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For details see post on 26-Feb-2015 at 6:00 am.

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GROUPS
Verify if the following vector addition & scalar multiplication satisfy the requirements of a group..

u+v=[{X1^(1/3)+X2^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)}^3]  AS PER DEFINITION

C*u=(C^3*X1,C^3*Y1)  ============= SCALARS BELONG TO A FIELD F OR REAL NUMBERS IN THIS CASE.VECTORS BELONG TO A VECTOR V. WE USE V(F) TO INDICATE VECTOR SPACE OVER A FIELD F.SOME TIMES IF THE FIELD IS UNDERSTOOD,WE MAY DROP (F) LEAVING V TO INDICATE VECTOR SPACE. THIS VECTOR IS DIFFERENT FROM WHAT WAS READ IN VECTOR ANALYSIS.THIS IS ONLY A PREDEFINED SET OF ELEMENTS,AS IN THIS CASE, IT IS AN ORDERED PAIR (X,Y) THE ADDITION ,MULTIPLICATION ETC.IN SCALARS IN THE FIELD ARE NORMAL OPERATIONS WITH NORMAL SYMBOLS USED. BUT THOSE OPERATIONS IN VECTORS ARE DIFFERENT IN DIFFERENT CASES AND ARE AS PREDEFINED,HENCE WE USE SPECIAL SYMBOLS FOR THE SAME LIKE *,0,ETC..OR EVEN NORMAL SYMBOLS ,BUT THEIR MEANING HAS TO BE UNDERSTOOD AS PER CONTEXT AND DEFINITION. THESE OPERATIONS IN VECTORS ARE CALLED COMPOSITIONS.THEY CAN BE INTERNAL THAT IS BETWEEN ELEMENTS OF VECTOR AS ADDITION HERE,OR COULD BE EXTERNAL BETWEEN ELEMENTS OF EXTERNAL FIELD AND VECTORS LIKE SCALAR MULTIPLICATION IN THIS CASE. I…WE HAVE TO PROVE THAT V HAS AN INTERNAL COMPOSITION CALLED ADDITION AND IT IS AN ABELIAN GROUP FOR THIS COMPOSITION….THIS REQUIRES  1.CLOSOURE..u+v IS AN ELEMENT OF V FOR ALL u&v ELEMENTS OF V 2.COMMUTATIVE...u+v=v+u……FOR ALL u&v ELEMENTS OF V 3.ASSOCIATIVE...u+(v+w)=(u+v)+w….FOR ALL u,v&w ELEMENTS OF V 4.IDENTITY ….THERE IS AN ELEMENT 0 IN V SUCH THAT 0+u=u FOR u AN ELEMENT OF V.0 IS THE ADDITIVE IDENTITY IN V 5.INVERSE...THERE IS -u IN V FOR EVERY u IN V SUCH THAT -u+u=0….-u IS ADDITIVE INVERSE IN V II.TPT V HAS AN EXTERNAL COMPOSITION OVER A FIELD CALLED SCALAR MULTIPLICATION ,THAT IS A*u IS AN ELEMENT OF V FOR ALL... A.. ELEMENT OF FIELD AND ALL u ELEMENT OF V. THAT IS V IS CLOSED WRT SCALAR MULTIPLICATION. III.TPT THE 2 COMPOSITIONS ADDITION  AND SCALAR MULTIPLICATION OF VECTORS SATISFY THE FOLLOWING.  1.DISTRIBUTIVE OVER VECTOR ADDITION….A*(u+v)=A*u+A*v…… FOR ALL A IN THE FIELD AND u&v IN V 2.DISTRIBUTIVE OVER SCALAR ADDITION...(A+B)*u=A*u+B*u….. FOR ALL A&B IN THE FIELD AND u IN V 3.ASSOCIATIVE….(AB)*u=A*(B*u)…….. FOR ALL A&B IN THE FIELD AND u IN V 4.UNIT……..1*u=u FOR 1 UNITY ELEMENT IN THE FIELD AND u IN V LET US USE SMALL CASE LETTERS..u,v,w FOR VECTORS AND CAPITAL LETTERS X,Y,Z ETC.FOR SCALARS. u=(X1,Y1) v=(X2,Y2) w=(X3,Y3) X1,Y1,Z1..ETC.ARE SCALARS. THE ABOVE ARE THE 10 POINTS TO BE PROVED.LET US PROCEED ONE BY ONE TO PROVE. I…WE HAVE TO PROVE THAT V HAS AN INTERNAL COMPOSITION CALLED ADDITION AND IT IS AN ABELIAN GROUP FOR THIS COMPOSITION….THIS REQUIRES  1.CLOSOURE..u+v IS AN ELEMENT OF V FOR ALL u&v ELEMENTS OF V u+v=[{X1^(1/3)+X2^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)}^3] EXPONENTIATION AND ADDITION OF X1,X2,Y1,Y2 BEING NORMAL ALGEBRAIC OPERATIONS ,THE RESULTANT ANSWER IS ALSO AN ORDERED PAIR IN THE REALFIELD AS THERE IS ALWAYS ONE REAL CUBIC ROOT FOR ANY REAL NUMBER.HENCE THE SUM BELONGS TO V =(X,Y)…SAY…WHERE X AND Y ARE SCALARS.HENCE u+v BELONGS TO V. 2.COMMUTATIVE...u+v=v+u……FOR ALL u&v ELEMENTS OF V u+v=[{X1^(1/3)+X2^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)}^3]=(X,Y)SAY v+u=[{X2^(1/3)+X1^(1/3)}^3,{Y2^(1/3)+Y1^(1/3)}^3]=(X,Y) EXPONENTIATION AND ADDITION OF X1,X2,Y1,Y2 BEING NORMAL ALGEBRAIC OPERATIONS ARE COMMUTATIVE AND THE SUM IS NOT ALTERED BY ORDER OF ADDITION. 