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Suppose 2kfor = some Dink's using that n is integer and even so Bible even = =2x2 ↳ n integer ↳n 2x n is is even, an is n from def + of even 2 in W2) = - -- this is even 2 as number 12 n2= (2k)2= 2(2xY=2k thus no is even Example & True for Ans:It An integers is n of 5P, odd no is n for 2kH) = integer some prooving do to what you must than YP, oddthen is i odd () is statement if P if true is integer Way own) your false? or all Stry on 0 3 Q -> prove it Similar statements Assumes P, Prove Q Q choose a is integer n is odd P, Prova a P 2k + 1 n = n2 and such (2k 1)2 then that meal either and + = n2 4k2 4k = n2 and itasfar and in 1 + 1 + 2(2k 2k)+) + = ----- ↳) some integer n2 2k + = ↳ prooved is n Is o n2= 2(2k2+2K)+) & True for negative Proof:There A prime Chat GPI choose numbers a number = to 4). Then a = is 41(H1 + 1 + 1) 41x43, = Thus, nothH or integer. Thus an is integer, 1 + 1) that n2+nthl so than 1 larger is a 41x43. So both = not positive integer odd he is divisors are 1 and itself so is n and n2+n+4) rith+hl has 41 as a divides 41x43=1763 4171763 412+41+ 4) = and 1241<n2+n+1 divisor prime. an integer me and integer, ntm= ntn ntur im :=n t n =an so that ntm Way an 2n = is even Choose where m n. = n is Then an m is an of is even prooving what you must statement do to integer, if P YP, than Q 0 3 -> Similar statements Assumes P, Prove Q 9 2k only than either haha be prime 21 and 43 true -> = some composite and it's equation with y interceptc an non-negative integer so is not = m +n for false? Proof:Suppose m R odd, n= 2kH negation For all integers n, there exists Ans:It is is n and a1 Dind's Bible + = it's integer quadratic a refer an 21 +4174) 41(41 & True since is prime. non-negative integer is told me, n 41 n (2K22K) where as n2th+4) We prove exists &> choose odd, false? or all an Answer: It i s false. def suppose his integer, suppose 5P, Q choose a P, Prova P prove it or 43 =1763 ask andki or such a h = that n = alx and k = 1 Proof:Suppose is Ans:It is n integer, an choose m (m+r 1) 6 -n = = true - ntm n + 6 = True n 2(3) 2(k) = 3 where = is integer, so an ntm is even false? (false) or There exists Negation for - 6 = integer an integers all m is an n is an m thatfor all integern, so m, there exists an ntm that so integern, is even that h i m is so odd integer integer m +3 n+1 2m 1 + 2(2m) +1 An integer ProofSuppose def odd () is n n is he an "FA 12 if I The does not converse Contrapositive for asand C, all integers bjc an integer and ntm-4m+1 where in is an integer, so - "If of as as original statement p a then not not truth values same truth values same if all then and als original statements Infact, they are logically equivalent al Way b at c as statement = = Some of if P integer attas YP, a(t t s) = = some integer -> = - and alb-c 5P, a(t s) as = - Q prooving do to what you must than t b -c at albtc integers are alb b+c is n "V A, if Q then " is have the have always = "If Q them P" PthenQ" always Then n 3m+1. false? or a integer, prooved 2x+ 1 integer choose of "If P then Q"is The contrapositive True some PthenQ"is of "If converse for 2kH) = 0 3 -> prove it Similar statements Assumes P, Prove Q Q choose a P, Prova P prooved True for all is the for false? or asande,if all integers, asand C, integers, a bc a lbtc alb b -c = 2 al = b albtc and alb-c then all and a c - c as = b aL b + = if int are b+c aL - aldtc and all- this true? of converse all and ale then -Ease as = for 2b a(Lts) odd integers we = b = false I L odd integers for we choose 0,5 1 = b ia = ->not a 2 12) an = b integer I - = a decrease = may of false c a = - = )a 2 = y a decreasing mag ot b false I want and a b to such integer that so we Is=false choose a to be even atm is odd We for choose example, Then negation:"There it's prove a,bic are btc a = albtc c that albtc and alb-e, but a by oral. 1 =- True a.0 True - a. = als and allow and and 1 = = = = so as and integers 2.0 0 = B -c 2 2.1 Thus, 2, b = bands integers are 2 XI because Example. For all This statement is true The "For is converse the is: and all proof is then aldbtc and al3btze easy. asand C, integers and ale if and alsbtze aldbtc them all and ak" true? converse Ans. The all and C, if as 6 integers true. is converse Working alabtc 3b 2c + = b 2B 21 - B = 3D = - = = B bax = i at = + 2b +c B c Y ak al3b 2 -at a(2k = - t) bat-bak aldt -3k) = = ↳I alb al Proof: let asbyc are integers. Theb 2(2b+c) = =Gat-bak=a (2t-3k) Thus Def I all and rational where - Suppose and alabic (3b 2) dak + = 2k-t and 2t-34a re number:A number that can be made as examples number:An irrational terminating decimal of ( 2(3b 2c) = + 3(2b some c) + irrational numbers number is as fraction of two integer b 0 3 or a(2k-t) and = - als - irrational -at for and 3b+2c-at 26+a= am integers. rational-a Def al3btes; thatis a real number 252/pi/culers that cannot be expressed as fraction of two integer integers I and t Quotient integers and g called a is that a when quotient I for Theorem so r the I 11 ~ Remainder - remainder and gdtr = divded (L 26 I 02 a and a aso, where there exists a unique v2d by ds - is a l integers any I1 4 - d the remainder a mod a dird:the a 20d = 20 3g 6r Proof thatfor el an in floors let so is we x3 and 0213 2 cases. have with is for and integer some with nen+1) = ax(n+1) = where indirect proof kinti) even. and nth n(n+1) integer some integer, an with so n(2k+ 2) = = is 22() = even Ceilings. and be real number a LX) the = x7 (-7) = +1 is 2 3 is even integer. Then a for n=2kH x 20 0223 odd is u(KA) n n, nt - by and even integer, Caseh is n max - is because all integers Proof:Suppose 20=6x4+ 2 divides 20 when Z mod3=1 caselh is = because = cased 2 = mode I by do divided is a by d divided is a quotientwhen = the remainder - when = the = floor of ceiling X of X Definitions (x) n a = Mx7 E) = a is is an an and n=x<n+1 integer integer and a-txn Ex (201) = L-201] True or =-3 because numbers the statement is Negation There I is because - an is integer an and integer 22012H1 and 1 - - 2012 - 2 = - 3H false? for all real Solution 2 exists a false, real x,(x) (Y) = we prove number, it's such negation that LxJF(x] - Proof choose 211 and Solution Negation I. Then X 23 the for all the statement is false. there exists choose numbers real x real a 1 is =3C2, an integer. Than Lx2)=14):a because I is an integer 2x4 2 +1 2x6. = and x 1 because (x)=LE1=1 = = y, Lxty and x numbers LyJ + = prove that to Easy 2x6 3 that Lxty/f(x)+Ly) such y, yE = = = 42 3) (3) 3 = = + ~ (x) + Lyb:(16 Proofchoose and 3 x = 31324. Then (x) and 111.322, There exists (1) 2 + = y = = 21.5) A = & is 1 131 is an 1 + = that so Lyty) where (x6+Lyb hence real number a Then 3. 3 where 3 2. (xd Thus, = is + hy) integer and an 121.5<2) Ly) integer and that so x = 21.5) = = 1 where I is an integer 2 x + y) 2 +3 = = (x2) (x)2 = Ans:True ol choose LOS L0012 = - 20001) 2 0 Proof L001 = 0 0 0 = 2x12 2001 There exists Ex Proof prove we likely use or when we Prove Suppose this a is an that so we kind of fail that to and integer an an and integer 0 0 <0.0kR =0.1). Thus (x2) = Lx) an integers it this a contradiction = x (27) and 2x4 2x12 = and then conclude p. - what we prove when means " " not directly. for all integers is proved 0 0 = fin is get suppose not proof prove integer. Suppose n not is x in, We even. odd. Since if it is n is even them o n is that prove - Suppose is O is o by contradictions contradiction, by because 0 = because = True real number a Indirect proof (proof P = 82 false? or Loo 20001) 0 = = = True To (x2) x=001, then choose then = = even odd. Since n a is even- by contradiction is odd, n 2kH. for = - - some is integer an 1. Then integer thus even er (2k+)2 4k2 = his CIn = odd which 2(2k27av)H +4kH where = contradicts the assumption that 222 it +2* is even,therefore, N is od& => Proof (by contradictions (TA) Suppose there exists an integer a so it that is even and i is odd. n is by if 31m...-- then it falsesiose netbut he n 3t a suppose there exist = ~ - - - r 3t S = n 3t - - = 35 2 or + where 3 An ) 3t 3 (x + = 3924733 + 3ab2 a (3x)3 211x3 + + 7 ak =- - - 3 n 272+1 C 31r = C contradiction 30 Thus 3 Definitions: A real - x where = A real of the Ex:SI r -> wir even I -> Thus in 221 2x2 II ! have no common m is t is r2 is have even dix, kE2 4x2 - = = use prive factors. da = n won't 2/nts n2 is rational. rational is m, in not proof (by contradiction] I a,m,at = is ESX numbers: = 2m no, proof of the Suppose 12 and irrational. is Sketch irrational rational Es 2-we irrational of Sketch is integers a re is x -> X number on number examples + + - - 3 3(3x)(1) 3(3k)1 + an = even irrational -> mis even n in tager =3kH 1 case 3Yt! n common factor of 2 is an them Ex is irrational is proof (by contradiction) Sketch of the suppose is 53 rational is ,nE2, a = n myn have mtm 3 = 31m2 o no prime factors. common 3r = 31m -> have m bkx,kx =129k) = factor of common 3 3r) = - n2 31n2+ 3x2- = Thus II 3 In irrational is ↳ How did you conclude? - I A for is, we use Suppose a Fuclidean Algorith: - is all an then integer, 3:kan is the 31n" integers n, ifbind Ei then gcd(6,9) 31n2 for that Homework: Prove Proof "if where that compute algorithm suppose n then 3 is 31m kn2 = some integer gad(asb) 3 = q2d(6,0) 6 = Divisors 6 of Divisors 0 of gcd(0,0) are are no = 11, 12, 13, 16 all common non-zero 2cd:Definition:Suppose 1) 2) integers divisor. asbe both so that n ot and glb, and gla For any Algorith de2,it dla and dlb uses following repeatedly lemmon lamuna:if a gbtr = then gcd(a,b) gcd(b,) = them dig a and b are zeros. Then ged(a,b) is an integer g so that Ex:find god (32)123) using Euclidean Algorithm. = + 21gcd(22,21) 48 27(1) = + + = 6(3) 3gad(6,3) + = 6 3(2) 0g2d(3,0) + = ↳ use R 123 YS 48 -> 6 Thus - -> 47 = 18 - y 47 + = +42(123) 3. = cuclidean the Use to 13 18 (8(32)) S 8 3 x method. 