Q. 12

# Consider the binary operations ∗: R × R → R and o: R × R → R defined as a ∗b = |a – b| 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 :

It is given that ∗: R × R → R and o: R × R → R defined as

a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R.

For a, b ϵ R, we get:

a*b = |a-b|

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

Therefore, a*b = b*a

⇒ the operation * is commutative.

We can see that

(1*2)*3 = (|1-2|)*3 = |1-3| = 2

1*(2*3) = 1*(|2-3|) = 1*1 = 1

Therefore, the operation * is not associative.

Now, consider the operation o:

We can observed that 1o2 = 1 and 2o1 = 2.

⇒ 1o2 ≠ 2o1(where 1, 2 ϵ R)

⇒ the operation o is not commutative.

Let a, b, c ϵ R. Then, we get:

(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 ϵ R, then we have:

a * (b o c) = a * b = |a –b|

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

Thus, 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)* (1o3) = 1 * 1 = |1 –1| = 0

Therefore, 1 o (2 * 3) ≠ (1 o 2)* (1o3)( where 1, 2, 3 ϵ R)

Therefore, the operation o does not distribute over *.

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