Solve {c}{1+1/1-1{1+frac{1/1-frac{1{3}}}}}{2-1/3/frac{3{4}+frac{1}{4}}{frac{3}{4}}*frac{9}{40}}

$$\left. \begin{array} { c } { 1 + \frac { 1 } { 1 - \frac { 1 } { 1 + \frac { 1 } { 1 - \frac { 1 } { 3 } } } } } \\ { \frac { 2 - \frac { 1 } { 3 } } { \frac { 3 } { 4 } + \frac { 1 } { 4 } } { \frac { 3 } { 4 } } \times \frac { 9 } { 40 } } \end{array} \right.$$

Sort

$$\frac{9}{32},\frac{8}{3}$$

Solution Steps

Convert 1 to fraction \frac{3}{3}.

$$sort(1+\frac{1}{1-\frac{1}{1+\frac{1}{\frac{3}{3}-\frac{1}{3}}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{3}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$sort(1+\frac{1}{1-\frac{1}{1+\frac{1}{\frac{3-1}{3}}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Subtract 1 from 3 to get 2.

$$sort(1+\frac{1}{1-\frac{1}{1+\frac{1}{\frac{2}{3}}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Divide 1 by \frac{2}{3} by multiplying 1 by the reciprocal of \frac{2}{3}.

$$sort(1+\frac{1}{1-\frac{1}{1+1\times \frac{3}{2}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Multiply 1 and \frac{3}{2} to get \frac{3}{2}.

$$sort(1+\frac{1}{1-\frac{1}{1+\frac{3}{2}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Convert 1 to fraction \frac{2}{2}.

$$sort(1+\frac{1}{1-\frac{1}{\frac{2}{2}+\frac{3}{2}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{2}{2} and \frac{3}{2} have the same denominator, add them by adding their numerators.

$$sort(1+\frac{1}{1-\frac{1}{\frac{2+3}{2}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Add 2 and 3 to get 5.

$$sort(1+\frac{1}{1-\frac{1}{\frac{5}{2}}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Divide 1 by \frac{5}{2} by multiplying 1 by the reciprocal of \frac{5}{2}.

$$sort(1+\frac{1}{1-1\times \frac{2}{5}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Multiply 1 and \frac{2}{5} to get \frac{2}{5}.

$$sort(1+\frac{1}{1-\frac{2}{5}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Convert 1 to fraction \frac{5}{5}.

$$sort(1+\frac{1}{\frac{5}{5}-\frac{2}{5}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{5}{5} and \frac{2}{5} have the same denominator, subtract them by subtracting their numerators.

$$sort(1+\frac{1}{\frac{5-2}{5}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Subtract 2 from 5 to get 3.

$$sort(1+\frac{1}{\frac{3}{5}},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Divide 1 by \frac{3}{5} by multiplying 1 by the reciprocal of \frac{3}{5}.

$$sort(1+1\times \frac{5}{3},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Multiply 1 and \frac{5}{3} to get \frac{5}{3}.

$$sort(1+\frac{5}{3},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Convert 1 to fraction \frac{3}{3}.

$$sort(\frac{3}{3}+\frac{5}{3},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{3}{3} and \frac{5}{3} have the same denominator, add them by adding their numerators.

$$sort(\frac{3+5}{3},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Add 3 and 5 to get 8.

$$sort(\frac{8}{3},\frac{2-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Convert 2 to fraction \frac{6}{3}.

$$sort(\frac{8}{3},\frac{\frac{6}{3}-\frac{1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{6}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$sort(\frac{8}{3},\frac{\frac{6-1}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Subtract 1 from 6 to get 5.

$$sort(\frac{8}{3},\frac{\frac{5}{3}}{\frac{3}{4}+\frac{1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Since \frac{3}{4} and \frac{1}{4} have the same denominator, add them by adding their numerators.

$$sort(\frac{8}{3},\frac{\frac{5}{3}}{\frac{3+1}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Add 3 and 1 to get 4.

$$sort(\frac{8}{3},\frac{\frac{5}{3}}{\frac{4}{4}}\times \frac{3}{4}\times \frac{9}{40})$$

Divide 4 by 4 to get 1.

$$sort(\frac{8}{3},\frac{\frac{5}{3}}{1}\times \frac{3}{4}\times \frac{9}{40})$$

Anything divided by one gives itself.

$$sort(\frac{8}{3},\frac{5}{3}\times \frac{3}{4}\times \frac{9}{40})$$

Multiply \frac{5}{3} times \frac{3}{4} by multiplying numerator times numerator and denominator times denominator.

$$sort(\frac{8}{3},\frac{5\times 3}{3\times 4}\times \frac{9}{40})$$

Cancel out 3 in both numerator and denominator.

$$sort(\frac{8}{3},\frac{5}{4}\times \frac{9}{40})$$

Multiply \frac{5}{4} times \frac{9}{40} by multiplying numerator times numerator and denominator times denominator.

$$sort(\frac{8}{3},\frac{5\times 9}{4\times 40})$$

Do the multiplications in the fraction \frac{5\times 9}{4\times 40}.

$$sort(\frac{8}{3},\frac{45}{160})$$

Reduce the fraction \frac{45}{160} to lowest terms by extracting and canceling out 5.

$$sort(\frac{8}{3},\frac{9}{32})$$

Least common denominator of the numbers in the list \frac{8}{3},\frac{9}{32} is 96. Convert numbers in the list to fractions with denominator 96.

$$\frac{256}{96},\frac{27}{96}$$

To sort the list, start from a single element \frac{256}{96}.

$$\frac{256}{96}$$

Insert \frac{27}{96} to the appropriate location in the new list.

$$\frac{27}{96},\frac{256}{96}$$

Replace the obtained fractions with the initial values.

$$\frac{9}{32},\frac{8}{3}$$

Evaluate

$$\frac{8}{3},\ \frac{9}{32}$$