THE TEACHING OF MATHEMATICS \noindent2010 commaTHE Vol TEACHING period XIII OF comma MATHEMATICS 1 comma pp period 5 1 endash 6 1 ON THE ANCIENT PROBLEM OF DUPLICATION \noindentOF A CUBE2010 IN HIGH , Vol SCHOOL . XIII TEACHING , 1 , pp . 5 1 −− 6 1 Andrea Grozdani acute-c and Gradimir Vojvodi c-acute \ centerlineAbstract period{ON The THE paper ANCIENT is devoted PROBLEM to exposition OF DUPLICATION of constructions} with straightedge and compass comma constructible numbers and their position with respect to all algebra hyphen \ centerline {OFTHE A CUBETEACHING IN HIGH OF MATHEMATICS SCHOOL TEACHING } ic numbers period2010 .. , Vol Although . XIII , the1 , pp large . 5 1 number – 6 1 of constructions may be accomplished with straightedge and compassON comma THE one ANCIENT of the known PROBLEM problems of this OF kind DUPLICATION dating from Greek \ centerline {Andrea Grozdani $ \acute{c} $ and Gradimir Vojvodi $ \acute{c} $ } era is duplication of a cube periodOF A.. The CUBE given IN proof HIGH in this SCHOOL paper is elementaryTEACHING and self hyphen contained period It is suitable forAndrea teachers Grozdani comma as wellc´ and as Gradimir for high school Vojvodi studentsc´ period \ centerline { Abstract . The paper is devoted to exposition of constructions with straightedge } ZDM Subject ClassificationAbstract : .. . GThe 45 paper comma is devoted H 45 to semicolon exposition of.. constructions AMS Subject with Classification straightedge : .. 97 G 40 comma \ centerline {and compassand compass , constructible , constructible numbers numbers and their positionand their with respect position to all algebra with - respect to all algebra − } 97 H 40 period ic numbers . Although the large number of constructions may be accomplished with Key words and phrases : .. Duplication of a cube semicolon construction with compass and \ centerline { i c numbersstraightedge . \ andquad compassAlthough , one of the the known large problems number of this kind of dating constructions from Greek may be accomplished with } straightedge semicolonera constructible is duplication of number a cube . semicolon The given algebraic proof in number this paper period is elementary and self - 1 period .. Introduction \ centerline { straightedgecontained and . It compass is suitable for , oneteachers of , as the well knownas for high problems school students of . this kind dating from Greek } Constructive problemsZDM Subject have alwaysClassification been the: favorite G 45 , H subject 45 ; AMS of geometry Subject Classification period The : 97 G 40 , traditional limitation of tools used for solving geometric constructions to j ust the \ centerline { era is duplication of a cube .97\ Hquad 40 . The given proof in this paper is elementary and self − } compass and the straightedgeKey words reaches and phrases far into: the Duplication past comma of a cube although ; construction the Greeks with compass had and also been using certain other instruments period .... The well known Euclidean geome hyphen \ centerline { contained . Itstraightedge is suitable ; constructible for teachers number ; algebraic , as number well . as for high school students . } try open parenthesis III B period C period closing1 . parenthesis Introduction was based on geometric constructions performed only by compass Constructive problems have always been the favorite subject of geometry . The \ centerline {ZDM Subject Classification : \quad G 45 , H 45 ; \quad AMS Subject Classification : \quad 97 G 40 , } and straightedgetraditional comma treated limitation as equal of tools instruments used for solving in constructions geometric period constructions .. In addition to j ust comma the the straightedge maycompass be used and only the as straightedge an instrument reaches for the far construction into the past of , althougha straight the Greeks had \ centerline {97 H 40 . } line comma butalso not been for measuring using certain theother lengths instruments period Although . the The large well number known of Euclidean constructions geome - may be accomplishedtry ( III this B . way C . comma ) was based we know on geometric of three problems constructions dating performed from Greek only era by compass \ centerline {Key words and phrases : \quad Duplication of a cube ; construction with compass and } that cannot beand solved straightedge in that way , treated : .. duplication as equal instruments open parenthesis in constructions doubling closing . In parenthesis addition , the of a cube endash to find a sidestraightedge may be used only as an instrument for the construction of a straight \ centerline { straightedge ; constructible number ; algebraic number . } of a cube whoseline volume , but is not twice for thatmeasuring of a given the lengths cube semicolon . Although .. tris the ection large of number an angle of constructions endash to find one third ofmay a given be accomplished angle semicolon this squaringway , we a know circle of endash three problems to construct dating the from square Greek that era has \ centerline {1 . \quad Introduction } the same area asthat a given cannot circle be solved period in that way : duplication ( doubling ) of a cube – to find a Unsolved problemsside of athat cube kind whose initiated volume a completely is twice that new of way a given of thinking cube ; endashtris ection of an angle – Constructive problems have always been the favorite subject of geometry . The how would it beto possible find one to third prove of that a given certain angle pro ; squaring b lems could a circle not– be to s construct o lved ? .. the The square that has traditional limitation of tools used for solving geometric constructions to j ust the answer is in modernthe same algebra area and as a group given theory circle . period .. The problem of solving algebraic equations dates farUnsolved back in the problems past and of forthat a kind long initiatedtime it was a completely the central new content way of thinking – how \noindent compass and the straightedge reaches far into the past , although the Greeks had of algebra periodwould Descriptions it be possible of solving to prove certain that certain simple algebraic pro b lems equations could not had be sappeared o lved ? The as early as 2000answer years B is period in modern C comma algebra for and example group in theory Egypt . comma The .. problem during the of solving Middle algebraicDynasty comma \noindent also been using certain other instruments . \ h f i l l The well known Euclidean geome − in the London papyrusequations known dates as far Ahmess back in calculation the past and comma for a .. long and time on Babylonian it was the tiles central comma content of approximately atalgebra the same . Descriptions time period of .. solving The Babylonians certain simple were algebraic able to equationssolve quadratic had appeared \noindent try ( III B . C . ) was based on geometric constructions performed only by compass equations commaas early while as in 2000 the XVI years century B . C , Girolamo for example Cardano in Egypt comma , during Nicolo the Tartaglia Middle comma Dynasty Lodoviko , in the London papyrus known as Ahmess calculation , and on Babylonian tiles , \noindent and straightedge , treated as equal instruments in constructions . \quad In addition , the approximately at the same time . The Babylonians were able to solve quadratic straightedge may be used only as an instrument for the construction of a straight equations , while in the XVI century Girolamo Cardano , Nicolo Tartaglia , Lodoviko \noindent line , but not for measuring the lengths . Although the large number of constructions may be accomplished this way , we know of three problems dating from Greek era

\noindent that cannot be solved in that way : \quad duplication ( doubling ) of a cube −− to find a side of a cube whose volume is twice that of a given cube ; \quad tris ection of an angle −− to find one third of a given angle ; squaring a circle −− to construct the square that has the same area as a given circle .

Unsolved problems of that kind initiated a completely new way of thinking −− how would it be possible to prove that certain pro b lems could not be s o lved $ ? $ \quad The

\noindent answer is in modern algebra and group theory . \quad The problem of solving algebraic equations dates far back in the past and for a long time it was the central content of algebra . Descriptions of solving certain simple algebraic equations had appeared

\noindent as early as 2000 years B . C , for example in Egypt , \quad during the Middle Dynasty , in the London papyrus known as Ahmess calculation , \quad and on Babylonian tiles ,

\noindent approximately at the same time . \quad The Babylonians were able to solve quadratic equations , while in the XVI century Girolamo Cardano , Nicolo Tartaglia , Lodoviko 52 .. A period Grozdani acute-c sub comma G period Vojvodi c-acute \noindentFerari comma52 \ ..quad ScipioneA .del Grozdani Ferro and many $ \acute others{ werec} dealing{ , } with$ G solving . Vojvodi cubic and $ \acute{c} $ quadratic equations period \noindentFor a longF time e r a commar i , \quad .. the questionScipione concerning del Ferro the possibility and many to solve others algebraic were dealing with solving cubic and quadraticequations by equations radicals remained . open in algebra period For an algebraic equation we shall say that it is solvable by radicals if its solutions may be obtained by using rational For a long time , \quad the question concerning the possibility to solve algebraic operations open52 parenthesisA . Grozdani additionc´ G . comma Vojvodi subtractionc´ comma multiplication comma and division closing equations by radicals remained, open in algebra . For an algebraic equation we shall parenthesis and theFerari operation , Scipione del Ferro and many others were dealing with solving cubic and say that it is solvable by radicals if its solutions may be obtained by using rational of taking n to thequadratic power ofequations th roots . comma under the assumption that those operations are applied a finite number of times ontoFor coefficients a long time or , onto the functions question ofconcerning coefficients the in possibility which only to the solve algebraic equa- \noindent operations ( addition , subtraction , multiplication , and division ) and the operation aforementionedtions operations by radicals appear remained period openThis wayin algebra comma . For the an quadratic algebraic comma equation cubic we and shall biquadratic say that equations are solvableit is solvable by radicals by radicals period if It its was solutions to be expected may be obtained that the by equations using rational of the \noindent of taking $ n ˆ{ th }$ roots , under the assumption that those operations are applied a finite fifth degree andoperations of higher degrees( addition would , subtraction be solvable , multiplication the same way , comma and division but it ) turned and the operation number of times onto coefficients or onto functions of coefficients in which only the out to be impossibleof taking periodnth roots , under the assumption that those operations are applied a finite aforementioned operations appear . This way , the quadratic , cubic and biquadratic The initial foundationsnumber of solvability times onto of coefficients algebraic equations or onto functions were est ablished of coefficients in which only the equations are solvable by radicals . It was to be expected that the equations of the by the French mathematicianaforementioned E operations period Galois appear comma . This by way connecting , the quadratic the solvability , cubic of and algebraic biquadratic fifth degree and of higher degrees would be solvable the same way , but it turned equations by radicalsequations with are group solvable theory by radicalsperiod The . It demand was to be that expected the roots that of the the equations algebraic of the fifth out to be impossible . equation f opendegree parenthesis and of x higher closing degrees parenthesis would = be0 may solvable be expressed the same by way coefficients , but it turned of that out equation to be comma .... by using impossible . The initial foundations of solvability of algebraic equations were est ablished rational operationsThe and initial taking foundations n to the power of solvability of th roots of isalgebraic expressed equations as a demand were est that ablished the field by the by the French mathematician E . Galois , by connecting the solvability of algebraic F has to be a componentFrench mathematician of a radical extension E . Galois field , by of connecting K period the.. When solvability this demand of algebraic is equations equations by radicals with group theory . The demand that the roots of the algebraic fulfilled one canby say radicals that the with given group algebraic theory equation . The demand is solvable that bythe radicals roots of period the algebraic Galois has determinedequation the criterionf(x) of solvability= 0 may of be algebraic expressed equations by coefficients that may of that be solved equation , by using \noindent equation $ f ( x ) = 0 $ may be expressed by coefficients of that equation , \ h f i l l by using only by radicalsrational comma operations and such criterion and taking is basednth roots on a is fact expressed that the as corresponding a demand that group the field F has of that equationto is be solvable a component period of a radical extension field of K. When this demand is fulfilled one \noindent rational operations and taking $ n ˆ{ th }$ roots is expressed as a demand that the field The general algebraiccan say equation that the given algebraic equation is solvable by radicals . Galois has determined $ F $ has to be a component of a radical extension field of $K . $ \quad When this demand is Line 1 n Line 2the sum criterion a sub k of x tosolvability the power of ofalgebraic k = 0 arrowdblleft-arrowdblright equations that may be solved a sub only 0 plus by radicals a sub 1 ,x plus a fulfilled one can say that the given algebraic equation is solvable by radicals . Galois sub 2 x to the powerand of such 2 plus criterion times times is based times on plus a fact a sub that n the x to corresponding the power of n group = 0 comma of that open equation parenthesis is n has determined the criterion of solvability of algebraic equations that may be solved greater 4 closing parenthesissolvable . Line 3 k = 0 only by radicals , and such criterion is based on a fact that the corresponding group of a degree higher than four with independentThe real general coefficients algebraic a sub equation k open parenthesis k = 0 comma 1 comma 2of comma that period equation period period is solvable comma n closing . parenthesis is not solvable by radicals period .. Many great mathematicians comma such as for examplen L period \ centerline {The general algebraic equation } Euler comma .. thoughtX that it was possible comma .. but Ruffini and Abel disputed that at the a xk = 0 ⇐⇒ a + a x + a x2 + ··· + a xn = 0, (n > 4) beginning of the XIX centuryk period .. This does0 not concern1 2 the issue ofn the existence of \ [ \abegin solution{ a ofl i g an n e algebraic d } n \\ equation of the n to the power of th degree period .. That wask = proved 0 by Gauss \insum 1 799 ina his{ PhDk } thesisx ˆperiod{ k } Abel= quoteright 0 \ si andf f Ruffinia { quoteright0 } + s problem a { was1 } whetherx + that equation a { 2 } of a degree higher than four with independent real coefficients a (k = 0, 1, 2, ..., n) is x ˆcould{ 2 be} s+ o lved\cdot by radicals\cdot and taking\cdot n to the+ power a of{ thn roots} ?x .. ˆ The{ n path} k= to the 0 solution , of ( that n > not solvable by radicals . Many great mathematicians , such as for example L . Euler 4problem ) \\ led to the development of modern algebra and group theory period , thought that it was possible , but Ruffini and Abel disputed that at the beginning k2 period = .. 0 Solving\end{ thea l i problem g n e d }\ of] duplication of a cube of the XIX century . This does not concern the issue of the existence of with compass and straightedge only a solution of an algebraic equation of the nth degree . That was proved by Gauss in According to a legend comma the problem of duplication of a cube arose when the 1 799 in his PhD thesis . Abel ’ s and Ruffini ’ s problem was whether that equation \noindentGreeks of Athensof a degree sought assistance higher from than the four Oracle with at Delphi independent in order to real appease coefficients $ a { k } could be s o lved by radicals and taking nth roots ? The path to the solution of that (k=0,1,2,...,n)$the Gods to grant relief from a devastating plague epidemic period The Oracle told them problem led to the development of modern algebra and group theory . isthat not todo solvable so they had by to radicals double the . size\quad of theMany alt ar greatof Apollo mathematicians which was in the , such as for example L . 2 . Solving the problem of duplication of a cube Eulershape of , a\quad cube periodthought .. Their that first it attempt was possible at doing that , was\quad a misunderstandingbut Ruffini of and the Abel disputed that at the with compass and straightedge only beginningproblem : .. of they the doubled XIX the century length of . the\quad sidesThis of the does cube period not concern .. That comma the howeverissue of comma the gave existence of According to a legend , the problem of duplication of a cube arose when the Greeks \noindent a solutionof Athens sought of an assistance algebraic from equation the Oracle atof Delphi the in $ order n ˆ{ toth appease}$ the degree Gods to. \quad That was proved by Gauss in 1 799 in hisgrant PhD relief thesis from a devastating . Abel ’ plague s and epidemic Ruffini . The ’ s Oracle problem told them was that whether to do sothat equation they had to double the size of the alt ar of Apollo which was in the shape of a cube \noindent could. be Their s firsto lved attempt by at radicals doing that and wastaking a misunderstanding $ n ˆ{ th of the}$ problem r o o t s : $ they ? $ \quad The path to the solution of that problem led todoubled the the development length of the of sides modern of the cube algebra . That and , however group , theory gave .

