For a, b ∊ Q, * is a binary operation on Q defined as: a * b = $\frac{3ab}{5}$ Now, b * a = $\frac{3ba}{5}$ As, ab = ba ⇒ $\frac{3ab}{5}$ = $\frac{3ba}{5}$ ∴ a *b = b*a So, the binary operation * is commutative Let a, b, c ∊ Q a * (b * c) = a * $\frac{3bc}{5}$ ⇒ a * (b * c) = $\frac{3a\frac{3bc}{5}}{5}$ ... (1) ⇒ a * (b * c) = $\frac{3abc}{25}$ Now, (a * b) * c = $\frac{3ab}{5}$ * c ⇒ (a * b) * c = $\frac{3\frac{3ab}{5c}}{5}$ ... (2) From equations (1) and (2): a * (b * c) = (a * b) * c So, the binary operation * is associative. Element e is the identity element on set A for the binary operation * if a * e = e * a = a ∀ a ∊ A Consider $\frac{5}{3}$ ∊ Q a * $\frac{5}{3}$ = $\frac{3a\frac{5}{3}}{5}$ = a And $\frac{5}{3}$ * a = $\frac{3\frac{5}{3a}}{5}$ = a Now, a * $\frac{5}{3}$ = $\frac{5}{3}$ * a = a Therefore, $\frac{5}{3}$ is the identity element of the binary operation * on Q.