Relations and Functions

Size: px

Start display at page:

Download "Relations and Functions"

Stephany Alexander
5 years ago
Views:

1 artesian product Relations and Functions Let and be two sets. Then the set of all ordered pairs ab where a and b is called the artesian Product or ross Product or Product set of and and is denoted b X. Thus X = {ab/a and b} Eample: = {12} = {JMPKV} then X={1J1M1P1K1V1 2J2M2P2K2V 2} Note:We can have the product of a set with itself and this product is defined as X = {ab/a and b}. This product is also denoted b Problems: 1. ={ab} ={123} Find X X X X X = {a1a2a3b1b2b3} X = {1a1b2a2b3a3b} X ={aaabbabb} X ={ } 2. If = {1234} = {25} = {347} write down the following X X X X X X X X X X X X X X X = { } = { } X = { } X = { } = {2} X = { } alculate the other quantities in the same manner Note: If and are finite sets then X = X = X = Eample:If has 8 elements and has 6 elements then X = 8 6 = 48 elements X = 6 8 =48 elements X = 6 6 = 36 elements Theorem:For an three sets prove the following results 1 X = X X 2 X = X X 3 X = X X 4 X = X X 5 X- = X-X Proof of 1 and 5 are given below others left to the reader to prove 2

2 Proof: 1 5 Relation: Let and be two sets. Then a subset of X is called a binar relation or a relation from to. Note: 1. If R is relation from to then R is a set of ordered pairs ab where a and b. 2. R is a set of ordered pairs ab where a and b then R is relation from to. 3. If abr we mean that a is related to b b R and is denoted as arb. 4. R is a relation from to i.e. R is a sub set of X then R is known as binar relation on 5. If a set has m elements and has n elements then the number of relations from to are 2 mn Hence or or or = Hence =

3 Problem: 1. Let = {123} and = {245} determine the following X Number of relations from to Number of binar relations on = = 9 No. of relations from to 2 mn =2 9 =512 No. of relations from to 2 mm =2 9 = onsider the sets = {012} and = {89}. Indicate the following sets of ordered pairs are relations from to. a R1 = {081829} b R2 = { } c R3 = {190880} a R1 is relation b R2 is not a relation ecause the ordered pair 22 does not belong to R2. c R3 is not a relation ecause the ordered pair 80 does not belong to R3. 3. Let = {1234} and R be the relation on defined b abr if and onl if a b. Write down R as a set of ordered pairs. First let us write the cross product = { } Now the relation R is R = { } 4. Let = {12346} and R be the relation on defined b abr if and onl if a is multiple of b. Write down R as a set of ordered pairs. First let us write the cross product = { } Now the relation R is R = { }

4 Function For nonempt sets a function or mapping f from to denoted as f: is a relation from to in which ever element of appears eactl once as the first component of an ordered pair. Note: 1. is called domain of f and is called codomain of f 2. The subset of consisting of the images of under f is called range of f and is denoted b f Problem: 1. Let = {1234}. Determine the following are functions. f = { } g = {314211} h = { } The element 2 in the domain appeared twice as the first element in the function. Hence f is not a function. The element 2 in domain has no image in the co domain. Hence g is not a function. ll the elements in the domain appeared once in the function hence h is a function. 2. If = {0±1±2±3} and f: R is defined b f = find the range of f. Given f = Let us calculate f0 = 1 f1 = = 1 f-1 = = 3 f2 = = 3 f-2 = = 7 f3 = = 7 f-3 = = 13 Therefore the range of f = f = {13713}

5 solution: 3. If f:r R is defined b f = 2 what is the range of f what is fz? What is f[- 21]? Range = [0 fz = { } f[-21] = [41] 4. If f:z R is defined b f = find f -1 {6} f -1 [67] f -1 [610] f -1 [-45] and f -1 [5 5. Let f:r R be defined b + 7 for 0 f = for 0< < 3 1 for 3 Find f f -1 0 f -1 4 f -1 6 f -1 7 f lso determine f -1 [-5-1] f -1 [-50] f - 1 [-24] f -1 [510] f -1 [1117] Tpes of functions Restriction of f Let and be nonempt sets and Φ 1. Given a function f: suppose we define a function f1:1 b f1 = f for all 1. Then f1 is called a restriction of f to 1 from Eample: Let ={12345} and a function f: R be defined b f={ }. If a function g:q R is defined b g = 3+7 for all Q show that f is a restriction of g to from Q Given g = Let us calculate g1 = = 10 = f1 g2 = = 13 = f2 g3 = = 16 = f3 g4 = = 19 = f4 g5 = = 22 = f5 Hence f is a restriction of g to from Q Etension of f Let and be nonempt sets and Φ 1. Given a function f1:1 suppose we define a function f: b f = f1 for all 1. Then f1 is called an etension of f1 to from 1 Eample Let ={12345} and a function f: R be defined b f={ }. If a function g:r R is defined b g = 5+5 show that g is an etension of f to R from

