Proof of inequality $\lceil x \rceil \le x+1$ I went through the Master Theorum extension for floors and ceiling section 4.6.2 in the book Introduction to Algorithms
It had the following statement:

Using the inequality $\lceil x \rceil \le x+1$

But I haven't seen the inequality anywhere and could not understand the verifiability of inequality.
Instead the Chapter Floors and ceilings defined floors and ceilings as:
$$x-1 \lt \lfloor x \rfloor \le x \le \lceil x \rceil \lt x+1 $$
Please clear my doubt over this.
On how to use this identity and which identity to be considered when because both of them define completely different inequalities.
Thank you.
The definition of $\lceil x \rceil$ is:

$\lceil x \rceil$ is the minimal integer $n$ such that $n \geq x$.

(The existence of such an integer makes the reals an Archimedean field.)
Let us assume, for the sake of contradiction, that $\lceil x \rceil \geq x + 1$. Then $\lceil x \rceil - 1 \geq x$. Since $\lceil x \rceil - 1$ is also an integer, this contradicts the definition of $\lceil x \rceil$. Thus $\lceil x \rceil &lt; x + 1$.
It is also easy to check that the inequality is tight, in the sense that $1$ cannot be replaced by any smaller $\theta$. Indeed, if $\theta = 1 - \epsilon$ for $\epsilon \in (0,1)$, then $\lceil \epsilon \rceil = 1 = \epsilon + \theta$.