Comment
Share
Q)

# For each operation $\ast$ defined below, determine whether $\ast$ is binary, commutative or associative: On $Z$, define $a \ast b = a-b$

Comment
A)
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.