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Unformatted text preview: Homework 2 Problems graded: 4.12 (10 points), 5.2 (10 points), 7.3 (15 points), 8.6 (15 points). The other problems were graded for completion to be worth five points each (and another five points were given to all students to make the total point value 100). 4.10. Suppose a > 0. By the Archimedean property there exists a natural m 1 with m 1 a > 1; dividing by m 1 , a > 1 /m 1 . However (by 3.2.vi), a 1 > 0; by the Archimedean property there exists a natural m 2 with m 2 a 1 > 1, i.e. m 2 > a . Letting n be the maximum of m 1 and m 2 , n N with a > 1 /m 1 1 /n (by 3.2.vii as n m 1 ) and n m 2 > a , i.e. 1 /n < a < n . 4.11. Suppose a < b are real. We construct an infinite increasing sequence of rationals { q i } in ( a,b ) by the density of Q as follows: Step 1: Let q 1 be a rational in ( a,b ). Step k (assuming step 1 ,...,k 1 are done): Let q k be a rational in ( q k 1 ,b ). After completing this procedure, we have that the q i are distinct rationals with a < q 1 < ... < q i < ... < b . 4.12. We begin by noting that if r is rational then r + 2 is irrational (as otherwise ( r + 2) r = 2, being the difference of two rationals, is rational; note Q forms a field). Now, if a < b , a 2 < b 2 so there exists a rational y with a 2 < y < b 2. Adding 2, we have that y + 2 is irrational with a < y + 2 < b . 4.14. (a) If is the supremum (which is an upper bound) of A and is the supremum (which is an upper bound) of B then for each s S there exist a A,b B with a + b = s so s = a + b + . ( and are finite as A,B are nonempty bounded). This tells us that sup S sup A + sup B . For equality we let > 0; there exist a ,b A,B respectively with  a ,  b both less than / 2. Therefore, a + b is an element of S with a + b > +  . Letting go to zero we have sup S sup A + sup B ; this indeed gives us sup S = sup A + sup...
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 Spring '08
 hitrik

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