DEAR STUDENTS

PLEASE GO THROUGH THE FOLLOWING WAY FOR DOING QUESIONS ON GROUPS

For proving the questions on Groups many students commits mistakes on the following questions.

**1. If Q _{1} is the set of rationall numbers other than 1 with binary operation _{* }defined by a*b=a+b-ab for all a,b εQ_{1}, Show that (Q_{1},*) is an abelian group and solve 5*x=3 in Q_{1}.**

**2.****If Q _{-1} is the set of all rational numbers except -1 and * is a binary operation defined on Q_{-1} by **

**a*b= a+b+ab, Prove that Q**

_{-1}^{ }is an abelian group.**For verifying ** closure law for the above the problems, many students simply do in the following way

**Wrong approach: **

Closure law: _{} a*b=a+b-ab is again a rational number and hence closure law is true. However Here student did not verified a*b is a rational number except 1. Hence it is very necessary to show that a*b is a rational number other than 1 . Other wise 1Mark will be reduced in the valuation. In april 2002, the 1 Mark is reduced for the students who did not verified the closure law correctly.

**The correct approach is as follows: **

**For question number 1: **

Let a, b _{}Q** _{1} ** such that a

_{}#1, b#1 a+b-ab is also a rational number and it cannot be equal to 1

**Because If a+b-ab=1, then a+b-ab-1=0 _{} (a-1)(b-1)=0 _{}a=1 or b=1 which is not true as a#1 and b#1. _{} a*b_{}Q_{1} _{} _{} Q_{1} is closed w.r.t * (GIVING REASONS IS IMP)**

**Similarly for verfying closure for 2 question**

Closure law: _{} or Q_{-1} a*b=a+b-ab is againa a rational number and hence closure law is true.

Let a, b _{}Q_{-1}such that a_{}# -1 b#-1 a+b+ab is also a rational number and it cannot be

equal to -1

**Because If a+b+ab=-1, then a+b+ab+1=0 _{} (a+1)(b+1)=0 _{}a= -1 or b= -1 which is not true as a# -1 and b# -1. _{} a*b_{}Q_{-1} _{} _{} Q_{1} is closed w.r.t * (GIVING REASONS IS IMP)**

** **