Solve 2*lfloor1-lceil1/1+0.0rceilrfloor

$$2 \times \lfloor 1- \lceil \frac{ 1 }{ 1+0.0 } \rceil \rfloor$$

Evaluate

$$0$$

Solution Steps

Add 1 and 0 to get 1.

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

Anything divided by one gives itself.

$$2\lfloor 1-\lceil 1\rceil \rfloor$$

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of 1 is 1.

$$2\lfloor 1-1\rfloor$$

Subtract 1 from 1 to get 0.

$$2\lfloor 0\rfloor$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 0 is 0.

$$2\times 0$$

Multiply 2 and 0 to get 0.

$$0$$

Factor

$$0$$