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

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 \(x=a+b\sqrt{2},y=c+d\sqrt{2};a,b,c,d∈ Q.\)
•  \(x+y=(a+c)+(b+d)\sqrt{2}∈ G,\) 
• 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 \(0=0+0\sqrt{2}∈ G\) such that for all \(x=a+b\sqrt{2}∈ G.\) 
•  \(x+0=(a+b\sqrt{2})+(0+0\sqrt{2})=a+b\sqrt{2}=x\) 
• Similarly, we have 0 + x = x.
• ∴ 0 is the identity element of G and satisfies the identity axiom.

(4) Inverse axiom:

• For each \(x=a+b\sqrt{2}∈ G\), there exists \(-x=(-a)+(-b)\sqrt{2}∈ G\)  such that \(x+(-x)=(a+b\sqrt{2})+((-a)+(-b)\sqrt{2})\) 
• = \((a+(-a))+(b+(-b))\sqrt{2}=0\) 
• Similarly, we have (-x) + x = 0.
• ∴ \((-a)+(-b)\sqrt{2}\) is the inverse of \(a+b\sqrt{2}\) and satisfies the inverse axiom.

(5) Commutative axiom:

•  \(x+y=(a+c)+(b+d)\sqrt{2}=(c+a)+(d+b)\sqrt{2}\) 
• = \((c+d\sqrt{2})+(a+b)\sqrt{2}\) 
• = y + x, for all x,y ∈ G.
• ∴ The commutative property is true.

Hence, Non-commutativity property fails for G