A binary operation (ast) is defined on the set of rational numbers by (m ast n = frac{m^{2} – n^{2}}{2mn}, m neq 0 ; n neq 0).
(a) Find (-3 ast 2).
(b) Show whether or not (ast) is associative.
Explanation
(a) (-3 ast 2 = frac{(-3)^{2} – (2)^{2}}{2(-3)(2)})
= (frac{9 – 4}{-12})
= (-frac{5}{12}).
(b) The operation (ast) is associative if given a, b and c which are real numbers, then
((a ast b) ast c = a ast (b ast c))
Let a = -3 ; b = 2 ; c = 1.
((-3 ast 2) ast 1 = (-frac{5}{12}) ast 1)
= (frac{(-frac{5}{12})^{2} – 1^{2}}{2(-frac{5}{12})(1)})
= (frac{frac{25}{144} – 1}{-frac{5}{6}})
= (frac{-frac{119}{144}}{frac{-5}{6}})
= (frac{119}{120})
(-3 ast (2 ast 1) )
(2 ast 1 = frac{2^{2} – 1^{2}}{2(2)(1)})
= (frac{3}{4})
(-3 ast frac{3}{4} = frac{(-3)^{2} – (frac{3}{4})^{2}}{2(-3)(frac{3}{4})})
= (frac{9 – frac{9}{16}}{-frac{9}{2}})
= (-frac{15}{8})
((-3 ast 2) ast 1 neq -3 ast (2 ast 1))
The operation (ast) is not associative.