6 Given that g = Let us calculate g1 = = 10 = f1 g2 = = 15 = f2 g3 = = 20 = f3 g4 = = 25 = f4 g5 = = 30 = f5 Hence g is an etension of f to R from Identit Function function f: such that fa = a for ever a is called the identit function on. In other words: If the image of ever element is itself in a function then it is identit function. Eample: f = { } onstant Function function f: such that fa = c for ever a is called the constant function on. In other words: If the image of ever element is same in a function then it is constant function. Eample : f = { } Onto Function or Surjective function function f: is said to be an onto function if for ever element b of there is an element a of such that fa = b. In other words: f is an onto function from to if ever element of has a preimage in. One-to-one Function orinjective function function f: is said to be a one-to-one function if different elements of have different images in under f. In other words: a1a2 with a1=a2 then fa1=fa2 In other words: a1a2 with a1 a2 then fa1 fa2 Eample : f = { } One-to-one orrespondence orijectivefunction function which is both one-to-one and onto is called one-to-one correspondence or a bijective function. In this function ever element of has unique image in and ever element in has unique preimage in. 1. Let = {a1a2a3} = {b1b2b3} = {c1c2} and D = {d1d2d3d4}. Find the nature of the following functions f1 = {a1b2a2b3a3b1} f2 = {a1d2a2d1a3d4} f3 = {b1c2b2c2b3c1} f4 = {d1b1d2b2d3b1d4b2} The givenfunction f1 is one to one and onto.hence it has one to one correspondence.

7 The given function f2 is one to one not onto. The given function f3 is not one to one but onto. The given function f4 is not one to one and onto. 2. If = {1234} = {vwz} f = {1v23z4} and g = {1v2v3w4} prove that f is one-to-one but not onto and g is neither one-to-one nor onto. Ever element in has unique image hence the function f is one to one. ut w in does not have preimage in hence it is not onto. The elements 1 and 2 in has same image hence the function g is not one to one. Similarl and z in does not have preimage in hence it is not onto. 3. Let = {123456} and = {678910}. If f : is defined b f = { }. Determine f -1 6 and f If 1 = {78} and 2 = {8910} find f and f f -1 6 = {4} f -1 9 = {56} f -1 1 = f ={123} and f -1 2 = f = {356} 4. In each of the following cases a function f:r R is given. Determine whether f is one-to-one or onto. If f is not onto find its range. i f = 2 3 ii f = 3 iii f = 2 iv f = 2 + v f = sin i f = 2 3 onsider f1 = f = = 22 1 = 2 Hence f is one to one Let f = 2 3 = X = + 3/2 + 3/2 R we get R such that f = Hence the given function is onto ii f = 3 onsider f1 = f2 1 3 = = 2 Hence f is one to one Let f = 3 =

8 X = 1/3 1/3 gives cube roots. We don t consider the comple numbers. R we get R such that f = Hence the given function is onto Floor and eiling Functions Let be an real number. Then is an integer or lies between two integers. Let denote the greatest integer that is less than or equal to and denote the least integer that is greater than or equal to. Then is called the floor of and is called ceiling of. Eample: = 3+4 = = 7.8 = = 8-9 = = 8.12 =9 Projection For sets and let D. Then the function π :D defined b π ab = a for all abd is called the projection of D on and the function π :D defined b π ab = b for all abd is called the projection of D on. Eample: Let = = R. Determine π D and π D for each of the following i D = { = 2 0 2} ii D = { = sin 0 π} iii D = { = 1} Let ==R. Determine π D and π D for each of the following sets D : 1. D = { = 2 0 2} π D = {R = 2 0 2} π D = { R 2 = 0 4} 2. D = { = sin 0 π} π D = {R = sin } π D = { R = sin 0 π} 3. D = { = 1} π D = {R =± } π D ={ R =± }

