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# Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that

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 $\forall a,b,c \in R$,
$a*(b\circ c) = (a*b)\circ (a*c)$. [If it is so, we say that the operation ∗ distributes
over the operation $\circ$]. Does $\circ$ distribute over ∗? Justify your answer.

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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 *.

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