Question

Prove that if x is a real number, then $\lfloor -x\rfloor =-\lfloor x\rfloor \text{and}\lfloor -x\rfloor =-\lfloor x\rfloor$

Short answer

The function is

$\lfloor -x\rfloor =-\lfloor x\rfloor \text{and}\lfloor -x\rfloor =-\lfloor x\rfloor$

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:$\lfloor -x\rfloor =-\lfloor x\rfloor \text{and}\lfloor -x\rfloor =-\lfloor x\rfloor$

PROOF

Let x be a integer. If x is an integer, then -x is the integer itself.

The ceiling function and floor function of an integer is the integer itself

$\begin{array}{r}\lfloor -x\rfloor =-x\\ \lfloor -x\rfloor =-x\\ \lfloor x\rfloor =x\\ \lfloor x\rfloor =x\end{array}$

Then we note:

$\begin{array}{r}\lfloor -x\rfloor =-x=-\lfloor x\rfloor \\ \lfloor -x\rfloor =-x=-\lfloor x\rfloor \end{array}$

We thus note that the equations hold for integer.

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.

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

Let us multiply the obtained inequality$m<x<m+1$ by -1.

$-m<-x<-(m+1)$

Or equivalently:

$-(m+1)<-x<-m$

The floor function of -x is then the smaller integer of the inequality, while the ceiling function is the larger of the two integers

$\begin{array}{r}[-x\rceil =-(m+1)\\ [-x]=-m\end{array}$

Then we note:

$\begin{array}{r}\lceil -x\rceil =-(m+1)=-\lceil x\rceil \\ \lceil -x\rceil =-m=-\lceil x\rceil \end{array}$

We thus note that the equation also holds for real numbers that are not integers.

Hence, the solution is

$\lfloor -x\rfloor =-\lfloor x\rfloor \text{and}\lfloor -x\rfloor =-\lfloor x\rfloor$