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 $$G$$​?

Correct option is D

​$$\begin{aligned}&\textbf{1. Closure Property:} \\&\text{Let } x = a + b\sqrt{2} \text{ and } y = c + d\sqrt{2}, \text{ where } a, b, c, d \in \mathbb{Q}. \\&\text{Then, } x + y = (a + c) + (b + d)\sqrt{2}. \\&\text{Since } \mathbb{Q} \text{ is closed under addition, } a + c \in \mathbb{Q} \text{ and } b + d \in \mathbb{Q}. \\&\text{Thus, } x + y \in G. \ \textbf{Closure holds.} \\[10pt]&\textbf{2. Associative Property:} \\&\text{Addition in } G \text{ inherits associativity from the real numbers.} \\&\text{For all } x, y, z \in G, \ (x + y) + z = x + (y + z). \ \textbf{Associativity holds.} \\[10pt]&\textbf{3. Identity Property:} \\&\text{The additive identity is } 0 = 0 + 0\sqrt{2} \in G. \\&\text{For any } x = a + b\sqrt{2} \in G, \ x + 0 = x. \ \textbf{Identity holds.} \\[10pt]&\textbf{4. Inverse Property:} \\&\text{For } x = a + b\sqrt{2} \in G, \text{ its additive inverse is } -x = -a - b\sqrt{2} \in G. \\&\text{Clearly, } x + (-x) = 0. \ \textbf{Inverse holds.} \\[10pt]&\textbf{5. Commutative Property:} \\&\text{For } x = a + b\sqrt{2} \text{ and } y = c + d\sqrt{2}, \\&\quad x + y = (a + c) + (b + d)\sqrt{2} = (c + a) + (d + b)\sqrt{2} = y + x. \\&\text{Thus, } G \text{ is commutative under addition.} \ \textbf{Commutativity holds.} \\[10pt]&\textbf{Conclusion:} \\&\text{All group axioms (closure, associativity, identity, inverse) are satisfied, and the operation is commutative.} \\&\textbf{No property fails for } G \text{ under addition.}\end{aligned}$$​​