3.ASSOCIATIVE...u+(v+w)=(u+v)+w….FOR ALL u,v&w ELEMENTS OF V u+(v+w)=(X1,Y1)+[{X2^(1/3)+X3^(1/3)}^3,{Y2^(1/3)+Y3^(1/3)}^3] (u+v)+w=[{X1^(1/3)+X2^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)}^3]+(X3,Y3) u+(v+w)=[[{X1^(1/3)+(((X2^(1/3)+X3^(1/3))^3)^(1/3))}^3],[{Y1^(1/3)+(((Y2^(1/3)+Y3^(1/3))^(1/3))}^3]] (u+v)+w=[[{X3^(1/3)+(((X1^(1/3)+X2^(1/3))^3)^(1/3))}^3],[{Y3^(1/3)+(((Y1^(1/3)+Y2^(1/3))^(1/3))}^3]] u+(v+w)=[{X1^(1/3)+X2^(1/3)+X3^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)+Y3^(1/3)}^3] (u+v)+w=[{X1^(1/3)+X2^(1/3)+X3^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)+Y3^(1/3)}^3] EXPONENTIATION AND ADDITION OF X1,X2,Y1,Y2 BEING NORMAL ALGEBRAIC OPERATIONS ARE COMMUTATIVE & ASSOCIATIVE AND THE SUM IS NOT ALTERED BY ORDER OF ADDITION. 4.IDENTITY ….THERE IS AN ELEMENT 0 IN V SUCH THAT 0+u=u FOR u AN ELEMENT OF V.0 IS THE ADDITIVE IDENTITY IN V 0=(0,0) 0+u=[{0^(1/3)+X1^(1/3)}^3,{0^(1/3)+Y1^(1/3)}^3]=(X1,Y1)=u WE CAN EASILY SEE THAT 0 VECTOR IS (0,0) WHICH SATISFIES THIS STIPULATION. 5.INVERSE...THERE IS -u IN V FOR EVERY u IN V SUCH THAT -u+u=0….-u IS ADDITIVE INVERSE IN V -u=(-X1,-Y1) -u+u=[{(-X1)^(1/3)+X1^(1/3)}^3,{(-Y1)^(1/3)+Y1^(1/3)}^3]=(0,0)=0 WE CAN EASILY SEE THAT -u IS (-X1,-Y1) IN V  WHICH SATISFIES THIS STIPULATION.CUBIC ROOT AND CUBING WILL NOT CHANGE - SIGN & ADDITION WILL RESULT IN GETTING 0.NOTE THAT THIS WOULD NOT HAVE BEEN THE CASE WITH SQUATRE ROOTS. II.TPT V HAS AN EXTERNAL COMPOSITION OVER A FIELD CALLED SCALAR MULTIPLICATION ,THAT IS au IS AN ELEMENT OF V FOR ALL a ELEMENT OF FIELD AND ALL u ELEMENT OF V. THAT IS V IS CLOSED WRT SCALAR MULTIPLICATION. A*u=(A^3*X1,A^3*Y1)=(X,Y) SAY WHERE X AND Y ARE SCALARS .HENCE A*u BELONGS TO V. WE FIND THAT ON MULTIPLICATION WE GET (A^3X1,A^3Y1) WHICH CLEARLY BELONGS TO V AS AN ORDERED PAIR. III.TPT THE 2 COMPOSITIONS  NAMELY ..ADDITION  AND SCALAR MULTIPLICATION OF VECTORS SATISFY THE FOLLOWING.  1.DISTRIBUTIVE OVER VECTOR ADDITION….A*(u+v)=A*u+A*v…… FOR ALL A IN THE FIELD AND u&v IN V A*(u+v)=A*[{X1^(1/3)+X2^(1/3)}^3,{Y1^(1/3)+Y2^(1/3)}^3]= [A^3*{X1^(1/3)+X2^(1/3)}^3,A^3*{Y1^(1/3)+Y2^(1/3)}^3] A*u+A*v= (A^3*X1,A^3*Y1)+(A^3*X2+A^3*Y2)= [{(A^3*X1)^(1/3)+(A^3*X2)^(1/3)}^3,{(A^3*Y1)^(1/3)+(A^3*Y2)^(1/3)}^3]= [A^3*{X1^(1/3)+X2^(1/3)}^3,A^3*{Y1^(1/3)+Y2^(1/3)}^3]=A*(u+v) 2.DISTRIBUTIVE OVER SCALAR ADDITION...(A+B)*u=A*u+B*u….. FOR ALL A&B IN THE FIELD AND u IN V (A+B)*u= (A+B)*(X1,Y1)= =((A+B)^3*X1,(A+B)^3*Y1) A*u+B*u= ((A^3)*X1,(A^3)*Y1)+((B^3)*X1,(B^3)*Y1)=[{(A^3*X1)^(1/3)+(B^3*X1)^(1/3))}^3,{(A^3*Y1)^(1/3)+(B^3*Y1)^(1/3))}^3]=((A+B)^3*X1,(A+B)^3*Y1)=(A+B)*u 3.ASSOCIATIVE….(AB)*u=A*(B*u)…….. FOR ALL A&B IN THE FIELD AND u IN V (AB)*u= ((AB)^3*X1,(AB)^3*Y1) A*(B*u)= A*(B^3*X1,B^3*Y1)=(A^3(B^3*X1),A^3(B^3*Y1))=((AB)^3*X1,(AB)^3*Y1)=(AB)*u 4.UNIT……..1*u=u FOR 1 UNITY ELEMENT IN THE FIELD AND u IN V 1*u=1*(X1,Y1)=(1^3*X1,1^3*Y1)=(X1,Y1)=u