Consider the binary operations $*$ : R × R $\rightarrow$ R and o : R × R$\rightarrow$ R defined as a $*$ b $= |a$ - $b|$ and aob $=$ a, $\forall a,b \in R$. Show that $*$ is commutative but not associative and o is associative but not commutative. Further, show that $\forall$a, b, c $\in$R, a $*$ ( boc ) $=$ (a $*$ b) o (a $*$ c). (If it is so, we sav that the operation $*$ distributes over the operation o]. Does o distribute over $*$? Justify your answer.

For commutativity :

a $*$ b $= |a$ - $b|$ and b $*$ a $= |b$ - $a| = |a$ - $b|$

$\therefore$ a $*$ b $=$ b $*$ a

Thus, the operation $*$ is commutative.

For associativity

Consider, a $*$ ( b $*$ c ) $=$ a $* |b$ - $c| = |a$ - $|b$ - $c||$

Also (a $*$ b) $* c$ = $|a$ - $b|$ * $c$ = $||a$ - $b|$ - $c|$

$\therefore$ a $*$ ( b $*$ c) $\ne$(a $*$ b) $*$ c

Thus, the operation $*$ is not associative.

For commutativity

aob $=$ a $\forall$a, b $\in$ R. Now, boa $=$ b $\Rightarrow$ aob $\ne$ boa

Thus, operation o is not commutative.

For associativity

ao ( boc ) $=$ aob $=$ a and (aob) oc $=$ aoc $=$ a

ao (boc ) $=$ ( aob ) oc. Thus, operation o is associative.

To prove : a $*$ (boc ) $=$ (a $*$ b) o (a $*$ c )

L.H.S. $= a$ * $( boc )$ = $a$ * $b$ = $|a$ - $b|$

R.H.S. $= (a$ * $b) o (a$ * $c )$ = $|a$ - $b| o |a$ - $c|$ = $|a$ - $b|$

Thus, a $*$ (boc) $=$ (a $*$ b ) o (a $*$ c ). Hence proved.

Another distributive law

ao( b $*$ c ) $=$ (aob ) $*$ (aoc)

L.H.S. $= a o (b$ * $c )$ = $a o ( |b$ - $c|)$ = $a$

R.H.S. $= (aob)$ * $(aoc )$ = $a$ * $a$ = $|a$ - $a|$ = $0 .$

As, L.H.S.$*$ R.H.S. Hence, the operation o does not distribute over $*$.