Question

The function INT is found on some calculations, where $INT\left(x\right)=\lfloor x\rfloor$when x is a nonnegative real number and$INT\left(x\right)=\lfloor x\rfloor$ when x is a negative real number. Show that this INT function satisfies the identity $INT(-x)=-INT\left(x\right)$

Short answer

The function is

$\mathrm{INT}\left(x\right)=\lfloor x\rfloor$

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:$\mathrm{INT}(-x)=-\mathrm{INT}\left(x\right)$

PROOF

Result previous exercise: if x is a real number,$\lfloor -x\rfloor =-x\text{and}\lfloor -x\rfloor =-x$

FIRST PART

Let x be 0. 0 is an integer. The ceiling function and floor function of an integer is integer itself:

Then we note:

$\mathrm{INT}(-x)=\mathrm{INT}(-0)=\mathrm{INT}\left(0\right)=\lfloor 0\rfloor$

SECOND PART

Let x be a negative real number, then -x is a negative real number.

$INT(-x)=\lfloor -x\rfloor =\lfloor -x\rfloor =-INT\left(x\right)$

THIRD PART

Let x be a negative real number, then -x is a positive real number.

$\mathrm{INT}(-x)=\lfloor -x\rfloor =-\lfloor x\rfloor =-\mathrm{INT}\left(x\right)$

Hence, the solution is

$\mathrm{INT}(-x)=-INT\left(x\right)$