1. **State the problem:** We want to find the multiplicative inverse of the expression $$2 + \frac{y}{S}$$ and show that it can be written in the form $$c + d \frac{y}{3}$$ where $$c, d$$ are rational numbers. 2. **Write the inverse expression:** Let the inverse be $$\frac{1}{2 + \frac{y}{S}}$$. 3. **Combine the terms in the denominator:** $$ 2 + \frac{y}{S} = \frac{2S + y}{S} $$ So, $$ \frac{1}{2 + \frac{y}{S}} = \frac{1}{\frac{2S + y}{S}} = \frac{S}{2S + y} $$ 4. **Assuming $$S = 3$$ (since the problem mentions denominator 3 in the form), substitute:** $$ \frac{S}{2S + y} = \frac{3}{2\cdot3 + y} = \frac{3}{6 + y} $$ 5. **Rewrite $$\frac{3}{6 + y}$$ in the form $$c + d \frac{y}{3}$$:** Start by rewriting the denominator: $$ \frac{3}{6 + y} = \frac{3}{6 + y} \times \frac{1/3}{1/3} = \frac{3 \times \frac{1}{3}}{(6 + y) \times \frac{1}{3}} = \frac{1}{2 + \frac{y}{3}} $$ 6. **Use geometric series expansion for $$\frac{1}{2 + \frac{y}{3}}$$ if $$\left|\frac{y}{6}\right| < 1$$:** $$ \frac{1}{2 + \frac{y}{3}} = \frac{1}{2} \cdot \frac{1}{1 + \frac{y}{6}} = \frac{1}{2} \sum_{n=0}^{\infty} \left(-\frac{y}{6}\right)^n $$ 7. **Express first two terms only for form $$c + d \frac{y}{3}$$:** $$ \frac{1}{2} - \frac{1}{2} \cdot \frac{y}{6} + \dots = \frac{1}{2} - \frac{y}{12} + \dots $$ Rewrite $$- \frac{y}{12}$$ as $$- \frac{1}{4} \cdot \frac{y}{3}$$, so $$ \frac{1}{2} - \frac{1}{4} \frac{y}{3} $$ 8. **Conclusion:** The inverse can be expressed as $$ c + d \frac{y}{3} $$ where $$c = \frac{1}{2}$$ and $$d = -\frac{1}{4}$$, both rational numbers. **Final answer:** $$ \frac{1}{2 + \frac{y}{S}} = \frac{1}{2} - \frac{1}{4} \frac{y}{3} $$ for $$S=3$$ and $$y$$ sufficiently small to ensure convergence. This proves the form with rational $$c, d$$ exists.