2) that also 2 is rational, which is a con- tradiction. The same argument in the more general case leads to the statement that n and n + 1 are rational. By problem 17 of 1.2 this implies that n = m 2 and n + 1 = k 2 for some positive integers m and k . As m 2 < k 2 we have m < k , so m + 1
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: k . This implies k 2 ≥ m 2 + 2 m + 1 > m 2 + 1 = n + 1 = k 2 , which is a contradiction. Page 29 Problem 5. a. We prove by induction that a n < a n +1 . The claim is true for n = 1 as we have that a 1 = √ 2 < p 2 + √ 2 = a 2 . Assume now that a n < a n +1 (induction hypothesis), then to prove a n +1 < a n +2 . Now a n +1 = √ 2 + a n < √ 2 + a n +1 = a n +2 . Hence ( a n ) is strictly increasing. b. Again we prove this by induction. The case n = 1 follows from a 1 = √ 2 < √ 4 = 2. Assume now a n < 2 (induction hypothesis), Then a n +1 = √ 2 + a n , √ 2 + 2 = 2 and the claim follows by induction. c. Let a = sup a n . Then also a 2 n ↑ a 2 , since a n and a are positive. Since a 2 n +1 = a n + 2, so we conclude a 2 = a + 2. Now a 2-a-2 = ( a-2)( a + 1) = 0 implies a =-1 or a = 2. From a > 0 we conclude that a = 2 is the only possibility. 1...
View Full Document

This note was uploaded on 02/05/2012 for the course MATH 554 taught by Professor Girardi during the Fall '10 term at South Carolina.

Ask a homework question - tutors are online