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'Ruler' printed from https://nrich.maths.org/
Andaleeb sent in this excellent solution.
The diagram shows some of the vertical lines drawn for values of x
between 0 and 1 as described in the question. The lines are of
height 1 unit at x = 0 and 1, of height ${1\over 2}$ units at $x=
{1\over 2}$ , of height ${1\over 4}$ units at $x= {k\over 4}$ and
${1\over 8}$ units at $x= {k\over 8}$ and so on... up to ${1\over
2^5}$ at ${k\over 2^5}$ where $k$ is a positive integer.
n |
$0$ |
$1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$\dots$ |
$n$ |
$n+1$ |
Height |
$1$ |
${1\over 2}$ |
${1\over 4 }$ |
${1\over 8}$ |
${1\over 16}$ |
${1\over 32 }$ |
${1\over 64}$ |
$\dots$ |
${1\over 2^n}$ |
${1\over 2^{n+1`}}$ |
Lines cut |
$2$ |
$1$ |
$2$ |
$4$ |
$8$ |
$16$ |
$32$ |
$\dots$
|
$2^{n-1}$ |
$ 2^n$ |
Thus if the height $h$ lies in ${1\over 2^n} > h> {1\over
2^{n+1}}$ then the number of lines cut is given by $$ 2 + 1 + 2 + 4
+ 8 + ... + 2^{n-1} = 2 + {{2^n - 1} \over {2 - 1}} = 2^n + 1.$$ As
$n$ tends to infinity the height of the lines tends to 0 and the
number of lines cut tends to infinity.