Solve 1divlceil4-1divlfloor2-1div(1+5/13)rfloorrceil=

$$1 \div \lceil 4-1 \div \lfloor 2-1 \div (1+ \frac{ 5 }{ 13 } ) \rfloor \rceil =$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

Convert 1 to fraction \frac{13}{13}.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 2-\frac{1}{\frac{13}{13}+\frac{5}{13}}\rfloor }\rceil }$$

Since \frac{13}{13} and \frac{5}{13} have the same denominator, add them by adding their numerators.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 2-\frac{1}{\frac{13+5}{13}}\rfloor }\rceil }$$

Add 13 and 5 to get 18.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 2-\frac{1}{\frac{18}{13}}\rfloor }\rceil }$$

Divide 1 by \frac{18}{13} by multiplying 1 by the reciprocal of \frac{18}{13}.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 2-1\times \frac{13}{18}\rfloor }\rceil }$$

Multiply 1 and \frac{13}{18} to get \frac{13}{18}.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 2-\frac{13}{18}\rfloor }\rceil }$$

Convert 2 to fraction \frac{36}{18}.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor \frac{36}{18}-\frac{13}{18}\rfloor }\rceil }$$

Since \frac{36}{18} and \frac{13}{18} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor \frac{36-13}{18}\rfloor }\rceil }$$

Subtract 13 from 36 to get 23.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor \frac{23}{18}\rfloor }\rceil }$$

Dividing 23 by 18 gives 1 and remainder 5. Rewrite \frac{23}{18} as 1+\frac{5}{18}.

$$\frac{1}{\lceil 4-\frac{1}{\lfloor 1+\frac{5}{18}\rfloor }\rceil }$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 1+\frac{5}{18} is 1.

$$\frac{1}{\lceil 4-\frac{1}{1}\rceil }$$

Anything divided by one gives itself.

$$\frac{1}{\lceil 4-1\rceil }$$

Subtract 1 from 4 to get 3.

$$\frac{1}{\lceil 3\rceil }$$

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of 3 is 3.

$$\frac{1}{3}$$

Factor

$$\frac{1}{3} = 0.3333333333333333$$