Page 56 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Thus 1y y y 1 y D y 1. (because multiplication is commutative) Therefore 1 is the unit
element of D.
,
Now 1 D and as such one of the elements aa aa 2 ,..., aa is equal to 1, i.e.,
1
n
aa 1 a a for some s such that 1 s n.
s s
Thus a D is the multiplicative inverse of the non-zero element a in D. Since a is arbitrary
s
element in D, we conclude that each non-zero element of D possesses multiplicative inverse.
Hence D is a field.

Illustrative Examples

Example 28: Prove that the set of complex numbers is a field with respect to addition and
multiplication operation.
or

Let C be the set of ordered pairs (a, b) of real numbers. Define addition and multiplication in C
by the equations

( , ) ( , ) = (a c b d )
a
d
,
b
c
b
( , ) ( , ) = (ac bd bc ad )
a
c
d
,
Prove that C is a field.
Solution: C is closed under addition and multiplication since a + c, b + d, ac – bd, bc + ad are all real
numbers.
Let ( , ), ( , ), ( , )a b c d e f C
then
e
f
f
,
e
a
b
[( , ) ( , )] ( , ) = (a c b d ) ( , )
c
d
= [(a c ) e ,(b d ) ] f
= [a (c e ),b (d f )]
= ( , ) (a b c , e d ) f
= ( , ) [( , ) ( , )]a b c d e f
Hence addition is associative
Since ( , ) ( , ) = (a c b d )
c
,
d
b
a
a
c
d
= (c a d b ) ( , ) ( , )
b
,
addition is commutative in C.
(0, 0) C is additive identity in C as
b
a
b
a
(0, 0) ( , ) = (0 a , 0 b ) ( , ) ( , ) C .
b
a
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