\ centerline {2 . \quad Solving the problem of duplication of a cube }

\ centerline { with compass and straightedge only }

According to a legend , the problem of duplication of a cube arose when the Greeks of Athens sought assistance from the Oracle at Delphi in order to appease the Gods to grant relief from a devastating plague epidemic . The Oracle told them that to do so they had to double the size of the alt ar of Apollo which was in the shape of a cube . \quad Their first attempt at doing that was a misunderstanding of the problem : \quad they doubled the length of the sides of the cube . \quad That , however , gave On the ancient problem of duplication of a cube .. 53 \ hspacethem eight∗{\ f t i imesl l }On the the original ancient volume problem since open of parenthesis duplication 2 x closing of parenthesisa cube \quad to the53 power of 3 = 8 x to the power of 3 period .. In modern notation comma in \noindentorder to fulfillthem the eight instructions t imes of the the Oracle original commaone volume must gosince from a $ cube ( of 2 side x x units ) ˆ{ 3 } = 8to x one ˆ{ of3 y} units. where $ \quad y to theIn power modern of 3 notation = 2 x to the , power in of 3 comma so that y = x surd of 3 2 sub periodorder Thus to comma fulfill essentially the instructions comma given a unit of the Oracle , one must go from a cube of side $ x $ u n i t s length comma they needed to construct a line segmentOn of the length ancient y = problem x surd of of duplication 3 2 sub period of a cube Now53 there are ways of doing thisthem but eight not tby imes using the only original the compass volume and since an (2 unmarkedx)3 = 8x3 straightedge. In modern notation , in \noindent to one of $ y $ units where $ y ˆ{ 3 } = 2 x ˆ{ 3 } , $ so that endash which wereorder the to only fulfill tools the instructions allowed in classical of the Oracle Greek , geometry one must period go from Constructive a cube of side pathx units $ y = x \surd{ 3 2 } { . }$ Thus , essentially√ , given a unit of solving this problem was known to ancient3 Greeks3 unless we would not demand to one of y units where y = 2x , so that y = x 32. Thus , essentially√ , given a unit limitation of constructionlength , they on needed use of to only construct compass a line and segment straightedge of length periody = Byx using32 Now hard there comma are ways \noindent length , they needed to construct a line segment of. length $ y = x 90 degree angleof and doing movable this but cross not hyphen by using shaped only the rectangle compass comma and an it unmarkedis possible straightedge to construct – a which \surd{ 3 2 } { . }$ Now there are side of a cube whosewere the volume only is tools twice allowed larger in than classical the volume Greek of geometry a cube with . Constructive unit path of solving ways of doing this but not by using only the compass and an unmarked straightedge side period Forthis detailed problem description was known see the to ancient book open Greeks square unless bracket we would 1 closing not square demand bracket limitation period of −− which were the only tools allowed in classical Greek geometry . Constructive path For the sake ofconstruction simplicity comma on use let of us only assume compass that and the straightedge given cube has . By the using side hard length , 90 degree angle of solving this problem was known to ancient Greeks unless we would not demand equal to the unitand of movable measurement cross - of shaped length rectangle period .. , Now it is thepossible problem to construct of duplication a side of of a cube whose limitation of construction on use of only compass and straightedge . By using hard , cube may be expressedvolume is in twice the following larger than way the : volume of a cube with unit 90 degree angle and movable cross − shaped rectangle , it is possible to construct a For a given cubeside with . For a s detailed ide of the description unit of measurement see the book find [ 1 the] . s ide side of a cube whose volume is twice larger than the volume of a cube with unit of a cube whose volumeFor the is twicesake of that simplicity of the given , let cubeus assume period that the given cube has the side length The problem isequal reduced to theto the unit solving of measurement of the cubic of equation length . x to Now the power the problem of 3 = of2 period duplication We shall of a \noindent side . For detailed description see the book [ 1 ] . show that this problemcube may cannot be expressed be s o lved in the b y following using compass way : and s traightedge only comma i period e period thatFor the rootsa given of cube polynomial with a sp ide open of parenthesis the unit of measurementx closing parenthesis find the = side x to the power of 3 For the sake of simplicity , let us assume that the given cube has the side length minus 2 are not constructible periodof a cube whose volume is twice that of the given cube . equal to the unit of measurement of length . \quad Now the problem of duplication of a To show this commaThe we problem shall define is reduced the term to the of constructible solving of the and cubic algebraic equation numbersx3 = 2 period. We shall show cube may be expressed in the following way : Then comma wethat shall this observe problem thecannot algebraic be s equation o lved b yx usingto the compass power of and 2 minus s traightedge 2 = 0 for only whose, root surd of 2 we shall i . e . that the roots of polynomial p(x) = x3 − 2 are not constructible . \ centerlineshow that is{ anFor irrational a given comma cube algebraic with a and s constructible ide of the number unit period of measurement Afterward we shallfind the s ide } To show this , we shall define the term of constructible and algebraic√ numbers . show that thereThen is no , analogy we shall for observe the cubic the algebraic equation commaequation i periodx2 − 2 e= period 0 for whose .. the root equation2 we x to shall the power \ centerline { of a cube whose volume is twice that of the given cube . } of 3 minus 2 = 0 show that is an irrational , algebraic and constructible number . Afterward we shall has one root thatshow is a that real there number is no and analogy two conjugate for the cubic complex equation roots , comma i . e . where the equation the realx3 − 2 = 0 The problem is reduced to the solving of the cubic equation $ x ˆ{ 3 } = 2 . $ root surd of 3 2has is irrational one root that comma is a algebraic real number comma and but two not conjugate a constructible complex number roots , where period the real We s h a l l √ 2 period 1 periodroot .. On32 constructible is irrational , numbers algebraic , but not a constructible number . show that this problem cannot be s o lved b y using compass and s traightedge only , As already mentioned comma the quest2 . for 1 . the answer On constructible concerning the numbers possibility to solve algebraic equationsAs already by radicals mentioned has ,led the us quest to the for answer the answer for the concerning question of the possibility to solve \noindent i . e . thatthe roots of polynomial $p ( x ) = xˆ{ 3 } − solving particularalgebraic geometric equations problems by by radicals using thehas compass led us to and the straightedge answer for only the question period of solving 2 $ are not constructible . In order to deriveparticular the proof geometric on the duplication problems by of using a cube the by compass applying and algebra straightedge comma it only is . In order necessary to convertto derive that the geometric proof on problem the duplication to the language of a cube of algebra by applying period algebra Each geo , it hyphen is necessary \ hspace ∗{\ f i l l }To show this , we shall define the term of constructible and algebraic numbers . metric constructionto convert may be that reduced geometric to the problem following to form the : language given a certain of algebra numbers . Each geo - metric of line segmentsconstruction a comma b comma may be c reduced comma period to the period following period form .. :and given looking a certain for one numbers or more of line line segments \noindent Then , we shall observe the algebraic equation $ x ˆ{ 2 } − 2 = 0 $ x comma y commasegments z commaa, period b, c, ... periodand period looking for one or more line segments x, y, z, ... Geometric for whose root $\surd{ 2 }$ we s h a l l Geometric constructionconstruction is than is than reduced reduced to solving to solving an algebraic an algebraic problem problem : : show that is an irrational , algebraic and constructible number . Afterward we shall bullet Determining• theDetermining connection the comma connection i period , i e . period e . equation equation between between the the wanted wanted measure measurex and x and the given measuresthe given a comma measures b commaa, b, c, c ...comma; period period period semicolon \noindent show that there is no analogy for the cubic equation , i . e . \quad the equation bullet Determining the unknown• Determining measure the x by unknown solving measure that equationx by solving semicolon that equation ; $ x ˆ{ 3 } − 2 = 0 $ bullet Determining• whetherDetermining that solution whether is that arrived solution at through is arrived a procedure at through that a corprocedure hyphen that cor - has one root that is a real number and two conjugate complex roots , where the real responds to theresponds construction to the performed construction by compass performed and by straightedge compass and period straightedge . Let us define the termLet us of define a constructible the term of number a constructible period .. number We shall . say We that shall a real say that a real number \noindent root $\surd{ 3 2 }$ is irrational , algebraic , but not a constructible number . number b is constructibleb is constructible comma, if it is is possible possible , comma in a definite in a definite number number of steps of , to steps construct comma , to construct comma with compass and straightedge , a segment of the length | b | . \ centerlinewith compass{2 and . 1straightedge . \quad commaOn constructible a segment of the numbers length bar} b bar period As already mentioned , the quest for the answer concerning the possibility to solve algebraic equations by radicals has led us to the answer for the question of solving particular geometric problems by using the compass and straightedge only . In order to derive the proof on the duplication of a cube by applying algebra , it is necessary to convert that geometric problem to the language of algebra . Each geo − metric construction may be reduced to the following form : given a certain numbers oflinesegments $a , b , c , . . .$ \quad and looking for one or more line segments $x,y,z,...$ Geometric construction is than reduced to solving an algebraic problem :

$ \ bullet $ Determining the connection , i . e . equation between the wanted measure $ x $ and thegivenmeasures $a , b , c , . . . ;$

\ centerline { $ \ bullet $ Determining the unknown measure $ x $ by solving that equation ; }

$ \ bullet $ Determining whether that solution is arrived at through a procedure that cor − responds to the construction performed by compass and straightedge .

