Defining Binary Operations?

Question

Determine whether the following binary operation is commutative and/or associative
* on Q defined by x*y = (x+y)/2

Could you please show answer in detail step by step.
Thanks...

Answer 1

We know addition is associative and commutative on R, so it is on Q.

If x and y are rational numbers, then, according to your definition, x*y = (x + y)/2. Since addition is commutative on Q, it follows that x*y = (y + x)/2 = y*x, so that x*y = y*x. Thefore, the binary operation * is commutative on Q.

Now, we have to verify if, for every rational numbers x, y and z, it's true that x*(y*z) = (x*y)*z. According to the definition, x*(y*z) = x*((y+z)/2) = (x + (y+z)/2)/2 = ((2x + y + z)/2)/2 = (2x + y + z)/4. Similarly, (x*y)*z = ((x + y)/2 + z)/2 = (x + y + 2z)/4. So, the 2 expressions give the same result if, and only if, 2x + y + z = x + y + 2z => 2x + z = x + 2z => x = z, which is a particular, but not a general condition. Therefore, in general, x*(y*z) is different from (x*y)*z. Therefore, * is not associative.

The operation * is the arithmetic mean of x and y. Our conclusions are not restricted to Q, but holds all over R, as well.

Answer 2

x*y = (x+y)/2
y*x = (y+x)/2= (x+y)/2

=> x*y = y*x, therefore the operation is commutative

x*(y*z) = x* (y+z)/2 = (x*y + x*z)/2
(x*y)*z = (x+y)/2 * z = (x*z + y*z)/2

=> x*(y*z) does not equal (x*y)*z, therefore the operation is not associative