If we have proved that $n$ cannot be greater than $9$, have we also proved that the most it can be is $9$?

Question

If we have proved that $n$ cannot be greater than $9$, have we also proved that the most it can be is $9$? I am conflicted because it could still be impossible for $n$ to be greater than, say, $5$, and it could never actually be $9$, but because we haven't proved this tighter bound, it seems it is valid to conclude that the possibility exists for $n$ to be $9$, as "can" expresses mere possibility. Which of these viewpoints is correct?

Edit: if we have shown $n \le 9$, is this logically equivalent to saying the most $n$ can be is $9$, or is that concluding too much; we don't know if it actually is possible for $n$ to be $9$?

Answer 1

If $n$ cannot exceed $9,$ then $n$ is indeed at most $9$ (in fact, these two claims are equivalent), which is not to say that it is not also at most $7$ nor a suggestion that $n$ in the context ever attains the value $9.$

"Every $n$ is at most $7$" is a stronger assertion that "every $n$ is at most $9$", and the two assertions are consistent with each other: they can—but needn’t—both be true.

Orthogonal point:

if we have shown that $n \le 9$, is this logically equivalent to saying that the most $n$ can be is $9$

Technically they are mathematically equivalent but not logically equivalent, because if we change the meaning of that inequality sign they may no longer have the same meaning.

Answer 2

To my ears, $n$ cannot be greater than $9$ is $n\not> 9$, and $n$ can at most be $9$ is $n\le 9$, and the two are equivalent. So I would say yes.

Both wordings leave open the possibility that $n$ is precisely $9$, but we do not know if equality holds.

Answer 3

$$\begin{align} \boxed{n \text{ cannot be greater than } 9} &\Longleftrightarrow n\notin (9, +\infty) \\ &\Longleftrightarrow n\in(\mathbb{R}/(9, +\infty)) \\ &\Longleftrightarrow n\in(-\infty, 9] \\ &\Longleftrightarrow \boxed{n \text{ is at most 9 }} \end{align}$$