Discrete Structures Lecture 11

Transcription

1 Introdution Good morning. In this setion we study funtions. A funtion is mpping from one set to nother set or, perhps, from one set to itself. We study the properties of funtions. A mpping my not e funtion. A funtion my or my not e invertile. We wnt ertin funtions to e invertile. For exmple, we wnt funtion tht enrypts messge to e invertile so tht we n derypt the messge. On the other hnd, we wnt our psswords to e impossile to derypt tht is we wnt the funtion tht mps our pssword into inry ode to e not invertile. DEFINITION EXAMPLE 0. Let A nd B e nonempty sets. A funtion f from A to B is n ssignment of extly one element of B to eh element of A. We write ff() = if is the unique element of B ssigned y the funtion f to the element of A. If f is funtion from A to B, we write ff: AA BB. Let AA = {,,, dd, ee}, BB = {0,,,,}, nd ff: AA BB is defined y the reltion RR = {(, 0), (, ), (, ), (dd, )} where (uu, vv) ff(uu) = vv in RR. Thus ff() = 0, ff() =, ff() =, ff(dd) =. Is ff: AA BB funtion? 0 d e Solution: Nope. Funtion ff fils to ssign element ee AA to ny element in BB. Rell, the definition A funtion f from A to B is n ssignment of extly one element of B to eh element of A. Sine ee AA, n element of BB must e ssigned to ee nd no suh ssignment is defined y ff: AA BB.

2 EXAMPLE 0. Let AA = {,,, dd, ee}, BB = {0,,,,}, nd ff: AA BB is defined y the reltion RR = {(, 0), (, ), (, ), (, ), (dd, ), (ee, 0)} where (uu, vv) ff(uu) = vv in RR. Thus ff() = 0, ff() =, ff() =, ff() =, ff(dd) =, ff(ee) = 0. Is ff: AA BB funtion? 0 d e Solution: Nope. Funtion ff fils to ssign extly one element of BB to element AA. Rell, the definition A funtion f from A to B is n ssignment of extly one element of B to eh element of A. Sine ff() = 0 nd ff() =, ff: AA BB is not funtion. Adms A Chou Goodfriend B C Rodriguez D Stevens F FIGURE Assignment of Grdes in Disrete Mthemtis Clss Remrk: DEFINITION Funtions re sometimes lso lled mppings or trnsformtions. If f is funtion from A to B, we sy tht A is the domin of f nd B is the odomin of f. If ff() =, we sy tht is the imge of nd is preimge of. The rnge of f is the set of ll imges of elements of A. Also, if f is funtion from A to B, we sy tht f mps A to B.

3 f =f() f A B FIGURE The Funtion ff Mps AA to BB EXAMPLE Wht re the domin, odomin, nd rnge of the funtion tht ssigns grdes to students shown in Figure. Solution: Let G e the funtion tht ssigns grde to student in our disrete mthemtis lss. Note tht GG(AAAAAAAAAA) = AA, for instne. The domin of G is the set {Adms, Chou, Goodfriend, Rodriguez, Stevens}. The odomin is the set {A,B,C,D,F}. The rnge is the set {AA, BB, CC, FF} euse eh grde exept D is ssigned to some student. EXAMPLE Let R e the reltion onsisting of ordered pirs (Adul, ), (Brend, ), (Crl, ), (Desire, ), (Eddie, ), nd (Felii, ), where eh pir onsists of grdute student nd the ge of this student. Wht is the funtion tht this reltion determines? Solution: The reltion defines the funtion f s given elow. f(adul)= f(brend)= f(crl)= f(desire)= f(eddie)= f(felii)= The domin of f is the set {Adul, Brend, Crl, Desire, Eddie, Felii} A odomin of f is the set of positive integers. The rnge of f is the set {,,}

