By Hugo D. Junghenn

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**Example text**

D) a, b, a, c, a, b, a, c . . (c) a, a, a, b, b, b, a, a, a . . (e) 1, 2, 3, 4, 1, 2, 3, 4, . . 2. Find a recursive formula for the sequence a, b, a, b, . . 3. Use the ε, N definition of limit to prove that 4n − 1 (a) lim = 2. n 2n + 7 (b) S √ 5 5 n+7 √ = . (c) lim n 3 n+2 3 2n2 − n lim 2 = 2. n n +3 1 n−1 = +∞. (e) S lim 2 + (d) lim √ n n n n+1 3 = 8. (f) limn n+2 = 1. n+1 4. Prove rigorously that the sequence {(−1)n n/(n + 1)} has no limit. rπ) for r ∈ Q. 6. Find limn 1 1 n+ n n p for all p ∈ R.

D) x · (y + z) = (x · y) + (x · z) (additivity). (e) |x · y| ≤ x 2 y 2 (Cauchy–Schwartz inequality). Proof. Properties (a) and (b) are immediate and parts (c) and (d) follow respectively from the calculations n n xj yj = t j=1 n n (txj )yj = j=1 xj (tyj ) and j=1 n xj (yj + zj ) = j=1 n xj yj + j=1 xj zj . j=1 The inequality in (e) holds trivially if y = 0. Suppose y = 0, so y By properties (a)–(d), 0 ≤ x − ty 2 2 = (x − ty) · (x − ty) = x 2 2 2 = 0. − 2t(x · y) + t2 y 22 . Setting t = (x · y)/ y 22 , we obtain 0≤ x 2 2 − 2(x · y)2 / y 2 2 + (x · y)2 / y which implies that (x · y)2 ≤ x 2 2 2 2 = x 2 2 − (x · y)2 / y 22 , y 22 .

2, Q is an ordered field under the algebraic operations and order relation inherited from R. The same is true for the set √ √ Q( 2) := {x + 2 y : x, y ∈ Q} (Exercise 19). This suggests that there are infinitely many ordered subfields of R. The property that distinguishes R from all other ordered fields is completeness, described in this section. 1 Definition. A nonempty subset A of an ordered field F is said to be bounded above if there exists a member u ∈ F, called an upper bound of A, such that a ≤ u for all a ∈ A.