Lecture 22 21 127 Concepts of Math 10 23 2012 Administrivia Prep Questions have been up Homework 6 Solutions are posted will be graded and returned tomorrow Updated Definitions sheet with Relations and Functions and some proof techniques Exam on Friday Bijections and Inverses Theorem f A B is bijective f has an inverse f 1 B A Proof Assume f is bijective WWTS f has an inverse To do this wee need to define a function F B A and show F f 1 Let b B be arbitrary and fixed Since f is surjective a A such that f a b Let such an a be given Since f is injective this a is unique because x 6 a f x 6 f a b i e PreImf b a Define F b a This is a well defined function where F B A because there is exactly one such a Next observe that f F b f a b so f F IdB Also observe that F f a F b a so F f IdA Thus F f 1 Assume f has an inverse function call it f 1 WWTS f injective and surjective First we will show f is injective WWTS that for every a1 a2 A we have f a1 f a2 a1 a2 Let a1 a2 A be arbitrary and fixed Suppose f a1 f a2 Since f a1 f a2 B and f 1 B A is a function we can apply f 1 to both sides This yields f 1 f a1 f 1 f a2 We can rewrite this as f 1 f a1 f 1 f a2 By the definition of f 1 we know that f 1 f IdA and so we can rewrite this as IdA a1 IdA a2 This allows us to conclude a1 a2 and this was our goal Second we will show f surjective WWTS b B a A such that f a b Let b B be given Since f 1 is a function a A such that f 1 b a Let such an a be given Since f 1 b a A we can apply f to both sides and deduce that f f 1 b f a Rewriting this we find that f f 1 b f a Since f f 1 IdB we deduce that b f a Thus we have shown that f is a bijection Since we have shown and we have proven Corollary If f is a bijection then f 1 exists and is also a bijection Furthermore f 1 f 1 Proof Suppose f is invertible i e f has an inverse This means f f 1 IdB and f f IdA by the definition of inverse These are exactly the conditions that show f 1 f again by the definition of inverse This shows f 1 is invertible and f 1 f 1 Example Let U y R 1 y 1 and I y R 6 y 12 and define g U I by g x 9x 3 Let s show this is a bijection by identifying g 1 Define G I U by G y x U and y I y 3 We can show that 9 G g x G g x G 9x 3 g G y g G y 9 y 3 9 9x 9x 3 3 x 9 9 3 y 3 3 y Thus G g 1 so g is indeed a bijection Note This shows that the intervals U and I are the same size We will talk about this more on Wednesday and next week when we discuss cardinality of sets Keep your minds open about this because in particular cardinality with infinite sets is very strange and counter intuitive Example Define a bijection f Z N Prove it is a bijection by identifying f 1 2z 2 if z 0 f z 2z 1 if z 0 Define F N Z by F n n2 1 n 1 2 if n is even if n is odd Let s show F f 1 Let z Z be given We have two cases Suppose z 1 Then f z 2z 1 Notice that 2z 1 N and 2z 1 is odd This means F f z F f z F 2z 1 2z 2z 1 1 z 2 2 Suppose z 0 Then f z 2z 2 Notice that 2z 0 so 2z 2 2 so 2z 2 N Also 2z 2 is even This means F f z F f z F 2z 2 2z 2 1 z 1 1 z 1 1 z 2 This shows F f IdZ Next let n N We have two cases 1 and so n2 1 1 0 This means n n 2n f F n f F n f 1 2 1 2 2 2 n 2 2 2 Suppose n is even Then F n n2 1 Notice that n 2 Notice that n 1 2 and so n 1 2 1 This means n 1 n 1 2n 2 f F n f F n f 2 1 1 n 1 1 n 2 2 2 Suppose n is odd Then F n n 1 2 This shows f F IdN Therefore F f 1 Note This shows that Z and N are also of the same size You might feel like there are twice as many integers as naturals but this is where your intution fails We can pair up the elements of these two sets 2 one by one so they must be of the same size Example Let p be a prime number Define Zp to be the set of equivalence classes of Z modulo p except for 0 p Let a 1 2 p 1 Define fa Zp Zp by f x p ax p Then f is a bijection by the MIRP Lemma since a has a multiplicative inverse so fa 1 Zp Zp is given by f 1 x p a 1 x p Note Zp is actually a special kind of mathematical object known as a group Example Find a bijection from the interval I y R 6 y 12 to R Let s do this by finding a bijection between I and the interval U y R 1 y 1 as well as a bijection between U and R Prove that f U R given by f x x 1 x is a bijection Here is the graph In recitation tomorrow you will prove it is a bijection Prove that g U I given by g x 9x 3 is a bijection We did this up above Use this to deduce that h f g 1 is a bijection from I to R Can you write down a rule for h Could you try to prove it is a bijection directly Isn t it much easier to use these two results to create h 3
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