Lecture 25 Enoch Cheung November 21 2013 Exam Review The biggest thing for this exam and most math exams is to make sure they re familiar with their formal definitions If they start off a proof by saying let a PreIm Y their next line should be to translate what that means i e b Y such that f a b Proofs on the test will rely mostly on definitions and some basic theorems Euclid s Lemma Fundamental Theorem of Airthmetic GCD Theorem MAL The other thing to work on is that they know how to read symbolic statements and how to begin proofs of universal existential statements especially universal quantifiers in front of a conditional statement A good list of the definitions and theorems that you need to know is on the review sheet on Blackboard The following is taken from it Number theory The Division Algorithm the definition of congruence modulo m the modular arithmetic lemma the definition of a gcd the gcd theorem Euclid s Lemma the fundamental theorem of arithmetic what the set Z mZ looks like which elements of Z mZ have multiplicative inverses how to solve linear diophantine equations or determine when there is no solution and the Chinese Remainder Theorem Functions Cardinality Know the definition of a function image and preimage of a set under a function how to prove disprove injectivity surjectivity and bijectivity how to prove two functions are inverses invertibile iff bijective cardinality CBS theorem Cantor s Theorem how to determine whether a set is countably or uncountably infinite countable union of countable sets is countable finite product of countable sets is countable the real numbers are uncountable and diagonalization arguments 1 Find all solutions to 4x 8 mod 12 Lemma If d m then xd yd mod m x y mod m d m Proof Certainly if x y mod m d then d x y so m x y Now suppose xd yd mod m m then m xd yd so mk x y d for some k N Therefore m d k x y so d x y Therefore since 4 12 4x 8 mod 12 x 2 mod 3 so the only solutions are x 2 mod 3 which in mod 12 are x 2 5 8 11 mod 12 2 Find all solutions to 4x 6 mod 12 There are no solutions Suppose x is such a solution then 12 6 4x and 4 12 so 4 6 4x so 4 6 which is false 3 Show that for all a b N d Z d a and d b if and only if d gcd a b Consider arbitrary a b N d Z such that d a and d b By gcd theorem there are x y Z such that gcd a b xa yb so by MAL since a b 0 mod d gcd a b 0 mod d Therefore d gcd a b Consider arbitrary a b N d Z such that d gcd a b By definition gcd a b a and gcd a b b so a b are multiples of gcd a b which itself is a multiple of d so d a and d b 4 For any a b N define a function f Z Z Z as f x y ax by What is Imf Z Z Let d gcd a b We know that there exists x y Z such that f x y ax by d so d Imf Z Z Recall the previous theorem that in fact d is the least positve integer in Imf Z Z Furthermore we know that for any k Z kd a kx b ky f kx ky so kd Imf Z Z In fact that is all of it meaning that Imf Z Z dZ This is because given any y Imf Z Z using division algorithm write y qd r for some q r Z and 0 r d Then since we know that qd Imf Z Z as well you can check that r qd y Imf Z Z However we know that d is the least positive interger in Imf Z Z so since 0 r d r 0 so y qd dZ 1 5 Show that the set S f n N f is a function n N which is the set of all finite sequences is countable Recall that we showed that A B is countably finite for any countably infinite set A B Inductively we can show that Nn N N for any n N is countably infinite z n times Base case N1 N is countably infinite Inductive case Assume for some n N that Nn is countably infinite then Nn 1 Nn N is countably infinite because it is the product of two countably infintie sets Therefore Nn is countably infinite for each n N For each n N let Sn f n N f is a function It is easy to verify that for each n N the following is a bijection sn Sn Nn f 7 f 1 f 2 f n Therefore each Sn is countably infinite and note that S countable sets so it is countably infinite 2 S n N Sn so S is a countable union of