1. **Stating the problem:** We have a binary operation defined on the rational numbers $\mathbb{Q}$ as $a \square b = a + b + ab$ for $a,b \in \mathbb{Q}$. We want to determine why $(\mathbb{Q}, \square)$ is not a group. 2. **Recall group properties:** For $(\mathbb{Q}, \square)$ to be a group, the operation $\square$ must be associative, have an identity element, and every element must have an inverse with respect to $\square$. 3. **Check commutativity:** $$a \square b = a + b + ab = b + a + ba = b \square a,$$ so $\square$ is commutative. Thus, option (a) is false. 4. **Check associativity:** We test if $a \square (b \square c) = (a \square b) \square c$. Calculate $b \square c = b + c + bc$. Then, $$a \square (b \square c) = a + (b + c + bc) + a(b + c + bc) = a + b + c + bc + ab + ac + abc.$$ Calculate $a \square b = a + b + ab$. Then, $$ (a \square b) \square c = (a + b + ab) + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc.$$ Both expressions simplify to $$a + b + c + ab + ac + bc + abc,$$ so $\square$ is associative. Thus, option (b) is false. 5. **Check identity element:** Suppose $e$ is the identity element such that for all $a \in \mathbb{Q}$, $$a \square e = a + e + ae = a.$$ Rearranged, $$e + ae = 0 \implies e(1 + a) = 0.$$ For this to hold for all $a$, $e$ must be zero and $1 + a \neq 0$ for all $a$ except $a = -1$. At $a = -1$, $1 + a = 0$, so the identity condition fails. Thus, $e=0$ acts as identity for all $a \neq -1$, but not for $a = -1$. Hence, there is only one identity element $e=0$, but it is not a global identity for all $\mathbb{Q}$. So option (c) is false. 6. **Check inverses:** For $a$ to have an inverse $a^{-1}$ under $\square$, we need $$a \square a^{-1} = 0,$$ which means $$a + a^{-1} + a a^{-1} = 0.$$ Rearranged, $$a^{-1}(1 + a) = -a.$$ If $a \neq -1$, then $$a^{-1} = \frac{-a}{1 + a}.$$ If $a = -1$, then $1 + a = 0$ and no inverse exists. Therefore, not every element has an inverse under $\square$. **Final conclusion:** The algebraic structure $(\mathbb{Q}, \square)$ is not a group because not every element has an inverse under $\square$. **Answer: d. not every element of Q has an inverse under □.**