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# ch11 - P1 PBU/OVY JWDD027-11 P2 PBU/OVY QC PBU/OVY T1 PBU...

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49 588 SECTION 11.1 CHAPTER 11 SECTION 11.1 1. lub = 2; glb = 0 2. lub = 2; glb = 0 3. no lub; glb = 0 4. lub = 1 , no glb 5. lub = 2; glb = 2 6. lub = 3; glb = 1 7. no lub; glb = 2 8. lub = 2; glb = 2 9. lub = 2 1 2 ; glb = 2 10. lub = 0; glb = 1 11. lub = 1; glb = 0 . 9 12. lub = 2 1 9 , glb = 2 1 9 13. lub = e ; glb = 0 14. no lub , glb = 1 15. lub = 1 2 ( 1 + 5); glb = 1 2 ( 1 5) 16. no lub , no glb 17. no lub; no glb 18. no lub; no glb 19. no lub; no glb 20. lub = 0; no glb 21. glb S = 0 , 0 ( 1 11 ) 3 < 0 + 0 . 001 22. glb = 1; 1 1 < 1 + 0 . 0001 23. glb S = 0 , 0 1 10 2 n 1 < 0 + 1 10 k n > k + 1 2 24. glb = 0; 0 1 2 n < 0 + 1 4 k for n > 2 k 25. Let > 0 . The condition m s is satisfied by all numbers s in S . All we have to show therefore is that there is some number s in S such that s < m + . Suppose on the contrary that there is no such number in S. We then have m + x for all x S. This makes m + a lower bound for S. But this cannot be, for then m + is a lower bound for S that is greater than m, and by assumption, m is the greatest lower bound. 26. (a) Let M = | a 1 | + · · · + | a n | . Then for any i , | a i | < M , so S is bounded (b) lub S = max { a 1 , a 2 , . . . , a n } ∈ S glb S = min { a 1 , a 2 , . . . , a n } ∈ S

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-11 JWDD027-Salas-v1 December 2, 2006 16:49 SECTION 11.1 589 27. Let c = lub S . Since b S, b c. Since b is an upper bound for S, c b. Thus, b = c . 28. S consists of a single element, equal to lub S . 29. (a) Suppose that K is an upper bound for S and k is a lower bound. Let t be any element of T . Then t S which implies that k t K . Thus K is an upper bound for T and k is a lower bound, and T is bounded. (b) Let a = glb S. Then a t for all t T. Therefore, a glb T. Similarly, if b = lub S, then t b for all t T, so lub T b. It now follows that glb S glb T lub T lub S. 30. (a) Let S = { r : r < 2 , r rational } . (b) Let T = { t : t < 2 , t irrational } . 31. Let c be a positive number and let S = { c, 2 c, 3 c, . . . } . Choose any positive number M and consider the positive number M/c . Since the set of positive integers is not bounded above, there exists a positive integer k such that k M/c . This implies that kc M . Since kc S , it follows that S is not bounded above. 32. (a) If S is a set of negative numbers, then 0 is an upper bound for S . It follows that α = lub S 0. (b) If T is a set of positive numbers, then 0 is a lower bound for T . It follows that β = glb T 0. 33. (a) See Exercise 75 in Section 1.2. (b) Suppose x 2 0 > 2. Choose a positive integer n such that 2 x 0 n 1 n 2 < x 2 0 2 . Then, 2 x 0 n 1 n 2 < x 2 0 2 = 2 < x 2 0 2 x 0 n + 1 n 2 = x 0 1 n 2 (c) If x 2 0 < 2, then choose a positive integer n such that 2 x 0 n + 1 n 2 < 2 x 2 0 . Then x 2 0 + 2 x 0 n + 1 n 2 < 2 = ( x 0 + 1 n ) 2 < 2 34. Assume that there are only finitely many primes, p 1 , p 2 , · · · , p n and let Q = p 1 · p 2 · · · p n + 1. Q has a prime divisor p . But Q is not divisible by any of the p i , so p = p 1 for all i a contradiction.
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ch11 - P1 PBU/OVY JWDD027-11 P2 PBU/OVY QC PBU/OVY T1 PBU...

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