If the operation *is defined on the set of real numbers by x*y=x^2-y

Question

1. Find (a) 4*5.    (b)5*4 and hence determine whether or not the operation *is commutative on the set of real numbers
2. Using one counter example, show that the operation *is not associative on the set of real numbers
3. Find the truth set of the equation x*4=0

Answer 1

1. 4*5=16-5=11; 5*4=25-4=21. Because x*y is not equal to y*x, the associative property does not apply.
2. Assume that a*b=b*a where a and b are real, but a is not equal to b, then a^2-b=b^2-a. a^2-b^2=b-a; (a-b)(a+b)=b-a=-(a-b); a+b=-1 (dividing through by a-b). But a+b=-1 forces a relationship between a and b, and that means the associative property does not apply generally. Therefore, a*b<>b*a.
3. x*4=0 implies x^2-4=0=(x-2)(x+2), so x=-2 or 2.