Question 44378
let R* be the set of all real numbers except 0. define * on R* by letting a*b = |a|b.
a) show that * gives an associative binary operation on R*.
TST   (A*B)*C=A*(B*C)
LHS=(|A|B)*C=||A|B|C......WHETHER A AND B ARE +VE OR -VE ,THE FINAL ANSWR IS
|AB|C 
RHS=A*(|B|C)=|A||B|C...HERE TOO THE FINAL ANSWER IS |AB|C
HENCE ASSOCIATIVE PROPERTY IS PROVED.
b)show that there is a left identity for * and a right inverse for each element in R*.
TST THERE IS X WHERE ....X*A=A
X*A=|X|A=A...THAT IS X=1 IS THE LEFT IDENTITY
BUT A*1=|A|1 IS NOT EQUAL TO A IF A IS -VE.HENCE X=1 IS ONLY LEFT IDENTITY AND NOT A RIGHT IDENTITY.
RIGHT INVERSE...LET A*A'=1
A*A'=|A|A'=1..SO A'=1/|A|WHICH EXISTS SINCE A IS NOT EQUAL TO ZERO AND IS  INDEPENDENT OF SIGN OF A 
LEFT INVERSE....A'*A=1=|A'|A=1....|A'|=1/A..WHICH EXISTS IF A IS POSITIVE.,BUT DOES NOT EXIST IF A IS NEGATIVE SINCE WE CANT FIND A' SUCH THAT |A'| IS NEGATIVE. 
c)is R* with this binary operation a group?
WE HAVE TO PROVE
1.CLOSOURE....A*B IS AN ELEMENT IN R*...WHICH IS OK.SINCE A/B CANNOT BE ZEROS AS GIVEN
2.ASSOCIATIVE...PROVED ALREADY
3.EXISTENCE OF IDENTITY...WE HAVE 1*A=A...BUT A*1 IS NOT EQUAL TO A .HENCE THIS IS NOT A GROUP.
d) explain the significance of this exercise.
IT SHOWS THAT THERE COULD BE AN IDENTITY ELEMENT WHICH IS PARTIAL IN THE SENSE THAT IT IS ONLY LEFT IDENTITY AND NOT RIGHT IDENTITY.