# Consider the Binary Operations*: R ×R → and O: R × R → R Defined as a * B = |A - B| and Ao B = A, andMnfore;A, B ∈ R - Mathematics

Consider the binary operations*: ×→ and o: R × R → defined as a * b = |a - b| and ab = a, &mnForE;ab ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that &mnForE;abc ∈ Ra*(b o c) = (ab) 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 *: ×→ and o: R × R → is defined as

a * b = |a - b| and a o b = a, &mnForE;ab ∈ R.

For ab ∈ 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 ab∈ R. Then, we have:

(b) o c = a o c a

a o (b o c) = a o b = a

⇒ b) o c = a o (b o c)

∴ The operation o is associative.

Now, let ab∈ R, then we have:

a * (b o c) = a * b = |a - b|

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

Hence, * (c) = (b) o (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 *.

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 12 Maths
Chapter 1 Relations and Functions
Q 12 | Page 30