# Explain this!

• December 28th 2010, 06:52 PM
Bruno J.
Explain this!
In a column, write the integer part of the integer multiples of $\sqrt{2}$. Then, in the second column, write the missing integers in increasing order. In the third column, write the difference between the second and first columns. You get :

1 3 2
2 6 4
4 10 6
5 13 8
7 17 10
8 20 12
9 23 14
11 27 16
12 30 18
14 34 20
15 37 22
16 40 24
18 44 26
19 47 28
21 51 30
22 54 32
24 58 34
25 61 36
26 64 38
28 68 40
29 71 42

...

Why does the third column consist of the even integers?
• December 29th 2010, 04:50 AM
DrSteve
Hmmm...do you have a proof that the third column is in fact the set of positive even integers? Any easy argument would seem to require that the first column follows a pattern - at first glance it is unclear to me if this is true.
• December 29th 2010, 12:25 PM
Bruno J.
Quote:

Originally Posted by DrSteve
Hmmm...do you have a proof that the third column is in fact the set of positive even integers? Any easy argument would seem to require that the first column follows a pattern - at first glance it is unclear to me if this is true.

Yes, it's true, and I'm challenging you to find the proof!
• December 29th 2010, 09:32 PM
simplependulum
HI , i think you are asking us to show that the form $2n + [n\sqrt{2}] ~,~ n \in \mathbb{N}$ represents all the missing integers . I am now going to prove this fact .

Before the proof , let me call the integers which cannot be expressed as $[k\sqrt{2}]$ happy integers . It is to avoid typing so much !

Proof :

Observation 1 :

Let $m$ be a positive integer , then $m$ is happy if and only if $\{ \frac{m}{\sqrt{2}} \} < 1 - \frac{1}{\sqrt{2}}$ .

Proof : It is easy to check that $m$ is happy if and only if there is no real number of the form $k\sqrt{2}$ lying between $m$ and $m + 1$ , since the common difference of the sequence $k\sqrt{2}$ is $\sqrt{2}$ , we can see that there must be such a number lying between $m - ( \sqrt{2} - 1)$ and $m$ . In other words , there exists $k \in \mathbb{N}$ such that $m - ( \sqrt{2} - 1) < k\sqrt{2} < m$ or $\frac{m}{\sqrt{2}} - (1 - \frac{1}{\sqrt{2}}) < k < \frac{m}{\sqrt{2}}$ . Therefore , $\{ \frac{m}{\sqrt{2}} \} < 1 - \frac{1}{\sqrt{2}}$

Now i am going to show all the integers of the form $2n + [n\sqrt{2}]$ are happy :

As $\frac{ 2n + [n\sqrt{2}] }{\sqrt{2}} = n \sqrt{2} + \frac{[n\sqrt{2}] }{\sqrt{2}}$

$= [n\sqrt{2}] + \{ n\sqrt{2}\} + \frac{ n\sqrt{2} - \{ n\sqrt{2} \} }{\sqrt{2}}$

$= n + [n\sqrt{2}] + \{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} )$

Since $\{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} ) < 1$ we have

$\left\{ \frac{ 2n + [n\sqrt{2}] }{\sqrt{2}} \right\} = \{ n\sqrt{2}\} ( 1 - \frac{1}{\sqrt{2}} ) < 1 - \frac{1}{\sqrt{2}}$ .

Therefore , $2n + [n\sqrt{2}]$ is happy for all positive integers $n$ .

Observation 2 :

If $m$ is happy , the next happy integer is greater than $m+2$ .

Proof :

$\{ \frac{m+1}{\sqrt{2}} \} = \{ \frac{m}{\sqrt{2}} \} + \frac{1}{\sqrt{2}}$ ( since $\{ \frac{m}{\sqrt{2}} \} + \frac{1}{\sqrt{2}} < 1$ )

$> \frac{1}{\sqrt{2}} > 1 - \frac{1}{\sqrt{2}}$ so $m + 1$ is not happy .

$\{ \frac{m+2}{\sqrt{2}} \} = \{ \frac{m}{\sqrt{2}} \} + \sqrt{2} - 1$ ( since $1 < \{ \frac{m}{\sqrt{2}} \} + \sqrt{2} < 2$ )

$> \sqrt{2} - 1> \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$ so $m+2$ is not happy !

We can see that $3 = 2 + [\sqrt{2}]$ is the first happy integer which is of this form . Moreover , $2(n+1) + [(n+1)\sqrt{2}] - (2n + [n\sqrt{2}] ) \leq 4$ . The conclusion is : there doesn't exist a happy integer lying between $2n + [n\sqrt{2}]$ and $2(n+1) + [(n+1)\sqrt{2}]$ exclusively as from the second observation , the separation between them will be greater than 4 which is impossible .

Therefore , $2n + [n\sqrt{2}]$ represents all the happy integers ! (Happy)
• December 30th 2010, 02:41 PM
Bruno J.
Good job simplependulum, you rock as always!

This fact is a consequence of the following more general and quite surprising theorem (a theorem which, obviously, you weren't expected to know off the top of your head) : the sets $\{\lfloor\alpha n\rfloor : n \in \mathbb{N}\}, \{\lfloor\beta n\rfloor : n \in \mathbb{N}\}$ form a partition of the integers if and only if $\alpha>1$ is irrational and $\frac{1}{\alpha}+\frac{1}{\beta}=1$.

So with $\alpha = \sqrt 2$, we find that $\beta = 2+\sqrt 2$, which solves the problem!