{VERSION 6 0 "SUN SPARC SOLARIS" "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 } {CSTYLE "" -1 256 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 33 "The extended Euclidean Al gorithm." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 128 "If a and b are positive integers, and d is the gcd, then we have \+ that the ideal (d) equals the ideal (a,b), so: Z*d = Z*a + Z*b." }} {PARA 0 "" 0 "" {TEXT -1 120 "In other words: a and b are Z-linear com binations of d (i.e. multiples of d) and d is a Z-linear combination o f a and b." }}{PARA 0 "" 0 "" {TEXT -1 112 "So d = s*a + t*b for some \+ integers s and t. These s and t can be computed with the Extended Eucl idean Algorithm." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "a:=72;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG\"#s" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "b:=50;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG \"#]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "d:=igcdex(a,b,'s',' t');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "s*a + t*b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "s;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#8" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 113 "Note that s and t are not uniquely determined. If you replace [s,t] by [s + b, t - a] you'll get the same result." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "s, t := s+b, t-a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"sG%\"tG6$\"#T!#f" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "s*a + t*b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 233 "Of course, by adding o r substracting a suitable multiple of [b,-a] to [s,t] we can get s in \+ the range 0..b-1, and then s,t will be uniquely determined. Or we coul d bring t in the range -(a-1) .. 0, that would give us the same result ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 "For \+ polynomials this all works very similar." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "a:=x^3-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,&* $)%\"xG\"\"$\"\"\"F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "b:=x^4-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG,&*$)%\"xG\"\"% \"\"\"F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "d:=gcdex (a,b,x,'s','t');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&\"\"\"!\" \"%\"xGF&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "s*a+t*b;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&%\"xG\"\"\",&*$)F%\"\"$F&F&F&!\"\" F&F+*$)F%\"\"%F&F&F&F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "e xpand(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\"\"%\"xGF$" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "If R is the ring Q[x], the ideal \+ (a,b)=R*a+R*b equals the ideal (d)=R*d, so d is an R-linear combinatio n of a and b." }}{PARA 0 "" 0 "" {TEXT -1 19 " d = s*a+t*b." }} {PARA 0 "" 0 "" {TEXT -1 86 "This equation will still hold if we repla ce [s,t] by [s,t] + some multiple of [b, -a]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t*b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\"\"%\"xGF$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ra:=randpoly(x,degree=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ra G,(\"#P!\"\"*&\"#&)\"\"\")%\"xG\"\"#F*F'*&\"#bF*F,F*F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "s, t := expand(s+ra*b), expand(t-ra *a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"sG%\"tG6$,.*&\"#a\"\"\"% \"xGF+F+*&\"#PF+)F,\"\"%F+!\"\"F.F+*&\"#&)F+)F,\"\"'F+F1*&F3F+)F,\"\"# F+F+*&\"#bF+)F,\"\"&F+F1,.\"#OF1*&F.F+)F,\"\"$F+F+*&F3F+F;F+F+*&F3F+F7 F+F1*&F:F+F/F+F+*&F:F+F,F+F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t*b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\" \"%\"xGF$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 425 "So if ra is some ar bitrary polynomial, we can add ra*b to s (and substract ra*a from t) a nd the equation s*a+t*b=d remains valid. If s is an element of Q[x], w e have seen that we can always find a unique polynomial q such that s- q*b will have degree < degree(b). Namely, q is the quotient of the div ision of s by b. So we can make the solution [s,t] of the equation s*a +t*b=d unique by requiring that degree(s,x) " 0 "" {MPLTEXT 1 0 1045 "GCDEX:=proc(a,b,x)\n # Produce d,s,t such tha t:\n # d = gcd(a,b)\n # d = s*a+t*b\n # degree(s,x)%&GCDEXGj+6%%\"aG%\"bG%\"xG6(%\"cG%\"rG%\"qG%\"dG%\" sG%\"tG6\"F1@%/9%\"\"!C%>8$-%(contentG6$9$9&@$/*&F<\"\"\"F8!\"\"FB>F8, $F8FB6%F@*&FAFAF8FBF5C%>8%-%$remG6&F6%8'8)8(-9!6%F4FIF=6%FQFS -%'expandG6#,&FRFA*&FSFAFNFAFBF1F1F16%F5F5F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG \"\"$\"\"\"F(F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG\"\"%\"\"\"F(F(!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "d,s,t:=GCDEX(a,b,x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"dG%\"sG%\"tG6%,&\"\"\"!