DCS Module 1 Part 3 FUNCTION Functions :- Injective, Surjective and Bijective functions - Inverse of a function- Composition ​ :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: A function is a special case of relation. ​ ​ Definition: Let X and Y be any two sets. A relation f from X to Y is called a function if for ​ ​ ​ every x∈ X, there is a unique element y ∈ Y such that (x, y) ∈ f. ​ ​ For every x there should be unique y .i.e. x has only one image. A relation must satisfy two additional conditions to become a function. These conditions are: ​ ​ 1. Every x∈ X must be related to y ∈ Y. ○ Domain of f must be X 2. Uniqueness . Every x has only one y image ○ (x,y)∈ f ∧ (x,z)∈ f ⇒ y=z Function f from set X to Y can be expressed as: f: X→Y or If (x,y) ∈ f then x is called an argument and Y is called image of x under f. ​ ​ ​ ​ (x,y) ∈ f is same as y=f(x) ​ f: x→y or Domain of function f from X to Y is denoted as D ​ ​f . Df=X ​ ​ Range of function f is denoted as Rf is defined as:- ​ ​ ​ Rf=f(X) ​ ​ Range of f, Rf is defined as:- ​ ​ Rf={y| ∃x∊X ∧ y=f(x) } ∃means “there exists” ​ ​ Rf ⊆Y ​ ​ Codomain f:X → Y. Here set Y is called codomain of function f ​ ​ E.g. Let X={1,3,4} Y={2,5,6} Check whether following are functions from X to Y or not ​ (1)R1={(1,2),(3,5),(4, 2)} (2)R2={(1,2)(1,5),(3,2)(4,6)} (3)R3={(1,2),(3,2)} R1 is a function because every x has unique image , Domain of R={1,3,4}=X R2 is not a function because one of the x value 1 has two images since (1,2) and (1,5) R3 is not a function because domain of R3={1,3}≠X E.g. Find the domain,codomain and range of the function ​ S={1,2,3} T={a,b} f:S->T f={(1,a),(2,b),(3,a)} Domain Df={1,2,3} ​ ​ Range Rf={a,b} ​ ​ Codomain=T={a,b} E.g. 1 Let X = {1, 2, 3}, Y = {p, q, r}, f : X → Y ​ and f = {(1, p), (2, q), (3,q)} then f(1) = p f(2) = q f(3)= r. Df={1,2,3} ​ ​ ​ Rf={p,q} ​ ​ ​ Codomain=Y={p,q,r} ​ E.g. 2: 2 If the function f is defined by f(x)=x +​ 1 on the set {−2, −1, 0, 1, 2}, find the range of f. ​ Ans. 2 f(−2) = (−2) +​ 1=5 2​ f(−1) = (−1) +​ 1 = 2 ​ f(0) = 0 + 1 = 1 f(1) = 1 + 1 = 2 f(2) = 4 + 1 = 5 f={(-2,5),(-1,2),(0,1),(1,2),(2,5) } So the range of f = {1, 2, 5}. Domain of f=x= {-2,-1,0,1,2} 2 E.g. Let X=Y=R and f(x)=x +​ 2 Df=R and Rf⊆R. Values are different and lie on a parabola. So ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ this is a function. E.g. Let X=Y=R set of real numbers(set of positive negative integers,rational numbers zero etc) ​ and let 2 f={(x,x )​ | x∊R} 2 ​ g={(x ,x)​ | x∊R} ​ 2 2 Here f is a function from X to Y f(x)= x ,​ because for every x there is only one x .​ ​ 2 ​ 2 But g is not a function from X to Y because for any real number a, we have (a ,a)​ and (a ,-a)​ in 2 2 ​ ​ g. This means that a ​ has two images a and –a. If a ​ i s 4 then a can be 2 and -2. But in a function ​ ​ for every x ϵX, there should be a unique image y ϵY. So g is not a function E.g. Let P is a set of all positive integers and f:P→P such that f(n)=n+1 where nϵP f(1)=2, f(2)=3 … … The function f is called Peano’s successor function. ​ ​ E.g. Let ​ f={(x,⌊x⌋)| x∊R ∧ ⌊x⌋ = the greatest integer less than or equal to x} ​ ​ ​ ​ g={{(x,⌈x⌉)| x∊R ∧ ⌈x⌉ = the least integer greater than or equal to x} ​ ​ ​ ​ Function f(x)=⌊x⌋is called floor of x. (whole number just before that ) ​ ​ ​ Function g(x)= ⌈x⌉ is called ceil of x. (whole number just after that ) ​ ​ ​ ​ f(3.6)=3 g(3.6)=4 f(-3.6)=-4 g(-3.6)=-3 E.g. If R is a set of real numbers identify which of the following relations define a function? 