Let us define the term of a constructible number . \quad We shall say that a real number $ b $ is constructible , if it is possible , in a definite number of steps , to construct ,

\noindent with compass and straightedge , a segment of the length $ \mid b \mid . $ 54 .. A period Grozdani acute-c sub comma G period Vojvodi c-acute \noindentLet us notice54 the\quad connectionA . Grozdani between some $ of\acute the simplest{c} algebraic{ , }$ operations G . Vojvodi $ \acute{c} $ and elementary geometric constructions comma .. where we shall assume that the given Letlengths us notice a and b arethe measured connection according between to the given some quotedblleft of the simplest unit quotedblright algebraic measure operations comma and that r and elementary geometric constructions , \quad where we shall assume that the given lengthsrepresents $ any a rational $ and number $ b period $ are measured according to the given ‘‘ unit ’’ measure , and that $ r $ 1 period .. Construction54 A . Grozdani of a linec´ segmentG . Vojvodi thatc´ has the length a plus b or a minus b represents any rational number, . Let us spot an arbitraryLet us point notice O the on connection an arbitrary between line period some .. of Construct the simplest the algebraic line operations and segment OA thatelementary has the length geometric a period constructions Construct , point where B on we that shall line assume comma that so the that given the lengths line a \ centerline {1 . \quad Construction of a line segment that has the length $ a + segment AB hasand theb lengthare measured b period according Then comma to the OB given = a “ plus unit b ” open measure parenthesis , and that Figurer represents 1 closing parenthesisany b $ or $ a − b $ } period rational number . Fig period 1 .. Fig period1 . 2 Construction of a line segment that has the length a + b or a − b Let us spot an arbitrary point $ O $ on an arbitrary line . \quad Construct the line The line segment aLet minus us spot b open an parenthesisarbitrary point a greaterO on anb closing arbitrary parenthesis line . Construct is constructed the line in a segment similar manner segment $OA$ that has the length $ a . $ Construct point $ B $ on that line , so that the line period .. On an OA that has the length a. Construct point B on that line , so that the line arbitrary line commasegment spotAB ahas point the O length periodb. ConstructThen ,OB the= linea + b segment( Figure OA 1 ) that . has the length a period \noindentConstructsegment point B in the $AB$ opposite hasthe direction on length that line $b comma .$ so that Then the line $ segment , OB = a + b ($ Figure1) . Fig . 1 Fig . 2 AB has the length b periodThe Then line comma segment OBa =− ab( minusa > b) b is open constructed parenthesis in a Figure similar 2 manner closing parenthesis . On an period 2 period .. Constructionarbitrary lineof a , line spot segment a point thatO. Construct has the length the line ra segment OA that has the length a. \ centerline { Fig . 1 \quad Fig . 2 } In order to constructConstruct ra we point simplyB in apply the opposite r times a direction plus a plus on timesthat line times , so times thatplus the line a comma segment where r is a natural numberAB periodhas the length b. Then ,OB = a − b( Figure 2 ) . \ hspace ∗{\ f i l l }The line segment $ a − b ( a > b ) $ is constructed in a similar manner . \quad On an 3 period .. Construction of a2 line . segmentConstruction that hasof a the line length segment a slash that b has the length ra In order to constructIn order a slash to b construct comma wera markwe simply OB = apply b andr times OA =a a+ ona + the··· arms+ a, where of anyr is a natural \noindent arbitrary line , spot a point $ O . $ Construct the line segment $OA$ angle with thenumber vertex in . point O comma and on line OB we mark the segment OD = 1 period that has the length $ a . $ Through D comma we construct3 . a Construction straight line of parallel a line tosegment line AB that that has intersects the length OAa/b at Construct point $ B $ in the opposite direction on that line , so that the line segment point C period Then OCIn will order have to the construct length aa/b, slashwe b mark openOB parenthesis= b and FigureOA = a 3on closing the arms parenthesis of any period Indeed comma fromangle the similaritywith the vertex in point O, and on line OB we mark the segment OD = 1. \noindent $AB$ hasthelength $b .$ Then $ , OB = a − b ( $ of triangles OABThrough and OCDD, we it followsconstruct that a straight OB : OD line = parallel OA : OC to linecommaAB ithat period intersects e periodOA bat : 1 = a : OC Figure 2 ) . comma point C. Then OC will have the length a/b( Figure 3 ) . Indeed , from the similarity wherefrom we canof triangles see thatOAB OC =and a slashOCD b periodit follows that OB : OD = OA : OC, i . e .b : 1 = a : OC, \ centerline {2 . \quad Construction of a line segment that has the length $ ra $ } Fig period 3 wherefrom we can see that OC = a/b. In order to construct $ ra $ we simply applyFig . 3 $ r $ times $ a + a + \cdot \cdot \cdot + a ,$ where $r$ isa natural number .

\ centerline {3 . \quad Construction of a line segment that has the length $ a / b $ }

\ hspace ∗{\ f i l l } In order to construct $a / b ,$ wemark $OB = b$ and $OA = a$ on the arms of any

\noindent angle with the vertex in point $ O , $ and on line $OB $ we mark the segment $ OD = 1 . $ Through $ D , $ we construct a straight line parallel to line $ AB $ that intersects $ OA $ at

\noindent point $C . $ Then $OC$ will havethe length $a / b ($ Figure 3 ) . Indeed , from the similarity

\noindent of triangles $OAB$ and $OCD$ it follows that $OB : OD = OA : OC ,$i.e$. b : 1 = a : OC ,$

\noindent wherefromwecan see that $OC = a / b . $

\ centerline { Fig . 3 } On the ancient problem of duplication of a cube .. 55 \ hspace4 period∗{\ ..f Construction i l l }On the of ancient a line segment problem that has of the duplication length ab of a cube \quad 55 As we did thus far comma we mark the line segments OA = a and OB = b on the \ centerlinearms of an angle{4 . whose\quad vertexConstruction is in point O period of a Online line segment OA comma that we mark has the the unit length segment $ ab $ } OC period .. Draw a straight line through points C and B comma and after that a straight line Aswethrough did A thus comma far that , is wemark parallel to the the straight line segments line which will $OA intersect = the a second $ and arm $OB = b $ on the of the angle in some point D period .... Then OD = abOn open the ancient parenthesis problem Figure of duplication 4 closing of a parenthesis cube 55 period arms of an angle whose vertex is in point $O . $ On line $OA , $ we mark the unit segment .... The proof of validity of 4 . Construction of a line segment that has the length ab $ OC . $ \quad Draw a straight line through points $ C $ and $ B , $ and after that a straight line this construction followsAs we did from thus the far similarity , we mark of triangles the line segments OAD andOA OCB= a openand parenthesisOB = b on theFigure arms 4 closing through $ A , $ that is parallel to the straight line which will intersect the second arm parenthesis periodof an angle whose vertex is in point O. On line OA, we mark the unit segment OC. Indeed commaDraw from the a straight mentioned line similaritythrough points it followsC and thatB, OCand : after OA = that OB a : straight OD comma line throughi period eA, period \noindent of the angle in some point $D . $ \ h f i l l Then $OD = ab ($ OD = OA timesthat OB is period parallel So to comma the straight it is indeed line which true that will intersectOD = a times the second b period arm Figure 4 ) . \ h f i l l The proof of validity of Fig period 4 of the angle in some point D. Then OD = ab( Figure 4 ) . The proof of validity of From the aforementionedthis construction constructions follows it from appears the similarity that .. quotedblleft of triangles notdef-aOAD and tionalOCB quotedblright( Figure 4 ) .. . lgebraic \noindent this construction follows from the similarity of triangles $ OAD $ and operations of addingIndeed comma , from .. the subtracting mentioned comma similarity .. multiplying it follows thatand dividingOC : OA of= knownOB : measuresOD, i . e . $OCB ( $ Figure 4 ) . may be conductedOD by= OA geometric· OB. So constructions , it is indeed period true that .. AOD set of= measuresa · b. that may be calculated in that way form the so hyphen called numerical field comma i period e period a set of numbers \noindent Indeed , from the mentioned similarityFig . 4 it follows that $ OC : OA = such From the aforementioned constructions it appears that “ notdef − a tional ” lge- OB : OD ,$ i.e. that the applicationbraic ofoperations rational operationsof adding , on subtracting two or more , members multiplying of that and set dividing comma of known mea- $ OD = OA \cdot OB . $ So , it is indeed true that $OD = a \cdot results in a numbersures that again belongs to that set period .. Rational comma real and complex b . $ numbers createmay such be fields conducted period by geometric constructions . A set of measures that may be Introduction ofcalculated the construction in that of way a square form the root so takes- called us numerical out the fields field arrived , i . e . a set of numbers such \ centerline { Fig . 4 } at in that mannerthat period the application of rational operations on two or more members of that set , results 5 period .. Constructionin a number of a that line again segment belongs that has to that the length set . surd Rational of a , real and complex numbers \noindent From the aforementioned constructions it appears that \quad ‘ ‘ $ notdef−a $ We claim that ifcreate the given such fieldsline segment . is of length a comma then the line of length surd of a t i o n a l ’ ’ \quad l g e b r a i c may be constructedIntroduction by compass of and the straightedge construction period of a square Apply root the line takes segments us out OAthe fields= a arrived at operationsand AB = 1 on of a adding straight line , \quad period ..subtracting Draw a circle whose , \quad diametermultiplying equals to the and line dividing of known measures in that manner . √ segment OB comma i period5 e . period Construction a circle whose of a line center segment is in the that midpoint has the length of line segmenta OB and \noindent may be conducted by geometric constructions . \quad A set of measures√ that may be whose radius equals OBWe slashclaim 2 that comma if the and given construct line segment a line perpendicularis of length a, tothen line the OB line from of length the a calculated in that way form the so − called numerical field , i . e . a set of numbers such point A that intersectsmay be theconstructed circle in pointby compass C period and .... straightedge Line segment . Apply AC = surdthe line of a segments sub periodOA ....= Thea proof that the application of rational operations on two or more members of that set , follows from theand similarityAB = of1 on triangles a straight OAC line and . ABC Draw period a circle .... Indeed whose comma diameter on equals the basis to of the line results in a number that again belongs to that set . \quad Rational , real and complex this similarity itsegment followsOB, that OAi . e : . AC a circle = AC whose : AB comma center is i period in thee midpoint period OA of line = AC segment to the powerOB and of 2 comma wherefromnumbers it create such fields . whose radius equals OB/2, and construct a line perpendicular to line√OB from the follows that ACpoint = surdA that of OA intersects sub period the Therefore circle in point commaC. ACLine = surd segment of a subAC period= a The proof Introduction of the construction of a square root takes us out the. fields arrived Note that eachfollows geometric from construction the similarity with of a triangles compassOAC and straightedgeand ABC. in a Indeed , on the basis of at in that manner . Euclid quoteright s plane always boils down to solving the following basic simpler2 t asks comma not this similarity it follows√ that OA : AC = AC√: AB, i . e .OA = AC , wherefrom it necessarily in followingfollows that orderAC : = OA Therefore , AC = a \ centerline {5 . \quad Construction. of a line segment. that has the length $\surd{ a }$ bullet .. ConstructNote a straight that each line geometrictrough two construction points semicolon with a compass and straightedge in a Euclid } ’ s plane always boils down to solving the following basic simpler t asks , not necessarily in following order : \ hspace ∗{\ f i l l }We claim that if the given line segment is of length $ a , $ then the line of length • Construct a straight line trough two points ; $\surd{ a }$

