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'Eyes Down' printed from https://nrich.maths.org/
Thank you to Alan of Madras College for
this solution.
If $x$ is a real number then $x = a + b$ where $a$ is an integer
and $b$ is a real number such that $0 \leq b < 1$. Here $a$ is
the integer part of $x$ and we write $a = [x]$. We have to consider
whether $[2x]$; $2[x]$ and $[x + 1/2 ] + [x - 1/2 ]$ can ever be
equal and whether they can take three different values.
If $1/2 \leq b < 1$ then $[2x]= 2a + 1$.
If $0 \leq b < 1/2$ then $[2x]= 2a$.
For any $b$, $2[x] = 2a$.
If $1/2 \leq b < 1$ then $[x+ 1/2 ] = a + 1$ and $[x - 1/2 ] =
a$ and so $[x + 1/2 ] + [x - 1/2 ] = 2a + 1$.
If $0 \leq b < 1/2$ then $[x+ 1/2 ] = a$ and $[x - 1/2 ] = a -
1$ and so $[x + 1/2 ] + [x - 1/2 ] = 2a - 1$.
$\bullet$
Case 1: $\; 0
\leq b < 1/2$
$[2x]= 2a = 2[x]$
but $[2x] \neq [x + 1/2 ] + [x - 1/2 ]$.
$\bullet$
Case 2: $\; 1/2
\leq b < 1$
$[2x]= 2a + 1 = [x + 1/2 ] + [x - 1/2]$
but $[2x] \neq 2[x]$.
Hence it is impossible for all of $[2x]$; $2[x]$ and $[x + 1/2 ] +
[x - 1/2 ]$ to be equal but they can never take three different
values.