Welcome to our community

Be a part of something great, join today!

Unsolved Challenge Challenge involving irrational number

  • Thread starter
  • Admin
  • #1


MHB POTW Director
Staff member
Feb 14, 2012
Let $x$ be an irrational number. Show that there are integers $m$ and $n$ such that $\dfrac{1}{2555}<mx+n<\dfrac{1}{2012}$.


Well-known member
Mar 31, 2013
We have $\frac{1}{2012} - \frac{1}{2555} = \frac{ 543}{2012 * 2555}$
Now to keep it simple let p = 2012 * 2555
divide the interval $[0\cdots 1]$ into p equal intervals from 1 to p the $k^{th}$ interval between
$\frac{k-1}{p}$ to $\frac{k}{p}$
Define the function $f(m) = mx - \lfloor mx \rfloor $
Let us find f(k) for k = 1 to p
We shall get p values and there is value in 1st interval.
We shall assert the above statement that there is value in 1st interval.
If there is no value in 1st interval then there are p values and p-1 intervals so 2 values must be in the same interval so say for a and b
So $ | f(a) - f(b) |$ must be in the 1st interval for (a-b)
So there is a value say c such that
$ | f(c) |= \frac{1}{p}$
Now because p = 2012 * 2555
So there exists integer s and t such that
$\frac{1}{2555} < st < rt < \frac{1}{2012}$
So if chose an integer m such that $sc <= m < rc$ then we have
$\frac{1}{2555} < f(c) < \frac{1}{2012}$
Because $f(c) < \frac{1}{p}$
And as $r < p$ we have
$f(mc) < 1$
So $f(m) = mx - \lfloor mx \rfloor $ and choosing $n= - \lfloor mx \rfloor $ we get the result
Hence proved