Question

Show that if x is a real number and m is an integer, then$\lceil x+m\rceil =\lceil x\rceil +m$

Short answer

The function is$\lceil x+m\rceil =\lceil x\rceil +m$

Step-by-step solution

Step 1:

Ceiling function [x] : smallest integer that is greater than or equal to x.

Floor function [x] : largest integer that is less than or equal to x.

Step 2:Given:

X is real number .

To prove:$\lceil x+m\rceil =\lceil x\rceil +m$

PROOF

Let x be a integer. Since m is an integer, then x+m is also an integer.

FIRST CASE

The ceiling function of an integer is the integer itself.

$\begin{array}{r}\lceil x\rceil =x\\ \lceil x+m\rceil =x+m\end{array}$

Thus we have derived:

$\lceil x+m\rceil =x+m=\lceil x\rceil +m$

Step 3:Part (b)

SECOND CASE

PROOF

Let x not be a integer. Any real number x is lies between two consecutive integers.

FIRST CASE

$\lceil x\rceil =n+1$

Thus we have derived$n<x<n+1$ .

Let us add m to each side of inequality:

$n+m<x+m<n+1+m$

The ceiling function of an larger integer of the inequality.

$\lceil x+m\rceil =n+1+m$

Then we have derived:

$\lceil x+m\rceil =n+1+m=\lceil x\rceil +m$

Hence, the solution is

$\lceil x+m\rceil =\lceil x\rceil +m$