Solve lceil2/lfloor3rfloorrceil

$$\lceil \frac{ 2 }{ \lfloor 3 \rfloor } \rceil$$

Evaluate

$$1$$

Solution Steps

The floor of a real number a is the largest integer number less than or equal to a. The floor of 3 is 3.

$$\lceil \frac{2}{3}\rceil$$

Dividing 2 by 3 gives 0 and remainder 2. Rewrite \frac{2}{3} as 0+\frac{2}{3}.

$$\lceil 0+\frac{2}{3}\rceil$$

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of 0+\frac{2}{3} is 1.

$$1$$

Factor

$$1$$