Question

Consider the binary operations*: R ×R → and o: R × R → R defined as a * b = |a - b| and a o b = a, "a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that "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

 ∗: 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
therefore, * is commutative.
We can see that (1*2)*3 = (|1-2|)*3 = |1-3| = 2
1*(2*3) = 1*(|2-3|) = 1*1 = 1
here, (1*2)*3 ≠ 1*(2*3)
Therefore, * is not associative.
Now, consider the operation o:
We can observed that 1o2 = 1 and 2o1 = 2.
⇒ 1o2 ≠ 2o1(where 1, 2 ϵ R)
 therefore, 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)
therefore, 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 *.