9 Number of onto functions Let and be finite sets with = m and = n where m n. Then the number of onto functions from to is given b the formula p n k n m n = 1 n k k = 0 n k Stirling number of the second kind The Stirling number of the second kind is defined as n p m n 1 k n m s m n = = 1 n k for m n n! n! k = 0 n k m Note: 1. Given that p64=1560 and p74=8400. We can prove that S64 = 65 and S74 = Prove the following: S53 = 25 S72 = 63 S85 = 1050 S54 = 10 S86 = 266 Sm1 = 1 Smm = 1 for all m 1. Unar and inar Operations Let be a nonempt set. Then the function f: is called a unar or monar operation on. For nonempt sets and a function f: is called a binar operation on. If = then the binar operation is said to be closed Result If = m then the number of closed operations on are Results If f: is a binar operation then 1. f is said to be commutative whenever fab=fba for all ab. 2. f is said to be associative whenever and fafbc=ffabc for all abc. 3. n element is called an identit for f whenever fa=fa=a for all a Problems 1. Determine whether the following closed operations f on Z are commutative and /or assosiative f = +- f = +-1 f = i f = +- = + since and are integers = f Hence f is commutative. ffz = faz where a = f = a + z - az

10 = f + z z f = + - z { + } = + z z + z.1 ffz = f b where b = fz = + b b = + fz fz = + + z z { + z z } = + + z z z + z.2 Hence 1 = 2 f is associative ii f = since and are integers f Hence f is not commutative ffz = faz where a = f = a z = f z = z ffz = f b where b = fz = b = fz = z Hence 1 = 2 f is associative Pigeonhole Principle If m pigeons occup n pigeonholes where m>n then at least one Pigeonhole must contain two or more pigeons in it. Generalized Pigeonhole Principle If m pigeons occup n pigeonholes then at least one pigeonhole must contain p+1 or more pigeons where p =[m-1/n]. Problems 1. is an equilateral triangle whose sides are of length 1cm each. If we select 5 points inside the triangle prove that at least two points are such that the distance between them is less than ½ cm. onsider the triangle DEF formed b the midpoints of the sides and of the given triangle as shown in the diagram. Then the triangle is partitioned into four small equilateral triangles each of which has sides equal to ½ cm. Treating each of these four triangles as pigeonholes and five points chosen inside the triangle as

11 pigeons we find b pigeonhole Principle that atleast on triangle must contain two or more points. Hence the distance between such points is less than ½ cm. 2. Shirts numbered consecutivel from 1 to 20 are worn b 20 students of a class. When an 3 of these students are chosen to be a debating team from the class the sum of their shirt numbers is used as the code number of the team. Show that if an 8 of the 20 are selected then from these 8 we ma form two different teams having the same code number. From the 8 of the 20 students selected the number of a teams of 3 students that can be formed is 8 3 = 56 ccording to the wa in which the code number of a team is determined we note that the smallest possible code number is = 6 and the largest possible code number is = 57. Thus the code numbers var from 6 to 57 and these are 52 in number. s such onl 52 code numbers are available consider them as pigeon holes and the 56 teams as pigeons. pigeon hole principle at least two different teams will have the same code number. 3. Prove that if 151 integers are selected from the set S = { } then the selection must include two integers where or. Let = { } Then ever integer n between 1 and 300 is of the form n = 2 k a where k is an integer 0 and a. Thus ever element of S corresponds to some a. The set has 150 distinct elements and therefore if 151 elements of S are selected then at least two of them sa and must correspond to the same a. Thus = 2 m a = 2 n a for some integers m n 0. Evidentl divides if m n and divides if n < m. This proves the required result. 4. Prove that if 101 integers are selected from the set S = { } then at least two integers are such that one divides the other. 5. Show that if an seven numbers from 1 to 12 are chosen then two of them will add to 13. Let us consider the sets These are the onl sets containing two numbers from 1 to 12 whose sum is 13. Since ever number from 1 to 12 belongs to one of the above sets each of the seven numbers chosen must belongs to one of the sets. Since there are onl 6 sets two of the seven chosen numbers have to belong the same set. onsidering the 7 sets as pigeons and 6 sets as pigeonholes b pigeonhole principle the sum of these two numbers must be equal to Show that if an 14 integers are selected from the sets = { } there are at least two integers whose sum is Show that if an n+1 numbers from 1 to 2n are chosen then two of them will have their sum equal to 2n+1.

