For a, b ∊ Q, * is a binary operation on Q defined as: a * b = $\frac{3\mathit{a}\mathit{b}}{5}$
Now, b * a = $\frac{3\mathit{b}\mathit{a}}{5}$
As, ab = ba
⇒ $\frac{3\mathit{a}\mathit{b}}{5}$ = $\frac{3\mathit{b}\mathit{a}}{5}$
∴ a *b = b*a
So, the binary operation * is commutative
Let a, b, c ∊ Q
a * (b * c) = a * $\frac{3\mathit{b}\mathit{c}}{5}$
⇒ a * (b * c) = $\frac{3\mathit{a}\frac{3\mathit{b}\mathit{c}}{5}}{5}$ ... (1)
⇒ a * (b * c) = $\frac{3\mathit{a}\mathit{b}\mathit{c}}{25}$
Now, (a * b) * c = $\frac{3\mathit{a}\mathit{b}}{5}$ * c
⇒ (a * b) * c = $\frac{3\frac{3\mathit{a}\mathit{b}}{5}\mathit{c}}{5}$ ... (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 * $\frac{5}{3}$ = $\frac{3\mathit{a}\frac{5}{3}}{5}$ = a
And $\frac{5}{3}$ * a = $\frac{3\frac{5}{3}\mathit{a}}{5}$ = 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.