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.

Answer 1

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 ∗.