2 (1) f={(x,x )|x​ ∈R} ​ 2 Every real number has a unique square eg: -1 =1​ ​ So f is a function 2 (2) g={(x ,x)|x​ ∈R} ​ Real number can have more than one square root e.g. square root of 4 is -2 and 2 So (4,-2) ,(4,2) is in R here 4 has two images(y values) -2 and 2 So g is not a function RESTRICTION AND EXTENSION If f : X → Y and A ⊆ X then f ∩(A ×Y) is a function g from A→ Y . g is called restriction of f ​ ​ ​ to A denoted as f/A. If g is a restriction of f then f is called extension of g. ​ ​ ​ ​ E.g. Let f : R → R. The function f(x)=|x| (Here |x| denote absolute value of x) ​ and R+ ⊆ R.and g:R+ → R ​ ​ R is a set of real numbers. R+ is a set of positive real numbers. Here g is a restriction of f. f is the extension of g. TYPES OF FUNCTION 1. Onto(surjective): ​ 2. One-to-one(Injection) 3. Oneto-one onto(bijective) 1.Onto(surjective) ​ A mapping f : X → Y is called onto or surjective or surjection if range Rf=Y ​ ​ ​ ​ ​ Otherwise i.e . Rf≠Y then function is not into. ​ ​ ​ ​ ​ E.g. X={1,2,3} Y={a,b} f : X → Y f={(1,a),(2,b),(3,a)} Here range Rf={a,b}=Y ​ ​ ​ So f is onto. ​ ​ E.g. X={1,2,3} Y={a,b,c} f : X → Y ​ g={(1,a),(2,b),(3,a)} Here range Rf={a,b}≠Y So g is not onto but into. ​ ​ ​ ​ ​ 2. One-to-one(Injection): A mapping f : X → Y is called one-to-one if distinct elements of X ​ ​ are mapped into distinct elements of Y , ​ i.e., f is one-to-one if x1 ≠ x2 ⇒ f(x1) ≠ f(x2) Or f(x1) = f(x2) ⇒ x1 = x2 for x1, x2 ∈ X. ⇒means “implies” or “”it means” ​ ​ ​ f(x1) = f(x2) ⇒ x1 = x2 is read as “if f(x1)=f(x2) then x1 =x2” E.g. Let X={a,b,c} Y={1,2,3,4} f : X → Y f={(a,2),(b,3),(c,4)} Here f is one-to-one(injection) f is not onto but f is into. 3. One-to-one onto(Bijective) ​ or one-to-one correspondence between two sets A mapping f : X → Y is called one-to-one onto(bijective) if it is both onto and one-to-one. ​ ​ ​ ​ ​ i.e. distinct elements of X are mapped into distinct elements of Y and Range of f, Rf=Y ​ ​ ​ ​ ​ E.g. Let X={1,2,3} Y={a,b,c} f :X→Y. f={(1,b),(2,a),(3,c}) Domain of f={1,2,3}=X Each element in x is mapped to different element in Y. Range of f=(a,b,c}=Y So f is bijective(one-to-one onto) E.g. Let X={a,b,c,k} and Y={1,2,3,4}.The function f :X→Y. Identify which of the following are ​ functions.Check whether they are onto,one-to-one or bijective.. Find the domain, range and co-domain of the function. 1)f={(a,1),(b,2),(c,3),(k,4) } Ans. f is a function from X→Y because here every element in X has unique image in Y and every element in X has image. Domain={a,b,c,k} Range={1,2,3,4}=Y Co-domain={1,2,3,4} Here distinct elements of X are mapped into distinct elements of Y. So this is one to one. ​ ​ ​ ​ Here Range set=Y So function is onto. So this function f is bijective(both one-to-one and onto). 2) g={(a,1),(c,2)} Here g is not a function since every element in X is not related to Y.Here b and k inX has no image in Y. 3) h={(a,1),(b,2),(b,3),(c,4),(k,2)} Here h is not a function since element b in X has more than one image in Y .(b,2 )and (b,3). 4) j={(a,1),(b,1),(c,1),(k,1)} j is a function. Here every element in X has unique image in Y. But a,b,c and k has same image inY i.e, 1 j is into.Not one-to-one Not onto. ​ ​ Domain={a,b,c,k} Range={1} Co-domain={1,2,3,4} Note :Range of a function may not be same as its co-domain. ​ Composition of function Let f : X → Y and g : Y → Z be two functions. ​ ​ ​ The composition relation gof is a relation from X→ Z ​ ​ gof is possible if and only if range of f is subset of domain of g.