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Question

# Consider the binary operations*: R × R → and o: R × R → R defined as and a o b = a , &mnForE; a , b ∈ R . Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mnForE; a , b , c ∈ 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.

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Solution

## The given binary operations ∗:R×R→R and o:R×R→R is defined as, a∗b=| a−b | and aob=a, ∀ a,b∈R. For a,b∈R, a∗b=| a−b | And b∗a=| b−a | =| −( a−b ) | =| a−b | =a∗b Thus, the binary operation ∗ is commutative. Consider, ( 1∗2 )∗3=( | 1−2 | )∗3 =1∗3 =| 1−3 | =2 And, 1∗( 2∗3 )=1∗( | 2−3 | ) =1∗1 =| 1−1 | =0 Here ( 1∗2 )∗3≠1∗( 2∗3 ). Thus, the binary operation ∗ is not associative. Take the binary operation o:R×R→R, 1o2=1 and 2o1=2 So, 1o2≠2o1. Thus, the binary operation o is not commutative. Consider a,b,c∈R, then, ao( boc )=aob =a And, ( aob )oc=aoc =a So, ( aob )oc=ao( boc ) Thus, the binary operation o is associative. Now, for a,b,c∈R, a∗( boc )=a∗b =| a−b | And, ( a∗b )o( a∗c )=| a−b |o| a−c | =| a−b | Therefore, a∗( boc )=( a∗b )o( a∗c ) Consider, 1o( 2∗3 )=1o( | 2−3 | ) =1o1 =1 And, ( 1o2 )∗( 1o3 )=1∗1 =| 1−1 | =0 This shows that 1o( 2∗3 )≠( 1o2 )∗( 1o3 ). Therefore, the binary operation o is not distributive over the operation ∗.

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