Page 44 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
Example 15: Prove that the set of residues {0, 1, 2, 3, 4} modulo 5 is using with respect to
addition and multiplication of residue classes (mod 5).
Solution: Let R = {0, 1, 2, 3, 4}.
Addition and multiplication tables for the given set R, are as under

+ mod 5 0 1 2 3 4 mod 5 0 1 2 3 4
0 0 1 2 3 4 0 0 0 0 0 0
1 1 2 3 4 0 1 0 1 2 3 4
2 2 3 4 0 1 2 0 2 4 1 3
3 3 4 0 1 2 3 0
4 4 0 1 2 3 4

From the addition composition table following is clear:
(i) Since all the elements of the table belong to the set, it is closed under addition (mod 5).
(ii) Addition (mod 5) is always associative.
(iii) 0 R is the identity of addition.
(iv) The additive inverse of the elements 0, 1, 2, 3, 4 are 0, 4, 3, 2, 1 respectively.

(v) Since the elements equidistant from the principal diagonal are equal to each other, the
addition (mod 5) is commutative.
Hence (R, +) is an abelian group.

From the multiplication composition table, we see that (R, .) is semi group, i.e., following
axioms hold good.
(vi) Since all the elements of the table are in R, the set R is closed under multiplication (mod 5).

(vii) Multiplication (mod 5) is always associative.
(viii) The multiplication (mod 5) is left as well as right distributive over addition (mod 5).
Hence (R, +, .) is a ring.

Example 16: Prove that the set of residue classes modulo the positive integer m is a ring
with respect to addition and multiplication of residue classes (mod m).
Solution: Let R = {0, 1, 2, …, r , …, r , … (m – 1) (mod m)}
1 2
R (R, +) is an abelian group.
1
(i) Let r , r R then
1 2
where r is the remainder obtained after dividing r + r by m.
1 2
R is closed under addition (mod m).
(ii) Addition is associative.
(iii) O R is the identity element for addition in R.

(iv) Since (m – r) + r = m = 0, the additive inverse of r R is (m – r) R.

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