4 EXAMPLE Let ff e the funtion tht ssigns the lst two its of it string of length or greter to tht string. For exmple, ff(00) = 0. Wht re the domin, odomin, nd rnge of ff? Solution: The domin of ff is the set of ll strings of length or greter. The odomin ff is the set {00,0,0,}. The rnge of ff is the sme s the odomin of ff, nmely the set {00,0,0,}. EXAMPLE Let ff: Z Z ssign the squre of n integer to this integer. Further define f nd identify the domin, odomin, nd the rnge of f. Solution ff(xx) = xx The domin of f is Z, the set of ll integers. The odomin of ff is Z, the set of ll integers. The rnge of f is the set of ll integers tht re perfet squres nmely, {0,,, 9, } DEFINITION Let ff nd ff e funtions from A to R. Then ff + ff nd ff ff re lso funtions from A to R defined y (ff + ff )(xx) = ff (xx) + ff (xx) (ff ff )(xx) = ff (xx)ff (xx) EXAMPLE 6 Let ff nd ff e funtions from R to R suh tht ff (xx) = xx nd ff (xx) = xx xx. Wht re the funtions ff + ff nd ff ff? Solution (ff + ff )(xx) = ff (xx) + ff (xx) = xx + (xx xx ) = xx (ff ff )(xx) = ff (xx)ff (xx) = xx (xx xx ) = xx xx DEFINITION Remrk: Let f e funtion from the set A to the set B nd let S e suset of A. The imge of S under the funtion f is the suset of B tht onsists of the imges of S. We denote the imge of S y ff(ss), so ff(ss) = tt ss SS tt = ff(ss). We lso use the shorthnd {ff(ss) ss SS} to denote this set. The nottion ff(ss) for the imge of the set S under the funtion f is potentilly miguous. Here, ff(ss) denotes set, nd not the vlue of the funtion f for the set S. EXAMPLE 7 Let AA = {,,, dd, ee} nd BB = {,,,} withff() =, ff() =, ff() =, ff(dd) =, nd ff(ee) =. Wht is the imge of the suset SS = {,, dd}? Solution: The imge of S under the funtion f is the set ff(ss) = {,}.

5 One-to-one nd Onto Funtions DEFINITION 5 A funtion f is sid to e one-to-one, or n injuntion, if nd only if ff() = ff() implies tht = for ll nd in the domin of f. A funtion is sid to e injetive if it is one-to-one. EXAMPLE 8 Determine whether the funtion f from {,,, dd} to{,,,,5} with ff() =, ff() = 5, ff() =, nd ff(dd) = is one-to-one. Solution: The funtion f is one-to-one euse f tkes on different vlues t the four elements of its domin. This is illustrted in Figure. d 5 FIGURE A One-to-One Funtion EXAMPLE 9 EXAMPLE 0 EXAMPLE DEFINITION 6 Remrk: Determine whether the funtion ff(xx) = xx from the set of integers to the set of integers is one-to-one. Solution: The funtion ff(xx) = xx is not one-to-one euse, for instne, ff() = ff( ) =, ut. Determine whether the funtion ff(xx) = xx + from the set of rel numers to itself is one-to-one. Solution: The funtion ff(xx) = xx + is one-to-one funtion. To demonstrte this, note tht xx + yy + when xx yy. Suppose tht eh worker in group of employees is ssigned jo from set of possile jos, eh to e done y single worker. In this sitution, the funtion ff tht ssigns jo to eh worker is one-to-one. To see this, note tht if xx nd yy re two different workers, then ff(xx) ff(yy) euse the two workers xx nd yy must e ssigned different jos. A funtion f whose domin nd odomin re susets of the set of rel numers is lled inresing if ff(xx) ff(yy), nd stritly inresing if ff(xx) < ff(yy), whenever xx < yy nd x nd y re in the domin of f. Similrly, f is lled deresing if ff(xx) ff(yy), nd stritly deresing if ff(xx) > ff(yy), whenever xx < yy nd x nd y re in the domin of f. (The word stritly in this definition indites strit inequlity.) A funtion tht is either stritly inresing or stritly deresing is gurnteed to e one-to-one. 5

6 Remrk: DEFINITION 7 A funtion ff is inresing if xx yy(xx < yy ff(xx) ff(yy). A funtion ff is stritly inresing if xx yy(xx < yy ff(xx) < ff(yy). A funtion ff is deresing if xx yy(xx < yy ff(xx) ff(yy). A funtion ff is stritly deresing if xx yy(xx < yy ff(xx) > ff(yy). A funtion f from A to B is lled onto, or surjetion, if nd only if for every element BB there is n element AA with ff() =. A funtion f is lled surjetive if it is onto. FIGURE 5 () One-to-one, not onto d FIGURE 5 () Onto, not one-to-one d FIGURE 5 () One-to-one nd onto d FIGURE 5 (d) Neither one-to-one nor onto FIGURE 5 (e) Not funtion 6

