Question

Consider the binary operations*: R ×R → and o: R×R →R defined as $a*b=\mid a-b \mid$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

It is given that *: R ×R → and o: R × R → R is defined as $a*b=\mid a-b \mid$ and a o b = a, ∀a, b ∈ R
For a, b ∈ R, we have:
$a*b=\mid a-b \mid$
$b*a=\mid b-a \mid=\mid-(a-b)\mid=\mid a-b\mid$
∴ $a*b=b*a$
∴ The operation * is commutative.
It can be observed that,
$(1*2)*3=(\mid 1-2\mid)*3=1*3=\mid 1-3\mid=2$
$1*(2*3)=1*(\mid2-3\mid)=1*1=\mid1-1\mid=0$.
Therefore $(1*2)*3\neq$ $1*(2*3)$ (where 1,2,3 ∈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
$\Rightarrow$ 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 =Ia-bI
(a * b) o (a * c) =(Ia-bI) o (Ia-cI)=Ia-bI
Hence, a * (b o c) = (a * b) o (a * c).
Now,1 o (2 * 3) =1o (I2-3I)=1o1=1.
(1 o 2) * (1 o 3) = 1 * 1 =I1-1I=0.
∴1 o (2 * 3) ≠ (1 o 2) * (1 o 3) (where 1, 2, 3 ∈ R)

The operation o does not distribute over *.