Page 58 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Since a 0 or b 0, a + b 2 0, i.e., C or d or both are non-zero real numbers.
2

Hence C is a field.
Note: The question could have been done by assuming the elements of C as a + ib etc. also.

Example 29: Show that the set of numbers of the form a b 2 with a and b as rational
numbers is a field.
a
Solution: Let R {a b 2 : , b Q }
F 1 ( , ) is a abelian group.
R
,
b
(F ) Let a b 2 R and a b 2 , R then a , , a b 2 are the elements of Q, the set of
11 1 1 2 2 1 1 2
rational numbers.
Now (a b 2) (a b 2) (a a ) (b b ) 2 R since a a , b b Q .
1 1 2 2 1 2 1 2 1 2 1 2
Hence closure axiom for addition is satisfied.
(F ) Addition is commutative for real numbers.
12
(F ) Addition is associative for real numbers.
13
(F 14 ) 0 0 2 0 R as 0 Q, hence 0 is the identity of addition in R because

(0 0 2) (a b 2) = (0 a ) (0 b 2)

= a b 2 , a b Q .

(F ) If a b 2 R then ( a ) ( b ) 2 R and also
15
[( a ) ( b ) 2] (a b 2)

= ( a a ) ( b b ) 2 0 0 2
= 0

each element of R possesses additive inverse .
(F ) Properties of field for (F,)
2
(F 21 ) (a 1 b 1 2)(a 2 b 2 2)

(a a 2b b ) (a b a b 2) R
1 2
1 2
1 2
2 1
Since a a 2b b a b a b Q for a a 2 , ,b 2 Q .
,
b
,
1
1 2
2 1
1
1 2
1 2
Thus R is closed under multiplication
(F ) Multiplication in R is commutative
22
(F ) Multiplication in R is associative
23
(F 24 ) 1 0 2 =1 R and 1 (a b 2)
= a b 2 , a b Q .
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