Answer
If x is an integer, then $\lceil{x}\rceil$-$\lfloor{x}\rfloor=0$
If x is not an integer, then $\lceil{x}\rceil$-$\lfloor{x}\rfloor=1$
Work Step by Step
Given : x is a real number
To Prove: if x is an integer, then $\lceil{x}\rceil$-$\lfloor{x}\rfloor=0$
If x is not an integer, then $\lceil{x}\rceil$-$\lfloor{x}\rfloor=1$
Proof:
Case 1- Let x be an integer
The ceiling function of an integer is the integer itself:
$$\lceil{x}\rceil=x$$
The floor function of an integer is the integer itself:
$$\lfloor{x}\rfloor=x$$
Then we obtain:
$$\lceil{x}\rceil-\lfloor{x}\rfloor=x-x=0$$
Case 2- Let x is not an integer. Any real number x lies between two consecutive integers (not inclusive, since x is not an integer).
$$\exists x \in Z:n
