# Consider the binary operations*: R ×R → and o: R × R → R defined as and a o b = a,

Question:

Consider the binary operations*: $\mathbf{R} \times \mathbf{R} \rightarrow$ and $0: \mathbf{R} \times \mathbf{R} \rightarrow \mathbf{R}$ defined as $a * b=|a-b|$ and $a \circ b=a, \& m n$ ForE; $a, b \in \mathbf{R}$. Show that * is commutative but not associative, o is associative but not commutative. Further, show that \&mnForE; $a, b, c \in \mathbf{R}, a^{\star}(b$ o $c)=\left(a^{*} b\right) \circ\left(a^{*} c\right) .$ [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.

Solution:

It is given that ${ }^{*}: \mathbf{R} \times \mathbf{R} \rightarrow$ and $0: \mathbf{R} \times \mathbf{R} \rightarrow \mathbf{R}$ isdefined as

$a * b=|a-b|$ and $a$ o $b=a$, \&mnForE; $a, b \in \mathbf{R}$.

For $a, b \in \mathbf{R}$, we have:

$a * b=|a-b|$

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

a * b = b * a

∴ The operation * is commutative.

It can be observed that,

$\therefore$ The operation * is not associative.

Now, consider the operation o:

It can be observed that $102=1$ and $201=2$.

$\therefore 1 \circ 2 \neq 2 \circ 1$ (where $1.2 \in \mathbf{R}$ )

∴The operation o is not commutative.

Let $a, b, c \in \mathbf{R}$. Then, we have:

$(a \circ b) \circ c=a \circ c=a$

$a \circ(b \circ c)=a \circ b=a$

$\Rightarrow a \circ b) \circ c=a \circ(b \circ c)$

$\therefore$ The operation o is associative.

Now, let $a, b, c \in \mathbf{R}$, then we have:

$a^{\star}(b \circ c)=a^{\star} b=|a-b|$

$\left(a^{*} b\right) \circ\left(a^{*} c\right)=(|a-b|) \circ(|a-c|)=|a-b|$

Hence, $a^{*}(b \circ c)=\left(a^{*} b\right) \circ\left(a^{*} c\right) .$

Now,

$1 \circ(2 * 3)=1 \circ(|2-3|)=1 \circ 1=1$

$(1 \circ 2)^{*}(1 \circ 3)=1^{*} 1=|1-1|=0$

$\therefore 1 \circ\left(2^{*} 3\right) \neq(1 \circ 2)^{*}(1 \circ 3)($ where $1,2,3 \in \mathbf{R})$

The operation o does not distribute over *.