\noindent may be constructed by compass and straightedge . Apply the line segments $ OA = a $ and $AB = 1$ ona straight line . \quad Draw a circle whose diameter equals to the line segment $ OB , $ i . e . a circle whose center is in the midpoint of line segment $ OB $ and

\noindent whose radius equals $ OB / 2 , $ and construct a line perpendicular to line $ OB $ from the

\noindent point $A$ that intersects the circle in point $C . $ \ h f i l l Line segment $ AC = \surd{ a } { . }$ \ h f i l l The proof

\noindent follows from the similarity of triangles $ OAC $ and $ ABC . $ \ h f i l l Indeed , on the basis of

\noindent this similarity it follows that $OA : AC = AC : AB ,$ i . e $ . OA = AC ˆ{ 2 } , $ wherefrom it

\noindent follows that $AC = \surd{ OA } { . }$ Therefore $ , AC = \surd{ a } { . }$

Note that each geometric construction with a compass and straightedge in a Euclid ’ s plane always boils down to solving the following basic simpler t asks , not necessarily in following order :

\ centerline { $ \ bullet $ \quad Construct a straight line trough two points ; } 56 .. A period Grozdani acute-c sub comma G period Vojvodi c-acute \noindentFig period56 5 \quad A . Grozdani $ \acute{c} { , }$ G . Vojvodi $ \acute{c} $ bullet From a given point comma as a center comma construct a circle semicolon \ centerlinebullet Find{ pointsFig representing. 5 } the intersection of two given circles semicolon bullet Find points representing the intersection of a given circle and a straight line \ centerlinedefined by two{ $ points\ bullet semicolon$ From a given point , as a center , construct a circle ; } bullet Find points56 representingA . Grozdani thec´ G intersection . Vojvodi c´ of two straight lines comma each defined by \ centerline { $ \ bullet $ Find, points representing the intersection of two given circles ; } two points comma Fig . 5 where the basic elements endash• From point a comma given point straight , as line a center and a , circle construct are treated a circle as ; known if $ \ bullet $ Find points representing the intersection of a given circle and a straight line they are specified at the beginning• Find points or if they representing were constructed the intersection within of a previous two given circles ; defined by two points ; step period • Find points representing the intersection of a given circle and a straight line defined Let us assume thatby two there points is only ; one given element comma unit segment 1 period As the set $ \ bullet $ Find points representing the intersection of two straight lines , each defined by of rational numbers• isFind closed points with representing regard to all the rational intersection operations of two comma straight i period lines e , period each defined operations by two p o i n t s , of adding commatwo subtracting points , comma multiplying and dividing two rational numbers open parenthesis excluding division by zerowhere closing the parenthesis basic elements comma – point again , straight result line in a and rational a circle number are treated comma as known it turns if outthey that all \noindent where the basic elements −− point , straight line and a circle are treated as known if rational are specified at the beginning or if they were constructed within a previous step . they are specified at the beginning or if they were constructed within a previous numbers may be constructedLet us assume by compass that there and is straightedge only one given period element .. Any , unit set segment of numbers 1 . As the set of step . with this characteristicrational numbersof being closed is closed in withregard regard to four to rational all rational operations operations is , i . e . operations referred to as aof numerical adding , fieldsubtracting period .., multiplying As we have and shown dividing that it two is possible rational to numbers construct ( excluding Let us assume that there is only one given element , unit segment 1 . As the set surd of k with compassdivision by and zero straightedge ) , again result comma in the a rational only thing number that , remains it turns is out to that check all whether rational of rational numbers is closed with regard to all rational operations , i . e . operations the extension ofnumbers the set may of rational be constructed numbers by Q compass open parenthesis and straightedge surd of . k Anysub closingset of numbers parenthesis with for any rational number kthis comma characteristic will of being closed in regard to four rational operations is referred to as \noindentinclude constructibleof adding numbers , subtracting only period , multiplying and dividing two rational numbers ( excluding √a numerical field . As we have shown that it is possible to construct Let us observe anyk with numerical compass field and F of straightedge constructible , the numbers only thing period that .. remainsLet us check is to check whether \noindentwhether itdivision would be also by possible zero )to construct, again numbers result in in the a form rational√ of p plus number q surd of , k it sub turns comma out that all rational the extension of the set of rational numbers Q( k) for any rational number k, will where p commainclude q and kconstructible come from the numbers field of only constructible . numbers F period \noindent numbers may be constructed by compass and straightedge . \quad Any set of numbers Let us pick a number kLet from us observefield F comma any numerical let us find field itF s squareof constructible root and constructnumbers . Let us check with this characteristic of being closed in regard to four rational√ operations is the field F to thewhether power it of would prime be comprising also possible of numbers to construct in the numbers form of p in plus the q form surd of ofp k+ subq k commawhere where p referred to as a numerical field . \quad As we have shown that it is possible, to construct and q are p, q and k come from the field of constructible numbers F. from F period It is easyLet to show us pick that a number adding commak from field subtractingF, let us comma find it multiplying s square root and and dividing construct two \noindent $\surd{ k }0 $ with compass and straightedge√ , the only thing that remains is to check whether numbers from thethe field F tocomprising the power of of numbers prime again in the results form inof ap + numberq k, where in thep formand pq are plus from q surdF. of k of where It is easy to show that adding , subtracting , multiplying and dividing two \noindent the extension of the set of rational numbers $ Q ( √\surd{ k } { ) }$ p and q are fromnumbers F period from Field the F field is aF subfield0 again ofresults the field in a F number to the inpower the form of primep + q periodk of If where we were to take for any rational number $ k , $ will that in p and q are from F. Field F is a subfield of the field F 0. If we were to take that in include constructible numbers√ only . the form p plus q surd of k sub comma q = 0 comma we conclude that all numbers from F are0 included in F the form p + q √k,q = 0, we conclude that all numbers from F are included in F , to the power of primeassuming comma that k is a number that does not belong to field F. \ hspace ∗{\ f i l l } Let us observe any numerical field $ F $ of constructible numbers . \quad Let us check assuming that surdFollowing of k is a allnumber those that considerations does not belong we are to ready field to F describe period the set of al l con - s tru Following all thosectible considerations numbers . Let we us are start ready from to describethe field theF0, sete . gof . al the l con field hyphen of rational numbers , \noindents tru ctiblewhether numbers period it would Let us be start also from possible the field F to sub construct 0 comma e period numbers g period in the the field form of rational of $ p +numbers q comma\surd{ k } { , }$ where $ p , q $ and $ k $ come from the field of constructible numbers $ F . $

\ hspace ∗{\ f i l l } Let us pick a number $ k $ from field $ F , $ let us find it s square root and construct

\noindent the field $Fˆ{\prime }$ comprising of numbers in the form of $ p + q \surd{ k } { , }$ where $p$ and $q$ are from $ F . $ It is easy to show that adding , subtracting , multiplying and dividing two

\noindent numbers from the field $ F ˆ{\prime }$ again results in a number in the form $ p + q \surd{ k }$ o f where

\noindent $p$ and $q$ are from $F . $ Field $F$ is a subfield of the field $ F ˆ{\prime } . $ If we were to take that in

\noindent theform $p + q \surd{ k } { , } q = 0 , $ we conclude that all numbers from $F$ are included in $Fˆ{\prime } , $

\noindent assuming that $\surd{ k }$ is a number that does not belong to field $ F . $

