For fixed $n_x \geq 1$ solution of the inequality $32N_x-31 \leq n_x \leq 32N_x$

Question

I have a very silly question about an inequality.

Let $n_x \geq 1$, integer, intuitively there's a unique integer $N_x \geq 1$ such that

$$ 32N_x - 31 \leq n_x \leq 32N_x $$

However I don't know why I'm struggling to prove rigorously that the statement is true. My attempt was to define the function

$$ N_x = N_x(n_x) = \left\lfloor \frac{n_x+31}{32} \right\rfloor $$

By the monotonicity of the floor function I have

$$ N_x \leq \frac{n_x+31}{32} \Rightarrow 32N_x-31 \leq n_x $$

which gives me the lower bound on $n_x$ for the upper bound

$$ N_x > \frac{n_x + 31}{32} - 1 = \frac{n_x - 1}{32} \geq 0 \Rightarrow N_x \geq \frac{n_x}{32} \Rightarrow n_x \leq 32N_x $$

Assuming everything is correct I'm puzzled if $N_x \geq \frac{n_x}{32}$, then I don't actually like the proof I just gave since it involves the guessing of the function $N_x(n_x)$, while instead I was specifically interested only in the existence of a unique solution, I'm not sure also that I've proved that the solution I gave is the unique one. I proved that there's a solution but not that such solution is eventually the unique one.

Is there a better way to prove the uniqueness of the solution without passing through the function I defined?

Answer 1

A simple argument for uniqueness: Consider two integers $N$ and $N'$ that both satisfy the two inequalities, that is: $$ 32N - 31 \le n_x \le 32N $$ and $$ 32N' - 31 \le n_x \le 32N'.$$ Multiplying the second inequality by $-1$ yields $$ -32N' \le -n_x \le -32N'+31, $$ and adding this new inequality to the first one yields the sum $$ 32(N-N')-31 \le 0 \le 32(N-N')+31. $$ For the integer $M:=N'-N$ the last inequality is equivalent to $$ -31 \le 32M \le 31.$$ The only multiple of $32$ between $-31$ and $+31$ is $0$. This implies $M=0$ and hence $N=N'$.

Answer 2

Note that if $k \in \mathbb{Z}$ and $x \in \mathbb{R}$ then $x \ge k \iff x \gt k-1\,$ so the inequality can be rewritten as:

$$ 32N - 32 \lt n \le 32N \;\;\iff\;\; N-1 \lt \frac{n}{32} \le N $$

Therefore $N$ is the smallest integer greater than or equal to $\cfrac{n}{32}\,$, since the next smaller integer $N-1 \lt \cfrac{n}{32}\,$, which proves both the existence and uniqeness of $N$. In fact, by the definition of the ceiling function it follows that $N = \left\lceil \cfrac{n}{32}\, \right\rceil\,$ (which is in fact equal to the posted $N = \left\lfloor \cfrac{n+31}{32}\, \right\rfloor\,$).