**Toolbox:**

- A binary operation $\ast$ on a set $A$ is a function $\ast$ from $A \times A$ to $ A$. Therefore, if $a,b \in A \Rightarrow a \ast b \in A\; \forall\; a,b, \in A$
- An operation $\ast$ on $A$ is commutative if $a\ast b = b \ast a\; \forall \; a, b \in A$
- An operation $\ast$ on $A$ is associative if $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in A$

Given On $Z \ast$ is defined by $a \ast b = a - b$.

$\textbf {Step 1: Checking if the operation is Binary}$:

It is given that $a\ast b = a - b$, which is $\in Z\; \forall\; a,b, \in Z$.

Therefore the operation is binary.

$\textbf {Step 2: Checking if the operation is Commutative}$:

For an operation $\ast$ to be commutative $a\ast b = b \ast a$.

Since $a \ast b = a - b$ and $b \ast a = b - a$ are not equal for all $a,b \in Z$, the operation is not commutative.

This can be verified with a simple example. Let $a=1, b = 2$.

$\Rightarrow a \ast b = 1 - 2 = -1$ and $b \ast a = 2 -1 = 1$ which are clearly not equal.

$\textbf {Step 3: Checking if the operation is Associative}$:

For an operation $\ast$ to be associative $a\ast ( b \ast c) = (a \ast b) \ast c\; \forall \; a, b, c \in Z$

We can see that the operation is not associative by applying the opertaion on the elements $a,b,c \in Z$ as follows:

$\Rightarrow a \ast (b \ast c) = a \ast (b - c) = a - (b-c) = a-b+c$.

$\Rightarrow (a \ast b) \ast c) = (a-b) \ast c = a-b-c$

This can be verified with a simple example. Let $a=1, b = 2, c = 3$.

$\Rightarrow a \ast (b \ast c) = a \ast (b - c) = a - (b-c) = a-b+c$ = 1-2+3 = 2$.

$\Rightarrow (a \ast b) \ast c) = (a-b) \ast c = a-b-c = 1 -2-3 = -4$.

These are clearly not equal, hence the operation is not assocative.