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Q. 12 Consider the binary operations $* : R \times R \rightarrow R$ and $\circ : R \times R \rightarrow R$ defined as
$a b = |a - b|$ and $a \circ b = a \;\forall a \in R$ . Show that ∗ is commutative but not
associative, $\circ$ is associative but not commutative. Further, show that $\forall a,b,c \in R$ ,
$a(b\circ c) = (ab)\circ (ac)$ . [If it is so, we say that the operation ∗ distributes
over the operation $\circ$ ]. Does $\circ$ distribute over ∗? Justify your answer.

Answers (1)

best_answer

Given $* : R \times R \rightarrow R$ and $\circ : R \times R \rightarrow R$ is defined as
$a *b = |a - b|$ and $a \circ b = a \;\forall a,b \in R$

For $a,b \in R$ , we have

$a *b = |a - b|$

$b *a = |b - a| = \left | -(a-b) \right |=\left | a-b \right |$

$\therefore a*b = b *a$

$\therefore$ the operation is commutative.

$(1*2)*3 = (\left | 1-2 \right |)*3=1*3=\left | 1-3 \right |=2$

$1*(2*3) = 1*(\left | 2-3 \right |)=1*1=\left | 1-1 \right |=0$

$\therefore (1*2)*3\neq 1*(2*3)$ where $1,2,3 \in R$

$\therefore$ the operation is not associative

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

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

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

Hence, $a*(b \circ c)=(a*b )\circ(a* c)$

Now,

$1\circ (2*3) = 1\circ(\left | 2-3 \right |)=1\circ1=1$

$(1\circ 2)*(1 \circ 3) =1*1=\left | 1-1 \right |=0$

$\therefore 1 \circ(2*3)\neq (1\circ 2)*(1 \circ 3)$ for $1,2,3 \in R$

Hence, operation o does not distribute over operation *.

.

Posted by

seema garhwal

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