GO Classes Test Series 2024 | Discrete Mathematics | Test 3 | Question: 19

Answer 1

• A. $x + y$ is closed for integer thus closure property exist addition is associative
      for addition identity exist i.e $0$
      for addition inverse exist
      and addition is commutative also
      Therefore it is abelian group
  B. $a \ast e = a = e \ast a$ identity element not exist thus not abelian group
  C. $x \times y$ multiplication is closed under integers, commutative associative and identity exist i.e $1$
      But for inverse $a \times a^{-1} =$ identity
      $a \times a^{-1} = 1$
      $a^{-1} = \dfrac{1}{a} $
     for $a = 0$ it will not be defined.
     $\therefore$ inverse not exist thus not abelian group.
  D.  $x \ast e = x$
    $x^{2} + e^{2} = x \Rightarrow e = \sqrt{x-x^{2}}$ does not belong to integers.
    $\therefore$ identity does not exist thus not abelian group.