Following all those considerations we are ready to describe the set of al l con − s tru ctible numbers . Let us start from the field $ F { 0 } , $ e . g . the field of rational numbers , On the ancient problem of duplication of a cube .. 57 \ hspacewhich∗{\ is definedf i l l } ifOn the the unit ancient line segment problem is given of period duplication Add surd of of k sub a cube0 where\quad k sub57 0 is from F sub 0 comma but \noindentsurd of k subwhich 0 is not is period defined By that if comma the unit we construct line segment the extended is field given F sub . 1Add of constructible $\surd{ numbersk { 0 }}$ commawhere $ k { 0 }$ i s from $ F { 0 } , $ but comprising of numbers in the form of p 0 plus q 0 surd of k sub 0 sub comma where p 0 and q 0 are arbitrary \noindent $\surd{ k { 0 }}$ is not . By that , we construct the extended field numbers from F sub 0 period .. We can define a new extensionOn the ancient of the problem field Fof subduplication 1 with of numbers a cube 57 $ Fp 1{ plus1 q}$ 1 surd of of constructible k sub 1 sub comma numbers where p 1 , and q 1 are arbitrary√ numbers from F sub 1 comma where which√ is defined if the unit line segment is given . Add k0 where k0 is from F0, but k sub 1 is from F subk 1is comma not . By that , we construct the extended field F of constructible numbers , \noindent comprising0 of numbers in the form of√ $ p 01 + q 0 \surd{ k { 0 }} { , }$ but surd of k subcomprising 1 is not period of numbers .. By in repeating the form thisof p0+ procedureq0 k where commap0 we and arriveq0 are to arbitrary the field numbers F sub n comma where $p 0$ and $q 0$ are arbitrary 0, after n from F . We can define a new extension of the field F with numbers numbers from $ F 0√{ 0 } . $ \quad We can define a new1 extension of the field $ F { 1 }$ additions of squarep1 + rootsq1 k periodwhere Allp1 constructible and q1 are arbitrary numbers numbers are those from andF only, where thosek thatis from F , with numbers √ 1, 1 1 1 may be arrivedbut at withk1 suchis not a sequence . By repeating of extended this fields procedure comma , in we fact arrive those to ones the that field Fn, after n are lo cated inadditions the extended of square field F roots sub .n All period constructible numbers are those and only those that may \noindent $ p 1 + q 1 \surd{ k { 1 }} { , }$ where $p 1$ and $q It would also bebe interesting arrived at to with mention such thata sequence if we assume of extended that we fields can , construct in fact those ones that are lo 1 $ are arbitrary numbers from $ F { 1 } , $ where $ k { 1 }$ i s from $ F { 1 } all numbers of acated certain in the numerical extended field field F commaF . by using only the s traightedge open parenthesis in fact , $ n connecting two pointsIt would within also that be field interesting or finding to the mention crossing that point if we of assume two lines that we can construct within that fieldall closing numbers parenthesis of a certain we shall numerical not exit field theF, fieldby in using questiononly periodthe s traightedge It is also a fact( in comma fact that \noindent but $\surd{ k { 1 }}$ i s not . \quad By repeating this procedure , we arrive to the field with connecting two points within that field or finding the crossing point of two lines within $ F { n } , $ a f t e r $ n $ only one use ofthat compass fieldopen ) we shall parenthesis not exit the the point field acquired in question as the. It intersection is also a fact of , two that circles with or a additions of square roots . All constructible numbers are those and only those that circle and a lineonly closing one parenthesisuse of compass we can ( the expand point acquiredth e field as F periodthe intersection .... The expansion of two circles of the or field a F would may be arrived at with such a sequence of extended fields , in fact those ones that be the field F tocircle the power and a ofline prime ) we composed can expand of th numbers e field inF. the formThe of expansion p plus q ofsurd the of field k subF commawould where are lo cated in the extended0 field $ F { n } . $ √ p and q are be the field F composed of numbers in the form of p + q k, where p and q are from F. from F period 2 . 2 . Duplication of a cube It would also be interesting to mention that if we assume that we can construct 2 period 2 period .. DuplicationAs we have of a already cube mentioned , we shall start from the algebraic equation all numbers of a certain numerical field $ F , $ by using√ only the s traightedge ( in fact As we have alreadyx2 − mentioned2 = 0. commaWe shall .. we show shall that start the from root the of this algebraic equation equation, 2 is an algebraic , connecting two points within that field or finding the crossing, point of two lines x to the powerconstructible of 2 minus 2 = and 0 period irrational .... We number shall .show Afterward that the we root shall of this observe equation the equation comma surdx3− of 2 sub commawithin is an that algebraic field comma ) we shall not exit the field in question . It is also a fact , that with 2 = 0 and we shall show that this equation has one root√ that is a real number constructible andand irrational two conjugate number complex period Afterwardroots , where we theshall real observe root the32 equation is an algebraic x to the , powerirrational of 3 minus \noindent only one use of compass ( the point acquired as the intersection of two circles or a 2 = 0 and we shallnumber show , but that not this a equation constructible has one number root .that is a real number and two conjugateLet complex us define roots an comma algebraic where number the real . rootWe shall surd sayof 3 that 2 is an a certain algebraic real comma or complex irrational \noindent circle and a line ) we can expand th e field $ F . $ \ h f i l l The expansion of the field number commanumber but notx ais constructiblealge braic if number it fulfills period a certain algebraic equation of the following form $ FLet $ us would define an algebraic number period We shall say that a certain real or complex n n−1 number x is alge braic if it fulfillsanx a+ certainan−1x algebraic+ ··· equation+ a1x + a of0 = the 0 following (n ≥ 1, a formn 6= 0), \noindenta sub n x tobe the the power field of n plus $Fˆ a sub{\ n minusprime 1 x}$ to the composed power of n of minus numbers 1 plus times in the times form times of plus $a sub p +1 x plus q a\ subsurd 0{ =where 0k open} {ak parenthesis,are}$ whole where nnumbers greater $p$ . equal 1 and comma $q$ a sub n are negationslash-equal 0 closing parenthesis √ 2 √ commafrom $ F . $As 2 is a root of the algebraic equation x − 2 = 0, we may conclude that 2 is where a sub k arean algebraic whole√ numbers number period . Based on the construction No . 5 above we may conclude√ \ centerlineAs surd of 2{ is2 athat . root 2 .of2 is\ thequad also algebraic aDuplication constructible equation number x ofto the a . powercube of} It 2 remains minus 2 to = be 0 comma shown that we may2 concludeis an that surd of 2 irratio nal number . We√ shall present an interesting geometric√ way of proving the \ hspace ∗{\ f i l l }As we have already mentioned , \quad we shall start from the algebraic equation is an algebraicirrationality number period of the .. Based number on the2. Let construction us suppose No the period contrary 5 above , that√ we may2 is conclude a rational that surd of 2 isnumber also a ,constructible i . e . it may number be written period in the .... form It remains of a real to fraction be shown2 that = m surdn ( ofm, 2 n is∈ anQ , 2 2 \noindentirratio nal number$ xlcd ˆ{ ( periodm,2 n}) =.... − 1) We. By2 shall squaring= present 0 this an interesting. equation $ \ h it f geometric i follows l l We that shall waym ofshow proving= 2n . that theNote the that rootthe of this equation $ ,irrationality\surd{ of2 theinequality} { number, }$n surd < ism of < an 22n sub algebraicis period valid . Let us , suppose the contrary comma that surd of 2 is a rational number comma i periodLet quadrangle e period ....ABCD it maybe written a square in with the form sides of equal a real to fractionm( Figure surd of 6 2 ) = and m sub let n open parenthesis\noindent mconstructible commaA1,A n2 in,C Q1,C comma2 andrepresent irrational the points number on the sides . Afterward of that square we , so shall that it observe is valid that the equation $ xlcd ˆ{ open3 } parenthesis − AA$ 1 = mAA comma2 = CC n closing1 = CC parenthesis2 = n. Let = us 1 notice closing points parenthesisE,F,G periodand H Byso squaring that thequad- this equation it follows$ 2 that = m 0 torangles $ the and powerAA we of1EA shall2 =2 2and n show toCC the1 that powerFC2 are ofthis 2squares period equation with Note sides that has equal the one to rootn, and that quadrangles is a real number inequality n less mA1 lessBC1 2G nand is validC2 periodDA2H are squares with sides equal to m − n. \noindentLet quadrangleand ABCDtwo conjugate be a square withcomplex sides equal roots to m , open where parenthesis the real Figure root 6 closing $\surd parenthesis{ 3 and2 }$ let isA an sub algebraic 1 comma A , sub irrational 2 comma C sub 1 comma C sub 2 represent the points on the sides of that square commanumber so that , but it is not valid a constructible number . that AA sub 1 = AA sub 2 = CC sub 1 = CC sub 2 = n period .. Let us notice points E comma F comma G andLet H us so define an algebraic number . We shall say that a certain real or complex numberthat the quadrangles$ x $ is square alge AA braic sub 1 if EA it sub fulfills 2 and square a CC certain sub 1 FC algebraic sub 2 are squares equation with sidesof the equal following form to n comma \ [and a quadrangles{ n } x squareˆ{ n } A sub+ 1 BC a sub{ n 1 G and− square1 } Cx sub ˆ{ 2n DA sub− 2 H1 are} squares+ \cdot with sides\cdot equal to m\cdot minus n+ period a { 1 } x + a { 0 } = 0 ( n \geq 1 , a { n }\not= 0 ) , \ ]

\noindent where $ a { k }$ are whole numbers .

As $\surd{ 2 }$ is a root of the algebraic equation $ x ˆ{ 2 } − 2 = 0 , $ we may conclude that $\surd{ 2 }$ is an algebraic number . \quad Based on the construction No . 5 above we may conclude

\noindent that $\surd{ 2 }$ is also a constructible number . \ h f i l l It remains to be shown that $\surd{ 2 }$ i s an

\noindent irratio nal number . \ h f i l l We shall present an interesting geometric way of proving the

\noindent irrationality of the $ number \surd{ 2 } { . }$ Let us suppose the contrary , that $\surd{ 2 }$ is a rational

\noindent number , i . e . \ h f i l l it may be written in the form of a real fraction $\surd{ 2 } = m { n } ( m , n \ in $ Q ,

\noindent lcd $( m , n ) = 1 ) . $ Bysquaring this equation it follows that $ m ˆ{ 2 } = 2 n ˆ{ 2 } . $ Note that the inequality $ n < m < 2 n$ is valid .

Let quadrangle $ABCD$ be a square with sides equal to $m ( $ Figure 6 ) and let $ A { 1 } ,A { 2 } ,C { 1 } ,C { 2 }$ represent the points on the sides of that square , so that it is valid that $ AA { 1 } = AA { 2 } = CC { 1 } = CC { 2 } = n . $ \quad Let us notice points $E , F , G$ and $H$ so that the quadrangles $ \ square AA { 1 } EA { 2 }$ and $ \ square CC { 1 } FC { 2 }$ are squares with sides equal to $ n , $ and quadrangles $ \ square A { 1 } BC { 1 } G $ and $ \ square C { 2 } DA { 2 } H $ are squares with sides equal to $m − n . $ 58 .. A period Grozdani acute-c sub comma G period Vojvodi c-acute \noindentFig period58 6 \quad A . Grozdani $ \acute{c} { , }$ G . Vojvodi $ \acute{c} $ Note that the quadrangle square FGEH is also a square period .. Let us mark the side of that \noindentsquare withFig k comma . 6 where k in N comma k less m period .... The area of the square square ABCD may be Notepresented that as the quadrangle $ \ square FGEH $ is also a square . \quad Let us mark the side of that P sub ABCD = P sub AA sub 1 EA sub 2 plus P sub A sub 1 BC sub 1 G plus P sub CC sub 1 FC sub 2 \noindent square with $k , $ where $k \ in $ N $ , k < m . $ \ h f i l l The area of the square plus P sub A sub 258 HCA sub . Grozdani 2 D minusc´, G P . sub Vojvodi FGEHc´ $ \onsquare the basis of ABCD which $ it follows may be that m to the power of 2 = n to the power of 2 plus open parenthesis m minus n Fig . 6 Note that the quadrangle F GEH is also a square . Let us mark the side of closing parenthesisthat to the power of 2 plus n to the power of 2 plus open parenthesis m minus n closing parenthesis to\noindent the power ofpresented 2 minus k to as the power of 2 comma i period e period square with k, where k ∈ N , k < m. The area of the square ABCD may be m to the powerpresented of 2 = 2 as n to the power of 2 plus 2 open parenthesis m minus n closing parenthesis to the power\ [P of{ 2ABCD = k to} the= power P of{ 2 periodAA { ....1 In} viewEA of{ the2 fact}} that+ m toP the{ powerA { of1 2} = 2BC n to{ the1 power} G of} 2 +comma P it{ followsCC that{ 1 } FC { 2 }} + P { A { 2 } HC { 2 } D } − P { FGEH }\ ] PABCD = PAA1EA2 + PA1BC1G + PCC1FC2 + PA2HC2D − PF GEH k to the power of 2 = 2 open parenthesis m minus n closing parenthesis to the power of 2 period .. That 2 2 2 2 2 2 comma however commaon the is basis in contradiction of which it follows with the that assumptionm = n + that (m m− n) + n + (m − n) − k , i . e . 2 2 2 2 2 2 \noindentis the smalleston the nonm = hyphen basis 2n + negative2( ofm − whichn) whole= k it. number follows forIn which view that of a naturalthe $ m fact ˆ{ number that2 }m n== exists 2n n, it so ˆ{ follows2 } that+ ( 2 2 m that− mn to the ) powerk ˆ{=2 2( ofm} 2− =+n 2) n. to nThat the ˆ{ power2 , however} of+ 2 period , ( is in m contradictionTherefore− surdn with of ) 2 theˆ is{ not assumption2 } a rational − thatk number ˆ{m2is period} , $ i .With e . this we didthe show smallest that non surd - of negative 2 is an whole algebraic number comma forconstructible which a natural and number irrationaln exists so 2 2 √ number period that m = 2n . Therefore 2 is√ not a rational number . \noindentLet us now$ observe m ˆWith{ the2 cubic} this= we equation did 2 show nx to that ˆ{ the2 power2} is an+ of algebraic 3 2 minus ( , 2 constructible = m 0 to− which andn the irrational problem ) ˆ{ 2 numberof} cube= k ˆ{ 2 } . $duplication\ h f i l l isIn reduced. view period of the .. Let fact us notice that that $mˆ this equation{ 2 } has= one 2 real n solution ˆ{ 2 } and , $ it follows that 3 two complex solutionsLetus period now .... observe Indeed the comma cubic weequation can writex − the2 = equation 0 to which x to the the problempower of of 3 minus cube 2 = 0 in\noindent the form $ kduplication ˆ{ 2 } is= reduced 2 . ( Let m us notice− thatn this ) ˆ equation{ 2 } has. $ one\ realquad solutionThat and , however , is in contradiction with the assumption that 3 $ mof $ open parenthesistwo complex x minus solutions surd of 3 . 2 sub Indeed closing , we parenthesis can write open the equation parenthesisx − x2 to = the 0 in power the form of 2 plus x √ 2 √ √ 2 surdis of the 3 2 smallest plus surdof ( ofx non− 3 2 to32−)( thexnegative+ powerx 32 of + 2whole sub32 closing) = number 0, wherefrom parenthesis for which we = can 0 comma asee natural that wherefrom the solutions number we can are $ see n that $ the exists so solutions are \noindent that $ m ˆ{ 2 } = 2 n ˆ{ 2 } . $ Therefore $\surd{ 2 }$ is not a rational number . we shall from and x suband 2x comma 3 to x sub√ 1 = surd√ of 3 2 and get = minus√ x sub 2 comma 3 x surd of 3 2 √2,3 2 . √ weshallx = 32andget=−x2,3x 321±i 3 2 seethatx1 = 32 √ 32±radical−minus−radical = √ 4322 .Bysortingoutthe expression 1 plusminux i surd of 3 2 see1 that x sub 1 = surd of 3 2 from 2 period to = to the power√ − of = surd of 3 2 to the− 32 radical−minus334. ,we1 further, == 32 andx2,3 3222 ± Finally powerWith of this minus we radicalbig did show sub that and x sub $\surd 2 comma{ 2 3} to$ the is power an algebraic of surd of 3 2 , plusminux constructible radical-minus-radical and irrational number . √ √ minus from = to 3 2 2As to the32 power is a root of 2 of surd the of algebraic 3 2 to the equation power ofx3 4− 32 sub = 0 plusminux, it follows to that the power32 is of an 2 to the power of 2 radical-minusalgebraic of number 3 3 4 sub . period Let us to show the that power is not of period a rational By sorting number sub . Finally Let us suppose sub comma the to the powerLet us of out now the observe sub we 1 the to the cubic power equationof expression $further x ˆ{ comma3 } − 2 = 0 $ to which the problem of cube duplicationAs surd of 3 2 is a reduced root of the . algebraic\quad equationLet us x notice to the power that of this 3 minus equation 2 = 0 comma has it one follows real that solution surd and of 3 2 is an \noindentalgebraic numbertwo complex period .. solutions Let us show that . \ h is f not i l l aIndeed rational number , we can period write .. Let the us suppose equation the $ x ˆ{ 3 } − 2 = 0$ intheform

