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.
(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.
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