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17 Jul 2023: Given constants b1, b2 such that b1 eb2 0 use the Lagrangian method to. ... Hint: There will be two cases to check depending the constants b1 and b2.].

7 Jan 2023: Exercise 3.2.20. (i) If a N+, then a = (a 1) 1 Z.(ii) If a, b Z, then a = a1 a2, b = b1 b2 with a j, b j ... Thus. a b = (a1 b1) (a2 b2) Z,a b = (a1b1 a2b2) (a1b2 a2b1) Z,a = a2 a1 Z.

20 Dec 2023: Chooseconstants M,D > 0 such that |θ(a)| 6 M|a| for all a A and |b1b2| 6 D|b1||b2|for all b1,b2 B; then for all a A ... 0|b|/r and choose constants M,D > 0 such that |f(a)|6. M|a| for all a A and |b1b2|6 D|b1||b2| for all b1,b2

23 Jan 2023: B(G,{µ}) = {[b] B(G) : κG(b) = µ,νb µ}.Note that for [b1], [b2] B(G,{µ}), we have [b1] = [b2] if and only if ν[b1] = ν[b2],. ... since [b1] and [b2] have common image µ under κG. Definition 2.2.4.