Question

1. Determine whether * is a binary operation on the given set. If it is a binary operation, decide whether it is associative and commutative. Justify your answers. a. Define * on Q+ by a *b = b. Define * on N by a*b = %.

Answer 1

(a) let, a = p/q and b = r/s belong to Q⁺

Then, a*b = a/b = ps/qr > 0

Hence, a/b belongs to Q⁺ for all a, b in Q⁺

So, Q⁺ is closed under the operation *

Also, a*b = a/b = ps/qr

&, b*a = b/a = qr/ps

So, a*b is not equal to b*a, in general. So, * is not commutative.

Take, a = 2 & b = 6. Then, a*b = 2/6 = 1/3 while, b*a = b/a = 6/2 = 3

So, a*b is not equal to b*a. Hence not commutative.

Similarly, * is not associative.

Take, c = 12

Then, b*c = b/c = 6/12 = 1/2

and, a*(b*c) = 2/(1/2) = 4

And, a*b = 1/3 then,

(a*b)*c = (1/3)/12 = 1/36

So, a*(b*c) $\neq$ (a*b)*c

So, * is not associative.

(b).take, a = 6 & b = 12

Then, a*b = 6/12 = 1/2 which does not belong to N

So, N is not closed under *

Hence, * is not a binary operation on N.