12 Let us consider the following sets: 12n22n 1 3 2n 2.n 1 n + 2 n n + 1 These are the onl sets containing two numbers from 1 to 2n whose sum is 2n + 1. Since ever number from 1 to 2n belongs to one of the above sets each of the n + 1 numbers chosen must belong to one of the sets. Since there are onl n sets two of the n + 1 chosen numbers have to belong to the same set. onsidering the 2n + 1 sets as pigeons and 2n sets as pigeonholes b pigeonhole principle the sum of these two numbers must be equal to 2n If 5 colours are used to paint 26 doors Prove that at least 6 doors will have the same colour. Let us consider 26 doors as pigeons and 5 colours as pigeonholes b generalized pigeonhole principle at least one of the colours must be assigned 26 1 or more doors. + 1= Prove that in an set of 29 persons at least five persons must have been born on the same da of the week. Let us consider 7 das of the week as pigeonholes and 29 persons as pigeons then b generalized pigeon hole principle at least one da of the week is assigned to or more persons. omposition of Functions onsider three non-empt sets and the functions f: and g:. The composition of these two functions is defined as the function gοf: with gοfa = g[fa] for all a. 1. Let f and g be functions from R to R defined b f = 2 and g=+5. Prove that gοf fοg gοf = g[f] for all R. = g 2 = fοg = f[g] for all R = f + 5 = Hence gοf fοg = Let fgh be functions from R to R defined b f=+2 g=-2 h=3 for all R. Find gοf fοg fοf gοg fοh hοg hοf fοhοg gοf = g[f] for all R = g + 2 = = fοg = f[g] for all R = f 2 = = = 5 7

13 fοf = f[f] for all R = f + 2 = = + 4 gοg = g[g] for all R = g 2 = 2 2 = 4 fοh = f[h] for all R = f3 = hοg = h[g] for all R = h 2 = 3 2 = 3 6 hοf = h[f] for all R = h + 2 = = fοhοg = f{h[g]} for all R = f{h 2} = f{3 2} = f{3 6} = = Let fgh be functions from R to R defined b f= 2 g=+5 and h=. Determine gοf hοg f 2 g 2 h 2 f 3 g 3 h 3 4. Let fgh be functions from R to R defined b f= 2 g=+5 and h=. Verif hοgοf = hοgοf 5. Let = {123} and f g h p be functions on defined as follows f={122331} g={122133} h={112231} p={112233}. Find gοf fοg gοp pοg fοp hοg hοf fοhοg gοf = g[f] = { } fοg = f[g] = { } Find the other quantities Invertible Functions functions f: is said to be invertible if there eists a function g: such that gοf = I and fοg = I where I is the identit function on and I is the identit function on. Problems 1. Let ={1234} and ={abcd}. Determine whether the following functions from to are invertible or not. f={1a2a3c4d} g={1a2c3d4d} gοf1 = g[f1] = ga = does not eist

14 gοf2 = g[f2] = ga = does not eist gοf3 = g[f3] = gc = does not eist gοf4 = g[f4] = gd = does not eist fοg1 = f[g1] = fa = does not eist fοg2 = f[g2] = fc = does not eist fοg3 = f[g3] = fd = does not eist fοg4 = f[g4] = fd = does not eist Hence f and g are not invertible. 2. Let = = R Show that f : defined b fa = a+1 for a is invertible. known theorem if f is one to one and onto then f is invertible. Let us consider f1 = f = = 2 Hence f is one to one Let f = + 1 = = 1R 1 R = there eists R =. Hence f is onto Thus f is invertible. 3. Let = = = R and f: and g: defined b fa= 2a+1 gb=b/3 for all a b. ompute gοf and show that gοf is invertible. What is gοf -1 gοf = g[f] = g2+1 = 2+1/3 It is evident that gοf is one to one and onto. Hence gοf is invertible. gοf = c 2+1/3 = c = 3c 1/2 gοf -1 = 3c 1/2 4. For the functions f and g from R to R given below verif that gοf -1 =f -1 οg -1 f = 2 g = 3 2 f = g = gοf = g[f] = g2 = 6 2 gοf -1 = + 2/ f -1 = /2 and g -1 = +2/3 f -1 οg -1 = f -1 [g -1 ] = f -1 [+2/3] =+2/ and 2 we get gοf -1 =f -1 οg -1 Theorem 1. Let f: and g: be an two functions. Then the following are true: If f and g are one-to-one so is gοf

15 i If gοf is one-to-one then f is one-to-one If f and g are onto so is gοf If gοf is onto then g is onto Theorem 2. Let f : g : and h: D be three functions. Then hοgοf = hοgοf Theorem 3 function f: is invertible if and onl if it is one-to-one and onto. Theorem 2 If f : and g : are invertible functions then gοf : is an invertible function and gοf -1 =f -1 οg -1 Theorem 4 If a function f: is invertible then it has a unique inverse. Further if fa=b then f -1 b=a Theorem 5 Let and be finite sets with = and f be a function to. Then the following statements are equivalent f is one to one ii f is onto iii f is invertible [For proof of all above theorems refer the tet Discrete and ombinatorial Mathematics b Grimaldi]

Section Summary. Definition of a Function.

Section Summary. Definition of a Function. Section 2.3 Section Summary Definition of a Function. Domain, Cdomain Image, Preimage Injection, Surjection, Bijection Inverse Function Function Composition Graphing Functions Floor, Ceiling, Factorial