Page 49 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Notes
),
= [a (c e b (d f )]
[addition is associative in real numbers]
= ( , ) (a b c , e d ) f

f
= ( , ) [(a b c d ) ( , )]
e
So the addition is associative in R
(R ) : (0, 0) + (a, b) = (0 + a, 0 + b) = (a, b) (a, b) R, so that (0, 0) is the additive identity in R.
3
b
a
,
(R ) : ( a b ) ( , ) ( a a b b ) (0,0) so the additive inverse of (a, b), is (–a, –b) (a, b) R.
,
4
(R ) : (a, b) + (c, d) = (a + c, b + d)
5
= (c a ,d b )
[because addition is commutative in real numbers]
= ( , ) ( , )c d a b (a, b), (c, d) R.
(R ): [(a, b), (c, d)] [e, f]
6
= ( ,ac bd )( , )
e
f
= {( ) ,(ac e bd ) }
f
= { ( , , (a c e b d f )}
[because ordinary multiplication is associative]

= ( , )( , ,a b c e d f )

f
a
b
d
c
e
= ( , )[( , )( , )]a b c d e f ( , )( , )( , ) R 
(R ) : ( , )[( , ) ( , )]a b c d e f
7
= ( , )(a b c e ,d ) f
= (a c a e b d b f ) (by distributive law of reals)
,
= (a c b d ),(a e b f )
,
a
c
f
= (a b ),( , ) ( , )( , ) .
e
d
b
Similarly
b
a
f
e
[(c d ) ( , )]( , )
= (c d ) ( , ) ( , )( , ) .
b
a
f
b
e
a
Hence R is a ring.
Now, in order to show that R is a ring with zero divisors we must produce at least two non-zero
elements whose product is zero. Clearly neither (a, 0) with a 0 nor (0, b) with b 0 is the zero
element (additive identity) or R yet their product
( ,0)(0, ) ( .0,0. ) (0,0)
b
b
a
a
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