# Proof involving floor and ceiling function

Problem: For all $x \in R$ $\lfloor x \rfloor \not= \lceil x \rceil$

I believe I need to utilize the definitions of both, whereas;

floor of x equals unique integer n s.t. n less than or equal to x less than 1;

ceiling of x equals unique integer n s.t. n-1 less than x less than or equal to n

I can quickly find an example to show there exists a real number (such as 1.2) that shows the floor and ceiling are not equal. However, I realize this is not a proof.

I do not believe (or at least found) a counter example.

I hope I have provided sufficient information. Any help would be greatly appreciated.

• If $x$ is an integer, then floor and ceiling are equal. – quasi Nov 6 '17 at 2:36

If $x$ is an integer, then $\lfloor{x}\rfloor = x = \lceil{x}\rceil$.

If $x$ is not an integer, let $n = \lfloor{x}\rfloor$.

Then $\lfloor{x}\rfloor = n < x < n+1 = \lceil{x}\rceil$.

Explanation:

Suppose $x$ is not an integer.

• Then $\lfloor{x}\rfloor \le x$, hence $n \le x$.
• Since $x$ is not an integer, we can't have $x = n$, hence we must have $n < x$.
• Since $n$ is the greatest integer less than or equal to $x$, it follows that $n+1$ is not less than or equal to $x$, hence $x < n+1$.
• Since $n < x < n+1$, it follows that $n+1$ is the least integer greater than or equal to $x$, hence $\lceil{x}\rceil = n+1$.