Nested Fraction D52815
1. **State the problem:** Simplify the complex nested fraction expression given:
$$\frac{\frac{1 + \frac{1}{3}}{1 + \left(2 - \frac{1}{3}\right)}}{\frac{1 - \left(2 - \frac{1}{2}\right)}{1 + \frac{1}{2}}} \div \frac{2 - \frac{1}{2}}{\frac{1 + \frac{1}{2}}{7 + 1}} \div \frac{1 - 1}{1 + \frac{7}{6}}$$
2. **Recall fraction rules:**
- To add/subtract fractions, find a common denominator.
- To divide fractions, multiply by the reciprocal.
- Simplify step-by-step from innermost parentheses outward.
3. **Simplify each part stepwise:**
- Calculate $1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3}$.
- Calculate $2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}$.
- Then $1 + \left(2 - \frac{1}{3}\right) = 1 + \frac{5}{3} = \frac{3}{3} + \frac{5}{3} = \frac{8}{3}$.
- Numerator of the big fraction is $\frac{4/3}{8/3} = \frac{4}{3} \times \frac{3}{8} = \frac{4}{8} = \frac{1}{2}$.
- Calculate $2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$.
- Calculate $1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}$.
- Calculate $1 - \left(2 - \frac{1}{2}\right) = 1 - \frac{3}{2} = \frac{2}{2} - \frac{3}{2} = -\frac{1}{2}$.
- Denominator of the big fraction is $\frac{-1/2}{3/2} = -\frac{1}{2} \times \frac{2}{3} = -\frac{1}{3}$.
- So the big fraction simplifies to $\frac{1/2}{-1/3} = \frac{1}{2} \times -3 = -\frac{3}{2}$.
- Next, simplify $\frac{2 - \frac{1}{2}}{\frac{1 + \frac{1}{2}}{7 + 1}}$:
- Numerator: $2 - \frac{1}{2} = \frac{3}{2}$ (from above).
- Denominator: $\frac{1 + \frac{1}{2}}{7 + 1} = \frac{\frac{3}{2}}{8} = \frac{3}{2} \times \frac{1}{8} = \frac{3}{16}$.
- So this fraction is $\frac{3/2}{3/16} = \frac{3}{2} \times \frac{16}{3} = 8$.
- Finally, simplify $\frac{1 - 1}{1 + \frac{7}{6}}$:
- Numerator: $1 - 1 = 0$.
- Denominator: $1 + \frac{7}{6} = \frac{6}{6} + \frac{7}{6} = \frac{13}{6}$.
- So this fraction is $\frac{0}{13/6} = 0$.
4. **Combine all parts:**
$$-\frac{3}{2} \div 8 \div 0$$
- Division by zero is undefined.
5. **Conclusion:** The entire expression is undefined because it involves division by zero.
**Final answer:** The expression is undefined due to division by zero.