Functions A function is a special type of relation In particular here are the rules for a relation to be a function for an arbitrary relation R A x B For each element in a A we must have EXACTLY 1 element in R such that a is the first term of the ordered pair In English that means for each element in the set A it MUST BE related to exactly one element in the set B Note that this means we must have A B when the sets A and B are finite Typically we call the set A the domain and the set B the co domain It is possible that all of the possible values of f a when a A form only a proper subset of B Thus the set of possible values of the function which can be more formally written as follows f A f a a A is known as the range of the function Here is an example of a function Let the set A the set of words in the English language For all a A define f a as follows f a the number of letters in the word a Thus we have that the domain is the set of all English words the co domain is the set of all positive integers and the range is all positive integers less than 29 assuming that antidisestablishmentarianism is the longest word in the English language However recall the relation I showed you in an earlier lecture Cocktails Orange Juice Vodka Cranberry Juice Vodka Coke Rum Orange Juice Peach Schnapps This is not a function because Orange Juice is related to more than one element of the range Note that you can define functions on multiple sets just as you can define relations on multiple sets Consider the following f a b a b This is a function of two real number variables Notice that different ordered pairs of this function map to the same element in the range For example f 2 3 f 4 1 We have already talked about composing relations together Now we will introduce function composition Consider the following example which uses the first function from this lecture f a the number of letters in the word a g b the sum of the digits in the integer b Consider inputting the output from f a into the function g Pictorially we get something like this Composing Functions produces a function Let f A B and g B C be two functions The composition of f and g as relations defines a function g f A C such that g f a g f a We need to show that for each element a A the expression or formula g f a assigns the unique element of C related to a based on the composition of f and g considered as relations By the definition of relation composition g f a c a A c C b B a b f b c g However a b f means b f a using the function notation and b c g means c g b by the function notation Thus the relation g f contains pairs a c with a A and c C such that c g b g f a by substitutions Further this element c that is related to element a via the composed relation g f is unique thus the relation g f defines a function g f A C Consider an example from math class Let f x x 3 and g x 2x 1 We have the following function compositions f g x f g x f 2x 1 2x 1 3 g f x g f x g x3 2x3 1 As can be seen from this example typically f g x g f x Injection Surjection and Bijection These definitions are often confusing but I think some pictures will help A function f is injective or is one to one if for all a b A a b f a f b Equivalently this means if f a f b then a b We also say f is an injection in this case f is surjective or is onto if f A B That is if for every b B there exists a A such that f a b We also say f is a surjection in this case f is bijective if it is both injective and surjective We also say f is a bijection in this case Consider these questions Is the function f x x2 injective where the domain and codomain are all real numbers Is the function f x sin x surjective where the co domain is the set of real numbers in the interval 1 1 Is the function f x x3 a bijection Consider functions f A B and g B C and the composition g f A C a If both f and g are injective then g f is injective b If both f and g are surjective then g f is surjective c If both f and g are bijective then g f is bijective To prove g f is injective we need to prove that for a b A if g f a g f b then a b by the definition of injection g f a g f b Since g is injective that means for these two to be equal we must have f a f b However we also know that f is injective thus this implies that a b Inverse Functions A bijection f A B maps each element of A to a unique element of B and conversely each element of B has a unique element of A as its pre image The following theorem makes a precise statement of this relationship Let f A B be a bijection Then the inverse relation of f defined from B to A as b a b B and a A and f a b is a function from B to A such that g f a a for all a A and f g b b for all b B The function g is called the inverse function of f denoted f 1 Note that since f A B is a bijection for each b B there exists one and only one element a A such that f a b Thus the relation from B to A relating b B to its pre image a A under f is a function by the definition of function That is we have defined a function g B A such that for b B g b a iff f a b g f a g f a g b a f g b f g b f a b Let f A B and g B A be two functions If g f a a for all a A and f g b b for all b B then both f and g are bijections and they are inverse functions of each other that is g f 1 and f g 1 First we will show that if g …
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