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Of elements in S with sn the least element of Sn = S − {s0 , s1 , . . , sn−1 } which is never empty. Notice in particular that s0 < s1 < · · · < sn < · · · , from which it easily follows that sn n. If s ∈ S, then for some m ∈ N0 must satisfy m by construction of the sn we must have s = sm0 for some m0 . Hence s, so S = {sn : n ∈ N0 }. Now define a function f : N0 −→ S by f (n) = n; this is easily seen to be a bijection. b) The simplest case is where X ∩ Y = ∅. Then given bijections f : N0 −→ X and g : N0 −→ Y we construct a function h : N0 −→ X ∪ Y by  n   f 2 h(n) = n−1   g 2 if n is even, if n is odd.

P2 More generally, show by Induction that for n pn−2 (1 + p) ≡ 1 + pn−1 , pn 2, the following congruences are true: n−1 (1 + p)p ≡ 1. pn What can you say about the case p = 2? 1-15. 14160, 51/11, 1725/1193, 1193/1725, −1193/1725, 30031/16579, 1103/87. In each case determine all the convergents. 1-16. If n is a positive integer, what are the continued fraction expansions of −n and 1/n? What about when n is negative? ] Try to find a relationship between the continued fraction expansions of a/b and −a/b, b/a when a, b are non-zero natural numbers.

Sn , . . of elements in S with sn the least element of Sn = S − {s0 , s1 , . . , sn−1 } which is never empty. Notice in particular that s0 < s1 < · · · < sn < · · · , from which it easily follows that sn n. If s ∈ S, then for some m ∈ N0 must satisfy m by construction of the sn we must have s = sm0 for some m0 . Hence s, so S = {sn : n ∈ N0 }. Now define a function f : N0 −→ S by f (n) = n; this is easily seen to be a bijection. b) The simplest case is where X ∩ Y = ∅. Then given bijections f : N0 −→ X and g : N0 −→ Y we construct a function h : N0 −→ X ∪ Y by  n   f 2 h(n) = n−1   g 2 if n is even, if n is odd.

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