{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Text Output" -1 6 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 2 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 269 "Try to find a field with \+ 4 elements, F4. I'm labelling the elements 0,1,2,3 with a=2, b=3 so t hat I can\nuse a 4 by 4 array for the addition and multiplication tabl es, indexed from 0 to 3.\nThis first try is with 1+1=0, i.e., char(F)= 2. So the addition table looks like" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "restart;\nMatrix([[0,1,a,b],\n [1,0,b,a],\n \+ [a,b,x,y],\n [b,a,y,z]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*Osza\"-%'MATRIXG6#7&7&\"\"!\"\"\"%\"aG%\"bG7&F-F,F/F .7&F.F/%\"xG%\"yG7&F/F.F3%\"zG%'MatrixG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "where x, y and z are yet to be determined.\nNow since th e characteristic is 2 we have x = a + a = 2 a = 0 and similarly z = b \+ + b = 2 b = 0." }}{PARA 0 "" 0 "" {TEXT -1 64 "( to prove this a + a = 1 a + 1 a = (1 + 1) a = 0 a = 0.) Hence" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 97 "Matrix( [[0,1,a,b],\n [1,0,b,a],\n \+ [a,b,0,y],\n [b,a,y,0]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*+tza\"-%'MATRIXG6#7&7&\"\"!\"\"\"%\"aG% \"bG7&F-F,F/F.7&F.F/F,%\"yG7&F/F.F2F,%'MatrixG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 229 "If y = 0 we have the additive inverse of a is a or \+ b which contractics the uniqueness of the inverse.\nIf y = a we have a + b = a ==> b = 0, a contradiction.\nIf y = b we have a + b = b ==> a = 0, a contradiction.\nThus y must be 1." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "a := 2: b := 3:" }}}{EXCHG }{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 119 "A := Array(0..3,0..3,\n [[0,1,a,b],\n \+ [1,0,b,a],\n [a,b,0,1],\n [b,a,1,0] ]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "In class we showed that th e multiplication table must be" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "M := Array(0..3,0..3,\n [[0,0,0,0],\n \+ [0,1,a,b],\n [0,a,b,1],\n [0,b,1,a]]):\nMatrix (M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*'>)za\"-%'MATRI XG6#7&7&\"\"!F,F,F,7&F,\"\"\"\"\"#\"\"$7&F,F/F0F.7&F,F0F.F/%'MatrixG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 253 "ASSOC := proc(T,n) local a,b,c;\n for a from 0 to n-1 do\n for b from 0 to n-1 do\n \+ for c from 0 to n-1 do\n if T[a,T[b,c]] <> T[T[a,b],c] t hen\n printf(\"Assoc %d %d %d\\n\",a,b,c);\n fi;\n \+ od;\n od;\n od;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ASSOC(A,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ASSOC(M,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 227 "DIST \+ := proc(A,M,n) local a,b,c;\n for a from 0 to n-1 do\n for b from 0 to n-1 do\n for c from 0 to n-1 do\nif M[a,A[b,c]] \+ <> A[M[a,b],M[a,c]] then\n printf(\"Dist %d %d %d\\n\",a,b,c);\nfi; \n od ; od; od;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "DIST(A,M,4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Thus the additio n table and multiplication " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "convert(A,Matrix);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6% \"*#\\(za\"-%'MATRIXG6#7&7&\"\"!\"\"\"\"\"#\"\"$7&F-F,F/F.7&F.F/F,F-7& F/F.F-F,%'MatrixG" }}}{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG }{EXCHG } {EXCHG }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "convert(M,Matrix);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*cvza\"-%'MATRIXG6#7&7 &\"\"!F,F,F,7&F,\"\"\"\"\"#\"\"$7&F,F/F0F.7&F,F0F.F/%'MatrixG" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Now to the characteristic = 3, i.e ., 1+1 is not zero so let 1+a=0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "a := 'a': b := 'b':\nMatrix([[0,1,a,b],\n [1,c ,0,d],\n [a,0,x,y],\n [b,d,y,z]]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\"*?wza\"-%'MATRIXG6#7&7&\"\"!\"\"\"%\"aG% \"bG7&F-%\"cGF,%\"dG7&F.F,%\"xG%\"yG7&F/F2F5%\"zG%'MatrixG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "Now d = 1 ==> 1+b = 1 ==> b = 0.\nAlso d = 0 ==> 1 + a = 0 and 1 + b = 0 ==> a = b.\nAnd d = b ==> a + b = b = => a = 0.\nThus d must be a and c must then be b. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "Now x=0 or y=0 are not possible thus z=0. Thus y =1 and x=b are forced.." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "a := 2: b := 3:\nA := Array(0..3,0..3,\n [[0,1,a,b],\n [1,b, 0,a],\n [a,0,b,1],\n [b,a,1,0]]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"AG-%'RTABLEG6%\"*%o(za\"-%&ArrayG6(;\"\"!\"\"$F,<./ 6$F-\"\"\"F2/6$F-\"\"#F5/6$F-F.F./6$F2F-F2/6$F2F2F./6$F2F.F5/6$F5F-F5/ 6$F5F5F./6$F5F.F2/6$F.F-F./6$F.F2F5/6$F.F5F2/%)datatypeG%)anythingG/%( storageG%,rectangularG/%&orderG%.Fortran_orderGF*" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 12 "DIST(A,M,4);" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 1 1" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 1 2" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 1 3" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 2 \+ 1" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 2 2" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 3 1" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 2 3 3" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 1 1" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 1 2" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 2 1" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 2 2" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 2 3" }} {PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 3 2" }}{PARA 6 "" 1 "" {TEXT -1 10 "Dist 3 3 3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "M[3,A[3,3 ]] <> A[M[3,3],M[3,3]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#0\"\"!\"\"$ " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "The distribute law does not h old." }}}}{MARK "10 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }{RTABLE_HANDLES 154797236 154797300 154798196 154797492 154797556 154797620 154797684 }{RTABLE M7R0 I6RTABLE_SAVE/154797236X,%)anythingG6"6"[gl!"%!!!#1"%"%""!"""%"aG%"bGF(F'F*F)F) F*%"xG%"yGF*F)F,%"zGF& } {RTABLE M7R0 I6RTABLE_SAVE/154797300X,%)anythingG6"6"[gl!"%!!!#1"%"%""!"""%"aG%"bGF(F'F*F)F) F*F'%"yGF*F)F+F'F& } {RTABLE M7R0 I6RTABLE_SAVE/154798196X,%)anythingG6"6"[gl!"%!!!#1"%"%""!F'F'F'F'"""""#""$F'F) F*F(F'F*F(F)F& } {RTABLE M7R0 I6RTABLE_SAVE/154797492X,%)anythingG6"6"[gl!"%!!!#1"%"%""!"""""#""$F(F'F*F)F)F* F'F(F*F)F(F'F& } {RTABLE M7R0 I6RTABLE_SAVE/154797556X,%)anythingG6"6"[gl!"%!!!#1"%"%""!F'F'F'F'"""""#""$F'F) F*F(F'F*F(F)F& } {RTABLE M7R0 I6RTABLE_SAVE/154797620X,%)anythingG6"6"[gl!"%!!!#1"%"%""!"""%"aG%"bGF(%"cGF'%" dGF)F'%"xG%"yGF*F,F.%"zGF& } {RTABLE M7R0 I6RTABLE_SAVE/154797684X,%)anythingG6"6"[gl!!%!!!#1!$!$""!"""""#""$F(F*F'F)F)F' F*F(F*F)F(F'F& }