Question

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

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

[If it is so, we say that operation * distributes over the operation o]. Does o distributes over *? Justify your answer.

Answer 1

We have: a * b = | a – b | and a o b = a.
(i) (a) a * b = | a – b |
and b * a = | b – a | = | a – b |
⇒ Operation * is commutative.

(b) (a * b) * c = | a – b | * c = || a-b | – c |
and a * (b * c) = a * | b – c | = |a- |b – c |
Clearly, ||a – b | – c | ≠ |a – |b – c||
⇒ Operation * is not associative.

(ii) (a) a o b = a, b o a = b.
Thus, a o b ≠ ⇒ b o a
⇒ Operation o is not commutative.

(b) (a o b) o c = a o c = a
and a o (b o c) = a o b = a
⇒ Operation o is associative.

(iii) For a * (b o c) = (a * b) o (a * c):
L.H.S. = a * (b o c) = a * b [∵ b o c = b]
= | a – b |

R.H.S. = (a * b) o (a * c)
= | a -b | o | a – c | = | a – b |
⇒ L.H.S. = R.H.S.
Hence, verified,
i.e., a * (b o c) = (a * b) o (a * c)

(iv) a o (b * c) = (a o b) * (a o c)
L.H.S. = a o (b * c) = a o | b – c | = a.
R.H.S. = (a o b) * (a * c) = a * a
= | a – a | = 0.
⇒ L.H.S. ≠ R.H.S.
o is not distributive over *.