Question

Consider the binary operations $*:R\times R\to R$ and $o:R\times R\to R$ defined as $a*b=\left| a-b \right|$ and $aob=a$ for all$a,b\in R$. Show that $'*'$ is commutative not associative and $'o'$ is associative but not commutative.

Answer 1

Hint: A binary operation is map which combines two values at the same time from two same or different sets and results in one value. Any binary operation $'\#'$ is commutative for some $a$ from some set $A$ and $b$ from some set $B$ if and only if $a\#b=b\#a$. The same binary operation $'\#'$ is associative with $a$ from $A$, $b$ from $B$ and $c$ from some set $C$ if and only if $\left( a\#b \right)\#c=a\#\left( b\#c \right)$.

Complete step by step answer:
The first given operation is $*:R\times R\to R$ defined as $a*b=\left| a-b \right|$. It means the operation $'*'$ takes two real numbers and returns a modulus of their differences.
Checking commutative property of $'*'$,
\[\begin{align}
  & a*b=\left| a-b \right|=\left\{ \begin{matrix}
   a-b & \text{if }a>b \\
   b-a & \text{if ba} \\
\end{matrix} \right. \\
 & b*a=\left| b-a \right|=\left\{ \begin{matrix}
   a-b & \text{if }a>b \\
   b-a & \text{if ba} \\
\end{matrix} \right. \\
\end{align}\]
So $a*b=b*a $.\[\]
From the above calculation we conclude that $'*'$ is commutative. Checking associative property of $'*'$,\[\]
\[\begin{gathered}
  \left( {a*b} \right)*c = \left| {\left| {a - b} \right| - c} \right| = \left\{ {\begin{array}{*{20}{c}}
  {\left| {a - b} \right| - c}&{{\text{if }}\left| {a - b} \right| > c} \\
  {c - \left| {a - b} \right|}&{{\text{if }}\left| {a - b} \right| < c}
\end{array}} \right. \\
  a*(b*c) = \left| {a - \left| {b - c} \right|} \right| = \left\{ {\begin{array}{*{20}{c}}
  {a - \left| {b - c} \right|}&{{\text{if }}a > \left| {b - c} \right|} \\
  {\left| {b - c} \right| - a}&{{\text{if }}a < \left| {b - c} \right|}
\end{array}} \right. \\
\end{gathered} \]
So, $\left( a*b \right)*c=a*\left( b*c \right)$.
From the above calculations we conclude that $'*'$ is not associative.\[\]

The second given operation is $o:R\times R\to R$ defined as $aob=a$. It means the operation $o$ takes two elements in order and returns the first element with respect to the order of operation.
Checking commutative property of $'o'$,
\[\begin{align}
  & aob=a \\
 & boa=b \\
\end{align}\]
So $ aob\ne boa$\[\]
From the above calculations we conclude that $'o'$ is not commutative.
Checking associative property of$'o'$,
\[\begin{align}
  & \left( aob \right)oc=aoc=a \\
 & ao(boc)=aob=a \\
\end{align}\]
So $\left( aob \right)oc=ao\left( boc \right)$
From the above calculations we conclude that $'o'$ is associative.

Note: The question tests the basic definition of commutative and associative property of any binary operation. The commutative property checks whether reversing the order of the elements returns the same value or not. The associative property checks whether the change in order of appliance of the operation changes the returning value or not. We need to be careful while testing the associative property because the calculation errors will result in erroneous conclusions.