Answer :

It is given that : R × R R and o: R × R R defined as

a b = |a – b| and a o b = a, a, b R.


For a, b ϵ R, we get:


a*b = |a-b|


b*a = |b-a| = |-(a-b)| = |a-b|


Therefore, a*b = b*a


the operation * is commutative.


We can see that


(1*2)*3 = (|1-2|)*3 = |1-3| = 2


1*(2*3) = 1*(|2-3|) = 1*1 = 1


Therefore, the operation * is not associative.


Now, consider the operation o:


We can observed that 1o2 = 1 and 2o1 = 2.


1o2 ≠ 2o1(where 1, 2 ϵ R)


the operation o is not commutative.


Let a, b, c ϵ R. Then, we get:


(a o b) o c = a o c = a


a o (b o c) = a o b = a


(a o b) o c = a o (b o c)


the operation o is associative.


Now, let a, b, c ϵ R, then we have:


a * (b o c) = a * b = |a –b|


(a * b) o (a * c) = (|a –b|) o (|a –c| = |a –b|


Thus, a * (b o c) = (a * b) o (a * c)


Now,


1 o (2 * 3) = 1 o (|2 –3|) = 1 o 1 = 1


(1 o 2)* (1o3) = 1 * 1 = |1 –1| = 0


Therefore, 1 o (2 * 3) ≠ (1 o 2)* (1o3)( where 1, 2, 3 ϵ R)


Therefore, the operation o does not distribute over *.


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