How to prove ceiling and floor inequality more 'formally'?

Question

The inequality in question is below:

$x - 1 < \lfloor x\rfloor \le x \le \lceil x \rceil < x + 1 $

Essentially, I must prove the above for every real number $x$. To begin this proof, I broke it into two cases: $x$ is an integer or $x$ is a real number.

Case 1 ($x$ is an integer):

I break this part up into two components to prove the left and the right side of the inequality.

For the left component,

For $x - 1 < \lfloor x\rfloor \le x$, we know that $\lfloor x\rfloor = x$ since $ x \in \mathbb Z$ therefore $ x - 1 < x \le x$ holds.

For the right component,

For $x \le \lceil x \rceil \le x + 1$, we know that $\lceil x \rceil = x$ since $ x \in \mathbb Z$ therefore $x \le \lceil x \rceil \le x + 1$ holds.

Case 2 ($x$ is a real number and non-integer):

Here is the part where I have trouble proving 'formally'.

For the left component,

For $x - 1 < \lfloor x\rfloor \le x$, we know that if $x = 4.5$ we know that $\lfloor x\rfloor = 4$, which is greater than $ x - 1 = 3.5 $. So, we know that $x - 1 < \lfloor x\rfloor \le x$ holds. This works for all $ x \in \mathbb {R} $.

For the right component,

For $x \le \lceil x \rceil \le x + 1$, we know that if $x = 4.5$ we know that $\lfloor x\rfloor = 5$, which is greater than $ x $. We know that $x + 1 > 5$, so $x \le \lceil x \rceil \le x + 1$ holds. This works for all $ x \in \mathbb {R} $.

Since both the left and the right components of the inequality hold, the whole inequality holds.

I am new to proofs, so I'm not sure if I'm doing this proof by induction or deduction (or something). All I know is that I was able to construct cases in which I know that this inequality works. However, I do not feel that my proof is 'all there'. I feel that Case 1 is sound enough to be considered a proof, but I feel my Case 2 is lacking. Is there any way I can prove this more formally if the above work is unacceptable as a proof?

EDIT:

Adding proof to show that the inequality holds:

The floor is defined as $\lfloor x \rfloor = n \le x < n + 1$.

If $x$ is a real number and $n$ is an integer, then $\lfloor x \rfloor$ is defined as the smallest integer less than or equal to $x$. (Credit to kccu) Since the smallest integer would be equivalent to $x$, we know that $x - 1 < \lfloor x \rfloor$ is less than $x$. Therefore, the left hand side of the inequality $x - 1 < \lfloor x\rfloor \le x$ holds.

Since $\lceil x \rceil$ is defined to be the smallest integer that is greater than or equal to $x$, $n + 1$ is an integer greater than or equal to $x$. Since $n$ is not $/re x$, we know that $n + 1$ is the smallest integer greater than or equal to $x$ (Again, credit to kccu). Since we know that $n + 1$ is the smallest integer greater than or equal to $x$, the right hand side of the equation holds since $\lceil x \rceil$ is equivalent to x since it is the smallest integer greater than or equal to $x$. Thus, the right hand side holds as well.

Answer 1

This all depends on your definition of the Floor and Ceiling functions... once you choose a definition you can show the different forms are synonymous, but you must choose a definition. Perhaps the easiest definition I have seen for you purposes comes from Wikipedia

Since there is exactly one integer in a half-open interval of length one, for any real x there are unique integers m and n satisfying $x-1 < m \leq x \leq n < x+1$

From this we can define $\lfloor x \rfloor = m$ and $\lceil x \rceil = n$ and your proof becomes trivial... a self evidence by a substitution

Answer 2

We can also define the fractional part of $x$, or $\{x\}$, to be a number between $0$ (inclusive) and $1$ (exclusive). Then, we can say $\lfloor x\rfloor = x - \{x\}$, and since $\{x\}$ is between $0$ and $1$, we have that $\lfloor x\rfloor$ is between $x-1$ and $x$. Likewise, we have $\lceil x\rceil = x + (1-\{x\})$, and $\{x\}$ is between $0$ and $1$, so $1-\{x\}$ is between $0$ and $1$, so $\lceil x\rceil$ is between $x$ and $x+1$.