Mathematical Foundation of Computer Science

DHARM

98 MATHEMATICAL FOUNDATION OF COMPUTER SCIENCE

Example 4.19. Test the following statements are true or false.

1. If x is an element of ring and m, n ∈ N, then (am)n = amn. [True]


2. Every subgroup of an abelian group is not necessarily abelian. [False]


3. The relation of isomorphism in the set of all groups is not an equivalence relation.
   [False]


4. If is a binary operation on any set X, then x x = x for all x ∈ X. [False]


5. If is any commutative binary operation on any set X, then
   x (y z) = (y z) x for all x, y, and z ∈ X. [True]


6. If is associative then x (y z) = (y z) x for all x, y, and z Î X. [False]
   Example 4.20. Test the following statements are true or false.


7. Every binary operation defined over a set of one element is both commutative and
   associative. [True]


8. A binary operation defined over a set X assigns at least one element of X to each or-
   dered pair of elements of X. [False]


9. A binary operation defined over a set X assigns exactly one element of X to each or-
   dered pair of elements of X. [True]
   Example 4.21. Prove that a group G is abelian, if for a, b ∈ G
   (i)a^2 = ê (ii)(a b)^2 = a^2 b^2
   (iii)b–1 a–1 b a = ê
   Sol.
   (i) Since, a^2 = ê ⇒ a–1 = a
   Further since a–1 = a and b–1 = b then a b = a–1 b–1 = (b a)–1 = b a (because (b a)–1 = b a).
   Hence G is abelian.
   (ii) Since (a b)^2 = a^2 b^2 ⇒ a b a b = a a b b
   ⇒ b a = a b [by cancellation law]
   Hence G is abelian.
   (iii) Given, b–1 a–1 b a = ê
   Since a b can be written as a b ê or
   a b = a b ê = a b b–1 a–1 b a [replace ê by b–1 a–1 b a]
   ⇒ a(b b–1) (a–1 b a)[∴ b b–1 = ê]
   ⇒ (a a–1) b a
   ⇒ b a [∴ a a–1 = ê]
   Hence G is abelian.
   Example 4.22. Examine whether the set of rational number (1 + 2p)/(1 + 2q), where p and q
   are integers, forms a group under multiplication.
   Sol. Assume two rational numbers are (1 + 2p)/(1 + 2q) and (1 + 2P)/(1 + 2Q), where p, q, P,
   and Q are integers. So we find, [(1 + 2p)/(1 + 2q)] • [(1 + 2P)/(1 + 2Q)] = (1 + 2p + 2P + 2pP)/(1

• 2q + 2Q + 2qQ) that returns again a rational number. Therefore set is closed under multi-
  plication. Associative property is also satisfied that can also verified. For identity,
  1 = (1 + 2 • 0)/(1 + 2 • 0)
  and the inverse of (1 + 2p)/(1 + 2q) is (1 + 2q)/(1 + 2p).
  Hence set of rational numbers of the form (1 + 2p)/(1 + 2q) forms an abelian group
  under multiplication.