3 D 5 - find integers a 344 = 344 235 109 and x y 3 235 = 25(1) +09 = 1092) = so compute that god largar alb - H7 i 3(2) ((3) = + 1 + 70.9cdC(,0] qcd (1,0) gcd(344,23S) = =I (344,235) (344,235) q(d(235,109) (7(6) + 7 = god smaller = = 3 to algorithm 7(2) 3 17 v andy x 2 = 2 -> integer N n 27 the table some ⑧ ⑧ 21 for = can zero. 3 gad(ab) axtby we 3 = remainder is when = = 3d) -> stop We This gad (32) (23) 3 gcd(2),6) 6 21 21)) Theorem g2d(48,27) 27 73 48(1) or gad (TS,48) 75(1) 48 + = = 21 gcd (123,75) + 25 321 2x123 123 and use n 2 23sy =344x NO 344 2350 109 17 ↑ M - 2 # -> 2 13 3 - A 69 69 = y c101 = 3 = 19 28 = x that + +01 + lemma:asbigs of a if , then bgtr = Proof:Suppose a b oth not g2 best It is ga easy r r is gcd(b,r) g1292, prove integers. Suppose easy exists. by proving that gally to prove Let both not a and b are eros, not by contradiction that god (asb) both exists. b and r g2 g(d(bs) = gi is and common divisor of 3 and r) galv. gcd(a,b) = and gala r to are It = prove cros gad(as) gcd(b,5) go=ged(ab). firstwe a re = let exists. Let and b a - q = gilb ba git- glIg = gx(t 48) - = Next, proof we prove gaegy by proving go is a common divisor a and 3, of and this is easy. Zeros, so ged(b,r) types Three proof Direct proof, of indirect proof, induction induction: iP 2 3 i +az a1 = ai=a 22 Ei 2 +2 4 Principle let ① ② a3 + ay s + + at + + + 32+ 4 s + + + (a (2) induction of + + 2 2 2 + = = 4 5 + + = P(n) be a where sentines n is integer an P(a), For all P(r) then - Ka, integers for 1,0,1,2, all if P(X) then D(k+1) integers nc a 5,5,5, 2,8,,xo, Template To prove Base 1) P(n) for case: Prove 2)inductive all integers is a we prove. P(a) step: Suppose KI a is an integer (IH) Suppose PCK) we want 4(k+) prore Thus 6 P(r) for Ex P(K+1) to prove Prove that all integers for nent's = 12a all a integers Solution: :Ii Basccase: 1 = z 4 4) = = = Inductive Step:let K:1 Eni we want to be an integer. Suppose k(IN) = (IA) 2 2) -i (1 x = prove 2 ii (i) = 1x1 + |x(kx 1) ((+) + + by (IA) 2 (x(1+71) + = 2((x+1) 2 2) (2x+ 2 I Thus Ex Prove for =m) that as+2n is divisible by all for 3 integers all ( rel. n= 0 integers by induction Solution: Base = Inductive step: want we 6 case:( n 0) - to 3x0 0 = divisible is is 8 an integer. (IH) by 3 divisible (k+1)+ 2(KH) prove where = (x+2Ki s Suppose (x+ 2k 3t 2x0 + by 3. 0 = Now, +1)" (kx 2(k+1) (x3 3(2 + 3k + + = =x3 + 3k 2 1 2(k+) + + 2 3k+ +2k + + (x 2k 31x2 - 3k2 + 3t = b(t xx2 = where Thus, Ex n8+2n Prove that Base + (t+k+K+1) is is an divisible 5" 4ns1 case: P(0) is kx 3 divisible be want Now an 1), (k+*+2(kH) so all by 1 = integers 3 - - is for all integers so 1 where 8(0) 0 = = IY-4(-) 5**! U(K+)-1 4k-4-1 5(54) - = 8(5t where integer thus integer. divisible 4k 5(8t) is some an by 8 (IH) by 5* -4k-1 bt = 8 4-5 = = is is divisible = 5(5* O is = 5th divisible by 3 480 integer. Suppose to prove 5(54) 3 + + for 50-4(0)-1 k2O Inductive step:let 3k integer, by + + + = we 3k 3 + + + - 1) - 16k + 16k 2) + 5"*- 4(xH)-1 is divisible by 8 for all integers n 0. Prove is it Solution Base induction by on for him mining that in all integers Basice na case 2 E= Jn+1 52 1 + = E > 52 52 ↳ - 52 ↳it because CLI 2 to Since 211 then 2 +C 2 I I I prooved KId Inductive step:let be an k integer. Suppose.I N , ist it /x+ it in It Thinking.- prove: Ei= if t * I >xxx see 1 by + fix+ then + e sixis x ! x - Thus, 1 0 for all integers + 2x 1 + 1 + Tx+ n72 + - Since Ext1) ( Fix Il Proof by tan n let divisible is be a -(r 0) by for 3 all integers no integer. an 3q a = example cases = n3+2n (39) 2(39) 4g3 6g = + = = + (r 1)n 3g 1 case 2 = = + 3 . . . . Case 3(3a2 3a) + 3(r 2) a = 3 + 3g + 2 = - - n 4 3g = n ifa 1 3 + = 24ifg 1 3a +) = + = 3a 2 35ifg 1 n = - + This take & all integers all cages covers Strong if = ... Principle Induction PCa), P(at), P(b) nz arb, where for all ixcb, if 4(m) and & for all integersm a Template Basecasa (n a, = Prove . . . . b) P(a), PCati), ...,P(3] Inductive step: let we to want kLb be an integer. Suppose P(m) for all integers D(K) prove Sprove PC17] Thus, Ex P(n) for all intergers R29 be let a 0,al92,...... a0=12-a1=29 by ak 5ax-1 - = Base - for and sequence (by cases 2 for strong induction (n 0,7) on of = B Next 0.16213 90 an 12 = 29 = 3x30 = 7x2 + 5x3'+ 7 = defined integers k22 that a n =5x34+ Tx24 Prove Proof bax the x 27 all integers n20 on where a zmck (IH) where a <mcK then P(KA) than P(n) for Now, definition by the 12 since of the sequences Jax-1-bak-2 ar= SISx3+ 2x2 * 25x3* + ")-6(5x** =wx2") 35x24 = (3") - + 13) 3 Thus, 1) - 1 )a" (7)2" + 5x3"+7x2" prove solution (n s 1 = 3 2 = 3x) 2x) = where 0 - co 1, 4 = 8 = and for 120 2= 2-1)0 be an - integer. Suppose CMCK that (= 3x2*- 2()* K22, by definition 2k (k -) + 2ck of sequences 2 +22k 2 - = 2 (x 3x2* - = M (x k prove Now, since = - = lat to want 3x20 = 3x2" 2 step m) Fol = putting n) = + = (k ( k -) on for all integers 2(-1)" = by = = inductive recursively 22n-2 3x2 = defined (n 0,1) 8 6 we + 1 2 n = the scavence (strong induction = = (n = that (n Base cases 20 for all integers be Co,Cl, C2 n=2 2)2 - an= Let 2 + 5(3*) x1 3 0)3"+ 15 - - 2* (2") - IoH by 42x2 - - + 1 0)3" Ex 30.3* + 3x3 = 2 * 1) * * 2(3x2*2 1"-) - * -26 - 4-l)" (m 3 = x2-2L-1)" for all integers integers A2L-1" x 