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Analytic Geometry and Calculus

by: Kavon Feest

Analytic Geometry and Calculus MATH 16B

Kavon Feest

GPA 3.93

J. Wilkening

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J. Wilkening
Class Notes
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This 3 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 16B at University of California - Berkeley taught by J. Wilkening in Fall. Since its upload, it has received 39 views. For similar materials see /class/226581/math-16b-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.


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Date Created: 10/22/15
H N 03 Let Zgt0 Math 113 Section 5 Fall 2010 Homework 1 Solutions Due Wednesday September 8 Section 01 Exercise 5 Determine whether the following functions f are well de ned a f Q a Z de ned by f a f is not well de ned Let 73 be another representative of that is a0 7 a and b0 y b but aob abo Then fg a0 7E a f QeQ mmW f is well de ned This is because the codomain is Q Let 57 be another representative of Then 1 aobab0 aob2 ab02 a3b2a2bg jig 10 f Prove Proposition 2 section 01 Let A be a nonempty set a Prove that if N determines an equivalence relation on A then the set of equivalence classes A N of N forms a partition of A Let a E A and E b E Alb N a We needto show 1 UaeAZL A and 2 W113 j a 8 1 Every element is in some equivalence class In particular a 6 7L 2 Assume 7 1 7 1 Then there exists a c E 7 1 Therefore 0 N a and c N 1 Because N is an equivalence relation a N b so a E b and b E 6 Further for any 0 E i c E I and vice versa Therefore 6 1 b Prove that if E I is a partition of A then there is an equivalence relation on A whose equivalence classes are precisely the sets Ahi E I De ne N on A by a N b if ab E A for some i We need to show for all a E A 1 aNa 2aNb gt bNa 3aNbandec gt aNc 1 is trivially satis ed 2 is also trivially satis ed because if a and b are in A then b and a are in A 3 is satis ed because if a b c E A then in particular a c E A Therefore N is an equivalence relation whose classes are the A 12 be the set of positive integers sometimes called the natural numbers and denoted N Strange but true there is a bijection of sets 4p Q a Zgt0 Determine 4p Hint you may describe Lp as a list of ordered pairs or as a formula or as any kind of diagram you like The de nition need not be unique so your answer may differ from your friends Most standard de nitions of 4p that I know of follow some kind of diagonalization argument The idea is to count the elements of Q q1q2q3 and let Apqi i Has 5 cm W my Km m law the mm awn m pmnble mum L x n w Mf39llrw 1r 4m 1 m J 7 7W 1339 7quot w m 7quot 71 H w m Kn m m 41 m neumbas uppibamgtdihadxepmbuveufthem mmmcbxmmaymwm Tmshnwsmmun uiQgtg Ymcm mmmmmmm ma Wecmcumbmed sgwmmunumgw 5mcmnum oa oimwnumbaufwayx gamma mama Mmwoum ebampmvmawwmmmmmm A UsetheEunlldeanakundsmm dmmmed egamcoxmdwlmxuf 12m 5 kmnOlecuu yumm pmemtd aedamma beziomkduza P9024 mnmmmba my yum v wmmmmm um mu 6 ag MM we Mix WWMWWWHWWM mcmmmnmm mmmhmab 5 9 c List the elements of Z12Z which are invertible Z12Zgtlt 17577171 because these Z are exactly those for which i712 1 Section 037 Exercise 3 Prove that the distinct equivalence classes in ZnZ are precisely 6127 n E 1 use the Division Algorithm For any a E Z7 use the division algorithm so a nqr for some q7 r E Z and 0 g r lt n ThenaEr mod n and7777 so e 612n71 To show that the classes are distinct7 let 0 S 071 f n such that a 7E 1 Assume 1970 gt 0 Since ab lt n 71 does not divide b E a and 6 7E b Prove Proposition 4 Section 03 Exercises 12 and 13 a Let n E Z with n gt 1 and a E Z with 1 g a S n Prove if am 3A 1 then there exists a b E Z with 1 g b lt n such that ab E 0 mod n and deduce that there cannot be an integer c such that ac E 1 mod Let 1 1771 and a a d Then is an integer and since 1 7 17 g lt n We have ga ga d na E 0 mod n so we let I g b Let n E Z with n gt 1 and a E Z with 1 g a g n Prove if am 1 then there is an integer c such that ac E 1 mod 71 use the fact that the gcd of two integers is a Z linear combination of the integers There exist z and y so that am 1 my 1 gt am 1 mod Let c m


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