Question
Consider the binary operations * : R x R → R and a : R x 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 * b). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Solution
a * b = | a – b |
= | – (b – a) = | b – a | = b * a
⇒ a * b = b * a ∀ a, b ∈ R
⇒ ‘*’ is commutative.
Let a, b, c ∈ R,
∴ (a * b) * c = | a – b | * c = || a – b | – c |
and a * (b * c) = a * | b – c | = | a – | b – c||
⇒ (a * b) * c ≠ a * (b * c)
∴ ‘*’ is not associative.
Again a o b = a and b o a = b
⇒ a o b ≠ b o a
⇒ o is not commutative.
Now, let a, b, c ∈R
∴ (a o b) o c = a o c = a and a o(b o c) = a o b = a ⇒ ‘o’ is associative.
Again a * (b o c) = a * b = | a – b | and (a * b) o (a * c) = | a – b | o | a – c |
= | a – b | ∴ a * (b o c) = (a * b) o (a * c)
Also a o(b * c) = a o | b – c | = a and (a o b) * (a o c) = a * a = | a – a | = 0 ⇒ a o (b * c) ≠ (a o b) *(a o c)
∴ ‘o’ is not distributive over ‘*’.