3(2) = 3(2)" aix = 4" - x2"-2(-)* 3 an - 2) 1 = = Thus, 4 - for all integers nzc Sets things elements Set collection of = of the set (members) "5" How do "means" is describe we Ans: Justfell the Using A Ex B 12 set? a what are it's us " 21 is elements. of the set all integers. curly brackets [3 [1,2,3, Dani] by listing 2 IR of element = = ↑ is an 2(x2 x >103. = x + rational set of all the the set of all is Complex = numbers. real numbers numbers subsets "I means" is Edition: A,B a of subset sets are A A C BL=) if x= then x = B (fx =A,x B) z %=the empty set let B [1,2,33 set that has - elements. no = all subsets list they ① P, [13,[23, 233, (123, (1,33,22,3) a re B E) if Definitions 0 than x= A This set (1,2,3] set 31,23 Chus):If it has Hlef: the has Afictions def 8 4 on subsets. They subsets. than elements an A, of sets:U, I, X = B X ARB = 2x 1x A = = 2x = A lB 3x = / 1 or XGA x EA B3 x = and and x= B3 B3 x = are 0, 913,[23 and 91,23 sat of all subsets of A. a"subsets. by PCA), is A,B sets A UB are and 21,2,3) They A has denoted Al BES Fx =A , operation 21,2133 B 31,23,22,3391,33 has set power xt and BEVxEA, XE B. I 8,313,323,333) This of B. the AUB E) AnB = E) x = x A-B CE we X= A E) sets see XEA o r xeB and xt A and X = using venn B xAB Diagram IIIIII E A AUB AnB It i An (AUBJ nC B A da Me B A t Mugsee B + - c we C 2 Au(Bre) W A 1B - A B - c At CA-B) - C C Jrue for or false? all sets This statement "There for is false. It We prove its sets A,B and so that example, choose Since So are A, B and C, AUCBrc) A, IDCAVB) so A 213 = I B c = Therefore point negation. A CAUB(nC 0 = but 10 Au(BRC) M - have Au(Brc) IEAUCBUC), RC. (AVB) MC. I = c 1CAUBSMC- Aure) Merce u We B 62AUB) nC True false? or For A, B and C, ifA BUC all sets A-B then EC W I IIIIImin - got A BCC - Se statement A C BUC true is suppose 6D + - A and *B X = - X GB W XC B A B t direc e C This B A True exist x I B A x XEC or F - G x Proof C letA,B,C be X= A and 2 *B. Thus, sets so Since that ALBUC. and XEA We ACBUg to want We know show that A-BCC. let XEBorXEC, but xAB A-BCC. True or false? for all sets A,Band C, ifAs Buc and BI CUA then - -Next*A_Buc > A / ↑ 1 - · choose and a cun R e point 2 c B A [a3 = = [3 E3 = C x = CCAUB A-B;ie, so xEC. for then 02 = True for false? sets, A, B or all ↑ but [13. = CUA, I AUB = = CAUB but = C, if Al(BUC) 0 and C = = and B 0 Buc ICC because [3, B 33 and A chose example A thn = AIC = B. 4 A((BRC) 4 Sketch = ↳ = x AK and X E A 4C x 9 This is No there means # Me) & x = A B Since I L ⑦ A =2 0 = & that means A T - and cB x o x BUC B or x*C B = x C always nullset is a subset every set. of : Suppose A,Band), thatA/C=B. Suppose Since - XE contradicts which June we such that AlCBRC) sets XEAK. We know XEA. the assumption that want Since to prove XEB, ALKBUC) 0. = sets, A, B and 2 C C, ifA/B, then - B A XE Brco Thus, XEB A - Ak Since and x=A. and Suppose prove XEB. XBnc, x= ALNC) AKCB. B. BEC Truc e C I contradiction C AlB= suppose AlC = I and XEA &X G A and - C x x . B A - - x - ↳ X by contradiction. - I f 55 want to v I -O XEB 0. We = false? or for all Alc, are G B I - ↑ ↑ As B and roof:Suppose prove and Since XEA thus True AUB x4C. Since XEAK, XEAIB Since therefore, that Alc-B. So x=B. know that we and we A or x XE or B and xEC xeBre y = [BrC) (AUB)nC, know we ye AUB and y = C. yEAUB, Since know we yGA or y EB. cases, ye ye A. B. Then y=AU (BUC) ytBnc yeB yeswe get (AuB(nc (AU(Bre) Since Thus, and and for so ye AU (BUC). - E(2,33 A all A,B, sets (and = B D 2 x1(x) (ADB) xCCID) = D [2,33 = 24,5,6) = [4193 ↳14↳(5),(ss) = ↑ ↓ ! " · W Es B - A - A B we true is ) <I (A- L - D - · & 2 >13 3 S => sats (Ax) ((BxD) I' true Ax2 & ↳ & 2. 2(x) ④ 8 · ⑤ (2x) 8 I ⑧ ⑳ ⑧ ~ &(2) biss L 4*s C There exists choose Not Y A,B, C and D 2x6) - ( - 6Ax(BXD such that (A1B) x(C(D) B LBXD= C] = 4). CAB ( xCCID) but (1,2) = CAXC) (KBXD) x=AK gat AlB2C, A/C proved CB ↓ = Au ProofSince casel: Suppose X-A and OC I cased we want to prove = AB and C, (AUB(nc Au(Brc) statemement This 2 sets all Suppose XCB. that XeAIB. know contradiction:XtC and false? or for a that ABC. We such by contradiction. XB; have we sat are that XeB we XE I Therefore we have x4C x = Counting When objects", count we A recipe looks of to way a which them, make is called a recipe. like: StepDo Step 2 think we this , DO done be can in ways. me n2 3) ~ " Step 1 Do A recipe makes A A recipe a when Ex How Ans 9x10 xoxo Ex goods is digit I digit -2 Choose 2nd digit = loways How to do object to the get make number of want. what we a does step in exactly the objects objects is depend not one (using nixhax...*nk. there? are 4 ways 1 (" digit [ loways positive are there? 