Eyes down
The symbol [ ] means 'the integer part of'. Can the numbers [2x];
2[x]; [x + 1/2] + [x - 1/2] ever be equal? Can they ever take three
different values?
Age
16 to 18
Challenge level
Problem
The symbol [ ] means 'the integer part of '.
Consider the three numbers
$$[2x];\ 2[x];\ [x + {1\over 2}] + [x - {1\over 2}]$$
Can they ever be equal?
Can they ever take three different values?
Getting Started
If the integer part of $x$ is $a$ then $x=a + b$ where $a$ is a whole number and $0\leq b < 1$.
Try some numerical values of $x$, evaluate the three functions and record the results. What do you notice? Can you prove that different values of $x$ will produce similar findings?
Student Solutions
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.