Consider the binary operations*: **R **×**R **→ and o: **R** × **R** → **R **defined as a * b = |a - b| and *a*o *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 *.