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

(A) Given $a \,^*\, b = |a-b|$ and $a \,o\, b = a$ for all $a, b \in R$.
For $^*$,$a \,^*\, b = |a-b|$ and $b \,^*\, a = |b-a| = |-(a-b)| = |a-b|$. Since $a \,^*\, b = b \,^*\, a$,$^*$ is commutative.
For associativity of $^*$,consider $(1 \,^*\, 2) \,^*\, 3 = |1-2| \,^*\, 3 = 1 \,^*\, 3 = |1-3| = 2$,while $1 \,^*\, (2 \,^*\, 3) = 1 \,^*\, |2-3| = 1 \,^*\, 1 = |1-1| = 0$. Since $2 \neq 0$,$^*$ is not associative.
For $o$,$1 \,o\, 2 = 1$ and $2 \,o\, 1 = 2$. Since $1 \neq 2$,$o$ is not commutative.
For associativity of $o$,$(a \,o\, b) \,o\, c = a \,o\, c = a$ and $a \,o\, (b \,o\, c) = a \,o\, b = a$. Since both sides equal $a$,$o$ is associative.
For distributivity of $^*$ over $o$,$a \,^*\, (b \,o\, c) = a \,^*\, b = |a-b|$ and $(a \,^*\, b) \,o\, (a \,^*\, c) = |a-b| \,o\, |a-c| = |a-b|$. Thus,$^*$ distributes over $o$.
For distributivity of $o$ over $^*$,consider $1 \,o\, (2 \,^*\, 3) = 1 \,o\, |2-3| = 1 \,o\, 1 = 1$,while $(1 \,o\, 2) \,^*\, (1 \,o\, 3) = 1 \,^*\, 1 = |1-1| = 0$. Since $1 \neq 0$,$o$ does not distribute over $^*$.