4-digitintegers are digit 4 number even x10x10x5 How many even positive there so that digitsare distinct? the X It - - 5 = 0,2,4,6,8 if "O" I choose W - then for my not So divide we choose 4th Choose st choose and choose god 1x9x8x2/4x8x0x7 Ans 1x9x8x7 + 4x8x8x7 se here options if them I not 1 10 options not a I have 17 over for 's". my worka s number independent. recipe into Hence two of have recipe steps on previous steps. way. the recipes and digits loways choose many we ways numbers choose 4 of ↑ every I ps choose Ans makes and works positive 4 many number works, the number of ways recipe a end at the the (i) iso it recipe When if sense works recipe Ex &2 * does are is n ixnax...... nk. Ex) How 2"because : A step - - in in of choose How many nx(n-1) x Arrange ways we a choices (chose have it or not) objects. a objects. can we of it arrange objects? a .x(n (x+) - ... .. 3 of 8 many ways let () be Ans] set, objects 8x7x6 As find we have? elements a be: can of a k setof a for each elementof the recipe a Arrange How does subsets many a can # the we of choose of to choose ways for (2) formula a objects? I objects from the by counting a objects? number of ways to arrange in of in objects. a(n-1).... (n kH) - differently, Now, count I of ((n=) objects stepl choose Step Arrange those chosen the (*) ! an k objects (k! answer: thus:(n)k! n(n-1) (k) h(n-is so (n -k + 1) ... = (n -k+1) .... = k! (9) letA How 98x = 84 = 31,2,3,4,5,6,7,8,9] = many subsetxof ↑x8x7 A that(x) so C |x(2) 3 and 3 eX S25,23 -> 1x 8x1 = number ·2 ↳ number i can of ways 2 - of ways choose 3 t hence i can you element choose next can make than this once setm ore not a good recipe. Counting:Recipe,makes A step in a can be: vecipe kof arrange sense, works, good. objects n -> A)) ways. 1x terms (2) How is atleast one are there I? using the digit1,2,3,4,5,6,7,8,9 9.x.9x9x9-8x8x8x8 Ans wrong popular 1:choose 2:fill !11 so not exactly 4x8 to onetoslot 0 st put ·He WI 11 fill 4: Ax9xax9 answer:- A 3. Fill Ans - 4-digitpositive integers many there ways. x)! meness...... k! - that (*) - = Example: so objects of I choose 'I you can a one good 11 /I make 1 in digl t 9 9 · a I 11 CL "9: this multiple ways recipe. exactly twol's exactly (24).+(8) three is 8x ()) + exactly x 4 four Is. 4-digit positive integers and at least one 1 atleast. a sample:How that there Answer: - what many is 94th One using the digits. 1,2,3,4,5,6,7,859 2? don't want is don't. No we there are I's No or ⑧ D - A NO (AUB1=(Al+IB)- want 84 for City the total, 84 + do the - not of set lANBD 44 9th have Whatw e digit. LetAbe - we 2's No Is B such want integers as thathave integers no a I's choices or for each of the nod's. is no B 25 what we /A1 have thathave intgers is we don'twant 8as = we &B1 8411 12 = PARB1=14 have 8 AUB is for choices 11 each 12 21 of the 13 1 4 digits. 111 ALBY-By- Now, AAUBI DAUBrc1 DAD = + DB1+1CD - DARBD-1Anc-1BreD+DAninc B A A M ↑ · A A C ↑ A UBUCUDD= - + - DAD+ ABD ADCD 6 doubles 4 tripples the AADA quadrable so Pascal identity (ii) (e) +(+) = Got(algebraic (i) + (in) - Easy is in nek-ll! - ! - - I - (x! ((x 1)!(n x)! = + - -(i) Binomial xty) idea Theorem (i) inyo = of the proof:just do it. (xty)=a() .....(xty) n times. functions from Def:A function every A XEA, domain : =co B there exactly is A one to a yet satis is a subset oft of AxB so that for (xy)ff. that so of f. domain of = (xy)=f set a f. y f(x) (= underf. the image of = = "f:AtB"means f is a function from A to B". 1:IR - IR f(x) x2 all XCIR for = V This f means 3xx,x)1x=IR] = which & A 31,2,3,43 Let and B = t is EC1,2)3 = [((2),(2,2) (3,4), (4)6)) [(1,4), (2) (, (3, 7, (4, a = - a = Recipe choose functions the image Choose the image choose the image choose the image of of are which there 17 2 + 31 of of a no may from be not what is a the image sat at A to B 3choices 43 choices. so from A to that (sy) It. Be Ans:Yes. 1) 3 choices 3 choices IRXIR. B? yet function set f(x) the image many by falling function A to exists yes How from = f function ⑦ function 1 define subset of [2,4,6] If We a = 2FA, butthere because Nos a is · of each is in the domain We visualize can : 2 0 - ⑧ a o f AxB subset using or diagram. ⑧ IR (i when X When A is rational is irrational X B I i arrow from Igoes to of 1 image 24 2 3 C 4 the diagram arrow of a functions 1 2 must have we 2 - - x4 3 function behaves A arrow ⑧ O x fex- In A to a s 234 W f:IR from ⑧ X O o X function a like exactly fo soucen 8 image X of X 0 a re o0 0 o 0 -> ⑧ 0 Let: f: AEB is Xajb => 1:A ESB Here exists one to one 7A, if one to feal f(b) = then a b. = many ajbt A, so that fat=f23) arrow from each x lights ource. a one but all. i n the element domain. f:At B E Xb= f() onto is " = - 08 B, FazA, O- b. = o -jo B everything A I 2 5 -> - B has to be used. in (co 2 46 domain). - - 810 12 A B f [(x2),(2,4),(3,6), (4,8)(5,12)] = How many from one XSx4x3 functions one = Ans:6 A to [1,2,3,4,53 A there are 22,4,6,3,10, 123. to x 2. [1,2,3,4] = I 2 - 30, 3 ~ IAM B IBI = DAME (=> 7 one 7 onto EC to functions from one from functions to B. A A to B. R 2x2 = ↳site offunction How sets many from A functions 21,2,3,43 = -2" >q Total whatwe ((2)(24 3" Let fXIR-112 - +1 3 x = a one is 22,43? ⑮ = - don't Ans f(x) B 2 = Ex) to there. a re - 2) 3] - + defined for to want. each one? is by X EIR fonto? Solution: - fi s one-to-onco foot: useact.Supposeateein f(x) f(b) = 2at 2 a f is 2bH = b = onto. Proof: Suppose b5117. Choose a b = then acIR, and 2(b) f(al=2a+1= +1 b01+1 = = f:2 2 - f(x) x2 on + = 5: Thinking. choose a so = one onto not 2 1 + f(x) that f(al to x = Not 2 + one to one not one b. 2at1 b = a EX 2-21 let 7: Yes to one defined one? to one is 5) = fx= f(b) E Xb= f() which fibl-1 is - 2 3. 