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 ∗.