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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

- Mar 31, 2013

- 1,334

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