235-= which is - onto is We must prove that · - is at 2 f:AtB 2. x = NO. that Then = each fonto Is 9,3t21. Suppose Suppose for = true one a b. f(x) 2xH) by -> definition. B, FazA, b. = B t1A (false) arto Negation:there exists 1 = fzal -b· for all acA By so roof exists There b 2200se b, 2. suppose = 2a then for all so a a ifa b = t2, E Xb= f() = B, FazA, b. f(a) b 7-1. 1 but = is I integer not for obo. 7: 2 Consider is Suppose show =22, asb that by xt2x one? to it one & given - a such that feal=f(b). We b.2a 2b = = implics to want a b= onto is it b 7 = G choose not b ]i =t such · Vaz that Suppose a II 21 fafb then Ga is is what an to we want integer and show thus hence onty. 7: Qt 21 Consider Ex, Show choose one? to it one is x f(11) Rs 1.1<2 XEx". but fext=fax; 10)= 1.2 bacause = No such that x= f(x) LX) by given (10) 1 21.2) and 1 because fax) = = = = but X*x 101+(.2. as 1.2<20 = it onto? is & 21. be suppose choose a Then b. = fix togiven by Consider is it feal=b. 7aGPsuch that *b=21, show is integer this x+2x one? to one d f(a)=La)=13). Because - such Suppose X, x=& 2x 2x" implies X x 1. that fax) f(x"). = to want show X x". = = = ontoh it is - X, x"G. Suppose Choose a Now = b = 2a f(a). = Ex f: let fone-to-one? IS defined 2+ 41 Is by tex) for each xn+2x x=21. Is = f one-to-one? Is forto? Solution: t (E) not one-toxone is roof:choose Ii s and x0 = f(x) f(y), xxyz -s - = y=- 2. (7yz2 x 2s onto = not Rof:choose y=-2. Then for any Ex f:4-1) let Then xyt2, xty) but and f(x) f(x) = ! f(x) +y) 0 = f( -z) f(y) defined by fix) = = = x y. Eab = x=1, f(x) x2 + 2x (x+12-11-12 = but = - 2 2x2 +x for each fexity. so y = f one-to-one? for any xc2, Is XE2. Is f onto solution: fi s to-one one- f(x) f(y) proof:Xxxy GA, = then 2x2 + x 2y +y. 2x2 2x2 X 2y2 - 2 - 2y 2(x2 yz) - x x - 2(x y)(x y) - x x - x y(2(x y) + xy + heace x = y y = Proof:(by f(x) completing the square) y=-1. Ten yet 2x2 +x = x 2 = +x 2 x2 x2 = + and +x + 2xx1)2|t)2 0 x2 = - y 0 = 1 = Y2 * x(x 1) + 0 - = + = = = y = = + 2(x y) y 0 y 0 1) + = = y + + = - + + y - x choose = + f onto 7 is Notinteger. so f(x) by. - - , fonto? is b. Composition fiAtB let Ex] functions: of and LetA S12,33) = 7 let g be git) B function. Then a function from defined to if got(x) by = g(f(x)) for [(2,1) (41), (6,2) Then 3. f is from function a A to B and is g a function from E 14 I 2 - 3 3 6 - > got I I fog? - = - xy IS Es-a E I f Ex g =) = They f:IR then g is same gilKeR fog(x) (2x+)2 = is Ars the S not is have A and B and x= = = I each [2,463, [1,2,3] = [CI, 2), (2,4), (3,6)3 = goti s domain and for every element = gof(x) 2x2H = False, because fog() 973 gof(l) = in X f(x) x2 by fog:got? = and onto is one to one onetoome defined and f onto fo I for the and each domain, g(x) = XEIR. f(x) g(x) = 2xH for each XER. B to 3 and got 3 (1,1), (2,1)((3,2)] = d IMPORTANT Ex a) f: Let AtB f and if and I then one-to-one are functions. be giBtC goti s True one false? or to one. An = ⑧ Co B A C gotTrue intratively proof:Suppose and and by and I want we to can you to show see that one XsYE where x=y = g(fx)) g(f(B)) one are g(x) gay) then = picture to f(a)=f(b) than is B. want We to prove be false. ab 9cb where = got is one 7 to A. Suppose one. 2,BEB. where B a= not it's possible g(f(x) g(f(b)) = f(x)=f(B) Since b) If because f(x):f(B) know we got is to one and of since f then one is is g to one 1-1 1-1 is x B = & one Not got o possible again intution will · - -eno - > o 6 - - 5 A 9 B ·C got mustprove got is one must one-to-one. We is f i s one-to-one. We pint not possible & - Jone: Suppose in 0 me C win innatever -- Suppose a to one, B I L = gof(a) g(f(a)) g(f(3)) gof(b) = get we - f(a) f(b). Suppose asGA. Since b. = prove f D that and = = ab got 2) If got This is statement is - got Choose A [1, = is 23, B f:A +B choose goti s g is as suppose g is one-to-onco f: AtB, 7 Negation: 7 sets A,B, C, 7 g: Btc one to fin):1 g(1) 1 one then one one [1,2,33) because not to = gof(l) to and not is g I) and f(2) g(2) 2 = g(z) and 1 as one gof(xl:gofcys hence I 2 = = -itse 18 32.52: - ( 21,23 = = 9:BtCas choose Now false: that such but one 2 = got(2)=2 g22) gC3) 2 = = case/ g(f(x)= 1 then casch g(f(x) then goti s 1-1 2 = x y1 = x = y 2. = = 3- I d) If one-to-one is got This statement is Proof and got that be B, Jaz A, fca) is is one-to-one. That xaP, 72x=A, y =Bs 7 g(feel=g(f(n) then = b. we g is one to one. show where 1 = g is is ec1=A. Suppose I is one-to-one. That onto gexi-gcy) we must snow X = y. X that such 6(xy) Y = = is want to 2 = g(f(x) g(f(y) got Since then that f(x)X such <pEA, have we onto is true. Suppose is f one to one we because have g(x) g(y) = [x=dy so x y=f(dy) f(x) = = showed we x=y. Inverse Definition f: A-B is g is invertible