Page 28 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Solution: Let the given set be denoted by Q . Then by group axioms, we have—
0
(G ) We know that the product of two non-zero rational numbers is also a non-zero rational
1
number. Therefore Q is closed with respect to multiplication. Hence, closure axiom is
0
satisfied.
(G ) We know for rational numbers.
2
a
(a b c a (b c ) for all , , c Q
)
b
0
Hence, associative axiom is satisfied.
(G ) Since, 1 the multiplicative identity is a rational number hence identity axiom is satisfied.
3
(G ) If a Q , then obviously, 1/a Q . Also
4 0 0
1/a . a = 1 = a . 1/a
so that 1/a is the multiplicative inverse of a. Thus inverse axiom is also satisfied.

Hence Q is a group with respect to multiplication.
0
Example 7: Show that C, the set of all non-zero complex numbers is a multiplicative
group.
Solution: Let C = {z : z = x + i y, x, y R}

Hence R is the set of all real numbers are i = ( 1) .

(G ) Closure Axiom: If a + i b C and c + id c, then by definition of multiplication of complex
1
numbers
(a i b ){(c i d ) (a c b d ) i (a d b c ) C ,

,
c
b
since a c b , d a d b c R for a , , , d R .
Therefore, C is closed under multiplication.
(G ) Associative Axiom:
2
(a i b ){(c i d ) (e i f )} (a c e a d f b c f b d ) e ( i a c f a d e b c e b d ) f

= {(a i b ) (c i d )} (e i f )
for a, b, c, d R.

(G ) Identity Axiom: e = 1 (= 1 + i ) is the identity in C.
3 0
(G ) Inverse Axiom: Let (a + i b) ( 0) C, then
4
1 a ib
(a + ib) –1 =
a ib a 2 b 2
a b
= i
a 2 b 2 a 2 b 2

a
= m + i n C, Where m = ,
a 2 b 2

22 LOVELY PROFESSIONAL UNIVERSITY