How to Show that $1+\left(\left\lceil\dfrac{x}{n}\right\rceil -1\right)n\leq x$?

Question

For $x$ integer in $\{1,\ldots,n^2\}$. I would like to show that $$ 1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \leq x. $$ I start by the property of the ceiling function which gives me $$ 1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n <x+1. $$

Answer 1

By the property of the ceiling function we have:

$$ \left\lceil\dfrac{x}{n}\right\rceil <\dfrac{x}{n}+1, $$ which gives me: $$ 1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n <x+1. $$

What I missed which seems some kind of obvious is that $\left(1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n\right)$ is a positive integer and since it is less than $x+1$, it must be less than or equal to $x$. Thus,

$$ 1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \le x. $$

Answer 2

Okay divide both sides by $n$, then: $$\left\lceil\dfrac{x}{n}\right\rceil-1 \le \frac{x-1}{n}$$ then we got: $$\left\lceil\dfrac{x}{n}\right\rceil = \frac{x+k}{n}$$ where $k$ some integer number. So we have inequality with $k$ and $n$ : $$n \ge k-1$$ , now think about what set of valid $k$ and you will prove it.