Consider the set G = {a + b : a, b ∈ Q, the set of all rational numbers} with respect to binary operation usual addition. Which condition fails for G?

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AAI ATC Junior Executive 27 July 2022 Shift 1 Official Paper
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  1. Inverse property
  2. Identity element 
  3. Associativity property
  4. Non-commutativity property

Answer (Detailed Solution Below)

Option 4 : Non-commutativity property
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Solution:

The set G = {a + b\(\sqrt2\) : a, b ∈ Q, the set of all rational numbers} with respect to binary operation usual addition. 

Satisfies closure, associative, identity, inverse, and commutative properties.

(1) Closure axiom:

  • Let x,y ∈ G. Then
  •   
  • Since (a + c) and (c + d) are rational numbers.
  • ∴ G is closed with respect to addition.

(2) Associative axiom:

  • Since the elements of G are real numbers, addition is associative.

(3) Identity axiom:

  • There exists  such that for all  
  •   
  • Similarly, we have 0 + x = x.
  • ∴ 0 is the identity element of G and satisfies the identity axiom.

(4) Inverse axiom:

  • For each , there exists   such that  
  •  
  • Similarly, we have (-x) + x = 0.
  • ∴  is the inverse of  and satisfies the inverse axiom.

(5) Commutative axiom:

  •   
  •  
  • = y + x, for all x,y ∈ G.
  • ∴ The commutative property is true.

Hence, Non-commutativity property fails for G

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