\noindent o f $ ( x − \surd{ 3 2 } { )( } x ˆ{ 2 } + x \surd{ 3 2 } + \surd{ 3 2 ˆ{ 2 }} { ) } = 0 , $ wherefrom we can see that the solutions are

\ [ we s h a l l ˆ{ and x { 2 , 3 }} { x { 1 } = \surd{ 3 2 }} and{ get { = − }} x { 2 , 3 }{ x \surd{ 3 2 }} 1 \pm i \surd{ 3 } 2 se e that x { 1 } = \surd{ 3 2 }ˆ{ 2 . } { = ˆ{ = }\surd{ 3 2 }ˆ{−{ \ sqrt }}ˆ{\surd{ 3 2 }\pm r a d i c a l −minus−r a d i c a l } { and x { 2 , 3 }} − ˆ{ = } {3{ 2 } 2 ˆ{ 2 }}\surd{ 3 2 }ˆ{ 4 3 } {\pm }ˆ{ 2 ˆ{ 2 }} r a d i c a l −minus{ 3 3 4 }ˆ{ . By s o r t i n g } { . } { F i n a l l y }ˆ{ out the } { , } { we }{ 1 }ˆ{ e x p r e s s i o n } f u r t h e r , }\ ]

As $\surd{ 3 2 }$ is a root of the algebraic equation $ x ˆ{ 3 } − 2 = 0 ,$ it follows that $\surd{ 3 2 }$ i s an algebraic number . \quad Let us show that is not a rational number . \quad Let us suppose the On the ancient problem of duplication of a cube .. 59 \ hspacecontrary∗{\ commaf i l l } thatOn the surd ancientof 3 2 is a rationalproblem number of duplication period It would of follow a cube that it\quad could be59 written in this to the power of the form that 2 to the power of of = to the power of real p q to the power of 3 to the power\noindent of 3 fractioncontrary comma , i period that e period$\surd 2{ q to3 the 2 power}$ of is 3 = a surd rational of 3 2 sub number p to the . power It would of 3 sub follow period that it could be written in to the power of = p sub q comma open parenthesis p comma Since q in the Q left sub hyphen to the power of comma\noindent sub hand$ tot h thei s ˆpower{ the of lcd form to the} power$ that of open $ parenthesis 2 ˆ{ o f } p comma= ˆ{ qr closinge a l { parenthesisp }} { q = ˆ side{ 3 of}}ˆ{ 3 } f r a c t i o n { , i . e . 2 q ˆ{ 3 }} = \surd{ 3 2 } { p ˆ{ 3 }}ˆ{ = } { . } to the power of 1 closing parenthesis sub the to the powerOn of the period ancient to problem the power of duplication of It would of a followcube equation59 is p { q } , ( p , {√Since } q \ in { the } Q { l e f t }ˆ{ , } { − }ˆ{ l c d } { hand }ˆ{ ( an from even contrary , that 32 is a rational number . It would follow that it could be written in p , } q ) = { s i d e } o f ˆ{ 1 ) }ˆ{ . } { the }ˆ{ I t } would follow { equation }$ number comma it followstheform that alsoof rightrealp hyphen3 hand side of√ equation= is also an even number, lcd(p, comma i period this that 2 = 3 fraction 3 = 32 3 .pq, (p, Sinceq ∈ theQ q) = ei s period an $ from { even }$ q ,i.e.2q p left− hand 1). It that p = 2 s commasideof sthe in Qwouldfollowequation period By including is that an from in theeven preceding equation comma equation 2 q to the power\noindent of 3 = pnumber tonumber the power , it , it of follows 3 that that also right also - hand right side− ofhand equation side is also of an equation even number is , also i . e an even number , i . e . 3 3 that$pbecomes 2 q to =. the that power2p = s 2 ofs, s3∈ = ,Q 8 s s. to By the including\ in power$ ofQ that 3 . comma in By the including preceding i period e equation period that q in , to equation the the power preceding 2q of= 3p = 4 equations to , equation 3 3 3 3 the$ 2 power q of ˆ{ 3 period3 }becomes= .. From p2q ˆ here={ 83s comma,}$i . e it.q turns= 4s out. thatFrom q here is also , it an turns even out that q is also an even becomesnumber comma $2number which qˆ is{ , inwhich3 contradiction} is= in contradiction 8 with s theˆ{ assumption with3 } the, assumption $ that i p . is e a that real $p . fractionis a q real ˆ comma{ fraction3 } i period ,= i . e 4.e period s ˆ{ 3 } . $ \quad From here , it turns out that $ q $ is also an even that the smallestthat common the smallest divisor common of p and divisor q is 1 ofperiod√p and q q is 1. q It remains to show that surdIt of remains 3 2 is not to show a constructible that 32 is number not a constructible period number . \noindentLet us assumenumber the opposite ,Let which us assume i period is inthe e period oppositecontradiction that i . the e . aforementioned that with the aforementioned the construction assumption construction is possi that hyphen is $ possi p $ - is a real fraction , i . e . ble period As weble saw . As in we the saw previous in the paragraph previous paragraph comma in , that in that case case x hasx has to belong to belong to a to certain a certain \noindentfield F subthat k whichfield the isFk an smallestwhich extension is an common extensionof a set of divisor of rational a set of numbers of rational $p$ comma numbers and arrived , arrived $q$ at by at consecutive by is consecutive $ 1 . q$ adding of squareadding roots of to square the set roots of rational to the numbers set of rational√ period numbers . \ centerlineTherefore comma{ It remains asTherefore we have to , shown as show we have that that surdshown of $ that\ 3surd 2 is32 not{ is3 a not rational 2 a rational}$ number is number not comma a , constructiblewe we may may conclude conclude number . } that x is not anthat elementx is not of the an rational element fieldof the F rational sub 0 comma field F but0, but instead instead belongs belongs to anotherto another \ hspaceextended∗{\ fieldf i l lFextended} subLet k us comma field assumeF wherek, where the k isk a oppositeis natural a natural number inumber . e period . Let that Let us us assume the assume aforementioned that thatk is k the is the smallest smallest construction is possi − natural numbernatural such that number x belongs such thatto thex extendedbelongs to field the F extended sub k period field F ....k. As weAs saw we that saw all that all \noindent ble . As we saw in the previous paragraph , in that case $ x $ has to belong to a certain elements of anelements extension of of an a set extension of rational of a numbers set of√ rational by a certain numbers square by a root certain may square root may be presented inbe the presented form√ of p in plus the q form surd of ofp k+ subq commak, it follows it follows that thatx may x may be written be√ written in the in the form form of \noindent f i e l d $ F { k }$ which is an extension of a set of rational numbers , arrived at by consecutive of x = p plus qx surd= p of+ wq subw, where commap, where q and pw commabelong q to and certain w belong√ field F tok− certain1, while fieldw Fdoes sub notk minus . 1 comma while surd of w does not periodLet us show now that if x = p + q w is a s o lution of the cubic \noindentLet us showadding nowequation that of .. square if x = p plus roots q surd to of the w is set a s o of lution rational .. of the ..numbers cubic .. equation . 3 √ x to the powerx of 3− minus2 = 0, 2then = 0 commay = p − thenq w yals = o p is minus its s qo surdlutio of n w. als We o is saw its that s o lutio by applying n period the .. We saw that\ hspace by applying∗{\ f i l the lbasic} Therefore rational operations , as we ( haveadding shown , subtracting that , multiplying $\surd{ 3 , dividing 2 }$ ) is , not as well a rational number , we may conclude basic rational operationsas open parenthesis adding comma subtracting comma multiplying comma .. dividing closing\noindent parenthesisthatapplying comma $ x .. $ the as squareis well not as root an , elementwe shall not of exit the a given rational field , then field we may $ F conclude{ 0 } that, $ but instead belongs to another 3 applying the squareas x belongs root comma to the we extended shall not field exitF ak, giventhen√ fieldx − comma2 also belongs then we to may that conclude field , and from \noindent extended field $ F 3 { k } , $ where $ k $ is a natural number . Let us assume that that as x belongsthere to theit follows extended that fieldx −√ F2 sub = k commaa + b w thenwhere x toa theand powerb are fromof 3 minus the field 2 alsoFk− belongs1. By to that $ k $ is the smallest field comma introducing√ x = p + q w√into the preceding equation , we get (p + q w3 − 2 = a + b w where , after cubing the equation , is follows that and from there it follows√) that x to the power√ , of 3 minus 2 =√ a plus b surd of w where a and b are from the \noindent natural3 number2 such2 that3 $ x $ belongs to the extended field $ F { k } field F sub k minusp + 3 1p periodq w + .. 3 Bypq w introducing+ q w w x− =2 = pa plus+ b qw surd. of wBy√ into grouping the preceding√ the corresponding equation comma .. . $ \ h f i l l As we saw that all 3 2 2 3 we get factors , it follows that (p + 3pq w − 2) + (3p q + q w) w = a + b w. Here it 3 2 2 3 open parenthesismay p plus be noticed√ q surd that of wa sub= p closing+ 3pq parenthesisw − 2 and tob = the 3p powerq + q ofw. 3 minusIn order 2 = to a show plus thatb surd of w sub\noindent comma ....elements wherey = commap − ofq w ....anis after aextension solution cubing of the cubic of equation a equation set comma ofy3 rational− ....2 = is 0, followswe numbers switch that values by a certain square root may 3 2 p to the powerof of q3 pluswith 3− pq toin the expressions power of for 2 qa surdand ofb. w plusIt follows 3 pq to that the powera = p of√+ 2 3 wpq plusw − q2 to and the power \noindent be presented2 in3 the form of $ p + q \surd3 { k } { , }$ it follows that of 3 w surd of w minusb = − 2(3 =p aq plus+ q w b) surd wherefrom of w sub it periodmay be .... noticed By grouping that y the− 2 corresponding = a − b w. $ x $ may be written in the form 3 factors comma it followsAs√ we have that assumed open parenthesis that x is a p solution to the power of cubic of equation 3 plus 3x pq− to2 the= 0, powerit follows of 2 that w minus 2 closingo f $ parenthesis x =a plus+ pb openw += parenthesis 0. qFrom\surd here 3{ itpw tofollows} the{ power that, }$a of= where 2b q= plus 0. Let q $p to us the assume power, theq$ of opposite 3 w and closing , $w$ i . parenthesis e . belong to certain field $ F { k − 1 } , $ while $\surd{ w }$ does not .√ surd of w = a plus√for b surd example of w that sub periodb 6= 0. ....Then Here , it from the equation√ a + b w = 0 it would follow that may be noticed thatw = a− =a/b, p towherefrom the power we of may 3 plus conclude 3 pq to that the powerw belongs of 2 w to minus the field 2 andFk b− =1, 3 p to the power \ hspace ∗{\ f i l l } Let us show now that \quad i f $ x = p + q \surd{ w }$ is a s o lution \quad o f the \quad cubic \quad equation of 2 q plus q to theto power which of both 3 wa periodand b ..belong In order , which to show is√ contrary to the assumption . Therefore , it that y = p minusmust q surd be valid of w thatis a solutionb = 0, and of cubic since equationa + b w y= to 0 theit follows power that of 3 minusa = 0. 2 =So 0 , comma with we switch \noindent $ x ˆ{ 3 } − 2 = 0 ,$then$y = p − q \surd{ w }$ values this , we have proved that a = b = 0. √ alsof o q with is its minus s q o in lutio expressionsHowever n . for\quad , a that andWe conclusion b period saw that.. takes It follows by us to applying that the claima = p thethat to they = powerp − q ofw 3 plusalso 3is pq a to the powerbasic of 2 rational w minus 2 andoperations ( adding , subtracting , multiplying , \quad d i v i d i n g ) , \quad as w e l l as b = minus open parenthesis 3 p to the power of 2 q plus q to the power of 3 w closing parenthesis wherefrom it\noindent may be noticedapplying that y to the thesquare power of 3root minus , 2 we = a shall minus b not surd exit of w sub a given period field , then we may conclude thatAs we as have $ assumed x $ belongs that x is a to solution the extended of cubic equation field x to $ the F power{ k of} 3 minus, $ 2 then = 0 comma $ x ˆit{ follows3 } − that2 $a plus also b surd belongs of w = to 0 period that From field here , it follows that a = b = 0 period Let us assume the opposite commaand from there it follows that $ x ˆ{ 3 } − 2 = a + b \surd{ w }$ where $a$i period and e period $b$ for example are from that b the equal-negationslash 0 period .. Then comma from the equation a plus b surdf i e of l d w = $ 0 F it would{ k follow− 1 } . $ \quad Byintroducing $x = p + q \surd{ w }$ intothat the surd preceding of w = minus equation a slash b comma , \quad wherefromwe get we may conclude that surd of w belongs to the field F sub k minus 1 comma \noindentto which both$ (a and p b belong + comma q \surd which{ w is contrary} { ) } toˆ{ the3 assumption} − 2 period = .. aTherefore + comma b \surd it { w } { , }$ \ hmust f i l l bewhere valid that , \ h b f = i l 0l commaafter and cubing since a the plus equation b surd of w , =\ 0h it f ifollows l l is that follows a = 0 period that .. So comma with \noindentthis comma$ we p have ˆ{ proved3 } + that a 3 = b p = ˆ 0{ period2 } q \surd{ w } + 3 pq ˆ{ 2 } w + q ˆHowever{ 3 } commaw \surd that{ conclusionw } − takes2 us = to the a claim + that b y =\surd p minus{ w q surd} { of. w}$ also\ ish a f i l l By grouping the corresponding \noindent factors , it follows that $ ( p ˆ{ 3 } + 3 pq ˆ{ 2 } w − 2 ) + ( 3 p ˆ{ 2 } q + q ˆ{ 3 } w ) \surd{ w } = a + b \surd{ w } { . }$ \ h f i l l Here i t

