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...

Answer Details

The operation is \(m\ast n=\dfrac{m^{2}-n^{2}}{2mn}\).

(a) Evaluate \(-3\ast 2\)

\[-3\ast 2=\frac{(-3)^{2}-2^{2}}{2(-3)(2)}=\frac{9-4}{-12}=\frac{5}{-12}=-\frac{5}{12}.\]

(b) Is \(\ast\) associative?

The operation is associative only if \((m\ast n)\ast p=m\ast(n\ast p)\) for all admissible values. Test the counterexample \(m=1,\;n=2,\;p=3\).

First \(m\ast n=\dfrac{1-4}{2(1)(2)}=-\dfrac{3}{4}\). Then

\[(m\ast n)\ast p=\left(-\tfrac{3}{4}\right)\ast 3=\frac{\left(-\frac{3}{4}\right)^{2}-3^{2}}{2\left(-\frac{3}{4}\right)(3)}=\frac{\frac{9}{16}-9}{-\frac{9}{2}}=\frac{-\frac{135}{16}}{-\frac{9}{2}}=\frac{135}{16}\times\frac{2}{9}=\frac{15}{8}.\]

Next \(n\ast p=\dfrac{4-9}{2(2)(3)}=-\dfrac{5}{12}\). Then

\[m\ast(n\ast p)=1\ast\left(-\tfrac{5}{12}\right)=\frac{1-\frac{25}{144}}{2(1)\left(-\frac{5}{12}\right)}=\frac{\frac{119}{144}}{-\frac{5}{6}}=\frac{119}{144}\times\left(-\frac{6}{5}\right)=-\frac{119}{120}.\]

Since \(\dfrac{15}{8}\neq -\dfrac{119}{120}\), we have \((m\ast n)\ast p\neq m\ast(n\ast p)\). Therefore \(\ast\) is not associative.