7 EXAMPLE Let ff e the funtion from {,,, dd} to {,,} defined y ff() =, ff() =, ff() =, nd ff(dd) =. Is ff n onto funtion? Solution: Beuse ll three elements of the odomin re imges of elements in the domin, we see tht ff is onto. This is illustrted in the figure elow. Note tht if the odomin were {,,,}, then ff would not e onto. d ff: n onto funtion EXAMPLE EXAMPLE EXAMPLE 5 Is the funtion ff(xx) = xx from the set of integers to the set of integers onto? Solution: The funtion ff is not onto euse there is no integer xx with xx =. Is the funtion ff(xx) = xx + from the set of integers to the set of integers onto? Solution: The funtion is onto, euse for every integer yy there is n integer xx suh tht ff(xx) = yy. To see this, note tht ff(xx) = yy if nd only if xx + = yy, whih holds if nd only if xx = yy. Consider the funtion ff in Exmple tht ssigns jos to workers. The funtion ff is onto if for every jo there is worker ssigned to this jo. The funtion ff is not onto when there is t lest one jo tht hs no worker ssigned to it. DEFINITION 8 The funtion f is one-to-one orrespondene, or ijetion, if it is oth one-to-one nd onto. 7

8 EXAMPLE 6 Let ff e the funtion from {,,, dd} to {,,,} with ff() =, ff() =, ff() =, nd ff(dd) =. Is ff ijetion? Solution: The funtion f is one-to-one nd onto. It is one-to-one euse no two vlues in the domin re ssigned the sme funtion vlue. It is onto euse ll four elements of the odomin re imges of elements in the domin. Hene, ff is ijetion. d We see grphil representtion of ff in the digrm ove. A funtion is one-to-one nd onto if it n e inverted. The funtion n e inverted if ll of the direted edges n e reversed nd ll of the elements in the odomin re mpped y the reversed direted edges to individul elements in the domin. d EXAMPLE 7 Let AA e set, the identity funtion on AA is the funtion ii AA : AA AA, where ii AA (xx) = xx for ll xx AA. In other words, the identity funtion ii AA is the funtion tht ssigns eh element to itself. The funtion ii AA is one-to-one nd onto, so it is ijetion. 8

9 Suppose tht ff: AA BB. To show tht f is injetive (one-to-one) Show tht if ff(xx) = ff(yy) for ritrry xx, yy AA with xx yy, then xx = yy. To show tht f is not injetive Find prtiulr elements xx, yy AA suh tht xx yy nd ff(xx) = ff(yy). To show tht f is surjetive (onto) Consider n ritrry element yy BB nd find n element xx AA suh tht ff(xx) = yy. To show tht f is not surjetive Find prtiulr yy BB suh tht ff(xx) yy for ll xx AA. Inverse Funtions nd Compositions of Funtions DEFINITION 9 Let f e one-to-one orrespondene from the set A to the set B. The inverse funtion of f is the funtion tht ssigns to n element elonging to B the unique element in A suh tht ff() =. The inverse funtion is denoted y ff. Hene, ff () = when ff() =. f() - =f () A f -() =f() f B f - FIGURE 6 The Funtion ff Is the Inverse of Funtion ff. EXAMPLE 8 Let ff e the funtion for {,, } to {,,} suh tht ff() =, ff() =, nd ff() =. Is ff invertile, nd if it is, wht is its inverse? Solution: The funtion ff is invertile euse it is one-to-one orrespondene. The inverse funtion ff reverses the orrespondene so ff () =, ff () =, ndff () =. EXAMPLE 9 Let ff: Z Z e suh tht ff(xx) = xx +. Is ff invertile, nd if it is, wht is the inverse? Solution: The funtion f hs n inverse euse it is one-to-one orrespondene, s hs een shown previously. To reverse the orrespondene, suppose tht y is the imge of x, so tht yy = xx +. Then xx = yy. This mens tht yy is the unique element of Z tht is sent to y y f. onsequently, ff (yy) = yy. 9

