Question

Show that if x is a real number and n is an integer, then

$\text{a)}x\le n\text{if and only if}\lceil x\rceil \le n$

$\text{b)}n\le x\text{if and only if}\mathrm{n}\le \lceil x\rceil$

Short answer

The function is

$\text{a)}x\le n\text{if and only if}\lceil x\rceil \le n$

$\text{b)}n\le x\text{if and only if}\mathrm{n}\le \lceil x\rceil$

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: (a)

X is real number .

To prove:$\text{a)}x\le n\text{if and only if}\lceil x\rceil \le n$

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x<n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

$\mathrm{\exists }n\in \mathrm{Z}:n\le x\le n+1$

The floor function of x is then the smaller integer of the inequality.

data-custom-editor="chemistry" $\lfloor x\rfloor =m+1$

Thus we have derived:

$\lfloor x\rfloor =m+1\le n\mid$

Thus if$x\le n,\text{then}\lfloor x\mid \le n$

Step 3:

SECOND PART

PROOF

Let x be a integer. If x is an integer, then$\lfloor x\rfloor =x\le n$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

$\mathrm{\exists }n\in \mathrm{Z}:n<x<n+1$

The floor function of x is then the smaller integer of the inequality.

$\begin{array}{r}\lceil x\rceil =m+1\\ x\le m+1=\lceil x\rceil \le n\end{array}$

Thus if$\left|x\right|\le n\text{, then}x\le n\mid$

Step 4:Given: (b)

X is real number .

To prove:$\text{b)}n\le x\text{if and only if}\mathrm{n}\le \lceil x\rceil$

PROOF

Let x be a integer. If x is an integer, then$n\le x=\lfloor x\rfloor$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

$\mathrm{\exists }n\in \mathrm{Z}:n\le x\le n+1$

The ceiling function of x is then the larger integer of the inequality.

$\lfloor x\rfloor =m$

Thus we have derived:

$n\le m=\lfloor x\rfloor$

Thus if$n\le x\text{, then}n\le \lfloor x\rfloor$

Step 5:

SECOND PART

PROOF

Let x be a integer. If x is an integer, then$n<\lfloor x\rfloor =x$

Thus let us assume that x is not an integer.

Any real number x is less between two consecutive integers.

$\mathrm{\exists }n\in \mathrm{Z}:n\le x\le n+1$

The ceiling function of x is then the largeer integer of the inequality.

$\lceil x\rceil =m$

Since$m<x<m+1$

$n\le \lfloor x\rfloor =m<x$

Thus if$n\le \lfloor x\rfloor \text{, then}n\le x$

Hence, the solution is

$\text{a)}x\le n\text{if and only if}\lceil x\rceil \le n$

$\text{b)}n\le x\text{if and only if}\mathrm{n}\le \lceil x\rceil$