Solve lfloor1/2+lceil1/2rceil

$$\lfloor \frac { 1 } { 2 } + \lceil \frac { 1 } { 2 } \rceil$$

Evaluate

$$1$$

Solution Steps

Dividing 1 by 2 gives 0 and remainder 1. Rewrite \frac{1}{2} as 0+\frac{1}{2}.

$$\lfloor \frac{1}{2}+\lceil 0+\frac{1}{2}\rceil \rfloor$$

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

$$\lfloor \frac{1}{2}+1\rfloor$$

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

$$\lfloor \frac{1}{2}+\frac{2}{2}\rfloor$$

Since \frac{1}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.

$$\lfloor \frac{1+2}{2}\rfloor$$

Add 1 and 2 to get 3.

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

Dividing 3 by 2 gives 1 and remainder 1. Rewrite \frac{3}{2} as 1+\frac{1}{2}.

$$\lfloor 1+\frac{1}{2}\rfloor$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 1+\frac{1}{2} is 1.

$$1$$

Factor

$$1$$