Given an irrational number $s>1$ and $n\in \mathbb{N}$ find a $w\in\mathbb{Q}$ such that $sI was reading a proof earlier which involved an irrational number $s>1$ and in which it was required to find, for each $n\in\mathbb{N}$, a rational number $w$ such that $s<w<s+\frac{1}{n}$. The authors give
\begin{align}
w&=\frac{[ (n+1)s]}{n+1}+\frac{1}{n+1}
\end{align}
without justification.
(The authors use the integer part  function $[\cdot]$ but since $s>1$, I will use the floor $\lfloor\cdot\rfloor$ isntead for notational clarity)
I see that this choice of $w$ works because
$\lfloor(n+1)s\rfloor+1>(n+1)s \Longrightarrow \frac{\lfloor (n+1)s\rfloor}{n+1}+\frac{1}{n+1}=w>s$ where the inequality is strict because $s$ is irrational and
\begin{align}\lfloor(n+1)s\rfloor<(n+1)s \Longrightarrow& \lfloor(n+1)s\rfloor+1<(n+1)s+1+\frac{1}{n}\\
\Longrightarrow &\frac{\lfloor (n+1)s\rfloor}{n+1}+\frac{1}{n+1}=w<s+\frac{1}{n+1}+\frac{1}{n(n+1)}=s+\frac{1}{n}
\end{align}
and while I appreciate the simplicity of this choice , I dont find it to be very  obvious. Can anyone please suggest alternate/more intuitive choice(s) for w?
 A: If you look at $s = \sqrt 2$ and $n=10$ you get $1.41\dots < w < 1.51\dots$ suggesting the choice of $w=1.5$ which generalizes to
$$s < \frac {\lceil sn \rceil}{n} < s + \frac 1n$$
$$sn < \lceil sn \rceil < sn + 1$$
Since $s$ is irrational, then $sn$ is also irrational, so the ceiling operation is not a fixed point.
A: Every irrational number has an infinite decimal expansion which represents the convergent limit of the infinite series
$$a + \frac{b}{10} + \frac{c}{10^2} + \cdots ~: ~a \in \Bbb{Z}, ~b,c,\cdots \in \{0,1,2,\cdots,9\}.$$
Each partial sum in the series is strictly below the irrational number, and if the series is truncated at any point and (for example) the coefficient $c$ is changed to $(c + 1)$, then the truncated sum is greater than the irrational number.
Simply extend the decimal expansion of $s$ out far enough so that truncating the series at a specific point and changing the pertinent coefficient from (for example) $c$ to $(c + 1)$ must result in a rational number $w$ that is greater than the irrational number $s$.
As long as the decimal expansion is taken far enough, before the truncation, you will routinely be able to guarantee that
$$w < s + \frac{1}{n}.$$
A: Let $s\in \mathbb R\setminus \Bbb Q,s>0$ and $n \in \Bbb N^*.$
Divide the segment $[s,s+\frac1n]$ into $n+1$ equal parts; each graduation on the drawing corresponds to $\frac{1}{n(n+1)}$. Then, if we enlarge $n+1$ times this drawing, we illustrate that :
$$\left\lfloor (n+1)s \right\rfloor<(n+1)s<\left\lfloor (n+1)s \right\rfloor +1<(n+1)(s+\frac1n)$$
The last inequality comes from the fact that $(n+1)(s+\frac1n)=(n+1)s+1+\frac1n>(n+1)s+1$
Then, if we now reduce by $n+1$, we get the desired inequalities.

A: There is two unnecessary complications in the given formula for
$w.$ Perhaps they are what make the result seem unintuitive?
The first complication is the use of $n + 1,$ rather than $n.$
The second complication is the use of the floor function, as in
$\left\lfloor{(n + 1)s}\right\rfloor + 1$ (or
$\left\lfloor{ns}\right\rfloor$ + 1) rather than
the ceiling function, as in
$\left\lceil{(n + 1)s}\right\rceil$ (or
$\left\lceil{ns}\right\rceil$).
I find the following argument intuitive.
Because $s$ is irrational, the real number $ns$ is not an integer,
so there exists an integer $m$ such that
$$
m - 1 < ns < m,
$$
whence
\begin{equation}
\label{4567634:eq:1}\tag{1}
w - \frac1n < s < w,
\end{equation}
where
$$
w = \frac{m}{n} =
\frac{\left\lceil{ns}\right\rceil}{n}.
$$
The only slight twist in this argument is the shift in perspective
needed to think of replacing the condition we were asked to satisfy,
$$
s < w < s + \frac1n,
$$
with the equivalent condition \eqref{4567634:eq:1}.
The only effect of replacing $n$ by $n + 1$ in the argument is to
ensure - quite unnecessarily! - that the stronger condition
$$
s < w < s + \frac1{n + 1}
$$
is also satisfied.
It is no wonder to me that you found the given formula puzzling.
