# Prove that spec of root 2 contains infinitely many powers of 2.

1. Oct 5, 2012

### sachin123

Spec of root 2 is that set of elements floor(k * (root 2)) ; k >= 0 .

I have no idea of how I can prove the statement in the question.

Prove that spec of root 2 contains infinitely many powers of 2.
I need ideas on how to proceed.

Thank you

2. Oct 5, 2012

### Ray Vickson

What is meant by the "spec" of root 2? I have never seen that term.

RGV

3. Oct 5, 2012

### sachin123

Here, it means Spec√2= {⌊k√2⌋ : k≥0}

4. Oct 5, 2012

Spectrum?

5. Oct 5, 2012

### Dick

It's little tricky. Here's big hint. Pick k=floor(2^n*sqrt(2)). Think about what binary representation of sqrt(2) looks like. Play around with that for a while.

6. Oct 5, 2012

### sachin123

I've been thinking about it but I can't seem to move ahead.
sqrt(2) is irrational. It's binary representation has no pattern.
It would look like 1.0110101000001001111... and multiplying by 2^n would left-shift it n times.
Taking floor of this irrational number loses a portion of the number. And inside the Spec definition, we have floor (k * √2) and this floor would lose some number too.
Finally if √2 * √2 * 2 ^ n = 2 ^(n+1), this attempt leads to an interger smaller than that.

7. Oct 6, 2012

### Dick

Yeah, I'm actually running into the same problem. It's pretty easy to show 2^n*sqrt(2)-floor(2^n*sqrt(2)) is less than 1/2 an infinite number of times and I thought that would cover it, but I keep getting an inequality pointing the wrong way.

Last edited: Oct 6, 2012
8. Oct 6, 2012

### Dick

Ok, I think you can do it. You can either pick 2^n*sqrt(2)-floor(2^n*sqrt(2)) to be in (0,1/2) or (1/2,1) (both happen an infinite number of times). Note it would also work fine if you could show the choice of k=floor(2^n*sqrt(2))+1 works. Just draw some careful diagrams of where all the numbers lie relative to each other.

Last edited: Oct 6, 2012