Page 42 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes (R ) Let a, b, c Q then
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(a + b) + c = a + (b + c)
because associative law for addition holds.
(R ) 0 Q and 0 + a = a + 0 = a a Q, i.e., 0 is the additive identity in Q.
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(R ) a Q, – a Q and a + (–a) = 0. Hence additive inverse in Q exists for each element in Q.
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(R ) Let a, b Q then a + b = b + a because addition is commutative for rationals.
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R (Q, .) is a semi group.
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(R ) Since the product of two rational numbers is a rational number, a, b Q a . b Q.
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(R ) Multiplication in Q is associative.
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R Multiplication is left as well as right distributive over addition in the set of rational
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numbers, i.e.,
a (b c ) = a b a c

)
(b c a = b a c a ,
for a, b, c, Q.
Hence (Q, +, .) is a ring.

Example 13: A Gaussian integer is a complex number a + ib, where a and b are integers.
Show that the set J (i) of Gaussian integers forms a ring under ordinary addition and multiplication
of complex numbers.

Solution: Let a + ib and a + ib be any two elements of J (i) then
1 1 2 2
(a + ib ) + (a + ib ) = (a + a ) + i (b + b )
1 1 2 2 1 2 1 2
= A + iB (say)

and (a + ib ) . (a + ib ) = (a a – b b ) + i (a b + b a )
1 1 2 2 1 2 1 2 1 2 1 2
= C + i D (say)
These are Gaussian integers and therefore J (i) is closed under addition as well as multiplication
of complex numbers.
Addition and multiplication are both associative and commutative compositions for complex
numbers.

Also, multiplication distributes with respect to addition.
0 (= 0 + 0i) J (i) is the additive identity.
The additive inverse of a + ib , J (i) is
(–a) + (–b) i J (i) is
(a + ib) + (–a) + (–b) i

= (a – a) + (b – b) i
= 0 + 0i = 0.
The Gaussian integer 1 + 0.i is multiplicative identity.

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