Ex There exists called fog-IB that giB -Aso of f inverse and got:IE. ft. g = Theorem: F.AtB f:2-21 Ex:Let f Is one-to-one? invertible is the function be fis Es defined by one to fixi-ex and onto. one for all x=4. f onto Is ~ Yes Thus, Ii s defined f:A not by gof=In No invertible, find a function 9:2- 21 I whenisever =2CI) X = so inverse of f in P this function and one-to-one. - onto B is can this case is both i y, 7 6 ↳ -ener 2 - fog2X B. = - 3 - 4 - - S 3 6 A - I ↳ 3 - - = i I -2 4 - g: B for example:let g:2+2 ii -> - - = that got:I gixs= - g but we - - 3 4 - S S - 6 9. be the function Relations Def:A relation to related is X sat A a on relation in the in g Ry x AXA when subset of is R is a relation on As we say (xxy) GR when (x,y)ER. > T 0 4 X 88 ⑧ 3 0 . I ⑳oo O - 734 Ex (etA 1 [1,2,3,4] = E(1,2), (1,1), (3,4)) is = How many 2 Ans= We can relation visualize on a A relation a 91,2,3,43 there? arc = A. on using its directed graph relation &· & 3 R Def:Let R R R Eb Fxxy, z EA, is 1, on = reflexive 1 = D and What relations is = antisymmetric:if121 for same then then 1 2 / and transitive:i f 122 relations what is? reflexive:1<A false antisymmetric:if then yRZ y. XRZ. 3, 43 if 4=4 then 4=4 true 1=1. True. 1-1. True "s"? than Symmetric:if122 x= True 21, 222, 313, 4-4. symmetric:112 and [1, 2, = and D=1 = 1 then yRx = 3,4 4 = then 1=1. D antisymmetric:if 1 = yRx. and xRy FxcA, XRx. reflexive ES "="? Symmetric:if 1=1 transitive:1 A Set 2 2, 3 = if is then Ry XxcyEA, xRy let R be relation what relations x if Ex is transitive A.R Sat on Axcy= A, if Ex antisymmetric Reflexive:1 3 the relation symmetric is is be and then I <1 transitive:if223 241 and and not symmetric. 124 213 true than then 1=1. True 113 · True false ASD. 3 C4 True. then Vaccous truth. 224 true. First "if weareashere 2 or it reflexive becaus not is iti s symmetric because not xcys2=2. Suppose and 4/2. 214 (1-1 Xly Suppose and yext have: then 2= 1 xtK but it. and -11 y/1. and then we prove x12. must and (E2. ykwhere the 2 where = nonce Ex X0. transitive:proof is we but 0 because iti s not antisymmetric it OC21 X/L. LetA E1, 2, 3, 43. = 3(k1), (2,2), (3,3), (4,4)3 [D,2), (2,4)) (3, 4), How many [4,4), (14)]. reflexive relations 2' because Ans: reflexive is 4 on are reflexive not 31, 2, 3, 43 A: fixed. arc because 11. there." (161), (212) (3,3), (4,4) I & 4 · O 3 0 O ⑧ O X X 2 0 I ⑧ X A 0 / ↳ ↳ · O L ⑧ O O 23Y Ans: 2k you Ex either choose or not choose from whatever is left. [1,2,3,43 LetA = S(1,2) (2,4), (2,1) (4,2)] is symmetric. S(1,2), (44), (2,1) 3 is symmetric. relations many symmetric How Ans: 20 4 X ⑧ & & 3 2 D X ⑳ ⑧ ⑧ A · a ⑳ * ·X ② 23 ⑳ 0 y are there on A 11,2,3,43? = so 2 k il yes this is reflexive. Ex 51,2,3,4] LetA = SC1,2)3 is antisymmetric Q [(1,2), (2,1) (1,1), (2,2)] a [(1,2)s(1,1), (2,2), 24,3] = & this -> vacuous is truthfirstcondition because not antisymmetric is : bacause 1Q2 antisymmetric. For does not not true. was but IF2. and dQ1 C1,2) and (4)3) case it is vacuously true. on to has - · 34 antisymmetric ( antisymmetric as not not antisymmetric How 24* 3) but x yRX fy. of having relation antisymmetric many and its there exists not as a re there 31,2,3,43? on AskTA. Ans: Ex Jxsyt A, XRY [1, 2, 3, 43. A Let = is transitive. * 2(12)3 R [(1,2)) (2,1), (1,1)] because T transitive. is is = 2R1 IRL and not transitive. 2Rd· but [(1,2),(2,4),(14), 434), = (3,4) 3 is transitive. ⑳ · 2 ↓i (12) DR2 and 2R4than 1R4· = (234) 2R4 and 4R4 then 2RL· (1323:AR4and RL them DR4 = (3)2):3R4 and URL than 3R4· CHSU)=GR4 and 4R4them 4R4. Ex let A [1,2,3,43 = Is there a relation R on [Islec 3,43 so that R not reflexive, symmetrics antisymmetric, but is Ans:No. Proof (by contradiction] Suppose there not transitive -> R is that x Ry that xRL Suppose is such a relation R. 7xcys2, xRy antisymmetric then xRy and and then XR2 yR2 and yRX yRZ butxR1. S and which x=y. is We a see contradiction. transitive? 1A reflexive, symmetric, antisymmetric, transitive? R IS 21,2,3,4].R relation = = R is R is R is R R 21 on defined R is [23 y then = = x yo2:223 +4 XRX. must snow we then 31x+2x B XJyt1. Suppose by + Suppose that xRy 3 is =3(x+2y) must we x+ x yRx by showing 3Dy+ 2x not antisymmetric. Suppose xsyt2. R is that is x 319H2=3121 Suppose transitive. 3k- dy and XRy yR2 choose Xt2y = x 9 and y 6. = - 2y = y = = * 6. We must = show - bk + - 4y 3y+ 6x 3(-yA2k) that XRML. - YA22=3t. is that yR and a y+2x yAd23k-2]) Suppose XRy = = but sys2t2. where 3K = 2 3t: and then 316H18:3124 and 3AXA2y, that is xRy = 2y 3 k 3k = · prove R is y - 2 Man 2L x + 3xx-2y + x + = 2k 3x x xy+37 y - - = 24 / + 2 =3kx 3+ + - 3y x722 3(x+ t- ~ = 3) A 31,2,3,45s is R is Choose and then Notations gal)=1 Relations:An equivalence = A is easy to A Transitive:Let gRf and on equivalence by f,gt F, fRg() than a set A is relation divides gaxs= faxs. g [(1,1), (2,2) = (332), (4, d) -79 but f(l)=g(I) Suppose number is = R that xyzPCA) fex)=gax) = relation I defined on fex)=fext. Suppose E1, 2, 3, 43. R relation · is Suppose xcy) 2EPCA) an Nx1=AX1, Reflexive:Let x=P(A). Since Symmetric:Let the relation is a and relation A into gal) KCI) · = on Since g2K K21) = it follows f(l) K21). that A equivalence classes. a = prove R of of Example:Let It class equivalence relation number A to A. 1xRa] (9.7 3x=A equivalence ktt-y cry. f [1,11, (1,1), 23, 1), (MD] and fig reflexive, symmetric and transitive. This The · figskff. transitive:Suppose Equivalence are fog ff antisymmetric. not falll=1 R them Suppose f. If symmetric. Suppose is where from = reflexive. R is y) f f set ofunctions = R XRY so ty. 2, xRY (b1x then 1 x = = symmetric. 