10 EXAMPLE 0 EXAMPLE Let f e the funtion from R to R with ff(xx) = xx. Is f invertile? Solution: Beuse ff() = ff( ) =, f is not one-to-one. If n inverse funtion were defined it would hve to ssign two elements to. Hene, f is not invertile. Show tht if we restrit the funtion ff(xx) = xx to the set of nonnegtive rel numers. Is f in exmple 0 to funtion from the set of ll nonnegtive rel numers to the set of ll nonnegtive rel numers, then ff is invertile. Solution: One-to-one: The funtion ff(xx) = xx from the set of nonnegtive rel numers to the set of nonnegtive rel numers is one-to-one. To see this, note tht if ff(xx) = ff(yy), then xx = yy, so xx yy = 0. xx yy n e ftored to (xx + yy)(xx yy) = 0. This mens tht xx + yy = 0 or xx yy = 0, so xx = yy or xx = yy. Beuse oth xx nd yy re nonnegtive, we must hve xx = yy. So, this funtion is one-to-one. Onto: Furthermore, ff(xx) = xx is onto when the odomin is the set of ll nonnegtive rel numers, euse eh nonnegtive rel numer hs squre root. Tht is, if yy is nonnegtive rel numer, there exists nonnegtive rel numer xx suh tht xx = yy, whih mens tht xx = yy. Invertile: Beuse the funtion ff(xx) = xx from the set of nonnegtive rel numers to the set of nonnegtive rel numers is one-to-one nd onto, it is invertile. Its inverse is given y the rule ff (yy) = yy. DEFINITION 0 Let g e funtion from the set A to the set B nd let f e funtion from the set B to the set C. The omposition of the funtions f nd g, denoted for ll AA y ff gg, is defined y (ff gg)() = ff(gg()) 0

11 ( f g)( ) g() f(g()) f g A B C f g FIGURE 7 The Composition of the Funtions ff nd gg. EXAMPLE Let g e the funtion from the set {,, } to itself suh tht gg() =, gg() =, nd gg() =. Let f e the funtion from the set {,, } to the set {,,} suh tht ff() =, ff() =, nd ff() =. Wht is the omposition of ff nd gg, nd wht is the omposition of g nd f? Solution: The omposition ff gg is defined y (ff gg)() = ff gg() = ff() =, (ff gg)() = ff gg() = ff() =, nd (ff gg)() = ff gg() = ff() = EXAMPLE Let f nd g e the funtions from the set of integers to the set of integers defined y ff(xx) = xx + nd gg(xx) = xx +. Wht is the omposition of ff nd gg? Wht is the omposition of g nd f? Solution: Both the ompositions ff gg nd gg ff re defined. Moreover, (ff gg)(xx) = ff gg(xx) = ff(xx + ) = (xx + ) + = 6xx + 7 nd (gg ff)(xx) = gg ff(xx) = gg(xx + ) = (xx + ) + = 6xx + Remrk: Note tht even though ff gg nd gg ff re defined for funtions f nd g in Exmple, ff gg nd gg ff re not equl. In other words, the ommuttive lw does not hold for the omposition of funtions.

12 Inverse Funtions: When the omposition of funtion nd its inverse is formed, in either order, n identity funtion is otined. To see this, suppose tht f is one-to-one orrespondene from the set A to the set B. Then the inverse funtion ff exists nd is one-to-one orrespondene from B to A. the inverse funtion reverses the orrespondene of the originl funtion, so ff () = when ff() =, nd ff() = when ff () =. Hene, (ff ff)() = ff ff() = ff () =, nd (ff ff )() = ff ff () = ff() =. The Grphs of Funtions DEFINITION EXAMPLE Let ff e funtion from the set AA to the set BB. The grph of the funtion ff is the set of ordered pirs {(, ) AA nd ff() = }. Disply the grph of the funtion ff(nn) = nn + from the set of integers to the set of integers. Solution: The grph of ff is the set of ordered pirs of the form (nn, nn + ), where nn is n integer. The grph is displyed in Figure 8. FIGURE 8 The Grph of ff(nn) = + from Z to Z. EXAMPLE 5 Disply the grph of the funtion ff(xx) = xx from the set of integers to the set of integers. Solution: The grph of ff is the set of ordered pirs of the form (xx, ff(xx) = (xx, xx ), where xx is n integer. The grph is displyed in Figure 9.