\noindent maybe noticed that $ a = p ˆ{ 3 } + 3 pq ˆ{ 2 } w − 2 $ and $ b = 3 p ˆ{ 2 } q + q ˆ{ 3 } w . $ \quad In order to show that $ y = p − q \surd{ w }$ is a solution of cubic equation $ y ˆ{ 3 } − 2 = 0 ,$ weswitch values

\noindent o f q with $ − q$ in expressions for $a$ and $b . $ \quad It follows that $ a = p ˆ{ 3 } + 3 pq ˆ{ 2 } w − 2 $ and $ b = − ( 3 p ˆ{ 2 } q + q ˆ{ 3 } w ) $ wherefrom it may be noticed that $ y ˆ{ 3 } − 2 = a − b \surd{ w } { . }$

As we have assumed that $ x $ is a solution of cubic equation $ x ˆ{ 3 } − 2 = 0 ,$ it follows that $ a + b \surd{ w } = 0 .$ Fromhere it follows that $a = b = 0 . $ Let us assume the opposite , i . e . for example that $ b \ne 0 . $ \quad Then , from the equation $ a + b \surd{ w } = 0 $ it would follow that $\surd{ w } = − a / b , $ wherefrom wemay conclude that $\surd{ w }$ belongs to the field $ F { k − 1 } , $

\noindent to which both $ a $ and $ b $ belong , which is contrary to the assumption . \quad Therefore , it mustbevalid that $b = 0 ,$ andsince $a + b \surd{ w } = 0 $ it followsthat $a = 0 .$ \quad So , with

\noindent this ,wehaveprovedthat $a = b = 0 .$

\ hspace ∗{\ f i l l }However , that conclusion takes us to the claim that $ y = p − q \surd{ w }$ a l s o i s a 60 .. A period Grozdani acute-c sub comma G period Vojvodi c-acute \noindentsolution of60 cubic\quad equationA . y Grozdani to the power $ of\ 3acute minus{ 2c =} 0{ comma, }$ keeping G . inVojvodi mind that $ y\ toacute the power{c} $ of 3 minus 2 = a minus b surd of w sub period \noindentLet us noticesolution that it is valid of cubic that x equal-negationslash equation $ y y ˆ comma{ 3 } i period − 2 e period = x minus0 , y equal-negationslash $ keeping in mind that 0$ comma y ˆ{ 3 since} x − minus2 y = = 2 q surd a of− w subb comma\surd and{ w } { . }$ 2 q surd of w = 0 only if q = 0 comma which consequentially leads to the conclusion that x = p comma \noindent Let us notice that it is valid that $ x \ne y , $ i . e $ . x i period e period60 xA would . Grozdani belongc´ toG . the Vojvodi fieldc´ F sub k minus 1 comma which is in contradiction with the − y \ne 0 , $ s i n c e, $ x − y = 2 q \surd{ w } { √, }$ and assumption period 3 3 solution of cubic equation y − 2 = 0, keeping in mind that y − 2 = a − b√ w. By this commaLet we ushave notice proved that the it claimis valid that that ifx 6= =y, p plusi . e q.x surd− y 6= of 0 w, since is a solutionx − y = of 2q cubicw and \noindent $ 2√ q \surd{ w } = 0 $ only if $ q = 0 , $ which, consequentially leads to the conclusion that equation x to the2q powerw = 0 of only 3 minus if q = 2 0, =which 0 comma consequentially then y = p leads minus to q the surd conclusion of w is also that itsx = solutionp, i . e comma $ x = p , $ where x equal-negationslash.x would belong y period to ....the But field commaFk−1, which is in contradiction with the assumption . iwe . are e thus $ . facing x$ a contradiction would belong comma to because the fieldsince p comma $F q{ andk surd− √ of w1 are} real, $ numbers which comma is in it contradiction with the assumption . By this , we have proved the√ claim that if x = p + q w is a solution of cubic follows that x andequation y arex real3 − numbers2 = 0, then commay = p which− q w is inis also contradiction its solution with , where the factx 6= thaty. But , \ hspace ∗{\ f i l l }By this , wehave proved the claim that if√ $x = p + q \surd{ w }$ the equation xwe to the are power thus facing of 3 minus a contradiction 2 = 0 has only , because one real since solutionp, q and periodw are real numbers , it is a solution of cubic Thus comma thefollows initial that assumptionx and y are has real led usnumbers to a contradiction , which is in comma contradiction i period with e period the fact the that solution of the equation x tothe the equation power ofx3 3− minus2 = 0 2has = 0 onlycannot one belong real solution to the field . F sub k comma therefore the duplication \noindent equation $ x ˆ{ 3 } − 2 = 0 ,$then$y = p − q \surd{ w }$ of Thus , the initial assumption has led us to a contradiction , i . e . the solution of is also its solution , where3 $ x \ne y . $ \ h f i l l But , a cube using athe compass equation andx straightedge− 2 = 0 cannot is impossible belong period to the field Fk, therefore the duplication of a 2 period 3 periodcube .. Theusing connection a compass between and straightedge constructible is impossible and algebraic . numbers \noindent we are thus facing a contradiction , because since $ p , q $ and $\surd{ w }$ Let us now notice2 some. 3 . of the The links connection between constructible between constructible and algebraic andnum algebraic hyphen numbers are real numbers , it bers period Let usLet show us that now al notice l constructible some of the numbers links between are alge constructible braic period and algebraic num - bers follows that $ x $ and $ y $ are real numbers , which is in contradiction with the fact that From the definition. Let of us algebraic show that numbersal l constructible it follows that numbers we claim are alge that braic .. each. constructible number isFrom a roo the t of definition an n to the of algebraic power of thnumbers degree it po follows lyn omial that period we claim that each \noindent the equation $ x ˆ{ 3 } − 2 = 0 $ has only one real solution . Let start from theconstructible field F sub number 0 comma is a the roo field t of ofan rationalnth degree numbers po lyn generated omial . by one line segment period .. NumbersLet start in from the thefield field F subF 1, arethe roots field of of a rational square equation numbers comma generated numbers by one from line \ hspace ∗{\ f i l l }Thus , the initial assumption0 has led us to a contradiction , i . e . the solution of F sub 2 are rootssegment of equation . Numbers of the fourth in the degree field F comma1 are roots and of comma a square generally equation comma , numbers numbers from ofF the2 field F sub k are rootsare of roots an equation of equation of 2 of to the the fourth power degree of k degree , and comma , generally with , numbersrational coefficients of the field period Let us \noindent the equation $ x ˆ{ 3 } − 2 = 0 $ cannot belong to the field $F { k } observe F are roots of an equation of 2k degree , with rational coefficients . Let us observe , $ thereforek the duplication of √ any number fromany F number sub 2 field from commaF2 field in , thein the form form of ofx =x = p plusp + q q surdw, where of wp, sub q, w commaare from whereF1, p comma q commaa cube w are using from aF sub compass 1 comma and straightedge√ is√ impossible√ . i . e . of the form p = a + b s,q = c + d s,w = e + f √s where a, b, c, d, e, f, s are i period e periodrational of the numbers form p = . a By plus squaring b surd the of s equation sub commax = qp =+ cq plusw we d surd arrive of to s sub the comma equation w = e plus \ centerline {2 . 3 . \quad The connection between constructible and algebraic numbers } f surd of s where a comma2 b comma2 2 c comma d comma e comma f comma s are x − 2px + p√= q w, where all the coefficients are in field F, which is the field rational numbersgenerated period By by squarings. Furthermore the equation , this x equation = p plus may q surd be written of w we in arrive the following to the equation form : Letx to us the now power notice of 2 minus some 2 px of plus the p to links the power between of 2 = q constructible to the power of 2 w and comma algebraic where all the num coefficients− bers . Let us show that al2 l constructible√ numbers√ 2 are√ alge2 braic√ . are in field F comma which is the fieldx − 2(a + b s)x + (a + b s) = (c + d s) (e + f s), generated by surd of s sub period Furthermore comma this equation may be written...... in the following form : \ hspaceLine 1∗{\ x tof ithe l l } powerFrom of the 2 minus definition 2 open parenthesis of algebraic a plus b numbers surd of s sub it closingfollows parenthesis that we x plus claim open that \quad each 2 2 2 2 2 parenthesis a plus b surd of s sub closing parenthesisx − 2ax + to (a the− powerc e + ofb s 2− =d openes − parenthesis2cdfs) = c plus d surd of s sub \noindent constructible number is a√ roo t of an $ n ˆ{ 2th }$2 degree po lyn omial . closing parenthesis to the power of 2 sub open= parenthesiss(2bx + e (− plus2ab f+ surd 2cde of+ sc subf + closingd fs)). parenthesis comma Line 2 period period period period period period Line 3 x to the power of 2 minus 2 ax plus open parenthesis a to the Keeping in mind that a, b, c, d, e, f, s are rational numbers , we come to the conclu - powerLet start of 2 minus from c to the the power field of 2 $ e Fplus{ b to0 the} power, $ of the 2 s minus field d to of the rational power of√ 2 numbers es minus 2 generated cdfs closing by one line sion that this equation may be presented in the form of x2 + ux + v = s rx + t) where parenthesissegment = . Line\quad 4 =Numbers surd of s sub in open the parenthesis field $ 2 Fbx plus{ 1 open}$ parenthesis are roots minus of 2 a ab( square plus 2 cde equation plus c to , numbers from u, v, s, r, t are all rational numbers . By squaring this equation , we get an the$ power F { of2 2 f}$ plus are d to theroots power of of equation 2 fs closingof parenthesis the fourth closing degree parenthesis , andperiod , generally , numbers of the field equation of the fourth degree (x2 + ux + v)2 = s(rx + t)2 with rational coefficients , as Keeping in mind that a comma b comma c comma d comma e comma f comma s are rational numbers comma it was claimed . we\noindent come to the$ conclu F { hyphenk }$ are roots of an equation of $ 2 ˆ{ k }$ degree , with rational coefficients . Let us observe It is worth noting that the opposite is not valid , i . e . that not all algebraic sion that this equation may be presented in the form of x to the power of 2 plus ux plus v = surd of s sub numbers are constructible ! We have shown that the roots of the third degree open\noindent parenthesisany rx number plus t closing from parenthesis $ F { 2 }$ field ,intheformof $x = p + q equation x3 − 2 = 0, i . e . algebraic numbers , are not constructible . \surdwhere{ w u comma} { , v}$ comma where s comma $p r comma , q t are ,all rational w$ numbersarefrom period $F .. By{ squaring1 } , this $ equation F is the algebraic extension of the field Q , if each and every element from F is comma we get an algebraic over Q . If that polynomial does not exist , than the number α is \noindentequation ofi.e.oftheform the fourth degree open parenthesis $p = x to thea power + of b 2\ plussurd ux{ pluss } v{ closing, } parenthesisq = to c the + powerd \surd of 2{ = s open} { parenthesis, } w rx = plus e t closing + parenthesis f \surd to{ thes } power$where$a of 2 with rational , coefficients b , comma c , das , it was e claimed , period f , s$are It is worth noting that the opposite is not valid comma .. i period e period that not all algebraic \noindentnumbers arerational constructible numbers ! .. We . have By shown squaring that the the roots equation of the third $ degree x = p + q \surd{ w }$ weequation arrive x to to the power equation of 3 minus 2 = 0 comma i period e period algebraic numbers comma are not con- structible period \noindentF is the algebraic$ x ˆ extension{ 2 } − of the2 field px Q comma + if p each ˆ{ and2 } every= element q ˆ{ from2 } w , $ where all the coefficients are in field $FF is algebraic , $ which over Q is period the .. field If that polynomial does not exist comma than the number alpha is \noindent generated $ by \surd{ s } { . }$ Furthermore , this equation may be written in the following form :

