document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=26758>Question 26758</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Show that it is a binary operation is a group and determine if it is Abelian.\r<BR>\n" );
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document.write( "Reals(R)\{-1} with operation * defined by <BR>\n" );
document.write( "a*b = a+b+ab </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #14616 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=14616\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Show that it is a binary operation is a group and determine if it is Abelian.\r<BR>\n" );
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document.write( "Reals(R)\{-1} with operation * defined by <BR>\n" );
document.write( "a*b = a+b+ab<BR>\n" );
document.write( "1.CLOSOURE..<BR>\n" );
document.write( "LET A AND B BE ELEMENTS OF REAL NUMBERS R.THEN <BR>\n" );
document.write( "A*B=A+B+AB IS ALSO REAL.SO IT IS IN R.<BR>\n" );
document.write( "2.ASSOCIATIVE...<BR>\n" );
document.write( "A*(B*C)=A*{B+C+BC)=A+B+C+BC+AB+AC+ABC<BR>\n" );
document.write( "(A*B)*C=(A+B+AB)*C=A+B+AB+C+AC+BC+ABC=A*(B*C)\r<BR>\n" );
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document.write( "3.IDENTITY ELEMENT<BR>\n" );
document.write( "LET A*E=A<BR>\n" );
document.write( "A+E+AE=A<BR>\n" );
document.write( "E+AE=0<BR>\n" );
document.write( "E=0<BR>\n" );
document.write( "LET US CHECK FOR..<BR>\n" );
document.write( "E*A=E+A+EA=0+A+0A=A...OK<BR>\n" );
document.write( "4.INVERSE..<BR>\n" );
document.write( "LET A*I=E=0<BR>\n" );
document.write( "A+I+AI=0<BR>\n" );
document.write( "I(1+A)=-A<BR>\n" );
document.write( "I=-A/(1+A)...THIS EXISTS SINCE A IS NOT EQUAL TO -1 AS GIVEN.R IS Reals(R)\{-1}<BR>\n" );
document.write( "LET US CHECK<BR>\n" );
document.write( "I*A=E=0<BR>\n" );
document.write( "-A/(1+A)*A=-A/(1+A)+A+(-A.A)/(1+A)=(1/(1+A)){-A+A(1+A)-A.A}<BR>\n" );
document.write( "=(1/(1+A)){-A+A+A.A-A.A}=0...PROVED<BR>\n" );
document.write( "5.ABELIAN TEST<BR>\n" );
document.write( "A*B=A+B+AB<BR>\n" );
document.write( "B*A=B+A+BA=A*B...HENCE IT IS ABELIAN GROUP.\r<BR>\n" );
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document.write( "SEE THE FOLLOWING EXAMPLE TO KNOW THE PROCEDURE ELABORATION.......<BR>\n" );
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document.write( "show that the binary operation is a group and determine if it is Abelian<BR>\n" );
document.write( "H = {(a,b)belong to R^(2): a can not equal 0} with product<BR>\n" );
document.write( "(a,b)(c,d) = (ac,ad + b)<BR>\n" );
document.write( "1 solutions\r<BR>\n" );
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document.write( "Answer 14520 by venugopalramana(860)   on 2006-02-07 03:41:54 (Show Source): \r<BR>\n" );
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document.write( "(a,b)*(c,d) = (ac,ad + b)<BR>\n" );
document.write( "TO SHOW THIS OPERATION * IS A GROUP,WE NEED TO PROVE...<BR>\n" );
document.write( "1..CLOSOURE...<BR>\n" );
document.write( "AC IS ALSO REAL NUMBER SINCE A AND C ARE REAL...AD IS ALSO REAL AND SO IS AD+B.SO (AC,AD+B)IS AN ELEMENT OF R^2<BR>\n" );
document.write( "2..ASSOCIATIVE....<BR>\n" );
document.write( "TPT...{(A,B)*(C,D)}*(E,F)=(A,B)*{(C,D)}*(E,F)}..<BR>\n" );
document.write( "LHS=(AC,AD+B)(E,F)=(ACE,ACF+AD+B)<BR>\n" );
document.write( "RHS=(A,B)(CE,CF+D)=(ACE,ACF+AD+B)=LHS<BR>\n" );
document.write( "3...EXISTENCE OF IDENTITY....LET IT BE (I1,I2)<BR>\n" );
document.write( "WE SHOULD HAVE (A,B)*(I1,I2)=(A,B) AND (I1,I2)*(A,B)=(A,B)<BR>\n" );
document.write( "AI1=A...SO...I1=1<BR>\n" );
document.write( "AI2+B=B...SO....I2=0....SO...(1,0) IS THE IDENTITY ELEMENT..IT IS AN ELEMENT OF R^2...OK....<BR>\n" );
document.write( "LET US CHECK...<BR>\n" );
document.write( "(1,0)*(A,B)=(1A,1B+0)=(A,B)...OK...<BR>\n" );
document.write( "4...EXISTENCE OF INVERSE....<BR>\n" );
document.write( "WE SHOULD FIND (X,Y)SO THAT (X,Y)(A,B)=(1,0)=(A,B)*(X,Y)<BR>\n" );
document.write( "XA=1....OR..X=1/A.....<BR>\n" );
document.write( "XB+Y=0.....Y=-XB=-B/A.....SO INVERSE IS (1/A,-B/A)..SINCE...A IS NOT ZERO ,WE HAVE THE INVERSE AS AN ELEMENT OF R^2.<BR>\n" );
document.write( "LET US CHECK...<BR>\n" );
document.write( "(A,B)(1/A,-B/A)=(A*1/A,((A*-B)/A)+B)=(1,0) ....OK....<BR>\n" );
document.write( "HENCE * AS A BINARY OPERATION IS A GROUP.<BR>\n" );
document.write( "WE FIND THAT<BR>\n" );
document.write( "(a,b)*(c,d) = (ac,ad + b)..WHERE AS<BR>\n" );
document.write( "(C,D)*(A,B) = (CA,CB+D) WHICH ARE NOT EQUAL.<BR>\n" );
document.write( "HENCE THIS IS NOT ABELIAN \r<BR>\n" );
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