4 Apr 2012

Here is the problem for tonight:

```
A set S contains 0 and 1, and the mean of every finite non-empty subset of S.
Prove that S contains all the rational numbers in the unit interval.
```

Ok….

Let's clarify what the rational number is. Maybe we'll get some ideas out of it.

```
A rational number is in the form of a/b where a and b are integer.
```

I still got nothing.

So, ``S`

` contains all the rational numbers in [0,1]. It implies that it might also contain non-rational numbers, as well, but we might not care about it.

I still got nothing.

So, ``S`

`contains 0 and 1, and the mean of a set, whose members are also the member of ```

S```.

It surely sounds like the barber paradox.

A mean of ``{ a, b, c, d }`

`is ```

(a + b + c + d) / 4``. A mean is guaranteed to be rational if`

`a```

, ``b`

`, ```

c``, and`

`d```

are all rational.

Walla! This is good shit.

``S`

` contains only rational numbers (Wrong!)Let's assume that ``S`

`contains an irrational number ```

a```.

```
Ok, this is a wrong direction. I cannot prove anything out of it.
Let's try again
--------------------
How about I claim that we can derive any ```a/b``` where ```a <= b```
### Base case
```

```
Still got nothing.
Let's try again
-----------------------
Ok, we are using a contradiction this time.
Let's assume that ```a/b``` is not a member of ```S```.
This means that there is no set ```{ c[1], ..., c[n] }``` that has its mean equal to ```a/b```.
```

Still got nothing.

Induction!

```
If we have 0/n, 1/n, ..., n/n, we can calculate 0/(n+1), 1/(n+1), ..., (n+1)/(n+1)
```

```
```0/(n+1)``` is ```0/n```. Ok!
```

Let's rephrase it to:

```
How do we calculate a/n from b/m where m > n?
```

If ``n = 2k`

`(an even number), assuming that ```

a = c + d (c != d)``, we get that`

`a/n = c/2k + d/2k = (c/k + d/k)/2```

. This case is ok.

If ``n = 2k+1`

`(an odd number), assuming that ```

a = c + d (c != d)```, we get nothing…

```
I gave up
-------------------
So.. The book starts with ```S``` contains all ```p/(2^n)```.
Ok, that's easy to see.
Basically, it claims that ```p/q``` can by the mean of ```x/(2^n)```.
Now this is getting a little weird.
```

We choose a very very large ``n`

` and replace 1 with 1-1/(2^{n), 1+1/(2}n), 1-2/(2^n), and so on.

For example, 1/7 = (1-1/(2^{n) + 0+1/(2}n) + 0-2/(2^{n) + 0+2/(2}n) + 0-3/(2^{n) + 0+3/(2}n) + 0) / 7

And that's it.

This is tooo creative.