\ [ \ begin { a l i g n e d } x ˆ{ 2 } − 2 ( a + b \surd{ s } { ) } x + ( a + b \surd{ s } { ) }ˆ{ 2 } = ( c + d \surd{ s } { ) }ˆ{ 2 } { ( } e + f \surd{ s } { ), }\\ ...... \\ x ˆ{ 2 } − 2 ax + ( a ˆ{ 2 } − c ˆ{ 2 } e + b ˆ{ 2 } s − d ˆ{ 2 } es − 2 c d f s ) = \\ = \surd{ s } { ( } 2 bx + ( − 2 ab + 2 cde + cˆ{ 2 } f + d ˆ{ 2 } f s ) ) . \end{ a l i g n e d }\ ]

\noindent Keepinginmindthat $a , b , c , d , e , f , s$ are rational numbers , we come to the conclu −

\noindent sion that this equation may be presented in the form of $ x ˆ{ 2 } + ux + v = \surd{ s } { ( } rx + t ) $ where $u , v , s , r , t$ areallrationalnumbers. \quad By squaring this equation , we get an

\noindent equation of the fourth degree $ ( x ˆ{ 2 } + ux + v ) ˆ{ 2 } = s ( rx + t ) ˆ{ 2 }$ with rational coefficients , as it was claimed .

It is worth noting that the opposite is not valid , \quad i . e . that not all algebraic numbers are constructible ! \quad We have shown that the roots of the third degree

\noindent equation $ x ˆ{ 3 } − 2 = 0 , $ i . e . algebraic numbers , are not constructible .

$ F $ is the algebraic extension of the field Q , if each and every element from $ F $ is algebraic over Q . \quad If that polynomial does not exist , than the number $ \alpha $ i s On the ancient problem of duplication of a cube .. 6 1 \ hspacetranscendental∗{\ f i l l over}On Q the period ancient .. Therefore problem comma of the duplication numbers that are of not a cubealgebraic\quad are called6 1 transcendental numbers period Liouville was the first to prove the existence of transcen hyphen \noindentdental numberstranscendental in 1 844 and in over 1 85 1 Q he . gave\quad the firstTherefore decimal representation , the numbers of such that are not algebraic are called transcendentala number comma the numbers so hyphen . Liouville called Liouville was constant the first : to prove the existence of transcen − dentalLine 1 infinity numbers Line in 2 sum 1 844 1 0 to and the in power 1 85 of minus 1 he k gave ! = 0 theperiod first 1 1 0 1 decimal 1000 period representation period period Line of 3 such a number , the so − called Liouville constant : k = 1 On the ancient problem of duplication of a cube 6 1 Some of the examplestranscendental of transcendental over Q . numbers Therefore are the , the number numbers pi comma that are as not well algebraic as the are called \ [ \ begin { a l i g n e d }\ infty \\ base of the naturaltranscendental logarithm e numbers period Transcendental. Liouville was nature the first of number to prove e wasthe existenceproved by of transcen - \sum 1 0 ˆ{ − k ! } =0 . 1 1 0 1 1000 . . . \\ Hermite open parenthesisdental numbers 1 873 in closing 1 844 and parenthesis in 1 85 1 comma he gave and the offirst number decimal pi representation comma by Lindemann of such open k = 1 \end{ a l i g n e d }\ ] parenthesis 1 882 closinga number parenthesis , the so - period called Liouville constant : REFERENCES open square bracket 1 closing square bracket R period Courant comma∞ H period Robbins comma I period \noindent Some of the examples of transcendental numbers are the number $ \ pi , $ Stewart comma What Is Mathematics ? comma OxfordX University Press comma 1 996 period 10−k! = 0.11011000... asopen well square as the bracket 2 closing square bracket M period Kac comma S period M period Ulam comma Mathe- matics and Logic comma Dover Publications comma 1 992 period k = 1 \noindentopen squarebase bracket of 3the closing natural square logarithm bracket V period $ e Peri .acute-c $ Transcendental sub comma Algebra nature II comma of IGKRO number Some of the examples of transcendental numbers are the number π, as well as the Svjetlost$ e $ comma was proved Sarajevo by comma 1 980 open square bracket in Serbian closing square bracket period base of the natural logarithm e. Transcendental nature of number e was proved by open square bracket 4 closing square bracket W period R period Knorr comma The Ancient Tradition of Hermite ( 1 873 ) , and of number π, by Lindemann ( 1 882 ) . Geometric\noindent ProHermite b lems comma ( 1 873Boston ) comma , and 1 of 986 number period $ \ pi , $ by Lindemann ( 1 882 ) . REFERENCES open square bracket 5 closing square bracket H period D dieresis-o rrie comma The Delian cube hyphen [ 1 ] R . Courant , H . Robbins , I . Stewart , What Is Mathematics ?, Oxford University Press , doubling\ centerline problem{REFERENCES comma in : 1} 0 Great Pro b lems of Elementary Math hyphen 1 996 . [ 2 ] M . Kac , S . M . Ulam , Mathematics and Logic , Dover Publications , 1 992 . ematics : Their History and Solutions comma New York : Dover comma 1 965 period [ 3 ] V . Peri c´ Algebra II , IGKRO Svjetlost , Sarajevo , 1 980 [ in Serbian ] . [ 1open ] R square . Courant bracket 6 , closing H . Robbinssquare, bracket , I B . period Stewart Bold comma , What The Is Delian Mathematics pro b lem comma $ ? in :, Famous $ Oxford University Press , 1 996 . [ 4 ] W . R . Knorr , The Ancient Tradition of Geometric Pro b lems , Boston , 1 986 . Problems[ 2 ] M of Geometry. Kac , andS . How M . to Ulam Solve ,Them Mathematics comma and Logic , Dover Publications , 1 992 . [ 5 ] H . D o¨ rrie , The Delian cube - doubling problem , in : 1 0 Great Pro b lems of New York : Dover comma 1 982 period Elementary Math - \ centerlineAndrea Grozdani{ [ 3 acute-c ] V . sub Peri comma $ \ 595acute Main{c Street} { comma, }$ Apt Algebra hash 1 104II comma, IGKRO New Svjetlost York comma , NY Sarajevo , 1 980 [ in Serbian ] . } ematics : Their History and Solutions , New York : Dover , 1 965 . 10044 USA [ 6 ] B . Bold , The Delian pro b lem , in : Famous Problems of Geometry and How to Solve \ centerlineE hyphen mail{ [ : 4 dominovic ] W . R at . y Knorr to the power , The ofa-h Ancient oo period Tradition com of Geometric Pro b lems , Boston , 1 986 . } Them , Gradimir Vojvodi acute-c sub comma Trg Dositeja Obradovi c-acute a 3 comma 2 1000 Novi Sad Serbia New York : Dover , 1 982 . \ hspaceE hyphen∗{\ mailf i l l :} gradimir[ 5 ] H period . D voj $ vodi\ddot c at{o dmi} $ period rrie uns , period The Delian ac period cube rs − doubling problem , in : 1 0 Great Pro b lems of Elementary Math − Andrea Grozdani c´,595 Main Street , Apt #1104, New York , NY 10044 USA \ centerline { ematics : Their HistoryE - mail and: dominovic Solutions@ya−h ,oo New . com York : Dover , 1 965 . } Gradimir Vojvodi c´, Trg Dositeja Obradovi c´ a 3 , 2 1000 Novi Sad Serbia \ hspace ∗{\ f i l l } [ 6 ] BE . - Bold mail : ,gradimir The Delian . voj vodi pro c b@ lemdmi . , uns in . : ac Famous . rs Problems of Geometry and How to Solve Them ,

\ centerline {New York : Dover , 1 982 . }

\ centerline {Andrea Grozdani $ \acute{c} { , } 595 $ Main Street , Apt $ \# 1 104 , $ New York , NY 10044 USA }

\ centerline {E − mail : dominovic $@ y ˆ{ a−h }$ oo . com }

\ centerline { Gradimir Vojvodi $ \acute{c} { , }$ Trg Dositeja Obradovi $ \acute{c} $ a 3 , 2 1000 Novi Sad Serbia }

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