13 (-,9) (,9) (-,) (,) (,) (0,0) FIGURE 9 The Grph of ff(xx) = xx from Z to Z. Some Importnt Funtions DEFINITION The floor funtion ssigns to the rel numer xx the lrgest integer tht is less thn or equl to xx. The vlue of the floor funtion t xx is denoted y xx. The eiling funtion ssigns to the rel numer xx the smllest integer tht is greter thn or equl to xx. The vlue of the eiling funtion t xx is denoted y xx.

14 FIGURE 0 () Grph of the Floor Funtion

15 FIGURE 0 () Grph of the Ceiling Funtion EXAMPLE 6 These re some vlues of the floor nd eiling funtions: = 0, =, =, = 0,. =,. =, 7 = 7, 7 = 7 5

16 EXAMPLE 7 Dt stored on omputer disk or trnsmitted over dt network re usully represented s string of ytes. Eh yte is mde up of 8 its. How mny ytes re required to enode 00 its of dt? Solution: To determine the numer of ytes needed, we determine the smllest integer tht is t lest s lrge s the quotient when 00 is divided y 8, the numer of its in yte. Consequently, 00 8 =.5 = ytes re required. 6

17 . Why is ff not funtion from R to R? Solution: Prt Funtion Explntion ) ff(xx) = xx The expression xx is meningless for xx = 0, whih is one of the elements in the domin; thus the rule is no rule t ll. In other words, f (0) is not defined. ) ff(xx) = xx The funtion ff(xx) does not mp vlues to the rel numers when xx < 0. In other words, when xx < 0, ff(xx) is undefined. ) ff(xx) = ± (xx + ) The rule for ff is miguous. We must hve ff(xx) defined uniquely, ut here there re two vlues ssoited with every xx, the positive squre root nd the negtive squre root of xx Find the domin nd rnge of these funtions. Solution: Prt Funtion Domin Rnge ) the funtion tht ssigns to eh pir of Z + Z + Z + positive integers the mximum of these two integers. ) the funtion tht ssigns to eh positive Z + {xx N 00 xx 99} integer the numer of the digits 0,,,,, 5, 6, 7, 8, 9 tht do not pper s deiml digits of the integer ) the funtion tht ssigns to it string (00 ) N the numer of times the lok ppers d) the funtion tht ssigns to it string the numeril position of the first in the string nd tht ssigns the vlue 0 to it string onsisting of ll 0s. (00 ) N 7

18 5. Determine whether ff: Z Z is onto if Solution: Rell the definition of onto. DEFINITION 7 A funtion f from A to B is lled onto, or surjetive, if nd only if for every element BB there is n element AA with ff() =. A funtion f is lled surjetion if it is onto. Prt Funtion Onto Explntion ) ff(mm, nn) = mm + nn Yes Given ny integer nn, we hveff(0, nn) = nn, so the funtion is onto. ) ff(mm, nn) = mm + nn No Clerly the rnge ontins no negtive integers, so the funtion is not onto. ) ff(mm, nn) = mm Yes Given ny integer mm, we hve ff(mm, 5) = mm, so the funtion is onto. (We ould hve used ny onstnt in ple of 5 in this rgument.) d) ff(mm, nn) = nn No Clerly the rnge ontins no negtive integers, so the funtion is not onto. e) ff(mm, nn) = mm nn Yes Given ny integer mm, we hveff(mm, 0) = mm, so the funtion is onto.. Determine whether eh of these funtions is ijetion from R to R. Solution: If we n find n inverse, the funtion is ijetion. Otherwise we must explin why the funtion is not one-to-one or not onto. Prt Funtion Bijetion Inverse or Explntion ) ff(xx) = xx + Yes One wy to determine whether funtion is ijetion, is to try to onstrut its inverse. Solve yy = + for xx, otining gg(yy) = yy Alterntively, we n rgue diretly. To show tht the funtion is one-to-one, not tht if + = xx +, then xx = xx. To show tht the funtion is onto, not tht yy + = yy, so every numer is in the rnge. ) ff(xx) = xx + No Not one-to-one: ff() = ff( ) = Not onto, sine ff(xx). ) ff(xx) = xx Yes Yes, this funtion is ijetion, sine it hs n inverse funtion ff(yy) = yy (otined y solving yy = xx for xx). d) ff(xx) = (xx + ) (xx + ) No Not one-to-one: ff() = ff( ) = Not onto euse the rnge is the set {yy R yy < } 8