Consider the binary operations*: R ×R → and o: R × R → R defined as a * b = |a - b| and ao b = a, &mnForE;a, b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mnForE;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.
Solution
It is given that *: R ×R → and o: R × R → R is defined as
a * b = |a - b| and a o b = a, &mnForE;a, b ∈ R.
For a, b ∈ R, we have:
a * b = |a - b|
b * a = |b -a| = |-(a-b)| = |a - b|
∴a * b = b * a
∴ The operation * is commutative.
It can be observed that,
`(1*2) *3 = (|1 - 2|)* 3= 1 * 3 = |1 - 3| =2`
1 * (2 * 3) = 1 *(|2 - 3|) = 1 * 1 = |1-1| = 0
:. (1*2)*3 != 1 * (2 * 3) (where `1, 2,3 in R`)
∴The operation * is not associative.
Now, consider the operation o:
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴1 o 2 ≠ 2 o 1 (where 1, 2 ∈ R)
∴The operation o is not commutative.
Let a, b, c ∈ R. Then, we have:
(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 ∈ R, then we have:
a * (b o c) = a * b = |a - b|
(a * b) o (a * c) =(|a-b|)o(|a-c|) = |a - b|
Hence, 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) * (1 o 3) = 1 * 1 =|1 - 1| = 0
∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ R)
∴The operation o does not distribute over *.
