# Thread: Getting two integers

1. ## Getting two integers

Please advise me how to prove the following assertion

Given two real numbers a , b, there always exists integers m, n such that a < m+n x sqrt(2)< b

It would be best if we can rely mainly on the basic theorems in number theory, such as the Archimedean Principle,
Density of Rationals and Density of Irrationals in proving the above assetion

2. ## Re: Getting two integers

Originally Posted by KLHON
Please advise me how to prove the following assertion

Given two real numbers a , b, there always exists integers m, n such that a < m+n x sqrt(2)< b

It would be best if we can rely mainly on the basic theorems in number theory, such as the Archimedean Principle,
Density of Rationals and Density of Irrationals in proving the above assetion
This is not true. Let a = 0, b = 1. There are no integers m, n that satisfy the inequality.

-Dan

3. ## Re: Getting two integers

$\displaystyle 0<3-2\sqrt{2}<1$

4. ## Re: Getting two integers

$\displaystyle a<b$

Find an integer $\displaystyle k\geq 0$

such that

$\displaystyle \left(3+2\sqrt{2}\right)^k(b-a)>1$

There is an integer $p$ such that

$\displaystyle \left( 3+2\sqrt{2} \right)^k a< p < \left( 3+2\sqrt{2} \right)^k b$

multiply by

$\displaystyle \left(3-2\sqrt{2}\right)^k$

to get

$\displaystyle a<p\left(3-2\sqrt{2}\right)^k<b$

5. ## Re: Getting two integers

This is the most interesting question we have had in months.
Idea's solution is equally interesting. Thank you.