# A proof needed

1. Jun 30, 2005

### Anzas

prove that between any two real numbers there is a number of the form
$$\frac{k}{2^n}$$
where k is an integer and n is a natural number.

2. Jun 30, 2005

### NateTG

This should be very easy.

Hint: Write the real numbers in binary.

3. Jun 30, 2005

### Hurkyl

Staff Emeritus
Sounds like homework. What have you tried?

4. Jul 3, 2005

### Icebreaker

Sounds interesting. I haven't been able to prove it, even with binary.

5. Jul 3, 2005

### HallsofIvy

Staff Emeritus
I did something similar in a proof showing that monotone convergence implies least upper bound property.

Let the two distinct real numbers be a and b with a> b. Then a- b> 0 so $$\frac{1}{a-b}$$ exists and is positive. By the Archimedian property of integers (given any real number x, there exist an integer m with m>x) there exist an integer n with $$n> \frac{1}{a-b}$$.

But it is easy to prove by induction that, for all positive integers n, 2n> n (Hint: first prove by induction that 2n>= n+1) so we have $$2^n> \frac{1}{a-b}$$. That means that 2n(a- b)= 2na- 2nb> 1.
Since the distance between those two numbers is greater than 1, there exist an integer, k, such that 2nb< k< 2na. Dividing through by 2n gives $$b< \frac{k}{2^n}< a$$.

Last edited: Jul 3, 2005
6. Jul 4, 2005

### NateTG

That's very elegant.

I was thinking more along the lines of:
WOLOG $a>b$.
Since $a$ and $b$ are real numbers, they can be written in binary notation in such a way that each of them has an infinite number of zeroes to the right of the binary (or is it still decimal?) point.

Now, since $a \neq b$ there is a first digit where they differ. At that position, $a$ will have a 1, and $b$ will have a zero. Now, since there is an infinite number of zeros, there is a next location where b has a zero. Let's say that that's the $2^{-k}$s place.

Then the integer part of
$$b*2^{k}+1$$
divided by
[tex]2^{k}[/itex]
Is between $a$ and $b$