21 x= where xdy x + 2x 3x = Xny:[1370 and Then = and xxy by Suppose reflexive. is y [1, 23,=323. YRX. so &2 0. X Letx=[1,23 yux=xnyto, than Xny*. i.e, = XRY. hence relation is XRY; = because xRYE6 xnyEG. d. dnd= since Proof:Let x [1 3 but X*2 antisymmetric. not xny1d ② xy GPCA). Suppose not transitive. R is by Axxy =PCA), Let symmetric. YRL, and defined 4GP(A), but$R0 because reflexive not PCA) on · equivalence classes. of on PCA) defined = P(A), xRY => NXD = AYD relation. XRX. XRY, i.e, NX1=1Y1. Then MYD= 1X, Suppose by:xxxy XRY and YR2, thati s so YRX. Ax1=MYD and MY1 12D. then Ax = 1Y1=121, = so XR2. = f(l) gCN). = where - yt2kzY. Example:Let A R Easy How to many Example 127A the relation is R that prove equivalence Two Ans 4. 31,2,3,43,7 = is classes functions related when defined PCA) on ↓x,y = fel b g(N) = relation. have the they B [1,2,3,4,5,6,7,8,93, = fig tf, fRg by there" = R relation from Ato A. functions all set of a defined on equivalence an are are = image same of So 1. the possible image number of of which 1, be can 1,2,3,4. [2,4,6,83 = by PCA), XRY DXUBD=DYUBD ES How there? classes are many equivalence r6 # of possible sizes of XVB' I = # Congruence Modelon on 21 where no 2 xcyt2, x=y(modn) tr to congruent x is nAx-y E) y modulus n. Proof Reflexive:Letx= 21. Since Let Symmetric Transitive and n xxy=2. X-x 0 nx0 where = = Suppose xEy(modn), i.e, and therefore y x (mode). 2) and suppose x=y(mode) site 2. Now, Lot xy, 2= for y-2=nt Of 21 so ndx-y, so (mode). nk for x-y- some key. Now, y -x =-(x-y) x-2 and y 2 (modn). = (x-y)+(y-2) = Then 1 2 +nt = adx-y n(stt) = and where ny- 2, 3 x 0 = (mod3)] (x= 2130x) = [d] 2x42139x 13 [ = - = - 5, - [ 6,- 3,0,3,63-remainder = - 2,14,7,....3-remainder, (2) (x 21131x- 3 Ea Esis = when n= 3 = we = have 3 congruence ↳ remainder d classes. 6 = I · S ! O B 2 4 S 38 4 S S ⑧ S 01 0 & 389 => 2 2 3 34 so SttE2, = = = - nk n(-xx) = = some [0] 3x =21 n x= x x-y so as = X= 2 (mode) where KEY, so aly-x. n 6 = X O 2 02 ↑ 3 03 4 0 S 2 34 4 02 4 O 3 3 O ·Y521 5 No n= ⑧ 00 O 0 R 5 234 I 0 U 2 0123x5? X 6 0 B 0 2 0 3 ⑧ 2345678 9 2 4680246 8 3 69 ↑ ⑧ 08 I 303 ! 6 25814 7 260 826 4 05 0S 0 S 628406284 07 41852963 ⑧ 7 8 ⑧ q D 642 64208 8 987654321 ad n 6 I ⑧ = 2 ↓ 0 N 2 I 3 - ac Theorem Modular Arthimatics if a b (mode) = Proof: and b a= (mode) and Stt 2. Then (atc) = d(modn) USCA but (moda) and - n(sc = bd ac= d(modn), c= (bAd) +bt), (modn) bd. = then atC = b+d(mode) and c= Easy Suppose 5 4 3 that is, (a b) A(c d) - - = where 8+tE2/ na-3 ns + = and ac bd (mode) = and nced. nt=n(stt) sctbtE2, and so Thus, az - arb=ns and for crd-nt bd acsbc Abc -bd (a b)c natc) = - = (b+d) and - nac-bd- thus, some + b(c d) - atc= bAd Def: An integer Theorem is a modulo invertible gcd(ayn) = Sketch of is a n Ex there exists an b integer that so ab=/(modn). n 1 = proof a invertible modulus 71 = Ib = a 21, (mode) ab=1 nbab Ex 7b E- 73jkty),ab-1 = modulo is invertible a 1 = nix = 7b,k= 2),abtnk 1 ( = = gcd(a,n)= 1 3 abtrk 1 = g gcd(a)n) = Sketch gla - of and gl gD = = gIn proof ab tak A = y gcd(a)n) gla = and + gen I gatg D = gak n - 9 3 - 8 Y = 7 - - 6 - 3 - 0123 whenever, we 4 0 have - 3 - 1 abtric: 1 2 2 we - 1 3 ⑧ 1 2 can 0. conclude ↑ b 34562 230 ↑ is inverse & of 2 ⑧ 3 9 O N mode. a 1000 -1(modll) = Ex Is 96 compute invertible modulo dT1? Find 21 - = 96 1x79+ 96 17 = + 11 29 17 Nx)) + 11 4x17 = = 11 1x6 = (x S A = 6 1 S 5x1 0 n - 5 - = 96 is invertible modulo 2410 A - 17 271x( k) - 2 = An 43 - G 19 S + 0 D 8 1D +S gcd(221,961=1. Thus, D 271 = 6 that 271x+96y 1 so gcd(271,as) 27 2x96 A79 79 xxy - 96x48 1 + 96 of inverse 271 is = modulus 48. 14 1] t 31 21117 28 E48 48 (mod = - 271) 48 (mod 271) t 48 27kc = + t - 48 27k = t 271 48 = x 1 = + if2a 23 (mode) this true is b(modn) then = a= modelo I is invertible when n ↓ this True a) ab = 3) (a) c 3 3 b if = al even ba=3b a · · a = 3 = = - = is a the remainder = 27 24 6(-4) = - = nence 3a = 3b 1, b 9 b 1 = (mode) = = a b a= ncyt. = and then a b 6, = 39-3b then (mode) exist ab, n (mode) bd but Proof:choose 3b = = false. statement is Negation:there if is = - What i 3a 36. and no2, if Fasb then 3a (mode) b a= then Ed = This 2, nt ifa b a when false? or X not true is where - 4E7 9 8 where-1-S #2 6(-).S) b 2243 when iti s divided of by 9? ab = 23 1 = = 80 * (23) =(-1)Y 273 x 9) then as (mode) 2 2 = =(2) a -> from theorem. qk b* = =( 1)9 - = So, the - 1 answer is 8. -END
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