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.

Answers (1)

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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