\"\"%\"xG F*,$F,F+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "expand(s*a+t* b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\"\"%\"xGF$" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a:=x^50-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG,&*$)%\"xG\"#]\"\"\"F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "b:=x^32-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG,&*$)%\"xG\"#K\"\"\"F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "d,s,t:=GCDEX(a,b,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"dG%\"sG%\"tG6%,&\"\"\"!\"\"*$)%\"xG\"\"#F*F*,0*$) F.\"#5F*F+*$)F.\"\"'F*F+F,F+*$)F.\"#9F*F+*$)F.\"#GF*F+*$)F.\"#CF*F+*$) F.\"#?F*F+,8F*F*F7F*F1F*F4F*F:F*F=F*F@F**$)F.\"#KF*F**$)F.\"#YF*F**$)F .\"#UF*F**$)F.\"#QF*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "e xpand(s*a+t*b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\"\"*$)%\" xG\"\"#F$F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"\"\"!\"\"*$)%\"xG\"\"#F$F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 210 "You can verify (for a proof you need to use the principle of m athematical induction because the algorithm uses recursion) that the c onditions on the degrees of s and t will hold when we use our algorith m GCDEX." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 395 "Consider the ring Q[x]/(f) where f is an element of Q[x]. Every e lement in this ring can be represented uniquely by a polynomial in x o f degree smaller than degree(f,x). We can add and multiply in this rin g, when we multiply and the result would get result >= degree(f,x) the n you just reduce modulo f, i.e. you replace your result by rem(result ,f,x) and it will have degree < degree(f,x) again." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "So we know how to add an d multiply in the ring Q[x]/(f). Basically all we need for that are th e Maple functions expand and rem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 478 "Now we want to divide as well. Suppose t hat g in Q[x]/(f), so we can represent g uniquely by a polynomial of d egree < degree(f,x). We want to do divisions, so we want to compute 1/ g if it exists. In other words, we want to find (if it exists) an elem ent t of Q[x]/(f) such that t*g = 1. In the ring Q[x]/(f) a polynomial equals 1 when its remainder modulo f is 1. So this means that when g \+ is given by a polynomial in Q[x], that we want to find a polynomial t \+ in Q[x] such that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " rem( t*g, f, x) = 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "In other words: t*g = q*f + r where q \+ = quo(s*g,f,x) and r=rem(s*g,f,x)=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "So: s*f + t*g = 1 where t = -q." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "This mean s: 1 is in the ideal (f,g), in other words gcd(f,g)=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Conclusion: There e xists a polynomial t in Q[x] such that t*g=1 in Q[x]/(f) if and only i f gcd(f,g)=1." }}{PARA 0 "" 0 "" {TEXT -1 63 "Furthermore: we can find t by the Extended Euclidean Algorithm." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f:=x^3-2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,&* $)%\"xG\"\"$\"\"\"F*\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "g:=x^2-5*x+3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG,(*$)% \"xG\"\"#\"\"\"F**&\"\"&F*F(F*!\"\"\"\"$F*" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 24 "d, s, t := GCDEX(f,g,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"dG%\"sG%\"tG6%\"\"\",&#\"#J\"#V!\"\"*(\"#AF)\"$H \"F.%\"xGF)F),(#\"#>F1F.*&#\"# " 0 "" {MPLTEXT 1 0 2 "t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(#\"#>\"$H\"!\"\"*&#\"# " 0 "" {MPLTEXT 1 0 4 "g*t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,(*$ )%\"xG\"\"#\"\"\"F)*&\"\"&F)F'F)!\"\"\"\"$F)F),(#\"#>\"$H\"F,*&#\"# " 0 "" {MPLTEXT 1 0 10 "expand(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&#\"#J\"#V\"\"\"* $)%\"xG\"\"$F(F(F(*&#\"#A\"$H\"F(*$)F+\"\"%F(F(!\"\"*&#\"#WF0F(F+F(F(# \"#>F'F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "rem(%,f,x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "One more example. Take g = x. So we're going to compute 1 /x." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "N ote that Q[x]/(x^3-2) is isomorphic to Q[ 2^(1/3) ], so computing 1/x \+ is just like computing 1/2^(1/3). And we know that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "1 / 2^(1/3) = 1/2 * ( 2^ (1/3) )^2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "f;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG\"\"$\"\"\"F(\"\"#!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "g:=x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "d,s,t := GCDEX(f,g,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6%%\"dG %\"sG%\"tG6%\"\"\"#!\"\"\"\"#,$*&F,F+%\"xGF,F)" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 2 "t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"# !\"\"%\"xGF%\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 125 "Every elem ent g in Q[x]/(f) is represented by a polynomial of degree < degree(f, x). Now 1/g exists if and only if gcd(f,g)=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 131 "Suppose now that f is an irreducible element of Q[x], i.e. there do not exist polynomials f1, \+ f2 of lower degree such that f1*f2=f." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 275 "Then you will see that every non-zer o polynomial g of degree < degree(f,x) will have no factors in common \+ with f, because f simply doesn't have any non-trivial factors of degre e < degree(f,x). In other words: every non-zero g in Q[x]/(f) will hav e an inverse 1/g in Q[x]/(f)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 194 "Q[x]/(f) is a commutative ring, i.e. we \+ can add and multiply and multipication is commutative. If f is irreduc ible then we can also divide by any non-zero element. Such a ring is c alled a field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "Note that if f is reducible, suppose f = f1*f2 with degr ees f1,f2 smaller than degree of f, then you could not divide by f1 in the ring Q[x]/(f). We conclude:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 71 "Q[x]/(f) is a field if and only if f is a n irreducible element of Q[x]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 749 "When f i s an irreducible polynomial we can multiply in the field Q[x]/(f) by e xpand and rem. We can divide by the Extended Euclidean Algorithm (call ed gcdex in Maple, for the help page type ?gcdex ). It would be easier to be able to just write * and / for products and quotients, and to h ave only one Maple command that will evaluate expressions in the ring \+ Q[x]/(f). This is done by the command evala. The element x modulo f in Q[x]/(f) should then be denoted by: RootOf(f,x). This RootOf should b e interpreted as: \"some unspecified root of the polynomial f\". So if alpha := RootOf(x^3-2,x) then you can not say that alpha equals 2^(1/ 3), you should think of alpha as an unspecified root of x^3-2, and not just one of the three roots in particular." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "f;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG\" \"$\"\"\"F(\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "A: =(x^4+2*x)/(x+1)+x^5/(x-2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG,&*&,&*$)%\"xG\"\"%\"\"\"F,*&\" \"#F,F*F,F,F,,&F*F,F,F,!\"\"F,*&F*\"\"&,&F*F,F.F0F0F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "n:=normal(A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG**%\"xG\"\"\",,*&\"\"#F')F&\"\"%F'F'*&F*F')F&\"\" $F'!\"\"*&F*F'F&F'F'F,F0*$)F&\"\"&F'F'F',&F&F'F'F'F0,&F&F'F*F0F0" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "n,d:=numer(A), denom(A);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>6$%\"nG%\"dG6$*&%\"xG\"\"\",,*&\"\"#F *)F)\"\"%F*F**&F-F*)F)\"\"$F*!\"\"*&F-F*F)F*F*F/F3*$)F)\"\"&F*F*F**&,& F)F*F*F*F*,&F)F*F-F3F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "g cdex(f,d,1,x,'s','t');" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "t; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&#\"\"\"\"\"$!\"\"*&\"\"'F'%\"xG \"\"#F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "t*n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&#\"\"\"\"\"$!\"\"*&\"\"'F(%\"xG\"\"#F(F&F+F &,,*&F,F&)F+\"\"%F&F&*&F,F&)F+F'F&F(*&F,F&F+F&F&F0F(*$)F+\"\"&F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "A_simplified := rem(%,f,x );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-A_simplifiedG,(*&#\"\")\"\"$ \"\"\"*$)%\"xG\"\"#F*F*!\"\"*&#F.F)F*F-F*F*#\"\"%F)F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Now we'll simplify A modulo f using Maple 's evala and RootOf." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "alp ha:=RootOf(f,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alphaG-%'RootOf G6#,&*$)%#_ZG\"\"$\"\"\"F-\"\"#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Ar:=subs(x=alpha,A);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ArG,&*&,&*$)-%'RootOfG6#,&*$)%#_ZG\"\"$\"\"\"F2\"\"#!\"\"\"\"%F2 F2*&F3F2F*F2F2F2,&F*F2F2F2F4F2*&F*\"\"&,&F*F2F3F4F4F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Ar:=evala(Ar);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ArG,(*&#\"\"#\"\"$\"\"\"-%'RootOfG6#,&*$)%#_ZGF)F*F* F(!\"\"F*F**&#\"\")F)F**$)F+F(F*F*F2#\"\"%F)F*" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "The n ame of the command evala stand for EVALuate Algebraic. This is Maple's command for handling algebraic numbers." }}}}{MARK "0 1 0" 1 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }